4.0 ac circuits - part 2

8/13/2019 4.0 Ac Circuits - Part 2

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4.0 AC Circuits• 4.1 Inductors and capacitors

• 4.2 Generation, frequency, mean values

• 4.3 Phasor concepts

• 4.4 Resistance, capacitance

and inductance in AC circuit

• 4.5 Power in AC circuits

• 4.6 Three phase supply star anddelta connections



VR

IR

w t

VC

IC

w t

VL IL

w t

RESISTOR CAPACITOR INDUCTOR

A phasor is an arrow whose length represents the amplitude of an AC voltage or current.

Phasor diagrams are useful in solving complex AC circuits.



R

L

C~

The impedance, Z, of a circuit relates max

current to max voltage:

Z V I m

m



R

L

C~

As in DC circuits,

I is same through all components.

: Voltages have different PHASES

they add as PHASORS.



Solve for the current:

Impedance:

Z V

X X R

V I m

Lc

mm

22 )(

Z R2

1

w C w L

2

R

L

C

~



• STEP

1. Replace the time descriptions of V and I with

correct corresponding phasors.

2. Replace inductances and capacitances by their

complex impedances, where;

3. Analyze the circuit by using any techniques

learned before.



• The instantaneous power dissipated in a

component is a product of the instantaneous

voltage and the instantaneous current

• In a the voltage and current are in

phase – calculation of p is straightforward

• In , there will normally be some

phase shift between v and i , and calculating the

power becomes more complicated



Current, voltage and power versus

time for a purely resistive load



.• Therefore, if a voltage is

applied across an inductance , the current will

be given by• Therefore

)2

2sin(

)sin(cos

sincos

)()()(

t I V

t t I V

t I t V

t it vt p

mm

mm

mm

w

w w

w w



• Half of the time the power is positive, showing

energy is delivered to the inductance, where it

stored in magnetic field, and other half is

negative, showing inductance returns energy

to the source

• The average power is zero, we can say that

reactive power flows from source to load




time for a purely inductive load



.

• Therefore, if a voltage is applied

across a capacitance , the current will be given by

• Then

)2

2sin(

)sin(cos

sincos

)()()(

t I V

t t I V

t I t V

t it vt p

mm

mm

mm

w

w w

w w



• The average power is zero (reactive powerflow)

• The power for capacitances carries oppositesign from inductance

• Thus, reactive power is positive for inductanceand is negative for capacitance.

• If load contains both inductance andcapacitance with reactive power of equal

magnitude, the reactive powers cancel




time for a purely capacitive load



• When a sinusoidal voltage is

applied across a circuit with resistance and

reactance, the current will be of the generalform

• Therefore, the instantaneous power, p is given

by

)cos(cos

)()()(

w w

t I t V

t it vt p

mm



• By using trigonometric identity,

is reduced to

cos

2/

2/

cos2

1

rmsrms

mrms

mrms

mm

I V P

I I

V V

I V P

)cos(cos

)()()(

w w

t I t V

t it vt p

mm

cosrmsrms I V P



• The dissipation given by

is termed the in the circuit and ismeasured in watts (W)

• The product of the voltage and current

is termed the . Toavoid confusion this is given the units of voltamperes (VA)

cosrmsrms I V P



• In other words, the active power is the

apparent power times the cosine of the phaseangle.

•This cosine is referred to as the

coscos

S I V P rmsrms

factor Power amperes)volt(inpower Apparent

watts)(inpower Active

iv

S

P

cosfactor Power



•When a circuit has resistive and reactive parts,the resultant power has 2 parts:

– The first is dissipated in the resistive element. Thisis the

– The second is stored and returned by the reactiveelement. This is the , which hasunits of

•

While reactive power is not dissipated it doeshave an effect on the system

– for example, it increases the current that must besupplied and increases losses with cables



• Consider an

RL circuit

– the relationship

between the variousforms of power can

be illustrated using

a power triangle



Active Power watts

Reactive Power var

Apparent Power S = V rmsIrms VA

S2 = P2 + Q2

cosrmsrms I V P

sinrmsrms I V Q



– measures power directly using a single meter which

effectively multiplies instantaneous current and

voltage

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• R is the real part of the impedancethrough which the current flows

• X is the imaginary part of theimpedance through which the

current flows• Reactive power Q is positive for

inductive loads and negative forcapacitive loads

• Vrms

is the rms voltage across theresistance

• VXrms is the rms voltage across thereactance

R

V P

Rrms2

R I P rms2

X I Q rms2

X

V Q

Xrms2



•

In resistive circuits the average power isequal to VI, where V and I are r.m.s. values

• In a the current leads the voltage

by 90 and the average power is zero

• In an the current lags the voltageby 90 and the average power is zero

•

In, the average power is VI cos

• The term is called the



3-Phase-Definition

37

• AC system that involves only one voltage source

is called one phase system. For this type of

generator, the emf is generated using 1

conducting coil that is rotated and strike throughthe magnetic field (north-south pole).

