chapter one introduction energy sources the various forms
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Chapter one Introduction
Energy sources
The various forms of energy available in nature are as follow.
Radient Energy:Energy obtained from sunlight
Thermal Energy: Energy obtained from earth’s interior
Chemical Energy: Energy store in wood,coal & oil in chemical form
Potential Energy: Energy obtained by virtue of position.eg:bulk of water stored in
high level pressure energy
Kinetic Energy: Energy obtained from moving bodies
Nuclear Energy: Energy release when atoms are broken down
Electrical energy
Production or generation of energy
The practical device used for the generation of electrical energy is generator.
Generator is rotating device which converts the mechanical energy into electrical energy.
In this process mechanical power is supplied to shaft in form of torque. The mechanical
torque produced by turbine is converted into electrical energy. The turbine used may be of
two type:Hydraulic and steam driven turbine. The electrical source may be thermal and
non thermal.
The various types of thermal and non thermal power generation schemes are as
follows:
Thermal power
Coal feed thermal power generation
Diesel power generation
Nuclear power generation
Solar power generation
Non-thermal power
Hydropower generation
Tidal power generation
Wind power generation
Coal feed thermal power generation
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It is the type of generator that uses coal for the generation of steam from the boiler.
This steam is then passed into steam turbine through super heater. The steam turbine
converts the energy of steam into mechanical rotation. The shaft of turbine is connected to
the generator which generates the electrical energy. The discharge from the steam turbine
is passed through a condenser in which steam is passed through a condenser. Here steam is
condensed into water and this water ia again supplied to boiler with help of field pump. In
this type of power generation the starting and warming up process is comparetively long
therefore frequently shutting up and starting up is not suggestable.
Diesel power generation
This type of power station uses diesel as fuel in diesel engine. The mechanical
rotation produced by diesel engine is coupled with generator by the help of a shaft. This
generator then converts the mechanical force into electrical energy. The engine has to be
cooled against the heat generation within the engine so, circulating water system is used
for cooling purpose.
Nuclear power generation
The nuclear fission of atomic material such as Uranium produces heat which is
utilized to produce steam to run steam turbine.Such power station is called nuclear power
generation. About 3000 netrictones of coal produce the same amount of heat as 1kg of
nuclear fuel when uranium 225 is bombarded with nutrons. Fission reaction takes place
releasing nutrons and heat energy. These neutrons then participate in chain reaction of
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emissioning atoms of 285 uranium. The nuclear reactor of this plant has to be carefully
shield against the radiation.
Solar power generation
In this type of power generation, solar energy is collected directly with the help of
solar panel which converts it into electrical energy. Solar panel consists of solar cell which
are semicinductor devices. Number of solar cells can be connected in series parallel
combination to generate electrical power at desired level. The energy generated by solar
pannel is used to charge battery which gives DC voltage output. Inverter may be used to
convert the voltage into ac if the user needs ac voltage.
Hydropower Generation
Hydropower is economical and pollution free source of energy. The initial capital
investment in dams, transmission and generation is quite high but operating cost is very
low. A tyical layout of storage type hydropower is given below
Hydropower scheme needs a supply of water and difference in water level so that
the water will possess potential energy. If certain amount of water can be collected at
certain height above a certain reference level. Then the potential energy possessed by this
amount of water can be converted into mechanical rotation and thus into electrical energy.
The power generated by a hydropower plant is given by,
P=Q x g x ρ x H =9.81x1000xQxH
∴ P = 9810QH
Where,
Q=discharge of water (in m3/sec)
H=height of water that falls through certain height (in meter)
The actual power is always less than this theoretical value because of various
power loss such as friction losses in pipe, losses in turbine, losses in generation etc.
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Tidal power energy
This type of power plant is used to convert the change in potential energy in
potential energy caused by tide levels into electrical form. A dam with large gates is made
across the mouth of bay in ocean and low head turbines are used. At the time of high tides,
gates are opened and after water reaches to electrical level the gates are closed. The water
so collected is used to drive the turbines that in turn rotates the generators converts it into
electrical energy.
Wind power energy
In this type of generation, wind is used to drive the air turbines which in turn
rotates the generator and the generator then converts the mechanical rotation into
electricity according to faradays laws of electromagnetic induction.
Generation, distribution and transmission of electric process
Above figure shows the layout of power distribution system consisting of generation
transmission and distribution section. The power system consist of two power station
system. Most of the power station are quite far from the city so the power generation in
power station has to be transmitted through a transmission line. In order to maintain lower
power loss the transmission is done at high voltage level. Set up transformer are used to
step up the voltage level generated by the power station and then transmitted through
transmitted line. The step down transformer are used at substation to step down the voltage
before distribution.
The 11kV distribution lines run around the city and 400V 4 wire and 220V single
phase system are trapped out from the distribution line at various points by using
11kV/400V step down transformer.
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Chapter two Passive elements and laws in electrical engineering
Electric current and emf
Electric current may be defined as time rate of net motion of an electric charge
across a cross sectional boundary.
Fig: 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦
Rate of transfer of electric energy =Quality of electric charge transfer
Time duration
=𝑑𝑄
𝑑𝑡 (c/sec) Ampere
Electromagnetic Force(emf) is the force that causes on electric current to flow in an
electric circuit while the potential between two point exist. Volt is unit of emf & P.d
(potential difference)
Resistance
It may be defined as the property of a substance due to which it opposes (or
restricts) the flow of electricity (i.e., electrons) through it. Commonly it is denoted by R.
The Unit resistance is ohm(Ω).
A conductor is said to have a resistance of
one ohm if it permits one ampere current to flow
through it when one volt is impressed across its
terminals.
Laws of Resistance
The resistance R offered by a conductor depends on the following factors :
(i) It varies directly as its length L. i.e. RαL ……………………………………..i
(ii) It varies inversely as the cross-section A of the conductor. i.e. Rα1
A……......ii
(iii) It depends on the nature of the material.
(iv) It also depends on the temperature of the conductor.
From Equation i & ii
RαL
A
∴R=ƍL
A ……………………………………………………………………..…iii
Where ƍ is a constant depending on the nature of the material of the conductor and is
known as its specific resistance or resistivity of conductor
If in Eq. (iii), we put L=1m and A =1m2, then R = ƍ
Hence, specific resistance or Resistivity of a material may be defined as the resistance
between the opposite faces of a meter cube (m3) of that material. Its unit is Ωm.
Conductance and Conductivity
Conductance (G)is reciprocal of resistance. Whereas resistance of a conductor measures
the opposition which it offers to the flow of current, the conductance measures the
inducement which it offers to its flow.
G= 1
𝑅 =
1
ƍ𝐿
𝐴
= 1
ƍ𝐴𝐿= σ 𝐴𝐿
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Where σ is conductivity or specific conductance of a conductor. The unit of
conductance is Siemens (S). Earlier, this unit was called mho.
Resistance variation with Temperature
Inferred zero temperature shows that at -t˚c is the temperature at which resistance
would be zero. If the rate of decrease between 100˚c & 0˚c were maintained constant at all
temperature. This doesn’t mean that resistance of the metal is actually zero at that
temperature.
From similarity of triangle
R2
R1=
t0+t2
t0+t1
Where R1 and R2 are resistance at which t1 and t2 respectively.
Temperature Coefficient of Resistance
Let a metallic conductor having a resistance of Ro at 0°C be heated of to 0°Cand let its
resistance
at this temperature be Rt, Then, Rt
R0=
t0+t
t0+0=1+
t
t0
∴Rt=R0(1+α0t)
Where α0=1
t0 is called temperature of coefficient of resistance of material at 0° C
And change in resistance is given by
∆R=Rt-R0
=R0(1+α0t)-R0
=R0α0t
Temperature Coefficient of Resistance is defined as the ratio of increase in resistance
per degree rise in temperature the original resistance.
It is found that the Temperature Coefficient of Resistance depends
(i) directly on its initial resistance
(ii) directly on the rise in temperature
(iii) on the nature of the material of the conductor.
Effect of Temperature on Resistance:
(i) For all pure metals temperature of coefficient is positive i.e. increase the temperature
increases resistance of pure metals.
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(ii) Increase in temperature increase the resistance of alloys, though in their case, the
increase is relatively small and irregular. For some high-resistance alloys like Eureka
(60% Cu and40% Ni) and manganine the increase in resistance is (or can be made)
negligible over a considerable range of temperature.
(iii)Increase in temperature decrease the resistance of electrolytes, insulators (such as
paper, rubber, glass, mica etc.) and partial conductors such as carbon. Hence, insulators
are said to possess a negative temperature-coefficient of resistance.
Ohm's Law , its application and limitation
Ohm’s law states that the ratio between potential difference applied across a
conductor and the current passing through it is always constant and is equal to value of
resistance ot that circuit element provided physical state remains constant.
i.e. V
I =constant=R
where R is known as resistance of conductor.
In figure the resistance R is supplied by
a voltage source. The voltmeter measures
voltage across the resistance and ammeter
measures current passingthrough the resistance.
When voltage across the resistance is increased,
the current through the resistance increases in
same proportion.
Application of ohm’s law
Basic tools for the electric circuit analysis
To calculate resistance of any circuit
To measure current through resistance
Limitation of ohm’s law
It doesn’t apply to the circuit whose temperature varies at different values of current
passing through it.
Resistance in Series
When some conductors having resistances
R1,R2and R 3 etc. are joined end-on-end so that
same current passes through all of them then they
are said to be connected in series
conductors.
V = V1+ V2+ V3= IR1 + IR2+ IR3
But V = IReqv
where R is the equivalent resistance of the series
combination.
IReqv= IR1 + IR2+ IR3
or Reqv= R1 + R2+ R3
Also1
𝐺𝑒𝑞𝑣=
1
G1+
1
G2+
1
G3
As seen from above, the main characteristics of a series circuit are:
1. Same current flows through all parts of the circuit.
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2. Voltage drop across different resistors have their individual voltage drops due to its
different resistance and is given by Ohm's Law
3. Voltage drops are additive.
4. Applied voltage is equals the sum of different voltage drops.
5. Resistances are additive.
6. Powers are additive.
Voltage Divider Rule
Since in a series circuit, same current flows through each of the given resistors, voltage
drop varies directly with its resistance. In above Fig.
Voltage drop across R1 is V1=IR1= V
ReqvxR1
Voltage drop across R2 is V2=IR2= 𝑉
Reqv𝑥R2
Voltage drop across R3 is V3=IR3= 𝑉
Reqv𝑥R3
Resistances in Parallel
When resistances,as joined in such way that current
divides into two or more paths from a node then the
resistors are said to be connected in parallel.
Here
I=I1+I2+I3=V
R1+
V
R2+
V
R3…………………..(i)
We have I= 𝑉
𝑅𝑒𝑞𝑣
Now equation (i) becomesV
Reqv=
V
R1+
V
R2+
V
R3
or1
Reqv=
1
R1+
1
R2+
1
R3
∴Reqv = R1R2R3
R1R2+R2R3+R1R3
The main characteristics of a parallel circuit are :
1. Same voltage acts across all parts of the circuit
2. Different resistors have their individual current.
3. Branch currents are additive.
4. Conductances are additive.
5. Powers are additive.
Current Divider Rule
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For fig:a
I1= IR1
R1+R2
I2= IR2
R1+R2
Power and energy in dc circuit
Let us consider a dc circuit as shown in figure which
consists of resistance R and dc voltage supply V. in the
circuit I flows through load resistance Rl and voltage drives
some charge q through circuit against opposition offered by
resistance Rl
. Energy delivered to load=V.q…………………………(i)
From defination of electric current q= I.t………………(ii)
Now from equation (i)&(ii)
E=VIt…………………………………………………..(iii)
Power is defined as time rate of energy delevering capacity of source.
Power (p)= E
T =
VIt
𝑇 = VI
Since V=IR
P=I2R
Active and passive elements
Active elements are those elements which supply energy to the networks.
Examples: voltage sources like batteries, dc generators etc.
The elements which dissipates or store energy are known as passive elements.
Example: resistors, capacitors etc.
Inductance
Inductance is the property of the circuit elements which oppose the rise or fall of
the current through it by inducing emf across the circuit elements. The circuit having these
property is known as inductor. It is capable of storing finite amount of energy. Generally
a wire wound in form of coil has these property.
Let us consider an inductive coil carrying a time varying
current ‘i’ as shown in fig.
The emf induced in the coil is given by E=Ldi
dt
Where L is inductance of coil and can be expressed as L=edi
dt
∴i=1
L∫ edt
Inductance can be defined as capacity to induce emf per unit time rate of change of
current. A coil having larger value of inductance can induce a large value of emf for given
value of 𝑑𝑖
𝑑𝑡
M=k√𝐿1𝐿2 where k= coupling coefficient
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Inductance in series
Fig:series adding connection Fig:series opposing connection
Let M = coefficient of mutual inductance
L1= coefficient of self-inductance of lst coil
L2= coefficient of self-inductance of 2nd coil
For series adding connection, the total emf induced in each of coils L1& l2 is due to coil
Self inductance & emf induced by other coil called mutual inductance
i.e. e1=L1di
dt+ M
di
dt
similarly
e2=L2di
dt+ M
di
dt
et= (L1 + L2 + 2M)di
dt
Total inductance =𝑒𝑡𝑑𝑖
𝑑𝑡
= L= L1 + L2 + 2M
For series opposing connection
Total inductance = L= L1 + L2 − 2M
Inductanc in parallel
When mutual flux helps the individual flux
L =L1L2 − M2
L1L2 − 2M
When mutual flux opposes the individual flux
𝐿 =L1L2 − M2
L1L2 + 2M
Capacitor
A capacitor is a circuit element which is capable of storing and delivering finite
amount of charges consisting of two conducting surfaces separated by a layer of an
insulating medium called dielectric. The conducting surfacesmay be in the form of either
circular (or rectangular)plates or be of spherical or cylindrical shape. The purpose of a
capacitor is to store electrical energy by electrostatic stress in the dielectric
Capacitance
The property of a capacitor to 'store electricity' may be called its capacitance.In
other word the capacitance of a capacitor is defined as "the ratio of magnitude of the total
charge Q on either side conductor to the potential difference V between the conductor.Its
unit is farad (F)
C=𝑄
𝑉
Q=CV
Differentiating both side with respect to t dq
dt =C
dv
dt i=C
dv
dt
∴C= i
dv
dt
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The voltage across the capacitor of any instant is given by
V=1
c∫ idt
Capacitance in series
V= Q
C
V1= Q
c1 V2=
Q
c2 V3=
Q
c3 (1)
Now,
V=V1+V2+V3
𝑜𝑟,Q
C=
Q
c1 +
Q
c2 +
Q
c3
∴1
C=
1
c1+
1
c2+
1
c3
∴ C = C1C2C3
C1.C2+C1.C3+C2.C3
We can also find the avlue of V1,V2 & V3.from (1) we get
Q=V1C1=V2C2=V3C3 =VC
or,V1C1= VC=VC1C2C3
C1.C2+C1.C3+C2.C3
∴ V1= V𝐂𝟐𝐂𝟑
𝐂𝟏.𝐂𝟐+𝐂𝟏.𝐂𝟑+𝐂𝟐.𝐂
𝟑
∴ V2= V𝐂𝟏𝐂𝟑
𝐂𝟏.𝐂𝟐+𝐂𝟏.𝐂𝟑+𝐂𝟐.𝐂
𝟑
∴ V3= V𝐂𝟐𝐂𝟏
𝐂𝟏.𝐂𝟐+𝐂𝟏.𝐂𝟑+𝐂𝟐.𝐂
𝟑
Capacitance in parallel
V= 𝑄
𝐶 → Q = CV
If Ceqv be equivqlent capacitance then
Q=Q1+Q2+Q3
Or Q=C1V+C2V+C3V
Or 𝑄
𝑉 =C1 +C2+C3
∴ Ceqv=C1 +C2+C3
Power and torque relation
Let us consider a rotating system rotating under a couple of force ‘f’ as shown
below.
Let,
r =radius of rotating disk in rpm
N=speed of rotating disk in rpm
Time for one revolution=60
𝑁
Distance covered in 1 sec= 2πr
Now,
Torque developed(T)=fxr
We know that,
Work done= force x distance
∴ 𝑤 = 𝑓𝑥2πr
∴ power(p) =work done
time taken=
f x 2πr60
N
=Tx2πN
60
∴ 𝑃 =Tx2πN
60
This equation shows that for a constant power of the system, more torque can be
developed at lower speed.
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Kirchoff’s Laws
These laws are more comprehensive than Ohm's law and are used for solving electrical
networks which may not be readily solved by the latter. Kirchoff’s laws, two in number,
are particularly useful
(a) In determining the equivalent resistance of a complicated network of conductors and
(b) For calculating the currents flowing in the various conductors. The two-laws are :
1. Kirchoff’s Point Law or Current Law (KCL)
It states as follows:
In any electrical network, the algebraic sum of the currents meeting at a point (or
junction) is zero.
In another way, it simply means that the total current
leaving a junction is equal to the total current entering
that junction.
Consider the case of a few conductors meeting at
a point A as in Fig. kcl 1 . Some conductors have
currents leading to point A, whereas some have currents
leading away from point A. Assuming the incoming
currents to be positive and the outgoing currents
negative, we have
I1+ (-I2) + (-I3) + (+ I4) + (-I5) =0
I1 - I2-I3+ I4-I5) =0
I1+ I4=I2+I3+ I5
or incoming currents =outgoing currents
Similarly, in Fig. KCL 2,
for node A
+I + (-I1) + (-I2) + (-I3) + (-I4)=0
or I=I1+ I2+ I3+ 14
We can express the above conclusion thus:
∑I= 0 at A junction
2.Kirchoff’s mess law or Voltage Law (KVL)
It states as follows :
The algebraic sum of the products of currents and resistances in each of the
conductors in any closed path (or mesh) in a network & the algebraic sum of the e.m.f’s.
in that path is zero.In other words,
∑IR + ∑e.m.f. = 0 ...round a mesh
It should be noted that algebraic sum is the sum which
takes into account the polarities of the voltage drops.
In Fig: KVL a closed circuit with resistor R1 ,
R2 & R3 are supplied with a voltage source V. I is
current in the circuit. The direction of the current is
always from negative to positive within the battery.
Sign convention
i. +ve to entering point of the resistance and –ve for the
leaving point
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ii.Potential difference measured from the –ve to +ve is called rise in voltage or +ve
voltage drop. Potential difference measured from the +ve to -ve is called fall in voltage or
-ve voltage drop.
Applying kirchhoff’s law
+V+(-V1)+(-V2)+(-V3)=0
Or V-IR1-IR2-IR3=0
∴ V=IR1+IR2+IR3
Note:It is important to note that
1.The sign of the battery e.m.f is independent of the direction of the current through that
branch.
2.The sign of voltage drop across a resistor depends on the direction of current through
that resistor but is independent of the polarity of any other source of e.m.f in the circuit
under consideration.
Use of kirchoff’s law
Determining equivalent resistance of a complecated network of conductor
Calculating current flowing in various conductor
Application of kirchoff’s law using mesh analysis
To perform mesh law analysis the total number of mesh in current is identified
then current for each mesh is given. The direction of all mesh currents are taken as
closewise. The KVL is applied on each mesh and finally the current at various branches
can be calculated by solving the KVL equations obtained for each mesh.
