Electricians Licence A (Domestic)

Licence (A)

General House Installation

(Lighting and Domestic Appliances)

1

Electrical Theory

Background Information on Units and Quantities

2

7 SI Base Units & Quantities

Unit

Name

Unit

Symbol

Base

Quantity

Quantity

Symbol

Dimension

Symbol

meter m length l L

kilogram kg mass m M

second s time t T

ampere A electric current I I

kelvin K thermodynamic

temperature

T Θ

mole mol amount of substance n N

candela cd luminous intensity Iv J

22 SI Special Derived Units

3

Derived quantity

Special

name

Special

Symbol

Expression in terms of

other SI units SI base units

plane angle radian rad m · m-1 = 1

solid angle steradian sr m2 · m-2 = 1

frequency hertz Hz s-1

force newton N m · kg · s-2

pressure, stress pascal Pa N/m2 m-1 · kg · s-2

energy, work, quantity of heat joule J N · m m2 · kg · s-2

power, radiant flux watt W J/s m2 · kg · s-3

electric charge, quantity of electricity coulomb C s · A

electric potential, potential

difference, electromotive force

volt V W/A m2 · kg · s-3 · A-1

capacitance farad F C/V m-2 · kg-1 · s4 · A2

electric resistance ohm Ω V/A m2 · kg · s-3 · A-2

electric conductance siemens S A/V m-2 · kg-1 · s3 · A2

22 SI Special Derived Units (cont.)

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Derived quantity Special name

Special

Symbol

Expression in terms of

other SI units SI base units

magnetic flux weber Wb V · s m2 · kg · s-2 · A-1

magnetic flux density tesla T Wb/m2 kg · s-2 · A-1

inductance henry H Wb/A m2 · kg · s-2 · A-2

Celsius temperature degree Celsius °C K

luminous flux lumen lm cd · sr cd · sr

illuminance lux lx lm/m2 m-2 · cd · sr

activity (of a radionuclide) becquerel Bq s-1

absorbed dose, specific energy

(imparted), kerma

gray Gy J/kg m2 · s-2

dose equivalent, et al. sievert Sv J/kg m2 · s-2

catalytic activity katal kat s-1 · mol

5

Electrical units and standards

The SI base unit for electrical measurements is the ampere (A), the unit of electric current.

It is defined in terms of a hypothetical experiment as that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 × 10-7 newton per meter of length.

6

The volt (V) is the unit of potential difference and of electromotive force. It is defined as the potential difference between two points of a conducting wire carrying a constant current of 1 ampere (A) when the power dissipated between these points is equal to 1 watt (W). From the ampere and the volt, the ohm (W) is derived by Ohm's law, and the other derived quantities follow in a similar manner by the application of known physical laws.

Ohm (W) is the unit of electrical resistance, equal to 1 volt per ampere. The ohm is defined as the resistance between two points of a conductor when a constant potential difference of 1 volt, applied to these points, produces in the conductor a current of 1 ampere

7

The remaining units of electrical and magnetic quantities

Coulomb (C): The unit of electric charge, equal to 1 ampere-second (As). The coulomb is the quantity of electricity carried in 1 second by a current of 1 ampere.

[C = As] Q = I * t

Farad (F): The unit of capacitance, equal to 1 coulomb per volt. The farad is the capacitance of a capacitor between the plates of which there appears a potential difference of 1 volt when it is charged by a quantity of electricity of 1 coulomb.

[F = C/V = As/V] C = Q/V

8

Henry (H): The unit of inductance, equal to 1 weber (Wb) per ampere. The henry is the inductance of a closed circuit in which an electromotive force of 1 volt is produced when the electric current in the circuit varies uniformly at the rate of 1 ampere per second.

[H = Wb/A = V/(A/s) = Vs/A] L = F/I

Siemens (S): The unit of electrical conductance (the reciprocal of resistance), equal to 1 ampere per volt.

[S = A/V] G = I/V

9

Tesla (T): The unit of magnetic flux density, equal to 1 weber per square meter.

[T = Wb/m2] B = F/A

Weber (Wb): The unit of magnetic flux, equal to 1 volt-second. The weber is the magnetic flux which, linking a circuit of one turn, would produce in it an electromotive force of 1 volt if it were reduced to zero at a uniform rate in 1 second.

[Wb = Vs]

10

DIMENSIONAL PREFIXESPrefix Symbol Magnitude

exa E 1018

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

hecto h 102

deka da 101

deci d 10-1

centi c 10-2

milli m 10-3

micro m 10-6

nano n 10-9

pico p 10-12

femto f 10-15

atto a 10-18

100 = 1 101 = 10 102 = 10 * 10 = 100 103 = 10 * 10 * 10 = 1000 10-1 = 1 / 10 = 0.1 10-2 = 1 / 10 / 10 = 0.01 10-3 = 1 / 10 / 10 / 10 = 0.001

Express 9.213 * 102 in decimal number

Solve 5 * 102 + 3.2 * 103

11

Ohm's law

Relationship between the potential difference (voltage), electric current, and resistance in an electric circuit. In 1827 Georg Simon Ohm discovered that at constant temperature, the current I in a circuit is directly proportional to the potential difference V, and inversely proportional to the resistance R, or I = V/R. Ohm's law may also be expressed in terms of the electromotive force Eof an electric energy source, such as a battery, or E= IR. In an alternating-current circuit, when the combination of resistance and reactance, called impedance Z, is constant, Ohm's law is applicable and V/I = Z.

12

RESISTANCE NETWORKS

There are basically three types of circuit -- series, parallel, and series and parallel circuit.

Series Circuit

The total voltage is the sum of the voltage on each component.VT = V1+ V2

The total resistance is equal to the sum of the resistance on each component.RT = R1 + R2

The total current is equal in every component.IT = I1 = I2

13

Example 1

What is the total voltage, resistance and current?

First, we have to find out the total voltage using equation V0 = V1+ V2 + V3 +...+ Vn, and then resistance using equation R0 = R1 + R2 + R3 +...+ Rn, and finally you can find out the current using equation I0 = I1 = I2= I3= I4 =...= In.

Total voltage is 9 + 1 + 16 + 4 = 30 VTotal resistance is 30 + 10 + 40 + 20 = 100 ohmUsing ohm's law, I = V / R, then we can find out the total current. I = 30 / 100 = 0.3 A

14

Example problem 1

What is the current through A and B?What is the voltage drop across A, B and C?What is the resistance of C?What is the total resistance?What is total current?

15

Parallel Circuit

The total voltage is equalin every component.VT = V1= V2

The resistance is equal to the product of resistance on each component divided by the sum of resistance of each component.RT = (R1*R2)/(R1+R2)

1/RT = 1/R1 + 1/R2

The total current is equal to the sum of current in each component.IT = I1 + I2

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Example 2

What is the total resistance and voltage? And voltage and current on A, B, and C?

In order to find out the total voltage, we have to find out the total resistance. Using equation 1/R0 = 1/R1

+ 1/R2 +...+ 1/Rn, we can find out the total resistance. 1/R = 1/15 + 1/15 + 1/30 = 5/30, R = 6 ohm

Then using ohm's law, V = I R, we can find out the total voltage. V = 5A * 6W = 30 V

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Using equation V0 = V1= V2= V3 =...= Vn, we now know the voltage on A, B, and C, which is 30 V each. Using ohm's law again, we can find out the current through A, B, and C.

IA = 30V/15W = 2 A,IB = 30V/15W = 2 A,IC = 30V/30W = 1 A .

When you add up all the current (using equation I0= I1 + I2 + I3 + I4 +...+ In), we get 5 A which is the total current.

18

Example problem 2

What is the total resistance, voltage, and current?What is the voltage across A, B and C? What is the current through A, B, C, and D?

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Series - Parallel Circuit

The total voltage is the voltage of series plus the voltage of parallel. VT = V1 + V2 = V1 + V3

The total resistance is the resistance of series plus the resistance of parallel. RT = R1 + [(R2 * R3) / (R2 + R3)]

The total current is equal to the current on series and to the sum of the current of parallel circuit. IT = I1 = I2 + I3

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Example 3

What is voltage across A, B, and D? What is current through A, B, C, and D? What is resistance of C? What is total current and resistance?

Using V0 = V1= V2= V3 =...= Vn, for parallel circuits, we know that the voltage on D is equal to C, which is 80 V. We also know A and B have the same voltage. Using the voltage law, we can find out the voltage on A and B, which is 230 - 80 = 150 V each.

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Now we get all the voltages on each component. Using ohm's law, we can find out the current through A, B, C, and D. IA= 150V/30W = 5 A; IB = 150V/30W

= 5 A; ID = 80V/40W = 2 A; IC = 10-2 = 8 A. The sum of the current on A and B is equal to that of C and D (I0 = I1 = I2= I3= I4 =...= In for serial circuit) IA+IB = IC+ID.

The resistance of A+B is 15 ohm RT =(R1*R2)/(R1+R2)The resistance of C is RC = 80V/8A = 10W. Therefore, the resistance of C+D is 8W.

Total resistance and current of this circuit are R = 15+8 = 23 W; I = 230V /23W = 10 A

22

ELECTRICAL POWER IN DC CIRCUITS

Power is the rate of doing work, or, of converting energy from one form to another. Its units are joules per second. One joule per second is called a watt W (symbol P).