• If the numbers of conducting coil increased, the

generator with multiple phase will be produced.



3-Phase-Definition

38

• 3-phase generation is based on the generation of

one phase system. Therefore, 3 conducting coil

should be arranged and separated 120° among

each other. This will lead to the 3 differentvoltages with the θ of 120°.

• Normally, these 3 conducting coil are termed

‘red’, ‘blue’ and ‘yellow’.

• The rotation of the rotor in the generation of

voltage can be positive (anti-clockwise) or

negative (clock-wise).



3-Phase-Definition

39



3-Phase-Definition

40



3-Phase-Definition

41

• Phasor form:

• The summation of VR, VB and VY =0



42

• Since the generator involved with three different

elements, logically we need 6 wires to connect

these 3 elements.

• However, one can be arranged so that all the

wires can be connected correctly. It’s via the

delta or wye (star) connection. Via these

connection, there is only 4 wires needed, 3 foroutput and one being connected among

themselves. The latter is called neutral.

• This configuration is also important when we are

dealing with the load connection.

• Note that this connection is in wye/star

connection.



43

3-Phase-Connection-Wye/Star

• There are two different voltages in 4 wire

y-connection: – Phase voltage-the voltage difference between appropriate

wire/line with neutral line (Vph)

– Line voltage-the voltage difference between line with other line

(VL)



44

Vph?

VL?

Vph VRN, VBN, VYN

VL VRB, VRY, VYB




45


VRN

VBN

VYN

-VYN

-VRN

-VBN

30° 30°

30°

VRY

VYB

VRB

VL= 3Vph

IL= Iph



• A 415 V, 50 Hz, 3-phase supply is connected to

a star-connected balanced load. Each phase of

the load consists of a resistance of 25 Ω and

inductance 0.1 H, connected in series.

• Calculate a) phase voltage,

b) the line current drawn from the

supply

46

Example 4.6.1



47

a) Phase voltage:

Vph=VL/√3= 415/√3=239.6 V

b) IL=Iph, so Iph= Vph/Zph

Zph

= ℎ

2 +

2

XL=2fL=2(50)(0.1)=31.42 Ω

Zph=(252+31.422)1/2 = 40.15

IL= 239.6/40.15=5.97 A



48

• The star-connection is based on KVL principle

which involved voltage while delta-connection

is based on KCL which involved current.

3-Phase-Connection-Delta



49


R

Y

B



50


IB

IR

IY

IBR -IYB

30°

IRY-IRY

IYB-IBR

30°

30°

IL= 3Iph

VL= Vph



• A balanced load of phase impedance 120

is connected in delta. When this load is

connected to a 600 V, 50 Hz, 3-phasesupply.

• Determine

a) the phase current, andb) the line current drawn



• a) Vph = VL = 600 V;Iph=Vph/Zph=600/ 120=5A

•

b) IL= 3Iph = 3 x 5=8.66 A



53

Power Dissipation in Star and Delta-connection Loads

• The power dissipation for both star and

delta connection are always the same.

The only difference is the value of VL andIL in the connection. The equation of

power dissipation is given by: Phase

power

factor



• A balanced load of phase impedance 100

and power factor 0.8 is connected

a) in star, and

b) in delta, to a 400 V, 3-phase supply.

Calculate the power dissipation in each case.



• a) IL=IpH=Vph/Zph;• Vph=VL/√3=400/√3

• IL=(400/√3)/100=4/√3

• P=√3ILVLcosф

= (√3)(4/√3)(400)(0.8)

= 1.28 kW



Star/Delta Supplies and Loads

• The same equation can also be applied for a

combination of supplies and loads with the

same arrangement.



• The star-connected stator of a three-phase, 50

Hz alternator supplies a balanced delta-

connected load. Each phase of the load

consists of a coil of resistance 15 Ω andinductance 36 mH, and the phase voltage

generated by the alternator is 231 V.

• Calculate (a) the phase and line currents, (b)the load power factor, and (c) the power

delivered to the load.



59



• a) Alternator: V1=Vph= 231 V;• V2=VL=√3Vph = (√3)(231)=400 V

• IL=Iph=I1

•

Load: VL=Vph=V2=400 V• XL= 2fL=2(50)(36 x 10-3)=11.3 Ω

• Zph=√(R2+XL2)=√152+11.32=18.78 Ω

• Iph=Vph/Zph=400/18.78=21.3 A

• IL=√3Iph = (√3)(21.3)=36.9 A; • So, I1=36.9 A



• b) Power factor = cos ϕ = (R/Z) = 15/

18.78=0.8

• c) P = Iph2R = (21.3)2(15) = 6.81 kW for 1 phase

• For 3 phase = 6.81 x 3=20.4 kW

or P = √3VLILcos

ϕ=√3(400)(36.9)(0.8)=20.4

kW

61



62

Conclusion

4.0 ac circuits - part 2

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