Using KVL in mesh I
V1 − I1R1 − (I1 − I2)R4 = 0
∴ V1 − I1(R1 + R4) + I2R4 = 0 ……………..(i)
Using KVL in mesh II
−(I2 − I1)R4 − I2R2 − (I2 − I3)R5 = 0
∴ I1R4 − I2(R4 + R2 + R5) + I3R5 = 0 ………………(ii)
Using KVL in mesh III
−(I3 − I2)R5 − I3R3 − 𝑉2 = 0
∴ I3(R3 + R5) + I2R5 − 𝑉2 = 0 ……………….(iii)
Solving equation (i),(ii)&(iii) we can find the current at various branches and loop.
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Application of kirchoff’s law using nodal analysis
In nodal analysis, the
number of nodes in an
electrical network is identified
then one of node is regarded
as reference node which has
zero potential. If there are n
nodes, the equation for (n-1)
nodes are obtained. These
nodal equation are then simplified in order to obtain various branches current.
Using KCL in node1
I1 = I2 + I4 V1−VA
R1=
VA−VB
R2+
VA
R4 ……(i)
Using KCL in node 2
I5 = I2 + I3 VB
R5=
VA−VB
R2+
V2−VB
R3 ……(i)
Solving equation (i)&(ii) we get the values of VA and VB and finally the value of currents
at the node can be obtained
Faradays law of electromagnetic induction
Frist Law:It states :
Whenever the magnetic flux linked with a circuit changes, an e.m.f. is always induced in
it.
or
Whenever a conductor cuts magnetic flux.. an e.m.f. is induced in that conductor.
Second Law. It states :
The magnitude of the induced e.m.f. is equal to the rate of change of flux-linkages.
Explanation. Suppose a coil has N turns andflux through it changes from an initial value
of Φ1webers to the final value of Φ2 webers in time t seconds. Then. remembering that by
flux-linkages is meant the product of' number of turns by the flux linked with the coil, we
have.
Initial flux linkages= NΦ1
Final nux linkages= NΦ2
∴induced emf (e) =NΦ2−NΦ1
t web/m or volt
Putting the above expression in its differential form, we get
E= d
dt(NΦ)
E =Nd
dtΦ volt
Induced e.m.f
Induced e.m.f can be either(i) dinamically induced or (ii) statically induced
1.Dynamically induced emf:
Let us consider a conductor AB placed in magnetic field as shown in the figure. When the
conductor is moved upward it will cut the magnetic flux.
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Therefore according to Faradays law of electromagnetic induction, emf will induced in
conductor AB
Let
L=length of conductor
V=velocity of conudctor
B=magnetic flux density
Then
Distance moved in small time dt =dx=Vdt
Area cut by conductor in time dt=𝐿 𝑑𝑥𝐵
Rate of change of flux= L dxB
dt
Hence emf =L dxB
dt=
L VdtB
dt =BLV
If conductor moves making angle θ with direction of magnetic flux then emf
induced is given by
E=BLVsin θ
Statically induced
There is no related relation between conductor and magnet.Coil A is supplied by
Dc voltage source with a variable resistance in series & coil B is connected with
galvanometer. When current through coil A is varied by variable resistance the
galvanometer will show some deflection indicating induced emf in coil B. flux linkage
in coil will change with respect to time hence emf will induced in B according to
faraday’s law of electromagnetic induction
Let coil has N numbers of turns. Magnetic flux changes from Ф1 to Ф2 in t second.
Initial flux linkage =NФ1
Final flux linkage =NФ2
∴ induced emf = rate of change of flux linkage
=NФ2−NФ1
t
E=NdФ
dt
Direction of induced emf
The direction of induced emf is determined by lenz law. According to lenz law the
statically induct ef will drive the current in such a way direction that magnetic flux
produced by induct current will oppose the cause by which emf was induct.
SOLVED NUMERICAL PROBLEMS
Q.No.1. A platinum coil has a resistance of 3.14 Ω at 40°C and 3.767 Ω at 100°C.
find the resistance at 0°C and the temperature – coefficient of resistance at 40° C.
soln,given
R40=3.14 Ω R100=3.767 Ω R0=? α0=?
We have, R100
R40=
1+100α0
1+40α0
or,3.767
3.14=
1+100α0
1+40α0 α0=0.00379per °C
now, α40 =α0
1+40α0=
0.00379
1+40x0.00379
∴ α40 = 3.29x10−3
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Q.No.2. What is the value of unknown
resistor R in fig if the voltage drop across the
500 resistor is 2.5 volts?All resistor are in Ω.
soln
By direct proportion method,
Voltage drop on 50 Ω =2.5 x50
500= 0.25 V
Voltage drop on CD=0.5+2.5=2.75 Ω
Voltage drop across 550 Ω = 12V – 2.75= 9.25 V
Current on 550 Ω resistor= V550
550=
9.25
550= 0.0168A
Current across 50 Ω resistor= 0.25
50= 0.005A
∴ Current across R=0.0168 – 0.005=0.0118A
Hence, R= V
I=
2.75
0.0118= 233 Ω
∴ R = 233 Ω
Q.No.3. Find the equivalent resistance between
point X & Y of the given circuit. Also find how
much current will pass through each
resistance?
soln
The given circuit can be re-arrange as follow
As 60 Ω ,40 Ω &12 Ω are in parallel
1
Reqv=
1
60+
1
40+
1
12=
40x12 + 60x12 + 50x60
60x40x12=
1
8
∴ Reqv=8 Ω
Total current through the circuit (I) = V
R=
24
8= 3A
I1 = 3x40x12
40x12+60x12+60x40=0.4A
I2 = 3x60x12
40x12+60x12+60x40=0.6A
I3 = 3x40x60
40x12+60x12+60x40=2A
Hence 0.4A,0.6A & 2A current will pass through 60 Ω ,40 Ω &12 Ω resistance
respectively.
Q.No.4. Find the value of total current and power in the circuit shown below.
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Reqv = 1 + 2 + 4||4 + 3 + 5 = 3 +4𝑥4
4+4+ 8 = 11 + 2
∴ Reqv = 13 Ω
Now, V=IR
0r,65= 13 x I
∴ I =65
13= 5A
P=VI=65 x 65
13=325w
Q.No.5. Find the resistance between the point A & B shown below.
Soln,
The equivalent resistance between point EF is
𝑅𝐸𝐹 = (6 + 10 + 6)||10 = 22||10 =22x10
22+10= 6.875𝛺
Now equivalent circuit became as in fig:A
The equivalent resistance between CD is given by
𝑅𝑐𝑑 = (6 + 6.875 + 6)||15 = 18.875||15 +18.875x15
18.875+15= 8.357𝛺
The circuit becomes as in fig:B
The equivalent resistance between AB is
RAB = (10 + 8.357 + 6) Ω = 24.357Ω
Q.No.6. If 20V be applied across AB shown in fig. calculate the total current,the
power dissipited in each resistor and the value
of series resistance to have total current.
Soln
As a&b are in parallel their equivalent resistance is
Rab= 2 // 4= 2x4
2+4=
4
3Ω
Also
Rcd= 6 // 8= 6x8
6+8=
24
7Ω
Again Rab & Rcd are in parallel hence
Rabcd= 4
3∥
24
7 =
4
3x24
74
3+
247
= 2425
Ω
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f & g are in parallel hence Rfg=3 ∥ 6 =3x6
3+6=
18
9= 2Ω
now,given circuit becomes as shown in fig beside
now,Reqv=(24
25+ 2) || 5 =
74
25 || 5 =
370
199 M
∴ Reqv= 370
199
Hence total current (Itotal)= V
Reqv=
20370
199
Itotal=10.76A
I1=𝑉
𝑅𝐴𝑀𝐵=
2074
25
=6.76A
I2=10.76 – 6.76=4A
Now voltage drop across Resistance a,b,c& d (Vabcd)=IRabcd=6.76X24
25= 6.48V
Ia=Vabcd
Ra=
6.48
2= 3.24A
Ib=Vabcd
Rb=
6.48
4= 1.62A
Ic=Vabcd
Rc=
6.48
6= 1.08A
Id=Vabcd
Rd=
6.48
8= 0.81A
Ie=V
Re=
20
5= 4A
If=I1𝑋𝑅𝑔
𝑅𝑓+𝑅𝑔= 6.76𝑋
6
3+6= 4.51A
Ig=I1𝑋𝑅𝑓
𝑅𝑓+𝑅𝑔= 6.76𝑋
3
3+6= 2.25A
Power dissipatation
Pa=𝐼𝑎2𝑥𝑅𝑎=3.242x2=20.99W
Pb=𝐼𝑏2𝑥𝑅𝑏=1.622x4=10.4W
Pc=𝐼𝑐2𝑥𝑅𝑐=1.082x6=7W
Pd=𝐼𝑑2𝑥𝑅𝑑=0.812x8=5.25W
Pe=𝐼𝑒2𝑥𝑅𝑒=42x5=80W
Pf=𝐼𝑓2𝑥𝑅𝑓=4.512x3=61W
Pg=𝐼𝑔2𝑥𝑅𝑔=2.252x6=30.4W
The value of series resistance to have total current is Reqv= 370
199 Ω
Q.No.7.Solve the following circuit using kirchoff’s laws(KVL or mesh analysis
method) and find current through each branch.
soln
Using KVL in ABED
+50 − 10I1 − 10I1 − (I1 − I2)10 −20 = 0
0r,30 − 30I1 + 10I2 = 0
∴ 3I1 − I2 = 3 …………………(i)
Using KVL in BCFE
20 − 10(I2 − I1) − 5I2 − 5I2 −40 = 0
0r,−2 − 2I2 + I1 = 0
∴ I1 − 2I2 = 2 ……………….(ii)
Solving equation (i)&(ii)
I1 = 0.8A
I2 = −0.6A
Hence current in resistor of 10Ω in branch AD & AB is given by
I1 = 0.8A
Current in resistor of 10Ω in branch BE is I1 − I2 = 0.8A + 0.6A = 1.4A
Currrent in resistor of 5Ω at branch BC & CF is given by I2 = −0.6A
i.e current flow from F-C & C-B
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Q.No.8. Determine the current supplied by each battery in the circuit shown below
by applying KVL( mesh analysis method).
Soln
Using KVL in mesh I
20 − 5I1 − 3(I1 − I2) − 5 = 0
∴ 8I1 − 3I2 = 15 ………….(i)
Using KVL in mesh II
5 − 3(I2 − I1) − 4I2 + 5 − 2(I2 − I3) + 5 = 0
∴ 9I2 − 3I1 − 2I3 = 15 ………… (ii)
Using KVL in mesh III
−5 − 2(I3 − I2) − 8I3 − 30 = 0
∴ 2I2 − 10I3 = 35 ……..(iii)
Solving equation (i),(ii)&(iii) we get
I1 = 2.56A
I2 = 1.82A
I3 = −3.13A Now,
Current supply by battery B1=I1 = 2.56A (discharge)
Current supply by battery B2=I1 − I2 = 2.56 − 1.82 = 0.74A(current)
Current supply by battery B3=I2 = 1.82A(discharge)
Current supply by battery B4=I2 − I3 = 1.82 + 3.13 = 4.95A(discharge)
Current supply by battery B5=I3 = −3.13A(discharge)
Q.No.9.Solve the given circuit by nodal
analysis method to find the current in
various resistor of the circuit shown in
fig.
soln
The given circuit can be redrawn as in fig
below.
At node 1
V1 (1
2+
1
2+
1
10) −
V2
2−
V3
10= 28
∴ 11V1 − 5V2 − V3 = 280……………(i)
At node 2
V2 (1
2+
1
5+ 1) −
V1
2−
V3
1= 0
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∴ 5V1 − 17V2 + 10V3 = 0……………(ii)
At node 3
V3 (1
4+ 1 +
1
10) −
V2
1−
V1
10= 0
∴ V1 − 10V2 − 12.5V3 = 20 …………………………(iii)
Solving equation (i),(ii) &(iii)
V1=37.32 V2=22.277 V3=19.20
Now,
I1=V1
2=
37.32
3=18.66A
I2=V1−V2
2=
37.32−22.277
2= 7.52A
I3=V1−V3
10=
37.32−19.2
10 = 1.812A
I4=V2−V3
1=
22.277−19.20
1= 3.077A
I5=V2
5=
22.277
5= 4.45A
Q.No.10. Using nodal analysis find the
different branch current in the given circuit.
All branch conductance are in siemen(i.e
mho). Also redraw circuit with values of
current.
soln
In frist node
V1(1 + 2) − V2X1 − V3X2 = −2
∴ 3V1 − V2 − 2V3 = −2………(i)
In second node
V2(1+4) – V1x1=5
∴ V1 – 5V2= – 5………(ii)
In third node
V3(2+3) – V1x2= – 5
∴ 2V1 – 5V3= 5……………(iii)
Solving equation (i),(ii) & (iii) we get
V1= – 3
2 = – 1.5
V2= 7
10 = 0.7
V3= – 8
5 = – 1.6
I1= (V1 – V2)X1=( – 1.5 – 0.7)x1 = – 2.2A
I2= (V3 – V1)X2=( – 1.6+1.5)x2= – 0.2A
I4= V2X4=0.7X4=2.8A
I3= (2A+ I4)=2A+2.8A=4.8A
Q.No.11.find the branch current in the circuit of given fig by using
a.Nodal analysis b.Loop analysis(mesh analysis)
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soln
let
R1=6 Ω R2=2 Ω R3=4 Ω R4=3 Ω R5=4 Ω
E1=6V E2=10V E3=5V
a. Nodal analysis
For node A
VA (1
R1+
1
R2+
1
R3) −
E1
R1−
VB
R2+
E3
R2= 0
VA (1
6+
1
2+
1
4) −
6
6−
VB
2+
5
2= 0
∴ 2VA – VB= – 3…………(i)
For node B
VB (1
R5+
1
R2+
1
R3) −
E2
R3−
VA
R2−
E3
R2= 0
VA (1
4+
1
2+
1
4) −
10
4−
VA
2+
5
2= 0
∴ VA – 2VB= – 10…………(ii)
Solving equation (i) & (ii) we get
VA= 4
3V VB =
17
3V
I1= 𝐸1−𝑉𝐴
𝑅1=
6−(4
3)
6=
𝟕
𝟗A
I2= 𝑉𝐴+𝐸3−𝑉𝐵
𝑅2=
(4
3)+5−(
17
3)
2=
𝟏
𝟑A
I3=𝐸2−𝑉𝐵
𝑅3=
10−(17
3)
4=
𝟏𝟑
𝟏𝟐A
I4=𝑉𝐴
𝑅4=
(4
3)
3=
𝟒
𝟗A
I5=𝑉𝐵
𝑅5=
(17
3)
4=
𝟏𝟕
𝟏𝟐 A
b. Loop analysis
In loop A
– 6I1 – 3(I1 – I2)+6=0
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∴ 3I1 – I2=2 …………….(a)
In loop B
5 – 2I1 – 4(I2 – I3) – 3(I2 – I1)=0
∴ 3I1 – 9I2+4I3= – 5……….(b)
In loop C
– 4I3 – 10 – 4(I3 – I2)=0
∴ 2I2 – 4I3=5………….(c)
Solving eqution (a),(b)&(c) we get
I1= 7
9A I2=
1
3A I3=−
13
12A
The – ve sign of I3 shows that the direction of current is opposite to that shown in fig.
Now the currents at various branch are as follow
Iab=I1= 7
9A
Ibf=I1 – I2= 7
9−
1
3=
4
9A
Ibc= I2= 1
3A
Ice=I2 – I3=1
3+
13
12=
17
12A
Idc= I3=−13
12A
Q.No.12.Three capacitor A, B &C have capacitance 10,50 & 25μF respectively.
Calculate
a) charge on each capacitance when connected in parallel to a 250V supply
b) Total capacitance when connected in parallel
c) P.D across each when connected in series
soln,
a.Q1=C1V
0r Q1=10x250
∴Q1=2500μf
Q2=C2V=50x250
0r, Q2=50x250
∴Q2=12500μf
Q3=C3V
0r, Q3=25x250
∴ Q3=6250μf
b. C= C1+ C2+ C3=10+50+25=85μf
c.1
c=
1
10+
1
50+
1
25
1
c=
25x50+10x25+50x10
10x50x25
C=6.25 μf
Then, total Q=CV=250x6.25=1562.5 μc
V1 =Q
C1=
1562.5
10= 𝟏𝟓𝟔. 𝟐𝟓𝐕
V2 =Q
C2=
1562.5
50= 𝟑𝟏. 𝟐𝟓𝐕
V1 =Q
C1=
1562.5
25= 𝟔𝟐. 𝟓𝐕
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Q.No.13.Find the charges in the capacitor shown below and P.d across them.
soln
Ceqv= 2x(3+5)
(2+(3+5))= 1.6
Now,
Total charge Q=CV=1.6x100=160 μc
V1 =Q
C1=
160
2= 80V
V2 = 100 − 80 = 20V = V3
Q1 = C1V1 = 2x80 = 160μc
Q2 = C2V2 = 3x20 = 60μc
Q3 = C3V3 = 5x20 = 100μc
Q.No.14. Find Ceqv of the given circuit
Soln
CBCD= 1𝑥2
1+2=
2
3 μF
CBE=4+2
3=
14
3μF
CABE=2𝑥
14
3
2+14
3
=1.4 μF
Ceqv=CAF= 3+1.4=4.4 μF
Q.No.15. Two capacitor A & B are connected in series across a 100V supply and it is
observed that the p.d.s across them are 60V and 40V respectively. A capacitor of 2μF
is now connected in parallel with A and pd across B rises to 90V. Calculate the
capacitance of A and B.
soln
As capacitor A & B are
connected in series charge
across them is same so
CAVA= CBVB 60CA=40CB
∴ CA =2
3CB………………(1)
After 2 μF capacitance
connected across capacitor A.
combine capacitor on parallel
is
(CA+2) μF
Again charge is same on series so
(CA+2)x 10=90CB
or (2
3CB+2)= 9 CB
∴ CB=0.24 μF
∴ CA =2
3CB =
2
3x0.24 =0.16 μF
Q.No.16.Two coils with a coeffcient of coupling of 0.5 between them are connected in
series so as to magnetize (a) in same direction (b) in different direction. The
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corresponding values of total inductances are for a) 1.9H and for (b) 0.7H. find the
self inductance of two coils and the mutual inductance between them.
soln
let mutual inductances be L1 & L2
for Case a) L1+L2+2M=L=1.9……..(i)
for Case b) L1+L2 – 2M=L=0.7……..(ii)
subtracting (ii) from (i) we get
M=0.3H
Putting this value in (i) we get
L1+L2+2x0.3=1.9
∴ L1+L2=1.3H ………….(iii)
We have M=k√L1L2
√L1L2 =𝑀
𝑘=
0.3
0.5= 0.6
∴ L1L2=0.36
From (iii) we get (L1 + L1)2 − 4L1L2 =
(L1 − L1)2
1.324𝑋0.36 = (L1 − L1)2
L1 − L1 =0.5……………(iv)
Solving (iii)&(iv)
L1= 0.9H L2=0.4H
Q.No.17.The combined inductance of two coils connected in series 0.6H or 0.1H
depending on the relative directions of the currents in the coils. If one of the coils
when isolated has a self inductance of 0.2 H calculate mutual inductance and
coupling co-efficient.
soln
L1+L2+2M=L=0.6……..(i)
L1+L2 – 2M=L=0.1……..(ii)
subtracting (ii) from (i) we get
M=0.125H
Let L1=0.2H then substituting value of M & L1 in equation (i) we get
L2= 0.15H
Coupling co-efficient= 𝑀
√L1L2=
0.125
√0.2𝑥0.15= 0.72
∴ k=0.72
Q.No.18.Two coils of inductances 4H & 6H are connected in parallel. If their mutual
inductances is 3H, calculate the equivalent inductance of combination if
i) mutual inductance assists the self – inductance
ii) mutual inductances opposes the self – inductances.
soln
i)L=(L1L2−M2)
(L1+L2 – 2M)=
4𝑥6−32
4+6−2𝑥3= 3.75H
ii)L=(L1L2−M2)
(L1+L2+ 2M)=
4𝑥6−32
4+6+2𝑥3= 0.94H
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Chapter three
Network theorems Superposition Theorem
“The current at any point in the linear circuit containing ore than one independent
source can be obtained by superimposing the currents at that point caused by each source
acting alone”
According to this theorem, if there are a number of e.m.fs. acting simultaneously in
any linear bilateral network, then each e.m.f. acts independently of the others i.e. as if the
other e.m.fs. did not exist. The value of current in any conductor is the algebraic sum of
the currents due to each e.m.f. Similarly, voltage across any conductor is the algebraic sum
of the voltages which each e.m.f would have produced while acting singly. In other words,
current in or voltage across, any conductor of the network is obtained by super imposing
the currents and voltages due to each e.m.f. in the network.