Example 1

If a machine converts 1,000 J of energy in 5 sec, what is its power?

The power, P, is given by:

P =energy converted / time = 1, 000 J/5 s = 200W

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When current flows through a wire, the wire gets hot: i.e., power is dissipated. (This heat is why the filament in a light bulb glows.)This leads to the definition of potential difference:when a current of one ampere flows through a resistor, one watt of power is dissipated by the resistor when a potential difference of one volt appears across it.In general the power, P, voltage and current are related by: P = V * I

Example 2If a current of 30A flows through a resistor to which a voltage of 100V is applied, what power is dissipated in the resistor?From P = V IP = 100V × 30A = 3, 000W (or 3 kW.)

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Multiple choice:If a current of 3 A flows along a wire with a potential difference of 4 V between the ends, how much power is dissipated along the wire?(a) 0W; (b) 7W; (c) 12W; (d) 4/3 W

There are other ways of writing the power P = V I.

Using Ohm‘s law P = I R I = I2 R (W)

or P = V V/R = V2/R (W)

Multiple choice:What is the power consumption of a 100W resistor if a 50mA current flows through it?(a) 0.25W; (b) 2.5x106W; (c) 2.5x10-4W; (d) 5x105W

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From Ohm‘s law, there are three equivalent expressions for the power dissipation in a circuit:

P = V I , P = V2/R, P = I2R

Exercise 1

(a) In the circuit if R = 6 and I = 3A, what is the power?(b) In the circuit if V = 8V and R = 2, what is the power?(c) Finally, what is the power if V = 8V and I = 0.25A?

RI

V

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Series and Parallel Circuits

In a series circuit:The same current flows through each resistor. Hence in the diagram the power dissipated in them areP1 = I2R1 , and P2 = I2R2 , respectively and the total power dissipated isPT = I2(R1 + R2)

By Ohm‘s law the voltage source is V = I(R1 + R2), the power can also be written as PT = V2/(R1 + R2).

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Example 3

In the series circuitsince 10W is twice as big as 5W, the power dissipated in the 10W resistor will be twice that dissipated in the 5W resistor.If I = 2A the power dissipation,P = I2R, will be 22 × 10 = 40W in the 10W resistor and 22 × 5 = 20W in the 5W resistor.

10W 5W

(a) If above R1 = 5W and R2 = 15W, how much more power is used in the 15W resistor?(b) If I = 0.8A, calculate the power dissipation in each resistor.(c) How much energy is dissipated over 30 minutes?

Exercise 2

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Example 4

If two resistors are connected in parallel, the effectiveresistance is less than either of the two individual resistors. (This is because there are more ways for the current to flow.)

The potential difference across the two parallel resistors is the same, V . Hence the total power in the resistors in parallel is

PT = V2/R1+ V2/R2= V2(R1 + R2)/R1R2 (W)

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Exercise 3Consider a 10W and a 5W resistor connected in parallel across a 2V source.

(a) What is the power dissipated in the 10W resistor?(b) What is the power dissipated in the 5W resistor?(c) How does the total power dissipated differ from the case if the same resistors were connected in series?

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Example 5The series circuit below represents a power source with an internal resistor Rs. If a load resistor R is connected across the terminals A and B, how does the power to load, PL, depend upon R?

RI

V RsA B

The current I is given by

V = I(Rs + R)I = V/(Rs + R)

Using PL = I2R, the power to load is thus

PL = V2R /(Rs + R)2

PL

RRs

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Maximum Power:

In the curve above it is shown that the maximumpower across the load resistance is when R = Rs, i.e., when the load resistance is equal to the internal resistance of the source (perhaps abattery or generator). This is called resistance matching.

Exercise 4

The power to load is given by PL = V2R/(Rs + R)2 .(a) What is PL when R = Rs?(b) What is PL when R = 100Rs?(c) What is PL when R = 0.001Rs?(d) The maximum power will be given when dPL/dR=0, use the rule of differentiation to show that the maximum is at R = Rs.

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Short Circuit:

If there is no resistance between the terminals,R = 0, the power to load is

PL = (V2 × 0)/(Rs + 0)2 = 0/Rs = 0 (W)

No power can be extracted from a short circuit: there must be a resistance to extract power.

Open Circuit:

If the terminals are disconnected then there is aninfinite resistance, R = , and no current flows. Again the power to load vanishes: a current must flow to extract power.

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ELECTRICAL ENERGY, COST OF ENERGY,TARIFFS RATE

Energy is capacity (ability) for doing work. Energy exists in various forms—including kinetic, potential, thermal, chemical, electrical (see electricity), and nuclear— and can be converted from one form to another. For example, fuel-burning heat engines convert chemical energy to thermal energy; batteries convert chemical energy to electrical energy. Though energy may be converted from one form to another, it may not be created or destroyed; that is, total energy in a closed system remains constant.

Units of energy is Joule (J)

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In the electricity supply industry the SI units of watts and joules are too small. Instead, the units used are:

power unit: kilowatt (1kW = 103W)

energy unit: kilowatt-hour(1kWh = 103×60×60 J= 3 600 000 J=3600 kJ)

This is such a large number that it is easy to understand why your electricity is not sold in joules!

Multiple choice:If a household electricity metre changes from 5732 to 5786 units, how much electrical energy has been dissipated in the house?(a) 2 × 108J (b) 2 × 1010J (c) 2 × 106J (d) 5.4×103J

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Multiple choice:If a current of 3A flows along a wire with a potential difference of 4V for one hour, how much energy is dissipated?(a) 12 J; (b) 720 J; (c) 4,320 J; (d) 43,200 J

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The cost of electricity, is expressed in terms of a unit cost (€ cent per kWh) delivered to the customer. This cost value, includes the capital costof the generating plant and equipment, transmission and distribution system; the cost of fuel burned and the cost of developing, operating and maintaining the system in whole.

Cost of consumed electrical energy= number of units * cost per unit

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Enemalta tariffs

Residential kWh

Band Cumulative Consumption 1 Apr 09 Tariff (€ cent)

1 2,000 0.1190

2 6,000 0.1340

3 10,000 0.1520

4 20,000 0.2090

5 60,000 0.2320

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Domestic kWh

Band Cumulative Consumption 1 Apr 09 Tariff (€ cent)

1 2,000 0.1610

2 6,000 0.1730

3 10,000 0.1890

4 20,000 0.2090

5 60,000 0.2320

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Band Cum1 Apr 09 Tariffs

(€ cents)

Day rates

(€ cents)

Night rates

(€ cents)

1 2,000 0.1040 0.1060 0.0990

2 6,000 0.1120 0.1140 0.1070

3 10,000 0.1250 0.1270 0.1200

4 20,000 0.1400 0.1420 0.1350

5 60,000 0.1570 0.1590 0.1520

6 100,000 0.1420 0.1440 0.1370

7 1,000,000 0.1290 0.1310 0.1240

8 5,000,000 0.1120 0.1140 0.1070

9 10,000,000 0.0860 0.0880 0.0810

Non-Residential kWh

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Band Cum1 Apr 09 Tariffs

(€ cents)

Day rates

(€ cents)

Night rates

(€ cents)

1 2,000 0.0960 0.0980 0.0910

2 6,000 0.1030 0.1050 0.0980

3 10,000 0.1150 0.1170 0.1100

4 20,000 0.1290 0.1310 0.1240

5 60,000 0.1440 0.1460 0.1390

6 100,000 0.1310 0.1330 0.1260

7 1,000,000 0.1190 0.1210 0.1140

8 5,000,000 0.1030 0.1050 0.0980

9 10,000,000 0.0790 0.0810 0.0740

Non-Residential kVAh

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Exercise 1

If factory‘s management decide to opt for the tariff system based on kVAh instead of kWh, what saving will be in term of € if factory total consumption is 75000 kWh annually and power factor is 0.99 after having installed power factor correction equipment?

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Maximum demand rate

Consumers being billed against this tariff are charged as

follows:

•A periodical maximum demand rate. (Maximum demand

is the highest power demand measured in KW or KVA. It

is measured within a definite period of time)

•A per unit rate of either kWh or kVAh.

kWh Billing € 20.50/kW

KVAh Billing € 19.20/kVA

kWh or kVAh billing on

reduced rates€ 17.20/kW or € 17.20/kVA

Enemalta Maximum Demand Charge As From 01.10.08

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RESISTANCE OF CONDUCTOR

Specific resistance is a constant for the type of conductor material being considered at specified temperature.

The resistance R of a conductor of uniform cross section can be calculated as

This formula relates the resistance of a conductor with its specific resistance ρ (Wm), its length l (m), and its cross-sectional area A (m2).

(W)

Conductor resistance increases with increased length and decreases with increased cross-sectional area, all other factors being equal.