Process to use superposition theorem in Numerical
Q. find I in AB
Let E2 be zero, then circuit become as shown in fig:a
RAB=R2 // R3= R2 .R3
R2+ R3
Total resistance = R1+RAB
Current due to E1, I1 =E1
R1+RAB
We know
IxR2=IyR3 → Iy = IxR2
R3
Ix+Iy = I1 →Ix+IxR2
R3= I1→Ix(
R3+R2
R3)= I1
∴Ix= I1R3
R3+R2
Let E1 be zero, then circuit become as shown in fig:b
RAB=R2 // R1= R2 .R1
R2+ R1
Total resistance = R3+RAB
Current due to E2, I2 = E2
R3+RAB
We know
Ix’R2=Iy’R3 → Iy
’= Ix’R2
R3
or, Ix’+Iy
’= I2→Ix’+
I’xR2
R3= I2→Ix
’(R1+R2
R3)= I2
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∴Ix’=
I2R1
R1+R2
Hence current in AB Is
I=Ix +Ix’
SOLVED NUMERICAL PROBLEMS
Q.No.1:By using Superposition Theorem.find the
current in resistance R shown in Fig:1 R1=0.005 Ω,
R2=0.004 Ω, R =1 Ω, EI=2.05 V, E2= 2.15 V
.Internal resistances of cells are negligible.
Solution
Removing E2 the circuit becomes
Resistances of 1Ω and 0.04 Ω are in parallel across poins A and C.
RAC=1//0.04 =1x0.04
1.04=0.038 Ω.
This resistance is in series with 0.05 Ω.
Hence, total resistance offered to battery E1=0.05+ 0.038 =0.088 Ω.
I= 2.05
0.088 = 23.3A.
Current through l Ω resistance,
I1=23.3 x 0.04
1.04 =0.896A from C to A.
When E1 is removed, circuit becomes
Combined resistance of paths CBA and CDA i.e
Reqv of AC =1 //0.05 ==1x0.05
1.05=0.048 Ω
Total resistance offered to E2 is =0.04 +0.048 =0.088 Ω.
Current I= 2.15
0.088 = 24.4 A.
Again, I2=24.4 x 0.05
1.05 = 1.16 A. from C to A
current through 1- Ω resistance when both batteries are present
I= I1+ I2=0.896 + 1.16 =2.056 A.
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Q.No.2:Use Superposition theorem to find current I in
the circuit shown in Fig:b. All resistances are in ohms
Solution.
Replacing voltage source by a short and the 40 A current
sources by an open. And considering only 120A current
source circuit becomes as shown in Fig.1
Using the current-divider rule, we get
I1=120 x 50
150+50 =30 A.
Replacing voltage source by a short and the 120 A current sources by an open. And
considering only 40A current source circuit becomes as shown in Fig.2
Again, using current-divider rule
I2=40 x 150
150+50 =30 A.
Again
Considering only voltage source circuit becomes as shown in Fig.3. Using Ohm's.law,
I3=10
150+50 =0.05 A.
Since I1 and I2 cancel out due to equal and opposite direction,
I=I3= 0.05 A.
Q.No.3: Compute the power dissipated in the 9-Ω resistor of given figure by
applying the Superposition principle. The
voltage and current sources should be treated
as ideal sources. All resistances are in ohms.
Solution.
An ideal constant-voltage sources has zero
internal resistances whereas a constant-current
source has an infinite internal resistance.
When Voltage Source Acts Alone
This case is shown in Fig. 1 where constant-current source has been replaced by an open-
circuit i.e. infinite resistance .
Further circuit simplification leads to the fact that total resistances offered to voltage
source is
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Rtotal =4 + (12//15) =4+ 12x15
12+15 =
32
3Ωas shown in Fig. 4
Hence current Itotal =3232
3
= 3
In fig 2 current is divided into two part. The part going alone AB is the one that also passes
through 9 Ω resistor. Hence current through 9Ω resistor is
r = 12x3
12+15= 4/3 A
When Current Source Acts Alone
As shown in Fig:a,
the voltage source is
replaced by a short-circuit
Further simplification gives
the circuit of Fig b. here 12Ω
and 4Ω are parallel so they
are replaced by their
eqivalent resistor 4x12
4+12 =3Ω
In fig:b, 4A current divides into two parts. By current divider rule . current through 9Ω
resistor is
r’=(3+6)x4
3+6+9 =2A
Since both r and r’flowin the same direction, total current through 9-Ω resistor is
I=r +r' = (4/3)+ 2 = (10/3)A
Power dissipated in 9 Ω resistor is
P=I2R=(10/3)2 x 9 =100 W
Q.No.4: Using Superposition theorem, find the value
of the output voltage Vo in the circuit of given
Fig:4.All resistances are in ohms.
Solution
(a) Replacing 4 A source by an open circuit and 6V
source by a short circuit , given circuit changes as in
fig:a.
Using the current-divider rule, we can find current i1 through the 2 Ω resistor
I1=6 x 1
(l + 2 + 3) = 1A
∴V0l= I1 x R2=1 x 2 =2 V.
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(b)Replacing 6A sources by an open-circuit and 6 Vsource by a short circuit, given circuit
changes as in fig:b. The current i2 can again be found with the help of current-divider rule
I2= 4x(2+1)
(2+3+1)=
12
6= 2A
∴ V02= I2 x R2 =2 x 2 = 4 V
(c)Replacing 6A sources by an open-circuit and 4A source by a open circuit, given circuit
changes as in fig:c.
Voltage drop overR2= 2 Ω resistor =VXR2
1+2+3=
6X2
6 = 2V
The potential of point B with respect to point A is= 6-2 = +4V. Hence. V03=- 4 V.
According to Superposition theorem, we have
V0= V01 + V02+ V03= 2 + 4-4 = 2 V
Q.No.5: Use superposition theorem to determine the voltage v in the network of
given fig.A.
soln
In fig A there
is one
dependent
source v
3
which cannot
be killed or
deactivate
during the
process of
using
superposition
theorem.
Let v1 be voltage across 3 Ω resistor due to 30V source only. So let 5A current source
replace by open circuit & 20V source by short circuit as in fig:B
Applying KCL to node 1
30−v1
6−
v1
3+
v13
−v1
2= 0
∴ v1=6V
Now , let 30V & 20V source replace by short circuit as in fig:C to find v2 . applying KCL
to node 1.
v2
6− 5 −
v2
3+
v23
−v2
2= 0
∴ v2= – 6V
Again let 5A current source replace by open circuit & 30V source by short circuit as in
fig:D to find v3. Applying KVL
– 2i – 20 – 2i – 1
3( – 2i)=0
∴ i=6A
Hence according to ohm’s law the component of v that correspond to 20V source is
v3=2x6=12V
∴v=v1+v2+v3=6 – 6+12=12V
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Q.No.6:Using the superposition theorem find the magnetude and direction of current
through each resistor in given circuit of fig:a
Let us short circuit 24V source then given circuit becomes like circuit as shown in fig:b.
then
Reqv = 3 + (12||6) = 3 +12x6
12+6= 7 Ω
Current through whole circuit (I)= V
R=
18
7A
Current through 3 Ω resistor = IBC=I= 18
7A
Current through 12 Ω resistor = ICA= Ix6
12+6=
18
7X
6
12+6=
6
7A
Current through 6 Ω resistor = ICD= Ix12
12+6=
18
7X
12
12+6=
12
7A
Let us short circuit 18V source then given circuit becomes like circuit as shown in fig:c
Reqv = 12 + (3||6) = 12 +3x6
3+6= 14 Ω
Current through whole circuit(I)= V
R=
24
14=
12
7A
Current through 12 Ω resistor = IAC=I= 12
7A
Current through 3 Ω resistor = ICB= Ix6
3+6=
12
7X
6
3+6=
8
7A
Current through 6 Ω resistor = ICD= Ix12
3+6=
12
7X
3
3+6=
4
7A
Now current due to both source is
Current through 12 Ω resistor = IAC – ICA = 12
7−
6
7=
6
7A(Direction A – C)
Current through 3 Ω resistor = IBC – ICB = 18
7−
8
7=
10
7A(Direction B – C)
Current through 6 Ω resistor = ICD + ICD = 12
7+
4
7=
16
7A(Direction C – D)
Q.No.7:Using the superposition theorem to find the current in R=8 Ω resistance in
branch AB for the circuit shown in fig:A.
soln for fig:A
Reqv I – A= 48||48= 48x48
48+48= 24 Ω
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Reqv E – F= 5||(8+12)=5||20=5x20
5+20= 4 Ω
Now, equivalant circuit of given circuit becomes like circuit as shown in fig:B
Let us short circuit 28V source then the circuit becomes as shown in fig:C
Now,
Reqv= 4 + 4||(6 + 8||24) = 4 + 4|| (6 +8x24
8+24) = 4 + 4||12 = 4 +
4x12
4+12= 7 Ω
I = V
R=
14
7= 2A
By applying current divider rule we get
Current across CA ICA= Ix4
4+(6+8||24)=
2x4
4+(6+8x24
8+24)=0.5A
Current across AB (IAB)= ICAx24
24+8= 0.5 x
24
24+8=
3
8A
Let us short circuit 14V source then the circuit becomes as shown in fig:D
Now,
Reqv= 24 + 8||(6 + 4||4) = 24 + 8|| (6 +4x4
4+4) = 24 + 8||8 = 24 +
4x4
4+4= 28 Ω
I = V
R=
28
28= 1A
By applying current divider rule we get
Current across AC IAC= Ix8
8+(6+4||4)=
1x8
8+(6+4x4
4+4)=
8
16=0.5A
Current across AB (IAB)= I – ICA=1 – 0.5=0.5A
Hence total current in 8 Ω resistor in branch AB is IAB+IAB= 3
8+ 0.5 = 0.875A
Q.No.8:Calculate the voltage drop across the 3 Ω resistor of fig:A using superposition
principle.
Let us make both 15A sourse open circuit to find voltage due to 20V only as shown in
fig:B
Reqv= 6||(1+2)+3= 6𝑥(1+2)
6+1+2+ 3 = 5 Ω
I= 𝑉
𝑅𝑒𝑞𝑣=
20
5= 4A
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∴ current on 3 Ω resistor = IBC=I=4A
∴ voltage across 3 Ω resistor(VBC)= 3x IBC=3x4=12V
Now, make one of 15A source open circuit and 20V source short circuit as in fig:C
The circuit can be re-arranged as in fig:D. now,
Reqv= 1||(6||3)+2=1||(6𝑥3
6+3+ 2) =1||4 =
4𝑥1
4+1=
4
5Ω
V= IReqv=15x4
5 =12V
As we know voltage is divided in series but remain same in parallel, by applying voltage
divider rule we get
Voltage at 2 Ω resistor =V2Ω = 𝑉𝑥2
2+3||6=
12x2
2+(3x6)
3+6
=6V
∴ voltage across 3 Ω resistor (VCB) = V – V2Ω =12 – 6=6V
Now, make another 15A source open circuit and 20V source short circuit as in fig:E
The circuit can be re-arranged as in fig:F. now,
Reqv= 2||(6||3)+1=2||(6𝑥3
6+3+ 1) =2||3 =
2𝑥3
2+3=
6
5Ω
V= IReqv=15x6
5 =18V
As we know voltage is divided in series but remain same in parallel, by applying voltage
divider rule we get
Voltage at 2 Ω resistor =V1Ω = 𝑉𝑥2
1+3||6=
18x1
1+(3x6)
3+6
=6V
∴ voltage across 3 Ω resistor (VBC)= V – V2Ω =18 – 6=12V
Hence net voltage on 3 Ω resistor = VBC – VCB+ VBC=12 – 6+12=18V
Q.No.9: For the circuit given in fig :A
a)determine current I1,I2,I3 when switch S is in position b
b)using the results of part a), determine the same current with switch ZS in position
a.
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a)When s is on b then circuit becomes as in fig:B. which can be redraw as in fig:C
As there’re more than two source let us use superposition theorem.
Frist let us find current due to 130 V source
only by short circuting other voltage
sources as in fig:E
Reqv=2||4+2=2𝑥4
2+4+ 2 =
10
3 Ω
Toal current due to 130V
I130=13010
3
= 39A
∴ I1=39A
∴ I2= Ix4
2+4=39𝑥
4
6= 26A
∴ I3=I – I2=39 – 26=13A
Secondly let us find current due to 120 V
source only by short circuting other voltage
sources as in fig:F
Reqv=2+2||4=2+2𝑥4
2+4=
10
3 Ω
Toal current due to 120V
I120=12010
3
= 36A
∴ I2=36A
∴ I1= Ix4
2+4=39𝑥
4
6= 24A
∴ I3=I2 – I=36 – 24=12A
As 20V is open circuted it doesnot play any role.
Now according to superposition theorem current due to both sources are as follow
I1=39 – 24=15A
I2=36 – 26=10A
I3=10+15=25A
b) When s is on b then circuit becomes as in fig:C. which can be redraw as in fig:G.
now let us find current due to 20V source only by short circuting other voltage sources. As
in fig H
Reqv=2+2||4=2+2𝑥4
2+4=
10
3 Ω
Toal current due to 20V
I120=2010
3
= 36A
∴ I2=6A ∴ I1= Ix4
2+4=6𝑥
4
6= 4A ∴ I3=I2 – I=6 – 4=2A
Hence net current I1,I2 & I3 are as follow
I1=15 – 4=11A
I2=10+6=16A
I3=25+2=27A
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Thevenin’s Theorem
It provides a mathematical technique for replacing a given network, as viewed
from two output terminals, by a single voltage source with a series resistance. It makes the
solution of complicated networks (particularly, electronic networks) quite quick and easy.
Statement
Any two points across a resistance of network can be replaced by an equivalent
voltage source together with an equivalent series resistance. The equivalent voltage
source is equal to open circuit voltage across two points & the equivalent series
resistance is equal to equivalent resistance across two point when looking back into the
network from two points with all source replace by their internal resistances.
In other word “the current IL flowing through a resistance RL connected across any two
terminals of the metwork containing one or more voltage source is given by
IL =Vth
Rth + RL
Where Vth= opern circuit voltage between two terminals with RL connected
Rth=equivalent resistance of network with RL disconnected and the source of
current/voltage replaced with their internal resistance if any. ”
How to Thevenize a Given Circuit?
Steps to thevenize a given circuit.
1. Temporarily remove the resistance (called load resistance RL whose current is required)
2. Find the open-circuit voltage Voc which appears across the two terminals from where
resistance has been removed. It is also called Thevenin voltage Vth.
3. Compute the resistance of the whose network as looked into from these two terminals
afterall voltage sources have been removed leaving behind their internal resistances (if
any) andcurrent sources have been replaced by open-circuit i.e. infinite resistance. It is
also called Thevenin resistance Rth
4. Replace the entire network by a single Thevenin source, whose voltage is Vthand
whose internal resistance is Rth.
5. Connect RLback to its terminals from where it was previously removed.
6. Finally, calculate the current flowing through RLby using the equation.
I = Vth
RL+Rth
Numerical process
For above fig:1
1.calculation of Vth
Removing load resistance RL given circuit became as fig:2
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Vth=VAB = R2
R2+r+R1X E where E=Vs
2. calculation of Rth
Short circuting voltage source(open circuit if current source is
given) the given Circuit became as fig:3
Rth=(R2//R1 + r) =R2XR1+r
R2+R1+r
3.Calculation of IL
Now replacing entire given network(circuit) by a single thevelin
source whose voltage is Vth and resistance is Rth, the thevenin’s equivalent circuit becomes
as shown in fig: 4
Now
IL = Vth
RL+Rth
SOLVED NUMERICAL PROBLEMS
Q.No.1:With reference to the network of Fig:a by applying Thevenin's theorem find
the following:
(i) The equivalent e.m.f of the network when viewed from terminals A and B.
(ii) The equivalent resistance of the network when looked into from terminals A
and.B.
(iii) Current in the load resistance RL of 15
Solution
(i) Current in the network before load resistance is connected [Fig.a]
I=V
∑R =
24
12+3+1 =1.5 A
:. voltage across terminals AB , Voc=Vth=IR= 1.5x 12 =18 V
Hence, so far as terminals A and B are concerned, the network has an e.m.f. of 18 volt
(and not 24 V).
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(ii)There are two parallel paths between points A and B. Imagine that battery of 24 V is
removed but not its internal resistance. Then, resistance of the circuit as looked into
from point A and B is Fig.b
Rth =12//(3+1)=12//4=12X4
12+4=3 Ω
(iii) When load resistance of 15 Ω is connected across the terminals, the network is
reduced tothe thevenin’s equivalent circuitas shown in Fig. c
I = Vth
RL+Rth=
18
15+3= 1A
No.2:Using Thevenin theorem, calculate the current
flowing through the 4 Q resistor of given Fig.
Solution.
(i) Finding Vth
If we remove the 4Ω resistor, the circuit becomes as
shown in Fig.a Since full 10 A current passes through 2Ω
resistor, voltage drop across it is
V=IR=10 x 2 =20 V.
Hence, VB=20 V
The two resistors of 3Ω and 6Ω are connected in series across the 12 V battery. Hence,
drop across 6Ω resistor =12 x 6/(3 + 6) =8 V. ∵ V6=V x R6/(R6+R3)
∴ VA = 8 V
∴ Vth = VBA =VB- VA =20 - 8 =12 V-with B at a higher potential
(ii) Finding Rth
Now, we will find Rth i.e. equivalent resistnace of the network as looked back into the
open circuited terminals A and B. For this
purpose, we will replace both the voltage by a
short circuit and current sources replaced by
an 'open' i.e. infinite resistance. In that case,
the circuit becomes as shown in Fig:b which
is equivalent to fig:c
Rth= 6//3 + 2 = 4Ω. Hence,
Thevenin's equivalent circuit consists of a
voltage source of12V and a series resistance
of 4Ω as shown in Fig:d
When 4Ω resistor is connected across
terminals A and B, as
shown in Fig:e
I= 12/(4 + 4) = 1.5 A-from B to A (∵ I=V/Reqv)
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Q.No.3:Use Thevenin's theorem to find the current
flowing through the 6Ω resistor of the network
shown in Fig:3. All resistances are in ohms.
Solution
For finding Vth
When 6 Ω resistor is removed given circuit reduced
as shown in Fig:a
Whole of 2A current flows along 2Ω in DC
producing a voltage drop
V=IR=(2 x 2) =4 V
As we go along BDCA,
the total voltage = - 4 + 12
=8 V -with A positive w.r.t. B. (∵
12V & 4V are in opposite direction)
Hence, Vth= 8 V
For finding Rth,
12 V voltage source is replaced by a short-circuit and the 2A current source by an
open-circuit, as shown in Fig:b.
The two 4Ω resistors are in series and are thus equivalent to an 8 a
resistance. However, this 8 a resistor is in parallel with a short of 0Ω.
Hence, their equivalent value is 0Ω. Now this 0Ω resistance is in series with
the 2Ω resistor.
Hence, Rth=2 + 0 =2Ω.
The Thevenin's equivalent circuit is shown in Fig:c
:. I = 8
2+6=1 Amp -from A to B
Q.No.4:The given circuit contains two voltage
sources and two current sources. Calculate
(a) Vth and (b) Rth between the open
terminals A and B of the circuit. All
resistance values are in ohms.
Solution.
For finding Vth
It should be understood that since terminals A and
B are open, 2 A current can flow
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only through 4Ω and 10Ω resistors, thus
producing a drop of
V1=(2x10)=20 V across the 10Ω resistor, as
shown in Fig:a
Similarly,
3 A current can
flow through its own closed circuit between A and C
thereby producing a drop
V2= (8x3)=24 V across 8Ω resistor as shown in
Fig.a
Also, there is no drop across 2Ω resistor because no
current flows through it.
Starting from point B and going to point A via points D and C, we get
Vth=V1+V2-V= 20+ 24 - 20 =24 V -with point A positive.
For finding Rth
short-circuit the voltage sources and open-circuit the current sources, as shown in
Fig.b
Rth =RAB =8 + 10 + 2 =20 Ω.
Q.No.5: Calculate the power which would be
dissipated in the 8Ω resistor connected across
terminals A and B of given Fig. All resistance values
are in ohms.
Solution.
For finding Vth
The Thevenin's voltage V this equals the voltage drop
across 10Ω resistor between points C and The with
AB on open-circuit,
120-V battery voltage acts on the two parallel paths EF and ECDF.
Hence,
current through 10Ω resistor is
I10= 120
20+10+20 =2.4 A
Drop across l0Ω resistor,
Vth= 10 x 2.4 =24 V
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For finding Vth
120 V battery is removed. It results in shorting the 40-Ω resistance since internal
resistance of the battery is zero as shown in Fig.a
Rth = 16+10//(20+20)=16+ 10x(20+20)
10+(20+20) +16=40Ω
Thevenin's equivalent circuit is shown in Fig:b
When 8-Ω resistoris connected in terminal AB as in fig:c then
Current through it is
I=24
40+8 =
1
2A
And power dissipated is
∴P = I2 R =(1
2)2x 8 = 2W
Q.No.6: Calculate the current in 8 Ω resistor of given fig:A by using thevenin’s
theorem.
soln
let us take out 8Ω resistor from the circuit then circuit became as shown in fig:B
As 6 Ω resistor is seen from AB the voltage of 6 Ω resistor is Vth
voltage on 6 Ω resistor due to 6V = 6x6
6+2+1=
6𝑥6
9= 4V
total voltage on 6 Ω resistor(Vth)= 12 –4= 8V
now for Rth let us make all voltage source short circuit as in fig:C then
Rth= 6||(2+1) = 6𝑥3
6+3=
18
9= 2 Ω
Hence current in 8 Ω resistor (I)= 𝑉𝑡ℎ
𝑅𝑡ℎ+𝑅𝐿=
8
2+8=
8
10= 0.8A
Q.No.7: Use thevenin’s theorem to calculate the p.d across terminals A & B shown in
fig:A below.
soln
let us take out 3Ω resistor on AB from the circuit then circuit became as shown in fig:B
To find Rth let us make all voltage source short circuit as in fig:C then
Rth= 3||6||6 = 3x6x6
3x6+6x6+6x3= 1.5 Ω
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As 3 Ω resistor is seen from AB the voltage of 6 Ω resistor is Vth
As there are more than one voltage source it’ll be easy to use superposition theorem to
find voltage on 3 Ω resistor.
Frist let us take only 6V source on HG then circuit become as in fig:D
Reqv= 6+6||3=6+ 6x3
6+3= 8 Ω
I = V
Reqv=
6
8=
3
4A
Current on 3 Ω resistor I3=
3
4x6
3+6=
1
2A
Voltage on 3 Ω resistor Vdc= I3x3 =1
2x3 =
3
2 V
Again take only 6V source on EF then circuit become as in fig:E
Reqv= 6||3+6 = 6x3
6+3+ 6 = 8 Ω
I = V
Reqv=
6
8=
3
4A
Current on 3 Ω resistor I3=
3
4x6
3+6=
1
2A
Voltage on 3 Ω resistor Vcd= I3x3 = 1
2x3 =
3
2 V
Again take only 4.5V source on DC then circuit become as in fig:F
Reqv= 3+6||6=3+ 6x6
6+6= 6 Ω
I = V
Reqv=
4.5
6=
3
4A
Current on 3 Ω resistor I3= 3
4
Voltage on 3 Ω resistor Vdc= I3x3 = 3
4x3 =
9
4 V
Now Vth=VDC= VDC – VCD– VCD=3
2−
3
2−
9
4= −
9
4
∴ VCD= 9
4V
Now
Current through 3 Ω resistor on AB (I)= Vth
Rth+RL=
9
4
1.5+3=0.5A
Hence p.d across terminal A&B is given by
VAB= IR= 0.5 x 3 =1.5V
Q.No.8: Compute the current flowing through the load resistance of 10 Ω connected
across terminals A & B in given fig A by using thevenin’s theorem.
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Frist load resistance 10 Ω is taken out from the given circuit as in fig:B. then to calculate
thevenin’s resistance shirt circuit the voltage source as in fig:C
Rth= 5+(10||10||10)+5= 10 +10𝑥10𝑥5
10𝑥5+5𝑥10+10𝑥10= 12.5 Ω
Now, for fig:B
VEF = 9 𝑥10
10+5= 6V
REF= 10||5 = 10𝑥5
10+5=
10
3 Ω
Now circuit changes to Fig:D
VCD = 6 𝑥10
10+10
3
= 4.5V
RCD= 10
3𝑥||10 =
10𝑥10
3
10+10
3
=5
2Ω
Again circuit changes to fig:E
According to thevenin’s theorem
Vth=Vcd=4.5V
Now
Current across 10 Ω load resistor = Vth
Rth+RL=
4.5
12.5+10=
1
5A = 0.2A
Q.No.9: Find thevenin’s equivalent circuit for terminal pair AB for the network
shown below in fig:A
soln
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here 4A current only passes through 6 Ω resistor . So voltage on 6 Ω resistor is
VCA=IR=4x6=24V
20V only regulates on path cfgh so
Voltage on 10 Ω resistor (Vcf)= 20 x 10
10+15= 8V
∴ Vth=VAB= VAc+Vcf = – 24V+8V= – 16V i.e VBA=16V
to find Rth short circuit voltage source and open circuit current source then circuit network
changes as in fig:B
Rth=6+15||10+4= 10+ 15𝑥10
15+10=16 Ω
The thevenin’s equivalent circuit for given network for terminal points A & B is shown in
fig:C
Delta/Star Transformation
Sometimes experiences great difficulty in solving networks (having considerable
number of branches) due to a large number of simultaneous equations that have to be
solved. However, such complicated network can be simplified by successively replacing
delta meshes by equivalent star system and vice versa.
Delta connection to Star connection
Suppose three resistance R12 , R23 and R31 are connected in delta connection
between terminals 1, 2 and 3 as in Fig.a. This delta connected resistances can be replaced
by the three resistances R1 , R2 and R 3 connected in star as shown in Fig.b so that
resistance measured between any pair of terminals is same in both cases.
First, taking delta connection:
Between terminals 1 and 2, there are two parallel paths; one having a resistance of RI2 and
the other having a resistance of (R12+R31).
Resistance between terminals 1 and 2 is =(R12//R23+R31)=R12X(R23+R31)
R12+(R23+R31)…..…….(i)
Taking star connection:
The resistance between the same terminals 1 and 2 is (R1+ R2)……………………….(ii)
As terminal resistances have to be the same
Equating (i) & (ii)
R1+ R2=R12X(R23+R31)
R12+(R23+R31) ……………………….(iii)
Similarly, for terminals 2 and 3 we get
R2+ R3=R23X(R31+R12)
R23+(R31+R12)……………………….(iv)
Similarly, for terminals 3 and 1 we get,
R3+ R1=R31X(R12+R23)
R31+(R12+R23)……………………….(v)
Adding equation (iii),(iv)&(v)
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R1+ R2+ R1+ R2+ R3+ R1= R12X(R23+R31)
R12+(R23+R31)+
R23X(R31+R12)
R23+(R31+R12)+
R31X(R12+R23)
R31+(R12+R23)
2(R1+ R2+ R3)= R12R23+R31R12
R12+R23+R31 +
R23R31+R23R12
R12+R23+R31 +
R31R12+R31R23
R12+R23+R31
2(R1+ R2+ R3) = 2(R12R23+R23R31+R31R12)
R12+R23+R31
∴R1+ R2+ R3 = R12R23+R23R31+R31R12
R12+R23+R31………………………….(vi)
Subtracting (iii) from (vi), we get
R3= R31R23
R12+R23+R31…………………………..………(1)
Subtracting (iv) from (vi), we get
R1= R31R12
R12+R23+R31…………………………………..(2)
Subtracting (v) from (vi), we get
R2= R23R12
R12+R23+R31 ……………..……………………(3)
Above equations are for converting Delta connection to Star connection. Resistance of
each arm of the star is given by the product of the resistances of the two delta sides that
meet at its end divided by the sum of the three delta resistances.
Star connection to Delta connection
This transformation can be easily done by using equations (iii), (iv) and (v) given
above. Multiplying (v) and (iv), (iv) and (v), (v) and (iii) and adding them together and
then simplifying them, weget
R12=R1R2+R2R3+R3R1
R3
R23= R1R2+R2R3+R3R1
R1
R31= R1R2+R2R3+R3R1
R2
The equivalent delta resistance between any two terminals is given by the sum of star
resistances between those terminals plus me product of these two star resitances divide by
the third star resistances.
SOLVED NUMERICAL PROBLEMS
Q.N.1.Calculate the current flowing through the 10Ω resistor of given Fig.
Solution.
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Given circuit has two deltas i.e.ABC and DEF. We can
convert them into their equivalent stars as shown in
Fig.a
Each arm of the delta ABC hasa resistance of 12Ω and
their equivalent star resistance are given by
RAo=ROB=Roc= 12X12
12+12+12 = 4Ω so, and each arm of the
equivalent star has a resistance of 4Ω.
Similarly, each arm of the delta DEF has a resistance of
30Ω and the equivalent star has a resistance
R= 30x30
30+30+30= 10Ωper arm.
The total circuit resistance between A and F is
Reqv=4 + (4+34+10) //(4+10+10) + 10
=14+48//24
=14+48x24
48+24 = 30Ω.
Hence Itotal=V
R =
180
30 =6 A.
Current through 10 Ω resistor as given by current-divider rule is
I10=Itotal xRopposite
Rtotal =6l x
(4+34+10)
(4+34+10) + (10+10+4) =4 A.
Q.N.2.A bridge network ABCD has arms AB, BC, CD
and DA of resistances 1, 1, 2 and 1 Ω respectively. If
the detector AC has a resistance of 1 ohm, determine
by star/delta transformation, the network resistance
as viewed from the battery terminals.
Solution
Delta connection DAC(part of given figure) shown in
fig:a can be reduces to its equivalent star connection as
shown in fig.b by following way
RD= 1x2
1+2+1 = 0.5Ω
RA= 1x1
1+2+1 = 0.25Ω
Rc= 1x2
1+2+1 = 0.5Ω
Hence, given figureis reduced to the one shown in fig.c.
There are two parallel paths between points N and B. Their combined resistance is
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RNB= (0.25+1)//(0.5+1)=(1.25)//(1.5) =1.25𝑋1.5
1.25+1.5=
1.875
2.75= 0.681818Ω
Reqv=RNB +0.5=0.681818Ω+0.5= 1.181818Ω
Q.N.3. Use delta-star conversion to find resistance between terminals 'AB' of the
circuit shown in Fig. 3 .All resistances are in ohms
Solution
First applying delta-star conversion in CGD loop
RC= 2X4
2+4+4=
8
10=0.8Ω
RD= 2X4
2+4+4 =
8
10= 0.8Ω
RG= 4X4
2+4+4 =
16
10=1.6Ω
Applying delta-star conversion in EGF loop
RE= 2X4
2+4+4 =
8
10 =0.8Ω
RF= 2X4
2+4+4 =
8
10 = 0.8Ω
RG= 4X4
2+4+4 =
16
10=1.6Ω
Now new circuit became as shown in fig:a which is Equivalent to another fig b
In fig:b 2Ω& 0.8Ω resistor are in series hence
RCJ=2Ω + 0.8Ω=2.8Ω
REJ=2Ω + 0.8Ω=2.8Ω
Fig:B Circuit became as shown in fig:c
Again We can change CHJ delta connection into Star
connection as follow
Rc=0.8X2.8
0.8+2.8+3.2 = 0.33Ω
RH=0.8X3.2
0.8+2.8+3.2 = 0.376Ω
RJ=3.2X2.8
0.8+2.8+3.2 = 1.32Ω
Circuit of fig:c changes to fig:d.
RAB=5+0.33+(1.32+2.8)//(0.378+0.8)
RAB=5.33+(4.12//1.178) =5.33+ 4.12X1.178
4.12+1.178=5.33 +
0.916
∴RAB=6.245
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Q.N.4. A network of resistances is formed as in given Fig. AB = 9Ω; BC = 1Ω; CA =
1.5 Ω forming a delta and AD = 6 Ω ; BD = 4 Ω and CD =3 Ω forming a star.
Compute the network resistance measured between (i) A and
B (ii) Band C and (iii) C and A.
Solution
The star of given Fig. may be converted into the equivalent delta
and combined in parallel with the given delta ABC.
RAB1=
6x3+3x4+6x4
3= 18Ω
RBC1=
6x3+3x4+6x4
6= 9Ω
RCA1=
6x3+3x4+6x4
4= 13.5Ω
Then given circuit becames changes to fig:a
In fig:a 9Ω &18Ω are parallel in AB,1.5Ω & 13.5Ω are parallel in AC And 9Ω & 1Ω are
parallel in BC. Hence
RAB2=
9x18
9+18= 6Ω
RCA2=
1.5x13.5
1.5+13.5=
27
20Ω
RCB2=
9x1
9+1=
9
10Ω
Again fig:a changes to fig:b
Here
RAB= 6//(27
20+
9
10)=6//2.25=
6x2.25
6+2.25=
18
11 =1.6363Ω
RBC= 9
10//(6+
27
20)= 0.9//7.35=
0.9x7.35
0.9+7.35= 0.8018Ω
RCA=27
20//(6+
9
10)= 1.35//6.9=
1.35x6.9
1.35+6.9= 𝟏. 𝟏𝟐𝟗Ω
Q.N.5. Find the current in 17Ω resistor in the network shown in fig:A by using Delta-
star transformation.
soln
RAC=2+4=6 Ω REG=11+4=15 Ω
Now we can arrange fig:A as in fig:B
let us apply star-delta transformation on ACH & DEG
As RAC= RAH= RCH on ACH then their corresponding delta will also be equal
RA=RC=RH= 6𝑥6
6+6+6= 2 Ω
As RDE= REG= RGD on DEG then their corresponding delta will also be equal
RD=RG=RE= 15𝑥15
15+15+15= 5 Ω
Now circuit will be as in fig:C
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Reqv= 2+(2+41+5)||(2+17+5)+5 = 7+48||24=7 +48𝑥24
48+24=23 Ω
I = 115
23= 5A
Current on 17 Ω resistor = I x 2+41+5
(2+41+5)+(2+17+5)=
5𝑥48
72=
10
3A
Q.N.6. find equivalent resistance between point A& B of given fig:A.
Soln
Given circuit can be re circuted as shown in fig:B. let us apply star-delta transformation in
ACD
RA=2RXR
2R+R+R= 0.5R
RC=2RXR
2R+R+R= 0.5R
RC=RXR
2R+R+R= 0.25R
Now circuit becameas shown in fig:C
Reqv=RAB=0.5R+(0.5R+R)||(0.25R+2R)=0.5R+ 1.5Rx2.25R
1.5R+2.25R= 1.4R
∴ Reqv=1.4R
Maximum power transfer theorem:
It states that “Maximum power is transferred from a source to load when the load
resistance is made equal to the internal resistance of the source as view from the load
terminals when load removed and all exit source are replaced by their interval resistance.”
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In figure Ri is equivalent to thevnin’s
resistance (Rth) and V is equal to
thevnin’s voltage (Vth) so maximum
power is transferred from the circuit to
load when RL is made equal to Ri .
Proof:
Let us consider a circuit shown above
then circuit current is given by
I =V
Ri+RL
Power deliver to load P = I2RL = (V
Ri+RL)
2
RL…….(i)
The value of V and Ri for a given source is constant so power deliver to the load depends
upon RL. To find the value of RL for which maximum value of power is obtained, we can
differentiate equation (i) with respect to RL and set to result zero, then
dp
dRL=
d
dRL((
V
Ri+RL)
2
RL) = 0
V2 [((Ri+RL)2)−2RL(Ri+RL)
(Ri+RL)4] = 0
((Ri + RL)2) − 2RL(Ri + RL) = 0
(Ri + RL)(Ri + RL − 2RL) = 0
Since Ri + RL cannot be zero (Ri + RL − 2RL) = 0 (Ri − RL) = 0
Ri = RL
Proved.
Maximum power transferred(Pmax)=I2RL = (V
Ri+RL)2RL
Pmax = (V
RL+RL)2RL = (
V
2RL)2RL
∴ Pmax =V2
4RL
The voltage across load at maximum power is given by,
Load voltage=IxRL =V
Ri+RLxRL =
V
RL+RLxRL =
V
2RLxRL =
V
2
Q.No.1:Find the value of load resistance RL in given fig below
i)for transfer of maximum power
ii)determine also maximum power
Solution,
Calculation of Vth from fig:A
Vth=120x20
20+40= 40
Calculation of Rth
Making voltage source zero circuit becomes as in fig:B
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Rth = 60 + 20||40 = 60 +20x40
20+40= 60 +
800
60= 73.33
For transfer of maximum power the value of load resistance is given by
RL=Ri=Rth=73.33Ω
The maximum power is given by
Pmax=V2
4RL =
402
4x73.33= 5.45watt
Q.No.2: Calculate the value of Are which will absorb maximum power from the
circuit of fig:A. Also compute the value of maximum power.
soln
Let us remove R and find thevenin’s voltage across A & B as shown in fig:B.