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Specific resistance at 20 deg C

Nichrome Alloy 112.2 (mW cm)Nichrome V Alloy 108.1Manganin Alloy 48.21Constantan Alloy 45.38Steel* Alloy 16.62Platinum Element 10.5Iron Element 9.61Nickel Element 6.93Zinc Element 5.90Molybdenum Element 5.34Tungsten Element 5.28Aluminum Element 2.650Gold Element 2.214Copper Element 1.678Silver Element 1.587* = Steel alloy at 99.5 % iron, 0.5 % carbon

Specific Resistance

(r) is a property of

any conductive

material, which is

defined as

resistivity, a figure

used to determine

the end-to-end

resistance of a

conductor given

length and cross-

sectional area.

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How much cross-sectional area of the aluminium conductor has to be bigger than the cooper one if both of them are of the same length and they should perform electrically the same?

Example 1

Rcu = Ral = R lcu = lal = l

R = rcu l/Acu = ral l/Aal

rcu /Acu = ral /Aal

Aal = ral Acu / rcu = 2.65 * Acu / 1.678 = 1.58 Acu

Cross-sectional area of the aluminium conductor has to be 58% bigger than the cooper one.

46

Calculate the resistance of a 2 km length of aluminium

overhead power cable if the cross-sectional area of the

cable is 95 mm2.

Exercise 1

A wire of length 8 m and cross-sectional area 3 mm2 has

a resistance of 0.16Ω. If the wire is drawn out until its

cross-sectional area is 1 mm2, determine the resistance

of the wire.

Exercise 2

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Temperature resistance dependence

In general, as the temperature of a material increases, most conductors increase in resistance, insulators decrease in resistance, whilst the resistance of some special alloys remain almost constant.

48

There are two types of temperature coefficient of resistance: positive and negative.

A positive coefficient of resistance increases resistance as the temperature rises, while a negative coefficient of resistance decreases the resistance as the temperature rises.

Positive temperature coefficient

Negative temperature coefficient

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Some typical values of temperature coefficient of resistance measured at 0°C are given below:

Copper 0.0043/°CAluminium 0.0038/°CNickel 0.0062/°CCarbon -0.00048/°CEureka 0.000 01/°C

(Note that the negative sign for carbon indicates that its resistance falls with increase of temperature.)

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Definition of temperature coefficient of resistance

Temperature coefficient of resistance is the ratio of the change of resistance per degree C change of temperature to the resistance of the same material at some definite temperature.

Definite temperature at 0oC

If the resistance of a material at 0°C is known the resistance at any

other temperature can be determined from:

R1=R0(1+a0t1)

R0 = resistance at 0°CR1 = resistance at temperature t1°Ca0 = temperature coefficient of resistance at 0°C

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A coil of copper wire has a resistance of 100Ω when its temperature is 0°C. Determine its resistance at 70°C if the temperature coefficient of resistance of copper at 0°C is 0.0043/°C

Exercise 3

A carbon resistor has a resistance of 1kΩ at 0°C. Determine its resistance at 80°C. Assume that the

temperature coefficient of resistance for carbon at 0°C is -0.0005/°C

Exercise 4

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Definite temperature at 200C

If the resistance of a material at room temperature (approximately 20°C), R20, and the temperature coefficient of resistance at 20°C, a20, are known then

the resistance R2 at temperature t2 is given by:

R2=R20[1+a20(t2-20)]

R2 is the resistance at t2

R20 resistance at 20oCt2 final temperature

a20 temperature coefficient of

resistance at 20°C

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A copper cable at 20oC has a resistance of 90Ω. The temperature is raised and the resistance measured reads 104Ω. If the temperature coefficient of resistance of copper at 20oC is 0.004/oC, calculate the final temperature.

Exercise 5

Exercise 6

An aluminum overhead cable has a resistance of 100Ω, when the effective daytime ambient temperature is 68oC. During night, the effective ambient temperature falls to 20oC. Calculate the night time resistance if the temperature coefficient of resistance of aluminum at 20oC is 0.0038/oC

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Resistance at 0oC is not known

If the resistance at 0°C is not known, but is known at

some other temperature t1, then the resistance at any temperature can be found as follows:

R1/R2 = (1+a0t1)/(1+a0t2)

R1 is resistance at temperature 1R2 is resistance at temperature 2t1 lower temperaturet2 upper temperaturea0 temperature coefficient of resistance at 0°C

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Exercise 7

Exercise 8

A nickel conductor has a resistance of 250Ω when its temperature is 25oC. If the temperature is raised to 120oC, calculate the value of the final resistance. Assume temperature co-efficient of resistance of nickel at 0oC is 0.0062/oC.

An aluminum wire has a resistance of 100Ω when the temperature is 10oC. A current flows through the wire and the temperature rises such that the resistance then reads 175Ω. If the temperature co-efficient of resistance of aluminum at 0oC is 0.0038/oC, calculate the temperature rise.

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VOLTAGE DROP AND POWER LOSS IN CABLES

The voltage drop is decline in voltage in an electrical circuit due to the resistance in the conducting line. This is why longer electrical runs in a building require thicker wire and why AC power is transmitted over high-voltage lines. Higher current requires thicker and more expensive wires, but higher voltage does not. The high-voltage lines are reduced by transformers near the end of the line

Voltage drop in cable (V) = Resistance of cable (R) x Current through cable (I)

Voltage drop% = (voltage drop/voltage at source)x100

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Whenever a voltage drop occurs in a cable, power is being lost across that voltage drop. This power must be paid for as well, so it is in the interest of the consumer to minimize power loss in cables.

Power loss in cable (W) = Resistance of cable (R) x Current through cable2 (I2)

orVoltage drop in cable (V) x Current through cable (I)

Some effects of voltage drop in cables:

Power wasted in cablesReduction in the efficiency of lamps and heatersDifficulty for fluorescent lamps to start upImproper speed attainment of motorsInability of motors to change position of centrifugal switches (lack of speed)

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Exercise 1

Exercise 2

The voltage at the terminals of a water pump motor is 234V whereas the voltage at the supply end is 240V. Calculate the percentage voltage drop.

A twin copper cable having a cross sectional area of 75mm2, supplies a 20kW load 1000m from source. If the terminal voltage is 400V when the winter ambient temperature is 20oC, calculate;

the resistance of the cablethe current absorbed by the circuitthe supply voltagethe percentage power loss in cable

continue on next page

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During the summer period, the effective ambient temperature of the cable is raised to 75oC and the supply voltage is kept constant. Calculate;

the resistance of the cablethe current absorbedthe voltage across the loadthe voltage drop across the cablethe percentage power loss in cable

Assume resistivity of copper as 0.017µΩm and temperature co-efficient of resistance of copper at 20oC is 0.004/oC

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INSULATION RESISTANCE IN CABLES

The purpose of insulation on electrical cables and equipment is to(1) isolate current-carrying conductors from metallic and structural parts and (2) insulate points of unequal potential on conductors from each other.The resistance of such insulation should be sufficiently high to result in negligible current flow through or over its surface.

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Testing insulation resistance

A low resistance between phase and neutral conductors, or from live conductors to earth, will result in a leakage current.This current could cause deterioration of the insulation, as well as involving a waste of energy which would increase the running costs of the installation. Insulation will sometimes have high resistance when low potential differences applyacross it, but will break down and offer low resistance when a higher voltage is applied.

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Required test voltages and minimum resistance

Nominal circuit voltage Test voltage (V)

Minimum insulationresistance (MW)

Extra-low voltage circuits supplied from a safety transformer

250 0.25

Up to 500 V except for above

500 0.5

Above 500 V up to 1000 V 1000 1.0

The insulation resistance tester must be capable of maintaining the required voltage when providing a steady state of current of 1mA.

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The insulation resistance test to earth must be carried out on the complete installation with the main switch off, with phase and neutral connected together, with lamps and other equipment disconnected, but with fuses in, circuit breakers closed and all circuit switches closed. Where two-way switching is wired, only one of the two wires will be tested. To test the other, both two-way switches should be operated and the system retested. If preferred, the installation can be tested as a whole when a value of at least 0.5 MW should be achieved, for the usual supply voltages. In the case of a very large installation where there are many earth paths in parallel, the reading would be likely lower. If this happens, the installation should be subdivided and retested, when each part must meet the minimum requirement.

Methods and equipment used for the insulation resistance tests

64

65

The testing equipment must be capable of delivering a

current of 1 mA at the minimum allowable resistance level,

which is:

250 kW for the 250 V tester

500 kW for the 500 V tester

1 MW for the 1,000 V tester

Basic instrument accuracy required is +/-5%

It must have a facility to discharge capacitance up to 5 mF

which has become charged during the test or may be

combined with the low resistance ohmmeter

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HEAT ENERGY AND MECHANICAL ENERGY

Energy is the ability to do work.

Energy can be found in a number of different forms. It can be chemical energy, electrical energy, heat (thermal energy), light (radiant energy), mechanical energy, and nuclear energy.

Definition: Heat energy (or just heat) is a form of energy which transfers among particles in a substance (or system) by means of kinetic energy of those particle. In other words, under kinetic theory, the heat is transferred by particles bouncing into each other.

Work is defined as the transfer of energy.

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Conduction occurs when energy is passed directly from one item to another. If you stirred a pan of soup on the stove with a metal spoon, the spoon will heat up. The heat is being conducted from the hot area of the soup to the colder area of spoon.