Now lwt us convert 120V source to current source. ∴ I=V
R=
120
10= 12A now circuit
become as in fig:C. by applying KCL we get Vth
10+
Vth
5= 12 + 6
∴ Vth=60V
Now to find Rth open circuit both current source in fig:C then circuit become as in fig:The
Rth= 10||5=10𝑥5
10+5=
10
3 Ω
The thevenins equivalent network is shown in fig:E
According to maximun power transfer theorem, R will absorb maximum power when it
equals to 10
3 Ω . In that case
I=V
Reqv=
6010
3+
10
3
= 9A
Maximum power (Pmax)=I2R=92x10
3=270W
∴ Pmax =270W
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Chapter four Ac fundamentals
Generation of Alternating current and voltage
AC emf may be generated by rotating a coil in a magnetic field. as shown In
Fig.(a) or by rotating of magnetic field within a stationary coil, as shown in Fig.(b). The
value of emf is depends upon the angle between magnetic field and conductor.
In fig.b conductor AB was kept around rotor (magnetic poles). When rotor is
rotated in particular direction the magnetic flux produced by rotor poles cuts conductor
AB continuously. Hence emf will induced in conductor. At various position in rotor poles
the magnitude and direction of flux linkage with conductor AB are different and
accordingly magnitude and direction of emf induced will also change.
Figure besides shows waveform of emf
generated in coil AB. This waveform generater Is
sine wave And can be describe by following
equation
E=Emsinωt
where
ω=Angular velocity of motor
Em=Maximum value of emf
t= time period to generate a cycle of emf
we know,
frequency of emf generated ‘f’ is defined as number of cycle generated in one second
∴ f =1
t
Angular displacement in t second =ωt
or, 2π = ωt
∴ ω = 2π
t= 2πf
Hence E=Emsin(2πft)
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Waveform and terms used in AC Phase :
Phase of an alternating waveform is defined as
the fraction of time period that has elapsed
since the waveform has passed through the
zero position. For example: the phase of
waveform at position A in figure beside is 𝑇
4
or 𝜋
2
Phase difference
In fig beside
ec leads ea by phase of 𝜋
2
eb lags ea by phase of 𝜋
2
If ea= Emsinωt then
ec= Emsin(ωt+ 𝜋
2)
eb= Emsin(ωt - π
2)
Amplitude
The maximum value, positive or negative, of an alternating quantity is known as its
amplitude.
Frequency
The number of cycle per second is called the frequency of the alternating quantity. lts unit
is hertz(Hz). It may be noted that the frequency is given by the reciprocal of the time
period of the alternating quantity.
f= 1
t or t=
1
f
Root-Mean-Square (R.M.S.) Value
The r.m.s. value of an alternaling current is given by that steady (d.c) current which when
flowing through a given time produces the same heat as produced by the alternating
current when flowing through the same circuit for the same time.
Consider an AC i=Imsinωt passing through resister of RΩ as in above fig.
the waveform of current is shown on above right side fig..
Since magnitude of current I is different at instant the heat generated is also
different at different instant. Let us divide time period T into n equal small
divisions/intervals so that magnitude of current during small time interval of t/n is nearly
constant.
Heat energy produced in 1st interval =i12R
T
n
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Heat energy produced in 2nd interval = i22R
T
n
Heat energy produced in 3rd interval = i32R
T
n
.
.
Heat energy produced in 1st interval = in2R
T
n
Total heat energy produced in T sec =( i12+ i2
2+ i32+ i4
2+…….+ in2)R
T
n
Suppose a steady current of magnitude I produced the same amount of heat energy in time
T, then
I2RT=( i12+ i2
2+ i32+ i4
2+…….+ in2)R
T
n
Or,I=√i12 + i2
2 + i32 + i3
2 ………………+ in 2 x
1
n = RMS value of I
Mathematically
Irms=√1
2π∫ i2dωt
2π
0
=√1
2π∫ (Imsinωt)2dωt
2π
0
=√Im
2
2π[ωt −
sin2ωt
2]0
2π
=√Im2
2π[2π −
sin4π
2]
∴ Irms =Im
√2
Where Im= Em
R Similarly, Vrms=
Vm
√2
Average value
The average value of an AC is given by that steady current which transfer across
any circuit the same amount of charge as transferred by that AC during that time. The
charge transferred at any instant is proportional to current at that instant.
Iav =i1+i2+i3………+in
n
or, Iav = 1
π∫ idωt
π
0=
1
π∫ Imsinωtdωt
π
0
or , Iav = Im
π[−cosωt]0
π = 2Im
π
∴ Iav = 2Im
π
Similarly
Vav = 2Vm
π
Phasor representation of sinusoidal waveform
A sinsoidal signal V = Vmsinωt can be reprasented by phasor rotating in
anticlockwise direction with constant angular velocity of ω and with a magnetude equal to
Vm so that vertical component of that rotating phasor at any instant represents instaneous
value of signal at that instant.
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Fig:b shows in waveform of signal V = Vmsinωt.. this signal can be represented by
rotating vector rotating in anticlockwise direction with angular velocity of ω & with a
magnetude equal to |OA|=|Vm|
At instant ωt=0, OA don’t have vertical component so that V=0 at ωt=0
After θ1 rotation the vertical component OA=Oasinθ which corresponds to instaneous
value of V at ωt1=θ1.
It is useful to note here that V=Vmsinωt=imaginary part of Vmejωt
ejωt=cosωt+jsinωt & j=√−1
∴ the equation V = Vmsinωt.. also can be written as
V = ImVmejcost.
Resistance exited by sinusoidal voltage(phasor representation)
Above fig:a shows an ac circuit with a pure resistance R exited by sinusoidal
voltage described by equation
V = Vmsinωt…………………………….. (i)
If I be the instanteneous value of current through circuit applying ohm’s law, we
can write,
i =V
R=
Vmsinωt
R = Imsinωt……………………………….(ii) where Im =
Vm
R
Comparing (i)&(ii)
We can say that I is also ac in nature and phase difference with V. The waveform
and phasor diagram of V and I is shown in fig:b and fig:c respectively.
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Instantaneous power
P=Vi=Vmsinωt x Imsinωt ∴ P = Vm Im(sinωt) 2
Average power:
Pav =1
2π∫ VmImsin2ωtdωt
2π
0
Pav =VmIm2π
∫ sin2ωtdωt
2π
0
Pav =VmIm2π
∫(1 − cos2ωt)dωt
2π
0
Pav =VmIm2π
∫(1 − cos2ωt)dωt
2π
0
=VmIm
2=
Vm
√2xIm
√2= Vrms Irms
∴ Pav =VmIm
2= Vrms Irms
Inductance exited by sinusoidal voltage(phasor representation)
Above fig:a shows an ac circuit with a pure inductance L exited by sinusoidal
voltage described by equation
V = Vmsinωt If e be the instanteneous value of emf induced then, we can write,
e =Ldi
dt…………………(i)
from figure e=Vmsinωt ……………………………….(ii)
From (i)&(ii) Ldi
dt= Vmsinωt
or,i =Vm
L∫ sinωtdt
or, i =Vm
Lx(−
cosωt
ω)
∴ i = −Vm
Lω x sin(ωt − 90) ………(iii)
Lω = 2𝜋𝑓𝑙 =inductance resistance
-ve sign implies waveform starts from –ve direction.
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Comparing equation (i) & (iii)
We can say that I is also ac
in nature and phase difference with
e.here I lags V by 90° The
waveform and phasor diagram of V
and I is shown in fig:b & c
Instantaneous power
P=Vi=Vmsinωt x Imsin (ωt − 90)
∴ P = −Vm Im
2cos2ωt
Average power:
Pav =1
2π∫ −
Vm Im2
cos2ωt
2π
0
= 0
Therefore pure inductance doesnot
consume power. The plot of instanteneous power versus ωt is shown in fig:b
Capacitor exited by sinusoidal voltage(phasor representation)
Above fig:a shows an ac circuit with a pure capacitor exited by sinusoidal voltage
described by equation
V = Vmsinωt The instanteneous value of voltage across the capacitor is given by
𝑉𝑐 =q
C …………………………….. (i)
At any instant,
q=VC
∴q=Vmsinωt.C
Differentiating both side with respect to time we get, dq
dt= C d
(Vmsinωt)
dt
or, dq
dt= CωVmcosωt
or,𝑖 =Vm1
ωC
sin (ωt + 90)
∴ i = Imsin (ωt + 90) ……….(ii)
Where Im =Vm
Xc
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and Xc =1
ωC=
1
2𝜋𝑓𝑐 is known as capacitance reactance of the capacitor.
Comparing (i)&(ii)
We can say that i is also ac in nature and phase difference with e.here i leads V by
90° The waveform and phasor diagram of V and I is shown in fig:b & fig:c respectively
Instantaneous power
P=Vi=Vmsinωt x Imsin (ωt + 90)
∴ P = −Vm Im
2sin2ωt
Average power:
Pav =1
2π∫
Vm Im2
sin2ωt
2π
0
= 0
Therefore pure capacitor doesnot consume
power. The plot of instanteneous power
versus ωt is shown in fig:11.64
R-L series circuit exited by Sinusoidal voltage(phasor representation)
Above fig:a shows an AC circuit with a resistance connected in series with an
inductance & supplied by an AC emf.
Suppose
V=RMS value of applied voltage
VR=voltage across R
VL=voltage across L
R=Resistance in ohm
L=Inductance in henry
I=RMS value of current through circuit
XL=Induca
nce
reactance=
2πfL
Now,
voltage drop in resistance ‘R’ is VR = IR , there is no
difference between VR & I
Voltage drop in inductance ‘L’ is VL = IXL , I lags V by 90
From kirchoff voltage law
V = VR + VL
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From phaser diagram
V = √VR2 + VL
2
Or, V = √I2R2 + I2XL2 = I√R2 + XL
2
Or,I =V
√R2+XL2
∴ I =V
Z where 𝑍 = √R2 + XL
2 known as impedence of circuit.
Here I lags V by same angle Ф
Where, tan∅ =VL
VR=
IXL
IR=
XL
R
∴ ∅ = tan−1(XL
R)
Instantaneous voltage and current
V = Vmsinωt
i = Imsin (ωt − ∅) where Im =Vm
Z
Waveform of V and I is shown in fig:c
Power in circuit
Let us divide current I into two
component in phasor diagram as
IcosФ=component of I in phase with V
IsinФ=component of I perpendicular to V
Now,
Active power(P)=V x IcosФ
=IZIR
z=I2R
It is responsible for producing heat.
Reactive power(s) =V x IsinФ
=IZI𝑋𝐿
𝑧= 𝐼2𝑋𝐿var
The product of V and total current I is called Volt
ampere of current & Is given by
𝑄 = 𝑉 𝑥 𝐼 = 𝐼2𝑧 = √𝑃2 + 𝑆2
It is also known as apparent power of the circuit.
Power factor
Among two component of power active power ‘p’ depends upon a factor cosФ
which is known as power factor of circuit.
If 𝑅 = 𝑋𝐿, ∅ = tan−1 (XL
R) = 45° & power factor=0.707
( Active power=reactive power)
If 𝑅 > 𝑋𝐿, ∅ = tan−1 (XL
R) < 45° & power factor>0.707
( Active power>reactive power)
If 𝑅 < 𝑋𝐿, ∅ = tan−1 (XL
R) > 45° & power factor<0.707
( Active power<reactive power)
If 𝑋𝐿 = 0, ∅ = tan−1 (XC
R) = 0° & power factor= 1(max)
Reactive power=0
If 𝑅 = 0, ∅ = tan−1 (XC
R) = 90° & power factor= 0(min)
Active power=0
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R-C series circuit exited by Sinusoidal voltage(phasor representation)
Above fig:a shows an AC circuit with a resistance connected in series with an
capacitance & supplied by an AC emf.
Soppose
V=RMS value of applied voltage
R=Resistance in ohm
VR=voltage across R
Vc=voltage across C
C=Capacitance in Farad
I=RMS value of current through circuit
Xc=Capacitive reactance=1
2𝜋𝑓𝑐 in ohm
Now,
voltage drop in resistance ‘R’ is VR = IR , there is no difference between VR & I
Voltage drop in capacitor ‘C’ is VL = IXc , I leads V by 90
From kirchoff voltage law
V = VR + Vc
From phaser diagram
V = √VR2 + Vc
2
Or, V = √I2R2 + I2Xc2 = I√R2 + Xc
2
Or,I =V
√R2+Xc2
∴ I =V
Z where 𝑍 = √R2 + Xc
2 known as impedence of circuit.
From phasor diagram we can say that I leads V by
same angle Ф
Where, tan∅ =Vc
VR=
IXc
IR=
XC
R
∴ ∅ = tan−1(XC
R)
Instantaneous voltage and current
V = Vmsinωt
i = Imsin (ωt + ∅) where Im =Vm
Z
Waveform of V and I is given besides
Power in circuit
Let us divide current I into two component in phasor diagram as
IcosФ=component of I in phase with V
IsinФ=component of I perpendicular to V
Now,
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Active power(P)=V x IcosФ
=IZIR
z=I2R
It is responsible for producing heat.
Reactive power(s) =V x IsinФ var
=IZIXC
z= I2XCvar
The product of V and total current I is called Volt
ampere of current & Is given by
Q = V x I = √P2 + S2
It is also known as apparent power of the circuit.
Power factor
Among two component of power active power ‘p’ depends upon a factor cosФ
which is known as power factor of circuit.
If R = Xc ,then ∅ = tan−1 (XC
R) = 45° & power factor=0.707
( Active power=reactive power)
If R > XC , then ∅ = tan−1 (XC
R) > 45° & power factor>0.707
( Active power>reactive power)
If R < XC , then ∅ = tan−1 (XC
R) < 45° & power factor<0.707
( Active power<reactive power)
If XC = 0, ∅ = tan−1 (XC
R) = 0° & power factor= 1(max)
Reactive power=0
If R = 0, ∅ = tan−1 (XC
R) = 90° & power factor= 0(min)
Active power=0
R-L-C series circuit exited by Sinusoidal voltage(phasor representation)
Above fig:a shows an AC circuit with a resistance connected in series with an
capacitance and inductance & exited by an AC voltage source V = Vmsinωt .
Suppose
V=RMS value of applied voltage
R=Resistance in ohm
VR=voltage across R
VL=voltage across L
VC=voltage across C
C=Capacitance in Farad
I=RMS value of current through circuit
XL=Inducance reactance=2πfL
Xc=Capacitive reactance=1
2πfc in ohm
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Case 1
Let, XL >Xc i.e VL>VC
For those phasor diagram will be as shown in fig below
Here I lags V by angleФ
Where,
∅ = tan−1(XL−XC
R)
Total voltage Is given by
V2 = (VL − VC)2 + VR
2
Or,V = √VR2 + (VL − VC)
2
Or, V = √I2R2 + I2(XL − XC)2
Or,𝑉 = I√R2 + (XL − XC)2
Or,I =V
√R2+(XL−XC)2
∴ I =V
Z where Z =
√R2 + (XL − XC)2 known as Impedence of circuit.
Case 2
Let, XC >XL i.e VC>VL
For those phasor diagram will be as shown in fig
below
Here I leads V by angle Ф
Where,
Ф = tan−1(XC−XL
R)
Total voltage Is given by
V2 = (VC − VL)2 + VR
2
Or,V = √VR2 + (VC − VL)
2
Or, V = √I2R2 + I2(XC − XL)2 =
I√R2 + (XC − XL)2
Or,I =V
√R2+(XC−XL)2
∴ I =V
Z where Z = √R2 + (XC − XL)
2 known
as Impedence of circuit.
Power in circuit
Let us divide current I into two component in phasor diagram as
IcosФ=component of I in phase with V
IsinФ=component of I perpendicular to V
Now,
Active power(P)=V x IcosФ
Reactive power(s) =V x IsinФ var
Summary of result of AC circuit
Types of impedance Value of impedance Phase
angle for
current
Power
factor
Resistance only R 0° 1
Inductance only ωL 90°lags 0
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Capacitance only 𝟏
𝛚𝐂
90° leads 0
Resistance+inductance √𝐑𝟐 + (𝛚𝐋)𝟐 0°<Ф<90°
lag
1>p.f>0
lag
Resistance+capacitance
√𝐑𝟐 + (𝟏
𝛚𝐂)𝟐
0°<Ф<90°
lead
1>p.f>0
lead
Resistance+inductance+capacitance
√𝐑𝟐 + (𝛚𝐋 −𝟏
𝛚𝐂)𝟐
Or
√𝐑𝟐 + (𝟏
𝛚𝐂− 𝛚𝐋)
𝟐
Between
0° and
90°
lags
Or
lead
Between
0 and
unity
lags
Or
lead
AC parallel circuit
Above figure show a typical ac circuit. Path 1 has a resistance ‘R’ in series with an
inductance ‘L’ and Path 2 has a resistance ‘R’ in series with an capacitor ‘C’. Total
combination is supplies by an AC voltage source.
Suppose,
V=RMS value of applied volage
I1=RMS value of current through path-1
I2=RMS value of current through path-2
XL=Inducance reactance=2πfL
Xc=Capacitive reactance=1
2πfc in ohm
Here,
In path-1 I1 lags V by Ф1,where Ф1 = Tan−1 (XL
R1)
In path-2 I2 leads V by Ф2,where Ф2 = Tan−1 (Xc
R2)
The magnitude of I1 & I2 are given by:
I1 =V
Z+∅1 where Z1=√R1
2 + XL2 and ∅1 = Tan−1 (
XL
R1)
I2 =V
Z+∅2 where Z1=√R1
2 + XC2 and ∅1 = Tan−1 (
Xc
R2)
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Phasor diagram is given below
I1cosФ1 = component of I1 in phase with V= active component of I1
I1sinФ1 = component of I1 which lags V by 90= reactive component of I1
I1cosФ2 = component of I2 in phase with V= active component of I2
I1sinФ2 = component of I2 which leads V by 90°= reactive component of I2
Net active component IX = I1cosФ1 + I1cosФ2
Net reactive component IY = I1sinФ1 + I1sinФ2
Then total current I is given by Phasor sum of Ix & IY as shown in phasor diagram. The
magnitude of total current I is given by
I = √(I1cosФ1 + I1cosФ2)2 + (I1sinФ1 + I1sinФ2)
2 = √IX2 + IY
2
Here I lags V by an angle Ф1. Where Ф1 = Tan−1 (IY
IX)
Total active power(P)=V x IcosФ
Total reactive power(s) =V x IsinФ
Resonance in R-L-C series circuit
Above figure shows R-L-C series circuit supplied by an ac voltage source.
Suppose,
L=inductance in henry
C=capacitance in farad
ω= Angular frequency in radian/sec
The reactance of inductor and capacitor depends upon the frequency of applied voltage
and are given by Inductive reactance, XL = 2πfL = ωL
capacitive reactance , XC =1
2πfC=
1
ωCΩ ………………(i)
Now net reactance of circuit=XL − XC &
Total impedence Z = √R2 + (XL − XC)2
From equation (i) it is clear that if frequency decreases then inductive reactance decreases
and capacitive reactance increases and vice versa.
At particular frequency net reactance becomes zero i.e.
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XL − XC = 0 & Z=R, power factor is unity,the current I is maximum and
I =V
R . Such condition is known as resonance in R-L-C circuit. The frequency at which
this condition occures is known as resonance frequency.
XL − XC = 0 → XL = XC → 2πf0L =1
2πf0C
∴ f0 =1
2π√LC
Above fig shows variation of XL, XC & Z with respect to
frequency f. At resonance frequency f0 net reactance
X=0
Figure given in right side shows variation of current in
circuit w.r.t frequency ‘f’. At resonance frequency ‘f0’
net reactance, x=0
i.e
XL = XC
At frequency f<f0 , XL > XC,net impedence is capacitive
& power factor is leading.
At frequency f>f0 , XL < XC,net impedence is inductive
& power factor is lagging.
f1 & f2 are two frequencies at which I =I0
√2
power at these two frequencies =(I0
√2)
2
R =1
2I02R=half the power at fo
These two frequencies f1 & f2 are known as half power frequency. The value of f1
& f2 are given below
f1 = f0 −R
4πL f2 = f0 +
R
4πL
The range of frequencies between f1 & f2 is known as bandwidth of circuit. This
indicates that the signal within these range of frequency will easily pass through this
circuit and signal of frequency below f1 & above f2 will be attenuated by this circuit
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Resonance in R-L-C parallel circuit
Above figure shows R-L-C parallel circuit supplied by an ac voltage source.
Suppose,
L=inductance in henry
C=capacitance in farad
R=resistance in ohm
ω= Angular frequency in radian/sec
Impedence of path I, Z1 = R + j(XL) = R + j(2πfL)
Current through path I, IL =V
Z1 =
V
R+j(2πfL)
Impedence of path II, Z2 = −j(XC) = −j1
(2πfC)
Current through path II, IC =V
Z2 = −j
V
(2πfC)
IL lags V by Ф=tan−1(XC
R) and IC leads V by 90°
The total current is phasor sum of IL and IC
At lower frequency XL is low and Xc is high.
∴ IL > IC
At higher frequency XL is high and Xc is low.
∴ IL < IC
At particular frequency Ic = ILsinФL so that total current I is in phase with V resulting
unity power factor as in fig c shown above. This condition is known as resonance in
parallel ac circuit.
At this condition
Ic = ILsinФL
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V
XC=
V
Zx
XL
Z → XLXC = Z2
or, 2πf0l x1
2πf0C= Z2 →
L
C= Z2 = R2 + XL
2
or,L
c= R2 + (2πf0l)
2 →L
C− R2 = (2πf0l)
2
or, √L
C− R2 = 2πf0l
∴ f0 =1
2π√
L
C− R2
Which is required equation for resonance frequency
Above figure shows resonance curve for parallel ac circuit. The current is minium at
resonance frequency ‘fo’ and equal to VR
Z2
Solved numerical problems
Q.No.1. A coil having Resistance of 4Ω and inductance of 9.55mH is connected in
series across a supply of 240V, 50Hz. Calcuate
a) Inductive reactance
b) Impedance
c) Phase angle
d) Total current
e) Power factor
f) Voltage across resistor
g) Voltage across inductor
h) Active power
i) Reactive power
j) Apparent power
k) Draw a phasor diagram
Soln,
Given
Resistance (R)=4Ω
Inductance (L)=9.55mh=9.55 x 10-3H
Voltage(V)=240V
Frequency (F)=50Hz
Now,
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a) Inductive reactance(XL)=ωL=2πfL= 2πx50 x 9.55 x 10-3=3Ω
b) Impedance(Z)=√R2 + XL2=√42 + 32 = √25 = 5 Ω
c) Phase angle( Ф)=tan−1 (XL
R) = tan−1 (
3
4) = 36.86°
d) Total current ( I )=V
Z=
240
5= 48A
e) Power factor = cos Ф=cos36.86° =0.80010 (lagging)
f) Voltage across resistor (VR)=IR=48 x 4=192V
g) Voltage across inductor(VL)=IXL=48 x 3 = 144 V
h) Active power (P)=I2R=(48)2x4=9216watt
i) Reactive power(Q)= I2XL=(48)2x3=6912VAR
j) Apparent power (S)= V x I=240x48=11520 or,√P2 + Q2 = 11520 VA
k) A phasor diagram is
Q.No.2. For a circuit below calculate
a) Power factor
b) RMS value of current
c) Frequency of supply voltage
d) Active power and reactive power
e) Expression for instantaneous
current in circuit
Soln,given
V=400sin314t
Vm=400
ω =314
R=15 Ω
L=0.1H
Now,
XL=ωL=314 x 0.1=31.4 Ω
Z=√R2 + XL2 = √152 + 31.42 = 34.8 Ω
Ф = tan-1(XL
R) = tan−1 (
31.4
15) = 64.47°
a) Power factor (pf)= cosФ=cos64.47° =0.43
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IM= Vm
Z=
400
34.8= 11.49A
b) RMS value of current(Irms) =Im
√2=
11.49
√2= 8.13A
We have
ω =2𝜋f
c) ∴ Frequency of supply voltage (f)=ω
2π=
314
2π= 50Hz
d) Active power(p)= I2R= (8.13)2 x 15 =991.4535watt
Reactive power (Q)= I2XL=(8.13)2x 31.4=8685.132VAR
e) i= IMsin( ωt – Ф)= 11.49sin( 314t – 64.47°)
Q.No.3.A coil takes a current of 6A when connected to a 24V d.c supply. To obtain
the same current with a 50-hz a.c supply the voltage required was 30V. calculate
inductance of coil and power factor of coil.
Soln
As coil offers resistance in direct voltage & impedance to Ac voltage
R=24
6= 4 Ω
Z=30/6=5 Ω
∴ XL=√𝑍2 − 𝑅2 = √52 − 42 = 3 Ω
Power factor= 𝑅
𝑍=
4
5= 0.8(lag)
Q.No.4. A capacitor is connected in series with a 40 Ω resistor across a supply of
frequency 60Hz. A current of 3A flows and the circuit impedence is 50 Ω .Calculate
a) The value of capacitance
b) Supply voltage
c) Phase angle
d) P.d across resistor and capacitor
e) Power factor
f) Active power ,reactive power and
apparent power
g) Draw the phasor diagram
soln, given
Resistance(R) =40 Ω
Current (I)=3A
Frequency(f)=60Hz
Impedence (Z)=50Ω
Now,
Z=√R2 + Xc2 ↔ Xc = √Z2 − R2 =
√502 − 402 = 30 Ω
Xc =1
ωc=
1
2πfc
∴ c =1
2πfXc=
1
2πx30x60= 88.42μf
a) ∴ The value of capacitance ( c)=88.42μf
b) Supply voltage(V)=IZ=3x50=150V
c) phase angle (Ф)=tan−1 (Xc
R) =
tan−1 (30
40) = 36.87°
d) p.d across resistor (VR)=IR=3x40=120V
p.d across capacitor (Vc)=IXc=3x30=90V
e) power factor = cos Ф =cos36.87°=0.8 (leading)
f) Active power(p)= I2R= (3)2 x 40 =360watt
Reactive power (Q)= I2XL=(3)2x 30=270VAR
A apparent power (S)= VI=150x3=450VA
g) The phasor diagram is shown on fig:1
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Q.No.5. A resistance of 100 Ω , inductance of 100mH and capacitance of 100 μf are
connected in series and supplied by an ac voltage source of 200 volts. Calculate
a) Resonance frequency
b) Half power frequency
c) Current at resonance
d) Bandwidth
soln,given
R=100Ω
L=100mH=100 x 10-3H
C=100 μf =100 x 10-6f
V=200V
Now,
a) Resonance frequency (f0)=1
2π√fc
=1
2π√100x10−3x100x10−6 = 50Hz
b) Half power frequency
f1 = fo −R
4πL= 50 −
100
4πx100x10−3 = −29.577Hz
f2 = fo +R
4πL= 50 +
100
4πx100x10−3 = 129.577Hz
c) Current at resonance (Io)=V
R=
200
100= 2A
d) Bandwidth =f2-f1=129.577 − (−29.577) = 159.154Hz
Q.No.6. A coil with resistance of 50 Ω and inductance of 0.318H is connected in series
with a 159 μf capacitor. The resulting circuit is connected across a 220V,50Hz ac
supply. Find
a) P.f of coil and p.f of circuit
b) Circuit current
c) Active ,Reactive & Apparent power
d) Voltage across the coil and capacitor
Soln,
XL=ωL=2πfL= 2πx50x0.318=100 Ω
Xc =1
ωc=
1
2πfc=
1
2πx50x159x10−6 = 20 Ω
∴ Фcoil=tan−1 (XL
R) = tan−1 (
100
50) = 63.43
p.f of coil=cosФcoil=cos63.43=0.447
∴ Фcircuit=tan−1 (XL−Xc
R) = tan−1 (
80
50) = 58°
p.f of circuit=cosФcircuit=cos58=0.57
a) ∴ p.f of coil and circuit are 0.447 & 0.57 respectively
b)Circuit current (I)=V
Z=
V
√R2+(XL−Xc)2=
220
√502+(100−20)2= 2.33A
c)Active power (P)=I2R=(2.33)2x50=271.445watt
Reactive power (Q)=I2(XL-Xc)= (2.33)2x (100-20)=434.312VAR
Apparent power (s)=VxI=220x2.33=512.6VA
Now,
VR=IR=2.33x50=116.5V
VL=IXL=2.33x100=233V
VC =IXc=2.33x20=46.6V
d)Voltage across coil(Vcoil) = √VR2 + VL
2 = √116.52 + 2332 = 260.509V
Voltage across capacitor(Vcapacitor) =Vc=46.64 V
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Q.No.7. Two impedance Z1=(10+j15) Ω & Z2 =(6-j8)Ω are connected in parallel.
The current supplied is 15A. what is the power taken by each branch?
soln
Z1=(10+j15) Ω =√102 + 152 = 18 Ω
Z2 =(6-j8)Ω = √62 + 82 = 10 Ω
∴ I1 =IxZ2
(Z1+Z2)= 15 x
10
18+10=
150
28= 5.357A
∴ I2 = IxZ1
(Z1+Z2)= 15 x
18
18+10=
270
28= 9.64 A
Power in branch 1 is P1
∴ P1= I12Z1 = 5.3572x18 = 516.55watt
Power in branch 2 is P2
∴ P2= I22Z2 = 9.642x10 = 929.296watt
Q.No.8. For given circuit, calculate
a) Power factor
b) Rms value of current
c) Impedence of the circuit
d) Active power and reactive power
e) Calculate the value of C for which p.f.=1
soln
XL=ωL=2πfL= 2πx50x100x 10-3=31.42 Ω
Xc =1
ωc=
1
2πfc=
1
2πx50x100x10−6 = 31.83 Ω
Impedence of whole circuit
Z =R + j XL − j Xc = 12 + j 31.42 − j 31.83 = 12 − j 0.41
Z=√122 + 0.412 = 12.007 Ω
The current through the circuit
I =V
Z=
200
12.007= 16.65 A
Now
a)power factor =cos Ф =cos1.957 (lead)
b)Irms=16.65A
c)Impedence =12.007
d) Active power =VI cosФ=200x16.65xcos1.957=3328.06 watt
Reactive power = VI sinФ=200x16.65xsin1.957=113.71VAR
e)For p.f=1
XL=XC
0r, XL= 1
2πfc
0r,C=1
2𝜋𝑓𝑋𝐿=
1
2𝜋𝑥50𝑥31.42= 101.32
∴ The value of C for p.f=1 is 101.32 μf
Q.No.9. For given circuit, calculate
a) Magnitude and phase of I1 and I2
b) Active power and reactive power of path-I
c) Active power and reactive power of path-II
d) Magnitude & power of total current I
e) Draw phasor diagram showing V,I,I1&I2
soln given,
R1=100 Ω L=100mH
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R2=100 Ω C=200 μF V=200V f=50Hz
Now,
XL=ωL=2πfL= 2πx50x100x10-3=31.415 Ω
Xc =1
ωc=
1
2πfc=
1
2πx50x200x10−6 = 15.915 Ω
Z1= √𝑅12 + 𝑋𝐿
2 = √1002 + 31.4152 = 104.818 Ω
Z2= √𝑅22 + 𝑋𝑐
2 = √1002 + 15.9152 = 101.258 Ω
a)Magnitude of I1=𝑉
𝑍1=
200
104.818= 1.908A
Phase of I1= Ф1=tan−1 (𝑋𝐿
𝑅) = tan−1 (
31.415
100) = 17.30°
I1 lags V by Ф1=17.30°
Magnitude of I2=𝑉
𝑍2=
200
101.258= 1.975A
Phase of I2= Ф 2=tan−1 (𝑋𝑐
𝑅) = tan−1 (
15.915
100) = 9.042°
I2 leads V by Ф2,where Ф2 = 9.042°
b)Active power for path I=𝑉𝐼1𝑐𝑜𝑠Ф1 = 200x1.908 cos(17.30) = 364.3367watt
Reactive power for path I= 𝑉𝐼1𝑠𝑖𝑛Ф1 = 200x1.908sin (17.30) = 113.478VAR
c)Active power for path II=𝑉𝐼2𝑐𝑜𝑠Ф2 = 200x1.975 cos(9.042) = 390.091watt
Reactive power for path II= 𝑉𝐼2𝑠𝑖𝑛Ф2 = 200x1.975sin (9.042) = 62.077VAR
d)I=I1+I2=1.908+1.975=3.883A
Q.No.10. A voltage V=100sin314t is applied to a circuit consisting of 25 Ω resistor
and an 80μF capacitor in series. Determine
a)an expression for the value of the current flowing at any instant
b)the power consumed & the p.d. across the capacitor at the instant when the current
is one-half of its maximum value.
soln given,
R=25Ω C=80 μF Vm=100
∴ Xc =1
ωc=
1
2πfc=
1
314x80x10−6 = 39.8 Ω
Z= √252 + 39.82 = 47 Ω
Im=Vm/Z=100/47=2.13A
Ф =tan – 1(39.8/25)=57°52’ 1.01 radian(lead)
Hence equation for instanteneous current
i= 2.13sin(314t+1.01)
power consumed =I2R=(𝐼𝑚
√2)2𝑥25 = (
2.13
√2)2𝑥25 = 56.7W
The voltage across the capacitor lags the circuit current by 𝜋
2 radians. Hence its equation is
given by Vc=Vcmsin(314t+1.01 – π/2) where Vcm=ImX xC = 2.13x39.8=84.8V
Now, when I is equal to half of maximum current then
i=0.5x2.13A
∴ 0.5x2.13=2.13sin(314t+1.01)
or, 314t+1.01=sin – 1(0.5)=π
6 or 5
π
6radian
∴ vc=84.8sin(𝜋
6−
𝜋
2)=84.8sin(−
𝜋
3)= – 73.5V
or, 84.8sin(5𝜋
6−
𝜋
2)=84.8sin(
𝜋
3)= 73.5V
hence p.d across the capacitor is 73.5V
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Q.No.11. It is desired to operate a 100-W,120Velectric lamp at its current rating from
a 240V,50Hz supply. Give a detail of simplest manner in which this could be done
using a)a resistor b) a capacitor c) an indicator having resistance of 10 Ω . what
power factor would be presented to supply in each case and which method is most
economical of power.
soln
Rated current of bulb = 100
120=
5
6A
a) by using resistance
we have p.d across R =240 – 120=120V
∴ R=
120
5
6= 144 Ω
Power factor of the circuit is unity.
Power consumed= 240x 5
6=200W
b)By using capacitor
we have
Vc=√2402 − 1202 = 207.5
Xc= 207.5 x 5
6 =249 Ω
∴ 1
314𝐶= 249 => c=12.8 μF
p.f= cos Ф = 120/240=0.5 (lead)
power consumed=240x 5
6𝑥0.5 = 100W
c) by using an indicator
VR=5
6𝑥10 =
25
3𝑉
∴ VL=√2402 − (120 +25
3
2) = 203V
314L x 5
6= 203
∴ L=0.775
Total resistive drop = 120+25
3= 128.3 V
Cos Ф = 128.3
240= 0.535(lag)
Power consumed = 240 x 5
6 x 0.535 = 107W
As method b) consumes less power , that is most economical.
Q.No.12. A resistance Are, an impedence L =0.01H and capacitance C are connected
in series. When voltage of V=400cos(3000t – 10 ° ) volts is applied to the series
combination, the current is 10√𝟐cos(3000t – 55 ° )A. Find R & C
soln
The phase difference between applied voltage and current circuit is (55° – 10°)=45° with
current lagging. The angular frequency=3000 redian/second. Since current lags XL>Xc.
Net reactance X= XL – XC. also XL= ωL= 3000x0.01=30 Ω
Tan Ф = X
R tan45=
X
R ∴ X=R
Now, Z= VM
IM =
400
10√2 = 28.3 Ω
Z2=R2+X2=2R2
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∴ R=𝑍
√2=
28.3
√2= 𝟐𝟎 Ω
X= XL – XC =30 – XC =20
∴ XC =10 Ω 1
𝜔𝐶= 10
∴ C=33μF
Q.No.13. find the values of current I.V1, V2 and p.f of given circuit A and also draw
phaser diagram.
soln
L=0.05+0.1=0.15H
XL=314x0.5=47.1 Ω
XC= 106/314 x 50=63.7 Ω
X=47.1 – 63.7= – 16.6
R=20+10=30 Ω
Z=√302 + (−16.6)2 = 34.3 Ω
I=𝑉
𝑍=
200
34.3= 5.83A
XL1=314x .05 =15.7 Ω
Z1=√102 + 15.72 = 18.6 Ω
V1=IZ1=5.83x18.6=108.4V
Ф1=cos – 1(10/18.6)=57.2° (lead)
XL2=314x0 .1 =31.4 Ω
Xc= – 63.7 Ω
X=31.4 – 63.7= – 32.2 Ω
Z2=√202 + (−32.2)2 = 38 Ω
V1=IZ2=5.83x37.905=221V
Ф2=cos – 1(20/38)=58.2° (lead)
Combined p.f= cos Ф =R/Z =30/34.3 =0.875
Phaser diagram is shown in fig:B
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Chapter six Transformer
Transformer & its constuction
A transformer is a state electrical device by means of which electrical energy is
transformed from one circuit to another circuit
without any moving path.
Basically a transformer consists of an iron core
on which two separated cores are wound.
The physical basis of transformer is mutual
induction between two circuit linked by a
common magnetic flux.
It consists of two inductive coils which are
electrically separated but magnetically linked
through the path of low reluctance.
The frist coil in which electric energy is supplied from AC supply is called primary
winding & another from which energy is drawn out is called secondary winding.
The voltage source of one circuit could be higher, lower or equal to voltage level of
other circuit as per need.
Working principle
Consider a transformer whose
secondary wiring is open circulated &
primary wiring is connected to
sinusuidal Alternating voltage V where,
V1=Vmsinωt……………………..(i)
If the coil is purely conductant
then the potential difference between
causes alternating current to flow in primary which lags voltage V1 by 90° i.e I0=
Imsin(ωt-90°)
As secondary winding is open, the primary winding draws a magnetising current
only whose funtion is to magnetise the core. Hence this current will magnetise the core
and It produce a flux Ф which will circulate in the core & will be alternating in nature and
in phase with I0
Ф =Ф msin(ωt-90°)………………………(ii)
As this flux is linked with secondary winding, according to faradays law of
electromagnetic induction emf E2 will be induce in secondary winding. This induced emf
E2 will be antiphase with V1. Also changing flux is linked with primary winding,emf E1
will also induced in primary winding and E1 is in phase with E2. Waveform of voltage,
flux and emf are as follow
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EMF equation of Transformer
Suppose,
N1=Number of turns in Primary
N2= Number of turns in secondary
Фm=maxium magnetic flux in core
F=frequency of applied voltage
V2=Terminal voltage across
secondary winding
V1=Primary voltage=Vmsinωt
Then,
Ф =Ф msin(ωt-
90°)…………………(i)
Differentiating equation (i) with respect to time
d∅
dt= ∅mωcos (ωt − 90°)
According to faradays law of electromagnetic induction, the instanteneous value of emf
induced in primary winding is given by
E1 = N1d∅
dt= N1∅mωcos (ωt − 90°)
∴ E1 = N1∅mωsinωt Comparing with V1 =Vmsinωt , we can say that the emf E1 has maxium value at N1∅mω
Hence the RMS value of E1 is given by
E1 =N1∅mω
√2 =
N1∅m2πf
√2= 4.44fN1∅m …………..(a)
Similarly RMS value of e.m.f induced in secondary winding is
E2 =N2∅mω
√2 =
N2∅m2πf
√2= 4.44fN2∅m …………..(b)
Here equation (a)&(b) are e.m.f equation of transformer.