Convection is the movement of gases or liquids from a cooler spot to a warmer spot. If a soup pan is made of glass, we could see the movement of convection currents in the pan. The warmer soup moves up from the heated area at the bottom of the pan to the top where it is cooler. The cooler soup then moves to take the warmer soup's place.

Heat energy moves in three ways:

Conduction; Convection; Radiation

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Radiation is the final form of movement of heat energy. The sun's light and heat cannot reach us by conduction or convection because space is almost completely empty. There is nothing to transfer the energy from the sun to the earth.The sun's rays travel in straight lines called heat rays. When it moves that way, it is called radiation. When sunlight hits the earth, its radiation is absorbed or reflected. Darker surfaces absorb more of the radiation and lighter surfaces reflect the radiation.

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As a form of energy, the SI unit for heat is the joule (J), though heat is frequently also measured in the calorie (cal), which is defined as "the amount of heat required to raise the temperature of one gram of water from 14.5 degrees Celsius to 15.5 degrees Celsius." Heat is also sometimes measured in "British thermal units" or Btu.

A piece of buttered toast contains about 315 kilojoules (315,000 joules) of energy. With that energy you could: Jog for 6 minutes Bicycle for 10 minutes Walk briskly for 15 minutes Sleep for 1-1/2 hours Run a car for 7 seconds at 80 km/hLight a 60-watt light bulb for 1-1/2 hours

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The amount of heat energy (Q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature -initial temperature)

Q = m x Cg x (Tf – Ti)

Specific Heat Capacity (Cg) of a substance is the amount of heat required to raise the temperature of 1kg of the substance by 1oC (or by 1 K). The units of specific heat capacity are J oC-1 kg-1 or

J K-1 kg-1

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aluminium Cg = 897 water Cg = 4181.3

carbon Cg = 720 ethanol (ethyl alcohol) Cg = 2440

copper Cg = 385 sulphuric acid (liquid) Cg = 1420

lead Cg = 127 sodium chloride solid Cg = 850

mercury Cg = 139.5potassium hydroxide

solidCg = 1180

Specific Heat Capacities of Some Substances

[Cg (J K-1 kg-1 or J oC-1 kg-1)]

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Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC.

Q = m x Cg x (Tf - Ti) m = 0.25 kg Cg = 4181.3 J oC-1 kg-1

Tf = 56oC Ti = 20oC Q = 0.25 x 4181.3 x (56 - 20) Q = 0.25 x 4181.3 x 36 Q = 37 632 J = 37.6 kJ

Example 1

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216 J of energy is required to raise the temperature of aluminium from 15o to 35oC. Calculate the mass of aluminium.

Example 2

Q = m x Cg x (Tf - Ti) Q = 216 J Cg = 897 JoC-1kg-1

Ti = 15oC Tf = 35oC 216 = m x 897 x (35 - 15) 216 = m x 897 x 20 216 = m x 17940 m = 216 ÷ 17940 = 0.012kg = 12g

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Example 3

Calculate the time needed to increase the temperature of 50 l of water from 20oC to 50oC, if the power of electrical heater is 1500W.

Q = m x Cg x (Tf - Ti) m = 50 kg Cg = 4181.3 J oC-1 kg-1

Tf = 50oC Ti = 20oC Q = 50 x 4181.3 x (56 - 20) Q = 50 x 4181.3 x 30 Q = 6271950 JQ = P x tT = Q/P = 6271950/1500 = 4181.3 s = 1.16 hAssuming 85% efficiency it will take 1.45 h

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Mechanical energy

There are two main types of mechanical energy. They are motion energy and stored mechanical energy.

Motion energy: This is the energy, that something has because it is moving. Motion energy is also called kinetic energy.

Stored mechanical energy: This is energy, that something has stored in it because of its height above the ground or because it is stretched or bent or squeezed (e.g. in a stretched rubber band). Stored mechanical energy is also called potential energy.

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Kinetic energy calculation

For an object that is moving the kinetic energy equals one half times the mass of the object times the square of the speed of the object. In symbols:

EK = (1/2)mv2

Example 4

How much kinetic energy does an object have if its mass is 5.0 kg and it is moving at a speed of 4.0 m/s

EK = (1/2)(5.0 kg)(4.0 m/s)2 = 40 J

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Potential energy calculation

Energy that is stored in the gravitational field is called gravitational potential energy, or potential energy due to gravity.

Since the work done on the object when it is lifted becomes the gravitational potential energy, the formula for gravitational potential energy equals the mass of the object times the acceleration due to gravity times the height that the object is lifted, as in:

Eg = m g h (J)m= mass (kg)g = gravitational acceleration (9.81 m/s2)h = displacement (m)

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Example 4

What is the gravitational potential energy for a 4 kg object that is lifted 5 m?

Eg = m g h (J)Eg = 4kg x 9.81m/s2 x 5m = 196.2 (J)

Example 5

How many energy is needed to move 10 m horizontally an object applying a force of 500 N?

E = F x m (J)E = 500 N x 10 m = 5000 (J) = 5 (kJ)

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Example 6

How many water would be raised from 40 m depth if 1.2 kW motor of water pump was running for 5 min? Assume 97% efficiency of the motor and 70% efficiency of the pump.

Effective power is 1.2 x 0.97 x 0.7 = 0.815 kWE = P x t = 815 x (5 x 60) = 24450(J)

E = m g h (J)m = E / (g h) = 24450 (J) / [9.81(m/s2) x 40(m)] == 62.31 (l)

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CONVERSION OF ELECTRICAL ENERGY INTO HEAT & MECHANICAL ENERGY

E = P t = U I t(VAs = Ws = J)

E = m Cg Dt(Kg J Kg-1 0C-1 0C )

(J)

E = P F t(Pa m3 s-1 s )(N m-2 m3 =J)

Heat

Mechanical energy

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ILLUMINATION, QUANTITY & UNITS, SYMBOLS

Luminous flux

Luminous flux describes the total amount of light emitted by a light source or received by a surface

[f] = Lumen (lm)

Quantity of light

The quantity of light, or luminous energy, is a product of the luminous flux emitted multiplied by time; luminous energy is generally expressed in klmh.

Q = F * t (lmh)

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Luminous efficacy

Luminous efficacy describes the luminous flux of a lamp in relation to its electrical power and is therefore expressed in lumen per watt (lm/W). The maximum value theoretically attainable when the total radiant power is transformed into visible light is 683 lm/W. Luminous efficacy varies from light source to light source, but always remains well below this maximum value.

h = F

P=

lm

W

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Luminous intensity

An ideal point-source lamp radiates luminous flux uniformly into the space in all directions; its luminous intensity is the same in all directions.

In practice, however, luminous flux is not distributed uniformly. This results partly from the design of the light source, and partly on the way the light is intentionally directed.

It makes sense, therefore, to have a way of presenting the spatial distribution of luminous flux, i.e. the luminous intensity distribution of the light source.

Luminous intensity I is the luminous flux Fradiating in a given direction per solid angle W.

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spatial angle

luminousintensity

luminous flux F

W

lm

srI = = cd

Luminous flux is power that is radiated from a source in all directions (lm)

Luminous intensity is power that is radiated from a source in specific direction (cd)

Typical values for luminous intensity

LED 0.005 cd

Candle 1 cd

100W incandescent bulb 150 cd

Automobile headlamp (high beam) 10000 cd

Lighthouse 300000 cd

Flash tube (peak value) 1000000 cd

;=

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The unit for measuring luminous intensity is candela (cd). The candela is the primary basic unit in lighting technology from which all others are derived.The candela was originally defined by the luminous intensity of a standardized candle.Later thorium powder at the temperature of the solidification of platinum was de-fined as the standard; since 1979 the candela has been defined by a source of radiation that radiates 1/683 W per steradian at a frequency of 540 ·1012 Hz.

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If a uniform point light source of 1 cd luminous

intensity (I) ―about the intensity of a normal wax

candle‖ is positioned at the center of a sphere of 1 m

radius, then every area of 1 m2 on the inside of that

sphere will receive a luminous flux of 1 lm.

I = 1 cd

W = A1/r12 = A2/ r2

2 (sr)

r1

r2

A1

A2

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Illuminance

Illuminance is the means of evaluating the density of luminous flux. It indicates the amount of luminous flux from a light source falling on a given area. Illuminance need not necessarily be related to a real surface. It can be measured at any point within a space. Illuminance can be determined from the luminous intensity of the light source. Illuminance decreases with the square of the distance from the light source (inverse square law).

E = F

A=

lm

m2= lx

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Luminance

Whereas illuminance indicates the amount of luminous flux falling on a given surface, luminance describes the brightness of an illuminated or luminous surface. Luminance is defined as the ratio of luminous intensity of a surface (cd) to the projected area of this surface (m2).

In terms of visual perception, weperceive luminance. It is anapproximate measure of how ―bright‖a surface appears when we view itfrom given direction.

L = I/Ap= [cd/m2 = nit]

luxnit

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Horizontal luminance Eh and vertical luminance Ev in interior spaces.

Average illuminance Em is calculated from the luminous flux F falling on the given surface A.

E = F / A [ lm/m2 = lx ]

E = I * W / A = I * (A/r2) / A

E = I / r2 [ cd/m2 ]

inverse square law

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The illuminance of a surface

E’ = F’/A

a

Fa

F`

The illuminance of a point

E’ = F*cos a/A

E‘ = E*cos a

E = Ig /r2

Ev = E*cosg = (Ig /r2)*cosg

Eh = E*sing = (Ig /r2)*sing

g

Eh

Ev

E

h

r

Ig

Mean value Em = SEi

ni=1

n

a is the angle between

illuminated surface and

the plane perpendicular

to light

COSINE LAW, POINT TO POINT METHOD

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I * cos3 g

h2

g

Ex

h

r

Light source

xy

=

I

Ey =I

h2

Ex =I * cos g

r2

cos g =h

r

r = h

cos g

Example:

I = 20000 cdH = 8 m

0

50

100

150

200

250

300

350

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71

illu

min

ati

on

(lx

)

angle

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Exercise 1

Example 5

A lamp of 1200 cd is placed 3.5 m above a surface. Find out the illumination at point ―X‖ 2 meters away from the centre line.

I * cos3 g

h2Ex = = 1200 cos3 29.75/3.52 = 64.1 lx

g = tan-1 (2/3.5) = 29.750

Two lamps being fixed 6 m apart on the height of 3 m above a surface. They are same with luminous intensity of 1500cd. Find out the illumination on the surface midway between the lamps and below each lamp.

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Exercise 2

A short, level driveway is illuminated by three 800cd lamps mounted in a straight line 6 m apart and 2.5 m above the road surface. Calculate the illuminance of the driveway halfway between two lamps.

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PHOTOMETRY AND LIGHTMETERS

Photometry is the science of measuring visible light in units that are weighted according to the sensitivity of the human eye.

For the visible part of the spectrum (380nm -780nm) a separate set of parameters is defined. These photometric values derive from the radiometric quantities by weighting them with the spectral response function for intensity of thehuman eye.

Luminous Flux [lm]Luminous Intensity [lm/sr = cd]Illuminance [lm/m² = lux]Luminance [cd/m²]

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Light meters

A photocell light meter is an instrument that directly measures the illuminance on a surface. The electrical resistance of some semiconductors, such as selenium, changes with exposure to light and this property is used in an electrical circuit connected to a galvanometer. This meter may be calibrated in lux or foot-candles.

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Measuring the lighting qualities of a lighting installation can serve a number of purposes. In the case of new installations measurements are taken to check that the planned values have been obtained. Measurements recorded on existing installations help the planner to decide what maintenance or renovation work is required.Measurements can also be taken during the planning process for the evaluation and comparison of lighting concepts. The factors that are measured are initially illuminance and luminance.To ensure that results of measurements taken are usable the measuring equipment must be of a suitably high quality.

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When measuring a lighting installation, a series of parameters have to be taken into account and documented in a report.This initially involves the recording of specific qualities of the environment, such as reflectance factors and colours of room surfaces, the time of day, the amount of daylight and the actual mains voltage.Features of the lighting installation are then recorded: the age of the installation, the lighting layout, the types of luminaries, the type and condition of the lamps and the overall condition of the installation. The type of measuring equipment and the class of accuracy of the measuring device has to be recorded.

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To record illuminance for an entire space, a floor plan is made of the space and has to include furniture. The arrangement of luminaries and the points at which measurements are to be taken are then entered. The measuring points are the central points on a 1-2 m grid, in the case of high rooms up to a 5 m grid.Measurements can also be taken at individual workplaces, in which case an overall light measuring grid is created for the area.Horizontal illuminance is measured at the individualmeasuring points at the height of the working plane of 0.85 mCylindrical illuminance for determining the formation of shadows on a 1.2 m plane of reference.

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Measuring illuminance on the working plane in empty or open furnished spaces ismade according to a regular grid of 1 to 2 meters.

Measuring points for the measurement ofilluminance at workplaces.

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LUMEN METHOD CALCULATIONS

The lumen method is used to acquire a rough estimation of the dimensioning of a lighting installation; it allows the designer to determine the number of luminaries required to produce the defined illuminance on the working plane, or viceversa, the illuminance on the working plane produced by a given number of luminaries.

This method does not provide exact illuminance at specific points in the space, which means that other methods must be applied to calculate the uniformity of a lighting installation or to determine illuminance levels at specific points.

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The deciding factor in this calculation is the utilisation, which is derived from the geometry of the space, the reflectance of the room surfaces and the efficiency and the distribution characteristics of the luminaries used.

To be able to calculate the appropriate utility in each individual case, there are tables available, which contain the utility of a standardised space with changing room geometry, changing reflection factors and luminaries with a variety of distribution characteristics.

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The lumen method formula is easiest to understand in the following form.

E = (n × N × F × UF × LLF)/A (lux)

E = average illuminance over the horizontal working plane (lx)n = number of lamps in each luminaryN = number of luminaryF = lighting design lumens per lamp, i.e. initial bare lamp luminous flux (lm)UF = utilisation factor for the horizontal working planeLLF = light loss factor (maintenance factor)A = area of the horizontal working plane

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Utilisation Factor

Utilisation factor (UF) is the proportion of the luminous flux emitted by the lamps and luminous flux which reaches the working plane. It is a measure of the effectiveness of the lighting scheme. Factors that affect the value of UF are as follows:

(a) light output ratio of luminary(b) flux distribution of luminary(c) room proportions(d) room reflectance(e) spacing/mounting height ratio

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The Utilisation factor (UF) can be read off the table from the column showing the corresponding room index and line showing the appropriate combination of reflectance factors of ceiling (rC), walls (rW) and floor (rF) or for greater accuracy, calculated through interpolation.

Room Reflectance

The room is considered to consist of three main surfaces:(a) the ceiling cavity,(b) the walls, and(c) the floor cavity (or the horizontal working plane).

The effective reflectance of the above three surfaces affect the quantity of reflected light received by the working plane.

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Room proportion

L = length of roomW = width of roomHm = mounting height, i.e. the vertical distance between the working plane and the luminaries.

RI =L*W

Hm * (L+W)

The room index RI describes the influence of the room geometry on the utilisation factor. It is calculated from the length and width of the room, and the height h above the working plane under direct luminaries and height Hm above the working plane under predominantly indirect luminaries.

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narrow-beamluminaries(A 60,DIN 5040)

Utilisation factor UF for typical interior luminaries

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wide-beamluminaries(A 40, DIN 5040)

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indirect luminaries(E 12, DIN 5040)

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Light output ratio of luminary (LOR) takes into account for the loss of light energy both inside and by transmission through light fittings. It is given by the following expression.

LOR = Output of luminary / Output of lamp

Flux Fraction of Various Luminaries

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Light Loss (maintenance factor) Factor

Light loss factor (LLF) is the ratio of the illuminance produced by the lighting installation at the some specified time to the illuminance produced by the same installation when new. It allows for effects such as decrease in light output caused by(a) the fall in lamp luminous flux with hours of use,(b) the deposition of dirt on luminaries, and(c) reflectance of room surfaces over time.

In fact, light loss factor is the product of three otherfactors:

LLMF = lamp lumen maintenance factorLMF = luminaries maintenance factorRSMF = room surface maintenance factor

LLF = LLMF x LMF x RSMF

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Lamp lumen maintenance factor (LLMF) is the proportion of the initial light output of a lamp produced after a set time to those produced when new. It allows for the decline in lumen output from a lamp with age. Its value can be determined in two ways:(a) by consulting a lamp manufacturer's catalog for a lumen depreciation chart, and(b) by dividing the maintained lumens by the initial lamps lumens.

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Luminaire maintenance factor (LMF) is the proportion of the initial light output from a luminaries after a set time to the initial light output from a lamp after a set time. It constitutes the greatest loss in light output and is mainly due to the accumulation of atmospheric dirt on luminaries. Three factors must be considered in its determination:(a) the type of luminaries,(b) atmospheric conditions, and(c) maintenance interval.

Room surface maintenance factor (RSMF) is the proportion of the illuminance provided by a lighting installation in a room after a set time compared with that occurred when the room was clean. It takes into account that dirt accumulates on room surfaces and reduces surface reflectance..

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Light loss factor LLF in relation to the degree of deterioration in the space.

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Example 1

Reflection factors:ceiling 80%walls 80%working surface 30%LLF = 0.8

Project data:

Eavg = 350 lxF= 6700 lm (2x36W TL36W)room length L=8 mroom with W=4 mmounting height Hm = 2.15 mheight of the working surface = 0.85 m

K 883 853 833 553 533 772 752 732 881 851 831 551 531 331

0.60 51 33 27 32 26 40 32 27 46 32 26 30 26 26

0.80 58 41 34 39 33 47 39 33 52 38 33 37 32 32

1.00 63 46 40 44 38 52 44 38 55 43 38 41 37 36

1.25 68 53 47 50 45 57 50 45 59 49 44 47 43 42

1.50 71 57 51 53 49 60 54 49 62 52 47 50 46 46

2.00 75 63 58 59 54 64 59 54 64 57 52 54 51 50

2.50 78 68 63 62 59 67 62 58 66 60 56 58 55 54

3.00 80 71 67 65 62 69 65 62 68 62 59 60 58 57

4.00 83 75 71 68 65 71 68 65 69 65 62 62 60 59

5.00 84 78 74 70 68 73 70 68 70 67 64 64 63 61

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k =L x W

Hm x (L+W)

room index

=8 x 4

2.15 x (8+4)= 1.24

E = (n × N × F × UF × LLF)/A (lux)

N = 350 x (8 x 4)

6700 x 0.68 x 0.8= 3.07 = 3

number of light fittings

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Optimal positioning of the light fittings

For optimal distance between the fittings (d) is given the following formula:

d = Hm xe

hHmh

e

e = 3.74 m

d = 2.15 x3.74

3.00= 2.68 m

d

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The illuminance at a point Ep is calculated from the luminous intensity I and the distance ―a‖ between the light source and the given point.