Voltage and current transformation ratio
If we divide equation (b) by equation (i) we get, E2
E1=
4.44fN2∅m
4.44fN1∅m=
N2
N1
∴E2
E1=
N2
N1 =K
neglect the internal resistance of a coil, then
E1≈ V1 E2≈ V2 E2
E1=
V2
V1=
N2
N1= K
Where K is known as Transformation ratio of transformer.
Note
If K>1; N2>N1 , then V2>V1 such transformer is called step up transformer
If K<1; N2<N1 , then V2<V1 such transformer is called step down transformer
If K=1; N2=N1 , then V2=V1 such transformer is called isolated transformer
For ideal loss less transformer
Input VA=Output VA
V1I1=V2I2
∴V2
V1=
I1
I2
∴E2
E1=
V2
V1=
N2
N1=
I1
I2= K
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No load operation of transformer
When no load is connected across the secondary winding , I0 current flow through the
primary winding.The ideal transformer cannot be made in real life which donot have any
flux leakage. The coils of real transformer will have some resistance.
The primary input current under no-load condition has to supply
i) Iron loss in the core
ii) A very small amount of cu loss in primary winding.
Hence primary input current I0 is not at 90° nehind V1 nut lags it by angle Ф0 which is
less than 90°.
The phasor diagram for no load operation is shown
above in right side. Here current I0 is resolved into two
components:
Iw=I0cos∅0=component of I0 in phase with V1.This
is known as active or working or iron loss
component.It is responsible for producing heat loss
due to heating of core.
Iμ=I0sin∅0=component of I0 in perpendicular with
V1.This is known as magnetising component.It is
responsible for sustaining alternating flux Ф in the
core.
The resolved component of current I0 has two distinc actions. The effect of this two
component can be represented by circuit model as shown in the figure. No load
equivalent circuit below.
The active power consumed by the transformer at no load is given by:
W0=V1I0cosФ0watts= No load power loss=Iron loss
Where cosФ0= nnoload power factor
It should be noted that no load current I0 is very small and no load copper loss is
negligible
Loaded operation of transformer
When the load is connected across the secondary winding ,secondary current I2
will flow through the load. Hence the secondary winding will produce its own magnetic
flux Ф2 which oppose the main magnetic flux Ф produced by primary winding.
For ideal transformer
Output power=Input power
For loaded real transformer
Output power=V2 I2
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Additional current I2′ will flow through winding so that the input power increases from V0
I0 to V1 I1 and there is power balance between primary and secondary winding.At this
stage primary winding produces the additional magnetic flux ∅2′ which is equal and
opposite to Ф2 so that net magnetic flux in the core is again Ф
In actual transformer , the primary and secondary winding will have some
resistance and also they will have some leakage inductance which donot help in the
process of emf inducing but cause reactive voltage drop.Hence equivalent circuit of an
actual transformer can be written as shown in fig below
Here,
V1= supply voltage to primary winding
I0=NO load primary current
Iw=Loss component of I0
Iμ=Magnetising component of Io
R0=Core loss resistance
X0=Magnetising reactant
R1=Resistance of primary winding
X1=Leakage resistance of primary winding
E1=Emf induced in primary winding
E2=Emf induced in secondary winding
R2=Resistance of secondary winding
X2=Leakage reactance of secondary
winding
V2=Terminal voltage across the load
The terminal voltage V2 decreases with the
load current because there is some voltage
drop in R2 & X2 and the terminal voltage is
given by
V2 = E2
− I2 R2 − I2 X2
The phaser diagram of above equation is
shown on right side
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Losses in transformer
In a static transformer, there are no friction or windage losses. Hence, the losses
occuring are Iron loss and coppor loss.
Iron loss
It is the summation of eddy current loss and hyteresis loss. The core flux in the
transformer remains practically constant for all loads(its variation being from no load to
full load). The core loss is practically the same at all load.These losses are minimized by
using steel of high silicon content for the core and by using very thin laminations. It is also
known as core loss. Iron loss for a transformer is constant.
Iron loss=Hysteresis loss+eddy current loss
Wi=Wh+We
Hysteresis current loss can be reduced by using high grade silicon steel and eddy current
loss can be reduced by using laminated iron core.
Copper loss
This loss is due to the ohmic resistance of the transformer windings.
cu loss=I2R
Total cu loss=I12R1 + I2
2R2 . Cu loss is proportional to I2.
Copper loss will vary according to load.
Efficiency of a transformer
Efficiency of transformer at a particular load and power factor is defined as the
output divided by input power in watts or kilowatts.
Efficiency =Output power
input power
Efficiency =Output power
Output power + losses=
Output power
Output power + Cu loss + iron loss
or, Efficiency =Input power − losses
Input power= 1 −
Cu loss + iron loss
Input power
It is denoted by ŋ
Condition for maximum efficiency in transformer
cu loss (Wcu)=I2R = I12R1 = I2
2R2 …………………………(i)
Wi=Wh+We
considering primary side
primary input = V1I1 cos ∅1
ŋ =V1I1 cos∅1−Wcu−Wi
V1I1 cos∅1
or,ŋ =V1I1 cos∅1−I21R1−Wi
V1I1 cos∅1
or,ŋ = 1 −I1R1
V1 cos∅1−
Wi
V1I1 cos∅1 …………………………(ii)
Differentiating equation (i) w.r.t I 1
dŋ
dI1= 0 −
R1
V1 cos∅1−
Wi
V1I12 cos∅1
For maximum efficienty R1
V1 cos∅1=
Wi
V1I12 cos∅1
→ Wi = I12R1
∴ Wi = Wcu …………………………(iii)
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Similarly
Wi = I22R2
I2 = √Wi
R2
The output current corresponding to maximum efficiency is I2 = √Wi
R2
Hence from above equation (iii) we can conclude that to obtain maximum efficiency in
the transformer iron loss=copper loss
Notes
If we’re given iron loss and full load copper loss then the load at which two losses will
be equal.
Max efficiency load = full load X √Iron loss
Full load copper loss
The efficiency at any load is given by
ŋ =x ∗ full load KVA ∗ power factor
(x ∗ full load KVA ∗ power factor) + Wcu + Wi
Reason for transformer rating in KVA
Copper loss of transformer depends on current and iron loss in voltage. Hence total
transformer loss depends on volt-ampere(VA) and not on phase angle between voltage and
current i.e. it is independent of load power factor. That is why rating of transformer is in
KVA and not in KW.
Solved numerical problems in transformer
Q.No.1. The maximum flux density in the core of the 250/3000V , 50 Hz single phase
transformer is 1.2wb/m2. If the emf induced per turn is 8V, determine
i)the primary and secondary turns ii)Area of core
Solution
Given
E1=250V
E2=3000V
f=50Hz
Βm=1.2wb/m2
Emf induced per turn=8V
Area of core=? Primary and secondary turns=?
We know,
E1=N1 x emf per turn
Or,250=N1 x 8 →N1=250
8
∴ 𝐍𝟏 = 𝟑𝟏. 𝟐𝟓 ≈ 𝟑𝟐
Similarly
E2=N2 x emf per turn
Or,3000=N2 x 8 →N2=3000
8
∴ 𝐍𝟐 = 𝟑𝟕𝟓
We have,E2=4.44fN2Фm → 300=4.44 x 50 x 375 x Βm x A →4.44 x 50 x 375 x 1.2 x A
∴ 𝐀 = 𝟎. 𝟎𝟑𝐦𝟐
Q.No.2. A 25KVA transformer has 500 turns in primary and 50 turns in secondary
winding. Primary winding is connected to 3000V,50HZ supply. Find the full load
primary and secondary current, secondary emf and maximum flux in the core.
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Soln,given
P=25KVA=25000VA
N1=500
N2=50
E1=3000V
f=50Hz
we have,
P=E1 x I1 →25000=3000 x I1 → I1=25000
3000 → I1=8.334A
Now, N2
N1=
I1
I2→ I2 =
N1
N2 xI1 =
500
50x 8.334 = 83.334
∴ 𝐈𝟐 = 𝟖𝟑. 𝟑𝟑𝟒𝐀 N2
N1=
E2
E1→ E2 =
N2
N1 x E1 =
50
500 x 3000 = 300V
∴ 𝐄𝟐 = 𝟑𝟎𝟎𝐕
E1=4.44fN1Фm→ Фm=E1
4.44x50xN1 =
3000
4.44x50x500 = 0.027wb=27.02mwb
∴ Ф𝐦 = 𝟐𝟕. 𝟎𝟐𝐦𝐰𝐛
Q.No.3. A 2200/200V transformer draws no load primary current of 0.6A & observes
400W. Find the magnetising & iron loss component of current.
Soln
V1=2200V
V2=200V
I0=0.6A
Power P=400w
We have
P=V1I0cosФ → V1Iw → Iw= 400
2200= 0.182A
∴ 𝐈𝐰 = 𝟎. 𝟏𝟖𝟐𝐀
I0 = √Iw2 + Iμ
2 → 0.62 = 0.1822 + Iμ2 → Iμ = √0.62 − 0.1822 = 0.572A
∴ 𝐈𝛍 = 𝟎. 𝟓𝟕𝟐𝐀
Q.No.4. A 2200/200V transformer takes 0.5A at power of 0.3 on open circuit. Find
magnetising & working component of no load primary current.
Soln, given
V1=2200V
V2=200V
I0= 0.5A
Power factor=cosФ= 0.3
We have
Iw=I0cosФ=0.5 x 0.3
∴ 𝐈𝐰 = 𝟎. 𝟏𝟓𝐀
I0 = √Iw2 + Iμ
2 → 0.52 = 0.152 + Iμ2 → Iμ = √0.52 − 0.152 = 0.476A
∴ 𝐈𝛍 = 𝟎. 𝟒𝟕𝟔𝐀
Q.No.5. In a 25KVA 2000/200V single phase transformer, the iron loss & full load cu
loss are 350 & 400watt respectively. Calculate the efficiency at unity power factor on
a)Full load
b)half load
soln given
We have,
For full load
Iron loss=350watt
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Cu loss=400watt
total loss=iron loss+cu loss=350+400=750watt
Full load output @ unity=25KVA x 1=25Kw=25000w
Input =750+25000 =25750W
Efficiency(ŋ) =output
inputx100%
ŋ =25000
25750 x 100%
∴ ŋ = 𝟗𝟕%
For half load
Iron loss=350watt
Half load Cu loss=400 x(1
2)2=100watt
Total loss=350+100=450 watt
Half load output @ unity=12.5 x 1=12.5 Kw= 12500w
Input=450+12500=12950w
Efficiency(ŋ) =output
inputx100%
ŋ =12500
12950 x 100%
∴ ŋ = 𝟗𝟔. 𝟓𝟐%
Q.No.6. A 5-KVA , 2300/230V, 50Hz transformer was tested for the iron losses with
normal excitation & cu losses at full load and these were found to be 40W and 112W
respectively. Calculate the efficiency Of transformer 0.8 power factor for the following
KVA outputs:
a)1.25
b)2.5
c)3.75
d)5.0
e)6.25
f)7.5
Soln, given
Full load cu loss=112W
Full load iron loss=40W
a)Cu loss @ 1.25 KVA=
112 x (1.25
5)2=7W
Total loss=40+7+47W
Output=1.25 x 0.8=1KW=1000W
Efficiency(ŋ) =output
output+lossx100%
ŋ =1000
1000+47x100% = 95.51%
∴ ŋ = 95.51%
b)Cu loss @ 2.5 KVA= 112 x (2.5
5)2=28W
Total loss=40+28=68W
Output=2.5 x 0.8=2KW=2000W
Efficiency(ŋ) =output
output+lossx100%
ŋ =2000
2000+68x100% = 96.71%
∴ ŋ = 96.71%
c)Cu loss @ 3.75 KVA= 112 x (3.75
5)2=63W
Total loss=40+63=103W
Output=3.75 x 0.8=3KW=3000W
Efficiency(ŋ) =output
output+lossx100%
ŋ =3000
3000+103x100% = 96.68%
∴ ŋ = 96.68%
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d)Cu loss @ 5 KVA= 112 x (5
5)2=112W
Total loss=40+112=152W
Output=5 x 0.8=4KW=4000W
Efficiency(ŋ) =output
output+lossx100%
ŋ =4000
4000+47x100% = 96.34%
∴ ŋ = 96.34%
e)Cu loss @ 6.25 KVA=112x (6.25
5)2=175W
Total loss=40+175=215W
Output=6.25 x 0.8=5KW=5000W
Efficiency(ŋ) =output
output+lossx100%
ŋ =5000
5000+215x100% = 95.88%
∴ ŋ = 95.88%
f)Cu loss @ 7.5 KVA= 112 x (7.5
5)2=252W
Total loss=40+252=292W
Output=7. 5 x 0.8=6KW=6000W
Efficiency(ŋ) =output
output+lossx100% =
6000
6000+292x100% = 95.36%
∴ ŋ = 95.36%
Q.No.7. A 200-KVA transformer has an efficiency of 98% at full load. If the maximun
efficiency occurs at three quarters of full load, calculate the efficiency at half load.
Assume negligible magnetizing current and p.f. 0.8 at all loads.
Soln, given
At full load efficiency=98%
Full load output= 200x0.8=160KW
Full load input=160/0.98=163.265KW
Full losses=163.265-160=3.265KW
Let Wcu be cu loss & Wfe be iron loss then
Wcu+ Wfe =3.265KW………………………………….(i)
loss at 75% of Full load= Wcu(3
4)2=
9Wcu
16………...(ii)
As Ŋmax occurs at three quarters of full load when Cu loss becomes equal to iron loss
∴ Wfe =9Wcu
16
Substituting value of Wfe in equation (i), we get
Wcu+ 9Wcu
16 =3.265KW=3265W
∴Wcu =2090W
∴ Wfe=1175w
Half load unity power factor
Cu loss=2090 x (1
2)2= 522W
Total loss=522+1175=1697W
Output=100x 0.8=80KW=80000
Efficiency(ŋ) =output
output+lossx100% =
80
80000+1697= 97.9%
∴ ŋ = 97.9%
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Chapter seven DC machines
DC Machines
DC machines are rotating electrical machines for continuous energy conservstion
which can be used as motor as well as generator. DC motor converts electrical energy into
mechanical rotation and DC generator converts mechanical energy into electrical energy.
Construction and Parts of DC machines
Field pole & field winding: Field poles are made of laminated silicon steel on
which the emamel insulated copper wire winding are provided.This winding is known as
field winding.
Armature: It is rotating part of machine.This is cylindrical in shape with a central
shaft. The armature is also made of laminated silicon steel sheets.In actual machines an
armature conductor represent many numbers of turns.
Commutator segments: End of armatures conductor are connected to segment
known as commutator segment which are electrically isolated by some insulator like
mica.They are made of copper. The function of the commutator is to facilate collection of
current from the armature conductor
Carbon brushes: The function of these brushes is to collect current from the
commutator & are usually made of carbon graphite. They are in shape of rectangular
block.These carbon blocks are fixed & donot moves while commutator segments rotates
along with the armature.
DC generator:
It is DC machine which converts mechanical energy into electrical energy.
Operating principal of dc generator
The simpler form of DC generator with two
field poles and an armature coil A-B is
given in figure shown beside . When the
armature is continuously rotated the
armature conductirs A and B will cut the
magnetic flux produced by the field poles.
Hence according to Faradays law of
electromagnetic induction, emf will induce
across the coil A-B.The nature of emf
induced will be ac as shown in figure
below
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If we could connect a load across coil A-B current will pass through the load. But it is very
difficult to connect the load directly across the rotating armature coil.in order to facilate the
connection of external load across the armature coil, carbon brushes and commutators
segments are used as shown in fig:Dc generator with carbon brush and commutator
segments. Carbon brushes and commutator also helps to convert the ac emf induced in the
armature coil into dc current across the load.
From zero position to 180° rotation, the direction current in the armature conductor ‘A’ is
going inside and the direction through the armature conductor ‘B’ is coming out as
determined by right hand fleming’s rule. Hence C2 collects current from the commulator
segment and delever to load.c1 receive the current back from the load. After 180° rotation
the situation will be as shown in fig @ right side.Here direction of current in the armature
conductors has changed but the C2 is still collecting the current from the armature and
delevering to the load hence the direction of current through the load is unidirectional DC.
Emf equation of dc generator(emf induced in dc generator)
Let us suppose
Ф=magnetic flux per pole
Ρ=number of magnetic poles
Z= Total number of armature conductor
N=Speed of armature in rpm
∴Average emf generator per condcutor=d∅
dt
Magnetic flux cut by each conductor in one revolution= dФ=Фp
Time for one revolution dt =60
Nsec
∴ Average emf generator per conductor=∅PN
60
Let A=number of parallel paths in armature winding
Then number of conductor in series =𝑍
𝐴
Total emf across the brushed are given by
E =∅ZNP
60A
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Method of excitation
The field winding needs some dc current to magnetise the field pole. This field current is
known as excitation for the dc generator.The excitation can be provided by various method
and accordingly the dc generator can be classified as follow.
1. Separately excited dc generator
2. Self excited dc generator.
Separately excited dc generator:In
this type of DC generator the field
winding is connected to an external
source of DC supply other than
armature of its own machine.
When the load is connected across
the armature terminals armature
currents Ia will flow which is equal to
the load current IL.The terminal voltage V fall from its open circuit emf E due to a
voltage drop caused by current flowing through armature resistance,Ra. Hence the
terminal voltage on separately excoted DC generator is given by:
V=E-IaRa
If there are voltage drop in brushes then
V=E-IaRa-voltage drop due to brushes
Self excited dc generator: In this type of generator the field winding is excited by
parts of the current generated by armature itself.No external DC source is required for
such generator.The field winding and armature winding have electrical connection self
excited DC generators can be classified into three types which are dscribed as follow.
DC shunt generator:
In DC shunt generator the field winding is
connected in parallel with the armature as shown
in fig:2.This generator shall started without load.
The field winding consisits of residual flux
because of which an emf is generated in
armature. Since no load is connected the total
armature current goes to the field winding and
increases flux which in turn increases the
generation of emf in armature. In doing so, the
production of flux at the field winding reaches to
saturation and load is connected to the generator . Then the field winding draws small
amount of current as it has relatively high resistance than armature resistance.The
current generated by armature Ia divides into two parallel path, one to the field winding
and other to the load. Hence
Ia=If+Ir
Now,
Field current ,If =V
Rf
Load current ,IL =V
RL
Terminal voltage across the load is V=E-IaRa……….(i)
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Above equation shows that with increase of current Ia, voltage V across the load
decreases. Fig:3.a shows how the terminal voltage decreases with increasing value of Ia.
this curve is known as load characteristics of DC shunt generator.