Ep = I / a2

@ 00 170 cd / 1000 lm

If TLD 36W gives flux of 2350lm, thenI = 0.17x 2350 = 400 cd on 2m distance there would be illuminance ofE = I / a2 = 100 lx.

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MOVING COIL AND MOVING IRON VOLTMETERS AND AMPERMETERS

Measuring instruments are classified according to both the quantity measured by the instrument and the principle of operation. Three general principles of operation are available:(i) electromagnetic, which utilizes the magnetic effects of electric currents(ii) electrostatic, which utilizes the forces between electrically-charged conductors(iii) electro-thermal, which utilizes the heating effect.

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The most common analogue instrument or meter is the permanent magnet moving coil instrument and it is used for measuring a dc current or voltage of a electric circuit.

Moving-iron instruments are generally used to measure alternating voltages and currents. In moving –iron instruments the movable system consists of one or more pieces of specially-shaped soft iron, which are so pivoted as to be acted upon by the magnetic field produced by the current in coil.

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Permanent Magnet Moving CoilInstruments (PMMC)

A moving coil instrument consists basically of a permanent magnet to provide a magnetic field and a small lightweight coil is wound on a rectangular soft iron core that is free to rotate around its vertical axis.

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PMMC instrument

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When a current is passed through the coil windings, a torque is developed on the coil by the interaction of the magnetic field and the field set up by the current in the coil. The aluminium pointer attached to rotating coil and the pointer moves around the calibrated scale indicates the deflection of the coil. A balance weight is also attached to the pointer to counteract its weight.

If the coil is carrying a current of I amps, the force on a coil side

F = B I l n (N)

n= number of turns of the coil B = flux density in (T)l = length of the vertical side of

the coil in (m) I= current in (A)

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In order to return the coil to its original position when there is no current through the coil, a hairsprings attached to each end of the coil. These hairsprings are not only supplying a restoring torque but also provide an electric connection to the rotating coil. With the use of hairsprings, the coil will return to its initial position when no current is flowing though the coil. The springs will also resist the movement of coil when there is current through coil. When the developing force between the magnetic fields (from permanent magnet and electro magnet) is exactly equal to the force of the springs, the coil rotation will stop.

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The resulting torque in a coil or motion of a coil in a magnetic field is due to the combined effect of deflecting torque, controlling torque and damping torque.

The interaction between the induced field and the field produced by the permanent magnet causes a deflecting torque, which results in rotation of the coil.

The value of control torque depends on the mechanical design of the control device. For spiral springs, the controlling torque is directly proportional to the angle of deflection of the coil.

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As the coil moves in the field of the permanent magnet, eddy currents are set up in the metal core. The magnetic field produced by the eddy currents opposes the motion of the coil. The pointer will therefore swing more slowly to its proper position and come to rest quickly with very little oscillation. This is a dumping torque.

When the moving system reached at steady state i.e. at final deflected position, the controlling torque becomes equal and opposite to the deflecting torque. The deflecting angle is directly proportional to the current in the movable coil.

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A multi-range ammeters and voltmeters

An ammeter is required to measure the current in a circuit and it therefore connected in series with the components carrying the current.

For higher range ammeters a low resistance made up of manganin (low temperature coefficient of resistance) is connected in parallel to the moving coil and instrument may be calibrated to read directly to the total current.

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Swamping resistance (manganin) which has a temperature coefficient practically zero is connected in series with the coil resistance in order to reduce the error due to the variation of resistance (temperature change) of the moving coil. The swamping resistance is usually three times that of coil thereby reducing a possible error of, say, 4% to 1%.

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A PMMC instrument has a coil resistance of 100Ω and gives a full-scale deflection (FSD) for a current of 500 μA. Determine the value of shunt resistance required if the instrument is to be employed as an ammeter with a FSD of 5 A.

Example 1

Rsh =100 * 0.0005

5-0.0005= 0.01 W

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Multi-range voltmeters is constructed by a connecting a resistor in series with a PMMC instrument. Unlike an ammeter, a voltmeter should have a very high resistance R and it is normally connected in parallel with the circuit where the voltage is to be measured.

The moving coil instruments can be suitably modified to act either as an ammeter or as a voltmeter.

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Example 2

A PMMC meter with a coil resistance 100Ω and a full scale deflection current of 100 μA is to be used as voltmeter. The voltmeter ranges are to be 50, 100 and 150V. Determine the required value of resistances for each range.

R50 =50 - 100 * 0.0001

0.0001= 0.4999 MW

R100 =100 - 100 * 0.0001

0.0001= 0.9999 MW

R150 =150 - 100 * 0.0001

0.0001= 1.4999 MW

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Moving-iron Instruments

There are two general types of moving-iron instruments:

1. Repulsion (or double iron) type2. Attraction (or single-iron) type.

The deflecting torque in any moving-iron instrument is due to forces on a small piece of magnetically ‗soft‘ iron that is magnetized by a coil carrying the operating current.

Controlling torque consists of1. Spring control (repulsion type)2. Gravity control (attraction type)

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The deflecting torque is proportional to the square of the current in the coil, making the instrument reading is a true ‗RMS‘. Rotation is opposed by a hairspring that produces the restoring torque.

Moving iron instruments having scales that are nonlinear and more dense in the lower range of calibration.

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In repulsion type moving–iron instrument consists of two cylindrical soft iron vanes mounted within a fixed current-carrying coil. One iron vane is held fixed to the coil frame and other is free to rotate, carrying with it the pointer shaft. Two irons lie in the magnetic field produced by the coil that consists of only few turns if the instrument is an ammeter or of many

turns if the instrument is a voltmeter. Current in the coil induces both vanes to become magnetized and repulsion between the similarly magnetized vanes produces a proportional rotation.

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Attractive type instrument consists of a few soft iron discs (B) that are fixed to the spindle (D), pivoted in jewelled bearings. The spindle (D) also carries a pointer (P), a balance weight (W1), a controlling weight (W2) and a damping piston (E), which moves in a curved fixed cylinder (F).

At equilibrium i.e. for steady deflection,Deflecting torque = Controlling torque.

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Shunts and multipliers for MI instruments

For moving-iron ampermeters it is difficult to design a shunt with the appropriate inductance, and shunts are rarely incorporated in moving iron ammeters. Thus the multiple ranges can effectively be obtained by winding the instrument coil in sections which may be connected in series, parallel or series-parallel combination which in turn changing the total ampere-turns in the magnetizing coil.

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For moving-iron voltmeters: Voltmeter range may be altered connecting a resistance in series with the coil. Hence the same coil winding specification may be employed for a number of ranges.

An ordinary arrangement with a non-inductive resistance in series with the fixed coil –results in error that increases as the frequency increases.

The change of impedance of the instrument with change of frequency introduces error in signal measurements. In order to compensate the frequency error, the multiplier may be easily shunted by the capacitor.

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INDUCTION WATTMETERS AND ENERGY METERS

It has two laminated electromagnets. One (series) is excited by the current in the main circuit, other (shunt) by current proportional to the voltage of the circuit.

Induction wattmeter

A thin aluminium disc cuts the AC fluxes and two eddy currents are produced with associated fluxes. The deflection torque is produced due to interaction of these eddy currents fluxes and inducing fluxes and is proportional to RMS of current and voltage.

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This instrument is spring controlled, which is fixed on spindle of the moving system which carries a pointer. The scale is uniform and extended over 3000

Normally it can handle around 100A current. In case of greater current, it should be connected to current transformer.

The main difference between energy meter and a wattmeter is that energy meter is equipped with registration mechanism, so that all instantaneous readings of power are summed over a definite period of time.

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Main parts of the induction energy meter

1.Current coil and

magnetic circuit

2.Voltage coil and

magnetic circuit

3. Rotating disk

4. Disk axis

5. Permanent magnet

6. Display

Permanent magnet is there to control the speed once the disc rotates. The speed of rotating disc can be controlled by changing magnet‘s position.

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Wiring of 3 phase energy meters

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PRIMARY AND SECONDARY CELLS

The term voltaic cell is defined as a combination of materials used to convert chemical energy into electrical energy. A voltaic or chemical cell consists of two electrodes made of different types of metals or metallic compounds placed in an electrolyte solution.

A battery is a group of two or more connected voltaic cells.

An electrode is a metallic compound, or metal, which has an abundance of electrons (negative electrode -anode) or an abundance of positive charges (positive electrode - cathode).

144

An electrolyte is a solution which is capable of conducting an electric current. The electrolyteof a cell may be a liquid or a paste. If the electrolyte is a paste, the cell is referred to as a dry cell; if the electrolyte is a solution, it is called a wet cell.

An ampere-hour is defined as a current of one ampere flowing for one hour. If you multiply the current in amperes by the time of flow in hours, the result is the total number of ampere-hours.Ampere-hours are normally used to indicate theamount of energy a battery can deliver.

145

The purpose of a battery is to store chemical energy and to convert this chemical energy through the chemical reaction into electrical energy when needed.

A voltaic cell develops a potential difference when electrodes of two different metals are immersed in an electrolyte. One electrode accumulates a positivecharge. The potential difference is due to the difference in charge between the two electrodes.

146

Primary Cells

Cells that cannot be returned to good condition, or recharged after their voltage output has dropped to a value that is not usable, are called primary cells. Dry cells that are used in flashlights and remote controllers e.g. Leclanché Cells ( AA, AAA, C, D) or mercury cells (button cells) are examples of primary cells.

Secondary Cells

Cells that can be recharged to nearly their original condition are called secondary cells. The most common example of a secondary, or rechargeable cell, is the lead-acid automobile battery.

147

Leclanché Cell

A electrolytic cell also known as a dry cell that uses a moist paste rather than a liquid as an electrolyte. Dry cells with a zinc cup for an anode, a carbon rod for a cathode, and a paste made of powdered carbon, Ammonium chloride, Zinc Chloride, and Manganese dioxide for an electrolyte.

148

The Leclanché Cell (carbon-zinc) cell is one of the oldest and most widely used types of dry cells. The carbon in the battery is in the form of a rod in the centre of the cell which acts as the positive terminal.The case is made from zinc and acts as the negative electrode. The electrolyte for this type of cell is a chemical paste-like mixture which is housed between the carbon electrode and the zinc case. The cell is then sealed to prevent any of the liquid in the paste from evaporating.The advantage of a carbon-zinc battery is that it is durable and very inexpensive to produce. It has a good shelf life.Disadvantages are high internal resistance and limitation of 1.5 volts.

149

Mercury Cell

Mercury cells come in two types; one is a flat cell that is shaped like a button, while the otheris a cylindrical cell that looks like a regular flashlight battery. Each cell produces about 1.35 volts. These cells are very rugged and have a relatively long shelf life. The mercury cell has the advantage of maintaining a fairly constant output under varying load conditions. For this reason, they are used in products such as electric watches, hearing aids, cameras, and test instruments.

150

New silver-oxide cell(zero-mercury, zero lead)

Silver-oxide cell(conventional)

151

The Lead Acid Cells consists of a series of cells, with each cell containing a lead peroxide positive plate and a lead negative plate immersed in a dilute sulphuric acid solution. This sulphuric

Lead Acid Cells – secondary cell

acid solution is known aselectrolyte. The whole arrangement is kept in a leak-proof casing. Each cell delivers around 2 volts and when six cells are connected in series there would be 12 V.

152

When a lead-acid battery is discharged, electrolyte and the active material on the plates of the battery are consumed to produce water and lead sulphate.

When a lead-acid battery is charged, electrical energy is added to the battery, causing the water and lead sulphate to be consumed and produce electrolyte and active material.

153

Voltage and Specific Gravity During Charge and Discharge

154

Nickel Cadmium Cell - secondary cell

The nickel-cadmium cell is a secondary cell, and the electrolyte is potassium hydroxide. The negative electrode is made of nickel hydroxide, and the positive electrode is made of cadmium hydroxide. The nominal voltage of a nickel-cadmium cell is 1.25 volts. The nickel-cadmium battery has the advantage of being a dry cell that is a true storage battery with a reversible chemical reaction (i.e., it can be recharged). The nickel-cadmium battery is a rugged, dependable battery. It gives dependable service under extreme conditions of temperature, shock, andvibration. Due to its dependability, it is ideally suited for use in portable communications equipment.

155

INTERNAL RESISTANCE, EMF, TERMINAL VOLTAGE

Internal resistance in a chemical cell is due mainly to the resistance of the electrolyte between electrodes. Any current in the battery must flow through theinternal resistance. The internal resistance is inseries with the voltage of the battery (EMF), causing an internal voltage drop.

With no current flow, the voltage drop is zero; thus, the full battery voltage (EMF) is developed across the output terminals (VB). If a load is placed on the battery, load resistance (RL) is in series with internal resistance (Ri).

156

When current flows in the circuit (IL), the internal voltage drop (IL * Ri) drops the terminal voltageof the battery. Thus, internal resistance reduces both the current and voltage available to the load.

VL = EMFB – IL * Ri

157

SERIES, PARALLEL AND SERIES-PARALLEL COMBINATION

When several cells are connected in series, the totalvoltage output of the battery is equal to the sum of the individual cell voltages. In the example ofthe battery where four 1.5V cells provide a total of6 volts. When we connect cells in series, the positive terminal of one cell is connected to the negativeterminal of the next cell. The current flow through a battery connected in series is the same as for one cell.

Ri =Ri1+Ri2+Ri3+Ri4+Ri5 (W)

E =E1+E2+E3+E4+E5 (V)

158

Cells connected in parallel, give the battery a greater current capacity. When cells are connected in parallel, all the positive terminals are connected together, and all the negative terminals are connected together. The total voltage output of a battery connected in parallel is the same as that of a single cell. Cells connected in parallel have the same effect as increasing the size of the electrodes and electrolyte in a single cell.

E=E1=E2=E3 (V)Ri = (W)

Ri=Ri1=Ri2=Ri3 (W)

Ri

3

159

The total voltage is equal to the sum of the voltages of each series connected cells. In our case 4.5 V.

Serial parallel combination

160

METHOD OF CHARGING

Classification by application

Main power source Stand by power source

Constant voltage

Constant voltage/Constant current

Two stepconstant voltage

Compensating(trickle/floating )

charge

161

Constant Voltage: A constant voltage charger is basically a DC power supply which in its simplest form may consist of a step down transformer from the mains with a rectifier to provide the DC voltage to charge the battery. The lead-acid cells used for cars and backup power systems typically use constant voltage chargers. In addition, lithium-ion cells often use constant voltage systems, although these usually are more complex with added circuitry to protect both the batteries and the user safety.

Constant Current: Constant current chargers vary the voltage they apply to the battery to maintain a constant current flow, switching off when the voltage reaches the level of a full charge. This design is usually used for nickel-cadmium and nickel-metal hydride cells or batteries.

162

Constant voltage / constant current method

This method charges the battery by controlling the current at 0.4 Ah and controlling the voltage at 2.45V/cell(unit battery) at room temperature of 200C to 250C. Proper charging time is 6 to 12 hours depending of the discharge rate.

163

Constant voltage / constant current method

Two step constant voltage charge control method uses two constant-voltage devices. At the initial stage, the battery is charged by the first constant voltage device of high setup voltage. When the charge current, the value of which is detected bythe current detection circuit, has reduced to the present value, the device is switched over to the second low setup voltage, for trickle charge voltage.

164

Flooded Lead-Acid battery charging typically have a IUI

charge profile of the following:

Bulk charge (I): Initially the battery is charged at a constant

current rate until the cell voltage reaches a preset value -

normally a voltage near to that at which gassing occurs.

Constant Voltage (U): When the preset voltage has been

reached, the charger switches into the constant voltage

phase and the current drawn by the battery will gradually

drop until it reaches another preset level.

Equalize (I): Finally the charger switches again into the

constant current mode and the voltage continues to rise up

to a new higher preset limit when the charger is switched off.

This last phase is used to equalize the charge on the

individual cells in a gassing process.

Method of charging lead-acid battery

165

is a measure (typically in Amp-hr) of the charge stored by the battery, and is determined by the mass of active material contained in the battery. The more electrolyte and electrode material there is in the cell, the greater the capacity of the cell. Thus a small cell has less capacity than a larger cell, given the same chemistry The battery capacity represents the maximum amount of energy that can be extracted from the battery under certain specified conditions. However, the actual energy storage capabilities of the battery can vary significantly from the "nominal" rated capacity, as the battery capacity depends strongly on the age and past history of the battery, the charging or discharging regimes of the battery and the temperature.

Battery capacity

166

Efficiency of a Lead Acid Cell

ampere-hour efficiency =

output(discharge), Ah x 100%

input (charge), Ah=

(around 85%, for good battery)

watt-hour efficiency =

average discharge, Wh x 100%

average charge, Wh=

(around 70%, for good battery)

167

Example 1

Discharged 12V battery is charged for 10h at 12A, at average charging terminal voltage of 15V. When connected to the load, a current of 10A for 8h at an average terminal voltage of 12V discharges the battery. Find out the ampere-hour and watt-hour efficiency.

ampere-hour efficiency = 10 x 8 x 100%

10 x 12=66.7%

watt-hour efficiency = 10 x 8 x 12 x100%

10 x 12 x 15=53.3%

168

MAGNETISM AND ELECTROMAGNETISM

There are two theories that explain magnetism. One states that magnet is is made up of a very large number of small, magnetized particles. When a bar of iron is not magnetized, the small magnetic particles are arranged in a random manner. When the bar of iron becomes a magnet, the magnetic particles are aligned so that their individual magnetic effects add together to form a strong magnet.