The open circuit characteristics of Dc shunt generator is shown in fig:3.b.It shows that the
emf across armature varies with field current.If OA is the magnitude of emf because of
residual flux. Whwn it increases, E also increases from A-B.When flux reaches its
saturation, then E because constant
DC series generator:In DC series generator the field winding is connected in series
with the armature winding as shown in fig:4.This type of generator shall be started
with the load. These generators are rarely used in practise. Here the figure shows that
the same current flows through
armatures, field and the load. The
terminal voltage ‘V’ across the load is
given by :
V=E-IRf – IRL
The load characteristic of series generator is shown in fig:5. When the current load
increases the emf will also increase. Hence series generator has rising voltage
characteristic but at overload condition the voltage starts decreasing.
DC compound generator: This type of DC generator have two sets of field winding
one of them is connected in series with armature winding and other is connected in
parallel with armature winding.The series field winding is made from thick wire with
few turns and shunt field winding is made from thin wire with many numbers of
turns.There are two types of DC compound generator as shown below
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Long shunt generator: In long shunt genrator the shunt field winding is
connected across armature together with series field winding. For long shunt
generator
𝐈𝐟 =𝐕
𝐑𝐬𝐡 𝐈𝐚 = 𝐈𝐟 + 𝐈𝐋
V=E-IaRsh-IaRse
Short shunt generator:In where as in the short shunt generator the shunt field
winding is connected across the armature winding only.
𝐈𝐟 =𝐕𝐬𝐡
𝐑𝐬𝐡=
𝑬−𝑰𝒂𝑹𝒂
𝐑𝐬𝐡 𝐈𝐚 = 𝐈𝐟 + 𝐈𝐋
V=E-IaRa-ILRse
The characteristic of compound generator lies between the characteristics of shunt
and series generators. Fig:7 shows the load characteristics of compound generator.
Armature reaction:Since a current carrying conductor produces magnetic field around
it,the armature conductor will also produce its magnetic field around it.The effect of
magnetic field produced by armature conductor on magnetic field poles is known as
Armature reaction.
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DC motor:It is a dc machine that converts electrical energy into mechanical rotation.The
electrical energy is given across the armature and field windings and the armature
produces the mechanical rotation.
Operating principle of DC motor
In DC motor the electrical energy is supplied across field winding and armature
winding’s as well. When field winding is supplied by DC voltage, the field current will
flow through the field poles winding this current will magnitize the field poles resulting a
magnetic field in the space betweeen two poles.
When the armature conductors are also supplied by DC current through carbon
brushes. These armature conductors interact with magnetic field produced by the field
poles and force will develop on the armature conductors. This force will produce
continuous rotation of armature.
Back emf in DC motor
When the armature of dc motor rotates the
armature conductor cuts the magnetic flux
produced by the field poles. Hence according to
faradays law of electromagnetic induction emf
will induced acros the armature conductors. The
direction of this emf is opposite to the applied
voltage ‘V’. This emf is known as back emf.
The field current is given by If =V
Rf
Using kifchhoff’s law we can write
V − IaRa − Eb = 0
∴ Eb = V − IaRa
The magnitude of back emf in DC motor is given by
Eb =∅ZNP
60A
The magnetude of back emf is always less than the applied voltage.
Role of back emf
Back emf protects the armature from short circuit during rumming condition
Back emf helps tge motor to prodce required amount of torque qccording to increased
or decreased external load torque (i.e it acts as current ontroling agent)
Back emf acts as energy converting agent. The back emf in dc motor acts as the
opposing agent due to which the Dc motor is able to convert electrical energy into
mechanical rotation.
Types of DC motor
DC shunt motor:In Dc shunt motor the field
wiring is in parallel with the armature across the
supply as shown in figure.
The armature current is flowing into the armature
against the opposition of back emf Eb and
armature develops the mechanical power to EbIa
watt.
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Let Ta=torque developed by the armature
Then, 2πNTa
60= EbIa
But Eb =∅ZNP
60A
or2πNTa
60=
∅ZNP
60A Ia
or, Ta =ZP∅Ia
2πA
∴ Ta ∝ ∅Ia
Using kifchhoff’s law we can write
V=Eb+IaRa
Ia =v−Eb
Ra
Armature torque- armature current characteristic
Since field current is constant in dc shunt motor, the magnetic flux is also constant.
Therefore Ta ∝ Ia .
Here torque increases proportionally with the armature current
Speed-torque characteristic
When external load on shaft of dc motor increases the speed will decrease. Them
back emf will be decrease resulting more armature current. Hence more torque will be
developed to overcome the increase load on shaft
DC series motor: In Dc series motor the field wiring is in series with the armature
across the supply as shown in figure.
In this motor current is flowing through both the armature and field winding.
If = Ia
𝑜𝑟, Ta ∝ ∅Ia
∴ Ta ∝ Ia2
Hence armature torque is propertional to the square of armature current.
Armature torque- armature current characteristic
The torque is directly proportionel to current over limited range before magnetic
saturation is reached.
Thus
Ta ∝ ∅Ia ∴ Ta ∝ Ia2
After magnetic saturation Ta ∝ Ia
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Speed-torque characteristic
At starting the speed is very low and back emf is also very small .Therefore the
motor draws very high current which produced strong magnetic flux. Since Ta ∝∅Ia the motor produce high torque at lower speed.
DC compound motor: These motor have both series and shunt field winding. There
are two types of dc compound motor. They are as follow:
Commulative compound:In this type of compound motor the series winding
produces the magnetic flux in same direction as produced by the shunt field
winding .
Differential compound: In this type of compound motor the series winding
produces the magnetic flux in the opposite direction to that produced by shunt
field winding.
Characteristic of compound generator
Armature torque- armature current characteristic
Speed-torque characteristic
Solved numerical examples
Q.No.1. An 8 pole wave connected armature has 600 conductor and is driver at 625
rpm. If the flux per pole is 20 μwb. Determine the generated emf.
Given,
P=8
Z=600
A=2
N=625
Ф=20μwb=20 x 10-3wb
E=?
We know that,
E =∅ZNP
60A=
20 x 10−3x 600x625x8
60x2= 500𝑣𝑜𝑙𝑡𝑠
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Q.No.2. A 4 pole generator has a lap wound armature with 50 slats with 16 conductor
per slot. The useful flux per pole is 30 μwb. Determine the speed at which the
machine must be driven to generate an emf of 240V.
Given,
P=4
A=4
Ф=30 μwb=30x10-3
Z=50 x 16=800
E=240V
N=?
We know that,
𝑁 =𝐸𝐴
∅𝑍𝑃=
240 𝑥4
30 𝑥10−3𝑥800𝑥4= 10𝑟𝑝𝑚
Q.No.3.Determine the terminal voltage of a separately excited generator which
develops an emf of 200V, and has an armature current of 30A on load. Assume that
armature resistance is 0.30Ω.
Soln,given
Ia=80A
E=200V
IL=30A
Ra=0.30Ω
V=?
We have
V=E-IaRa=200 -30x 0.30= 200 – 9.0 =191V
∴ V = 191V
Q.No.4. A separately excited generator is connected to a 60Ω load and a current of
8A flows. If the armature resistance is 1 Ω. Determine
a)terminal voltage b) generator emf
soln, given
Ia=IL=8A
RL=60Ω
Ra=1Ω
V=?
E=?
We have
V=ILRL=8x60
∴ V = 480V
Again
E=V+IaRa=480+(8x1)
∴ E = 488V
Q.No.5. A dc shunt generator delivers 50A to the load at 220 volts. Calculate the emf
generator by the armature. Given that the armature winding resistance is 110 Ω.
Soln, given
IL=50A
Ra=0.8Ω
Rf=110Ω
E=?
Now,
If =v
Rf=
220
110= 2A
𝐼𝑎 = 𝐼𝑓 + 𝐼𝐿 = 2 + 50 = 52𝐴
𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 = 220 + 52𝑥0.08 = 220 +4.16
∴ E = 224.16V
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Q.No.6. A long shunt compound generator delivers a load current of 40A at 400V
and has a armature, series field and shunt field resistances of 0.07Ω,0.04 Ω & 200 Ω
respectively. Calculate the generated voltage and armature current.
Soln, given
V=400V
IL=40A
Rse=0.04 Ω
Rsh=200 Ω
Ra=0.07 Ω
Ia=?
E=?
We have
If =V
Rsh=
400
200= 2A
Ia = If + IL = 2 + 40 = 42𝐴
𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 + IaRse = 400 + 42𝑥0.07 + 42𝑥0.04
∴ E = 404.62𝑉
Q.No.7. A short shunt compound generator delivers a load current of 28A at 220V
and has armature , series and shunt field resistance of 0.06Ω,0.25Ω & 240Ω
respectively. Calculate the induced emf and armature current.
S0ln,given
V=240V
IL=28A
Rse=0.25 Ω
Rsh=240 Ω
Ra=0.06 Ω
Ia=?
E=?
We have
Vsh = V + ILRse = 220 + 28x0.25
∴ Vsh = 227V
If =Vsh
Rsh=
227
240
∴ If = 0.95
𝐼𝑎 = 𝐼𝑓 + 𝐼𝐿 = 0.95 + 28 = 28.95𝐴
𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 + IaRse
= 220 + 28.95𝑥0.06 + 28𝑥0.25
∴ E = 228.73𝑉
Q.No.8.A short shunt cumulative compound dc generator supplies 7.5Kwat 230V.
the shunt field , seris field and armature resistances are 100 Ω,0.3 Ω and 0.4 Ω
respactively. Calculate the induced emf and the load resistance.
soln, given
p=7.5Kw=7500w
Rsh=100Ω
Ra=0.4 Ω
Rse=0.3 Ω
V=230V
E=?
RL=?
Now,
P=VxIL
0r,7500=230x IL
∴ 𝐼𝐿 =7500
230= 32.61Ω
We know,
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V=IL xRL
Or,230=32.61 x RL
∴ 𝑅𝐿 =230
32.61= 7.053Ω
Vsh = V + ILRse = 230 + 32.61x0.3
∴ Vsh = 239.78V
If =Vsh
Rsh=
239.78
230
∴ If = 2.39
𝐼𝑎 = 𝐼𝑓 + 𝐼𝐿 = 2.39 + 32.61 = 35𝐴
𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 + IaRse
= 230 + 35𝑥0.07 + 32.61𝑥0.3
∴ E = 253.78𝑉
Q.No.9. A dc motor operates from a 240V supply, the armature resistance is 0.2Ω.
determine the back emf when the armature current is 50A.
Given,
V=240V
Ra=0.2 Ω
Ia=50A
Eb=?
We know
Eb=V-Ia Ra=240 – 50x 0.2=240 – 10
∴ Eb = 230V
Q.No.10. A dc machine has an armature resistance of 0.5 Ω. If the full load armature
current is 20A. Find the induced emf when machine acts as
1)generator 2)motor
Given
V=220V
Ra=0.5 Ω
Ia=20A
Eb=?
We have
1)when machine acts as generator
Eb = V + IaRa = 220 + 20x0.5
∴ Eb = 230V
2)when machine acts as motor
Eb = V − IaRa = 220 − 20x0.5
∴ Eb = 210V
Q.No.11. A 240V shunt motor takes a total current of 30A.If the field winding
resistance Rf =150 Ω and armature resistance 0.4V. Determine the back emf and
current in armature.
Given
V=240V
I=30A
Ra=0.4 Ω
Rf=150 Ω
Eb=?
Ia=?
We know
𝐼𝑓 =𝑉
𝑅𝑓=
240
150= 1.6𝐴
𝐼𝑎 = I − If = 30 − 1.6 = 28.4𝐴
𝐸 = 𝑉 − 𝐼𝑎𝑅𝑎 = 240 − 28.4𝑥0.4 = 240 − 11.36
∴ E = 228.64V
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Chapter Eight AC machines
Induction motor
It is an electrical motor which operates from AC voltage source and operates under
the principle of electromagnetic induction. It is also known as asynchronous motor.
Construction of Induction motor
An induction motor consists of three parts: a)Strator b)Rotor c)yoke
Stator: It is stationary part of motot that is made up of circular stamping.Stator winding
are placed on the inner circumference of strator core.Generally three phase winding are
provided on these slots and each phase windings are spaced 120° electrically apart.
Rotar: It is the central rotating part of induction motor which is cylindrical in space with a
central shaft. The shaft is supported by bearing at the both ends so that the rotor rotates
freely. It is made up of laminated silicon steel. There are two types of rotor.
1. Squiral cage rotor: It is made up of cylindrical laminated core with parallel slots
nearby the outer circumference. These parallel slots carry rotor conductors and the end
of these conductors are short circulated by coppering known as ring.
2. Phase wound rotor: It is also made up of laminated core but it has open slots along
the circumference on which three phase winding are provided with same number of
pole as that in the stator winding. The three ends of rotor winding are connected to the
three separate slip rings and the slip rings are short circuted by the carbon brushes with
or without external resistance.
Yoke:It is the outermost frame of the machine. It houses the stator core and provides
mechanical protection for the whole machine.the outer surface of the yoke have many
number of fins to cool the machine.
Operating principal of induction motor
When the three phase stator are supplied by the three phase voltage source, three
phase current will flow through stator windings. These three phase current will magnetize
the strator core. Each winding will produce their own magnetic flux which are 120° apart.
The net flux of the induction motor will be equal to 1.5Фm
Where, 1.5Фm = maximum flux. The stator winding produces rotating magnetic field and
the speed of the rotating magnetic field is given by,
Ns =120f
P
Where, f=frequency of voltage applied across the stator winding
P=number of magnetic poles for which stator winding is wound
This speed is known as synchronous speed(Ns)
The rotating magnetic field produced by the stator cuts the rotor conductor and
hence emf will induce on the rotor conductor according to faradays law of electromagnetic
induction. As the rotor conductor are short circulated current will circulate within the rotor
conductor. Now the current carrying conductor are lying in the magnetic field produced by
strator. Hence force will developed on the rotor conductorand therefor the rotor starts
rotating under the action of this force
The rotor will try to catch up the speed of rotating magnetic field(Ns) but it never
success to do so and always runs at speed less than the synchrinous speed(Ns). Therefor an
induction machine is also called asynchronous machine.
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Slip
The fraction by which the speed of the rotor is less than the synchronous speed is
known as slip(s). it is given by
S =Ns−N
Ns
Slip is usually expressed in percentage.
Synchronous speed Ns =120f
p
Synchronous Generator
It is an ac rotating machine which has to be driven at constant speed equal to
synchronous speed. They are also called alternators as it produce alternate voltage.
Construction of synchronous machine(both generator and motor)
The main parts of synchronous machine are described as follow:
i) Strator: It is stationary part of motot that is made up of circular stamping.Stator
winding are placed on the inner circumference of strator core.Generally three phase
winding are provided on these slots and each phase windings are spaced 120° electrically
apart.
ii) Rotor: It is rotating part of machine with number of magnetic poles excited by dc
source from excitor. There are two types of rotor.
a. Salient pole rotor: Its construction is easier and cheaper than cylindrical rotor. It
is mainly used in generator driven by low speed prime movers such as water
turbine, diesel engine etc
b. Cylindrical type rotor: It has smooth magnetic poles in form of a closed cylinder.
Its construction is more compact and robust with compare to salient pole rotor.
They are generally used in generator driven by high speed prime movers like steam
turbine, gas turbine etc
iii) Exciter :Exciter is a felf exciteed dc generator mounted on the shaft of the alternator. It
is supplied dc current to field winding of rotor.
Operating principal of synchronous generator
In synchronous geberator the field poles are rotating and armature conductor are
stationary. The shaft of the machine is driven by prime mover at a constant speed equal to
synchronous speed. The exciter built uo its voltage byself excitation and supply dc current
to the field winding of alternator. The magnetic flux produced bythe rotor poles will cut
the stationary three phase stator winding. Hence according to faradays law of electro
magnetic induction, three phase emf will be induced in stator winding.
Advantages of rotating magnetic system and stationary armature system
The magnetic field system in synchronous generator is opposite to that in dc
generator. Following are the advantages of rotating magnetic system and stationary
system:
The output current can be led to the load directly from the fixed terminals on the stator
without slip rings and brushes.
It is easier to insulate stationary armature winding for high voltage (usually 11kv or
higher) rather than rotating armature
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The field winding deals with low current at low voltage. Therefor the rotating field
winding can be easily insulated. Also slip ring and brushes do not have to handle large
current so that the aparking problem at the slip rings minimum.
Parallel operation and syncronization
In power system , two or more alternators in parallel is known as synchronization.
The numbers of alternators are connected through bus bar(usually infinite bus bar). An
infinite bus bar is the bus bar whose voltage and frequency is independent and constant
with load. The following conditions have to be satisfied for synchronizing an alternator
The terminal voltage of both alternator should be equal.
The frequency of both alternator must be equal
The waveform of emf generated by both alternators should be in phase
The percentage impedance of both alternators should be same.
The phase sequence of both alternators must be same
When two alternators are operating ,so that all above requirements are fulfilled.
They are said to be in synchronism. The proces of connecting them in synchronism called
as synchronizaton.
Synchronous motor
It is an ac motor which always which rotates at constant speed equal to
synchronous speed. Some characteristic features of synchronous motor are as follow.
It ran either at synchronous speed or not at all. i.e. while running it maintains a
constant speed equal to the synchronous speed.
As it is not self starting. Some auxillary means has to be used to start thr motor.
The motor can be operated at wide range of power factors both lagging and leading
Operating principle
When the stator winding are supplied by three phase voltage , rotating magnetic
field will produce. At starting the rotor of field winding are unexcited and the rotor is
driven at synchronous speed by some auxilliary means. Then the rotor field winding are
supplied by dc current and the auxilliary means axis disconnected. The rotor pole and
stator pole will get engaged with a strong force and the rotor continuously rotates with
synchronous speed.
Solved numerical problems
Q.No.1.A 6 pole, 50Hz squirrel cage induction motor runs at the speed of 970rpm.
Calculate
i)Synchronous speed
ii)slip
iii)the frequency of rotor current
Soln
Given
P=6
f=50Hz
N=970rpm
Ns=?
S=?
f’=?
now,
Ns =120xf
p=
120x50
6
Ns = 1000rpm
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s =Ns−N
Ns=
1000−970
1000= 0.03
𝑠 = 3%
f’ = sf = 0.03x50
f’ = 1.5Hz
Q.No.2. A 6 pole 3-Ф 60Hz induction motor runs at 4% slip at certain load.
Determine
i)synchronous speed
ii)rotor speed
iii) frequency of rotor current
soln
p=6
f=60Hz
s=4%=0.04
Ns=?
N=?
f’=?
now,
Ns =120xf
p=
120x60
6
Ns = 1200rpm
𝑠 =𝑁𝑠−𝑁
𝑁𝑠=
1200−𝑁
1200= 0.04 ↔ 1200 − 𝑁 = 48
𝑁 = 1152𝑟𝑝𝑚
f’ = sf = 0.04x60
f’ = 2.4Hz
Q.No.3. A 4 pole 3-Ф 50Hz induction motor runs at 1440 rpm. Determine the
percentage slip of induction motor.
Soln
p=4
f=50Hz
N=1440rpm
S=?
Now,
𝑁𝑠 =120𝑓
𝑝=
120𝑥50
4= 1500𝑟𝑝𝑚
𝑠 =𝑁𝑠−𝑁
𝑁𝑠=
1500−1440
1500=
60
1500= 0.04
𝑠 = 4%