Demagnetized iron Magnetized iron (saturated)

169

The other theory of magnetism is associated with the electron. The orbiting electrons cause circulating currents and form microscopic magnetic dipoles. In addition, both the electrons and the nucleus of an atom rotate (spin) on their own axes with certain magnetic dipole moments.

orbital motion spin of an electron

170

In the absence of an external magnetic field the magnetic dipoles of the atoms of most materials (except permanent magnets) have random orientations, resulting in no net magnetic moment. The application of an external magnetic field cause both an alignment of magnetic moments of the spinning electrons and an induced magnetic moment due to a charge in orbital motion of electrons.

171

A fundamental law of magnetism state that unlike poles attract each other and like poles repel each other.

172172

Magnetic flux

The direction of a line of magnetic flux at any point in a non magnetic medium, such as air, is that of the north seeking pole of a compass needle placed at that point.

Each line of magnetic flux forms a closed loop.

Lines of magnetic flux never intersect

Lines of magnetic flux are like stretched elastic cords, always trying to shorten themselves.

Lines of magnetic flux which are parallel and in the same direction repel one other and vise versa.

Characteristics of lines of magnetic flux are:

Magnetic flux is given symbol F, and is measure of the magnetic field: its unit is the weber [Wb]

173

Magnetic Field

Currents produce magnetic fields, a phenomenon described mathematically by the Biot-Savart Law and Ampère's Law. The magnetic field generated by a current travels in a circular path around the current in a plane perpendicular to the flow of charge i.e. current.

It is a unique and fundamental property of magnetic field that, unlike electric field, does not begin on a charge and end on a charge. On the contrary, magnetic fields close in on themselves, forming a circular field path.

174

Right hand rule

The magnetic field in space around an electric current is proportional to the electric current which serves as its source, just as the electric field in space is proportional to the charge which serves as its source.

(T)

m0 permeability of free space or vacuum=4p 10-7 (H/m)

175

Magnetomotive force

I

N

F

Magnetomotive force symbol F, is the force which establish magnetic flux in the magnetic circuit (its analogy in the electric circuit is e.m.f. which establish a current in the electrical circuit).

F = N x I (Ampere-turn)

176

Magnetic field intensity (strength)

Magnetic field intensity, symbol H is defined as m.m.f. per unit length of the magnetic circuit

I

l H =F

l

N x I=

l[

A

m]

F = H x l A[ ]

= H x l A[ ]N x I

177

Magnetic flux density

Magnetic flux density symbol B, is the amount of flux passing through unit area perpendicular to the ―direction‖ of the flux.

The unit of flux density is the tesla.

T=Wb

m2B =

F[ ]

A

F

A

178

Permeability

Permeability symbol (m) is the specific measure of a material's acceptance of magnetic flux, analogous to the specific resistance of a conductive material (ρ), except inverse (greater permeability means easier passage of magnetic flux, whereas greater specific resistance means more difficult passage of electric current).

Permeability of free space or vacuum and non magnetic materials, symbol m0 is defined as

m0 =B

H= 4 p 10-7 [

H

m]

179

For ferromagnetic material permeability increases by

factor mr, called relative permeability, where mr > 1

(up to 7000 and even more).

B(T)

1.4

0

0 10000

H(A/m)

cast iron

cast steelmr

800

0

0 10000

H(A/m)

cast iron

cast steel

Absolute permeability

=B

H=mr m0m B = mr m0 HX

180

Example 1

A coils of 200 turns is wound uniformly over a wooden ring having a mean circumference of 600 mm and a uniform cross sectional area of 500 mm2. If the current through the coil is 4 A, calculate:

(a) the magnetic field strength,(b) the flux density, and(c) the total flux

H =N x I

l=

200 x 4

0.6= 1333 A/m

B =mr m0 H = 1 x 4p10-7 x 1333 = 1.675 mT

F = B x A = 1675 x 10-6 x 500 x 10-6 = 0.8375 mWb

181

E = I * RF = F * S

Resistance, R [W]Reluctance, S [A / Wb]

Current, I [A]Flux, F [ Wb]

Electromotive force, E [V]Magnetomotive force, F [A]

Electric circuitMagnetic circuit

Permeability, m [H / m] Conductance, G [S]

Comparison of magnetic and electric circuit

Reluctance S

F = B * A [ T * m2 = Wb ]

F = H * l [ A / m * m = A ]S =

F

F=

H*lB*A

=1

m*

lA

1

mr* mo*

lA

=

In electrical circuit resistance = r *lA

[ A

Wb

]

[ W ]R

182

Exercise 1

An iron circuit with a small air gap cut in it. A 6000 turn coil carries a current I=20 mA which sets up a flux within the iron and across the air gap. If the iron cross section is 0.8×10-4 m2, the mean length of flux path in iron is 0.15 m, mr=800 in iron and air gap length is 0.75 mm, calculate the air gap flux density. It may be assumed that the flux lines flow straight across the air gap, i.e. air gap cross section is also 0.8×10-4 mm2.

Answer: 0.16 T

183

Electromagnetic induction

Faraday and Henry have discovered that a voltage can be induced in a conductor which is moving relative to an external magnetic field. A current will flow if a complete circuit is present.

Whenever the magnetic field in the region of a conductor is moving, or changing in magnitude, such that magnetic field lines are moving across the conductor, an electric current is induced in the conductor, if the conductor is part of a complete circuit.

184

Faraday's law of electromagnetic induction states that:

E = N-dF

dt

N is number of turns

The emf induced in an electric circuit always acts in such a direction that the current it drives around the circuit opposes the change in magnetic flux which produces the emf.

dF

dtis rate of change of flux linkages

Lenz‘s gives the direction of the induced emf saying:

[ V ]

E is the electromotive force (emf) in volts

F is the magnetic flux in weber

185

Three principal methods of inducing an e.m.f.

1. Self induction

dt

tdNte

)()(

f-=

186

2. Induction by motion

vBdt

de =-= l

f

187

2. Induction by motion (rotation)

A’ = A cos a

A

)cos()( tABdt

dN

dt

dNte w

f-=-=

tABNte ww sin)( =

188

3. Mutual induction

N2 dt

tdNte

)()(

22

f-=

189

Force on a current-carrying conductor in a magnetic field

When a current-carrying conductor is placed in a magnetic field, there is an interaction between the magnetic field produced by the current and the permanent field, which leads to a force being experienced by the conductor.

F = B l I (N)

190

Force between parallel current-carrying conductors

If currents pass along two parallel wires, each wire will set up a magnetic field and the fields will interact

The force is directlyproportional to the currents I1 and I2and length l, andinversely proportional to distance d.

F = m0

I1 I2

2 p dl [N ]

191

AC and DC

DC stands for ―Direct Current,‖ meaning voltage or current that maintains constant polarity or direction, respectively, over time.

AC stands for ―Alternating Current,‖ meaning voltage or current that changes polarity or direction, respectively, over time.

192

An alternating current is thus one which rises in one direction to a maximum value, before falling to zero and repeating in the opposite direction. Instead of drifting steadily in one direction, the electrons forming the current move backwards and forwards in the conductor.

The time taken for an alternating quantity to complete its pattern (to flow in both directions and then return to zero) is called the periodic time (symbol T) for the system, which is said to complete one cycle in this time.

193

The number of complete cycles traced out in a given time is called the frequency (symbol f ), usually expressed in hertz (Hz), which are cycles per second (c/s). If there are f cycles in one second, each cycle takes 1/f seconds, so that

)(1

HzT

f =)(1

sf

T =

A frequency of 50 Hz is the standard for the supply system in many parts of the world, including the Malta, but 60 Hz systems are also common for mains supplies.

)(02.050

11s

fT ===

194

Advantages of AC systems

(a) An alternating-current generator (often called an alternator) is more robust, less expensive, requires less maintenance, and can deliver higher voltages than its DC counterpart.(b) The power loss in a transmission line depends on the square of the current carried(P = I2R). If the voltage used is increased, the current is decreased, and losses can be made very small. The simplest way of stepping up the voltage at the sending end of a line, and stepping it down again at the receiving end, is to use transformers, which will only operate efficiently from AC supplies.(c) Three-phase AC induction motors are cheap, robust and easily maintained.

195

(d) Energy meters, to record the amount of electrical energy used, are much simpler for AC supplies than for DC supplies.(e) Discharge lamps (fluorescent, sodium, mercury vapour etc.) operate more efficiently from AC supplies, although filament lamps are equally effective on either type of supply.(f) Direct-current systems are subject to severe corrosion, which is hardly present with AC supplies.

196

Peak, average and rms values of sinusoidal waves

The alternating current or voltage changes continuously, so that it is not possible to state its value in the same simple terms that can be used for a direct current.

Instantaneous values are the values at particular instants of time, and will be different for different instants. Symbols for instantaneous values are small symbols,v(t) for voltage, i(t) for current

Maximum or peak values are the greatest values reached during alternation, usually occurring once in each half-cycle. Maximum values are indicated by Umfor voltage, Im for current and so on.

197

Average or mean value is the average value of the current or voltage. If an average value is found over a full cycle, the positive and negative half-cycles will cancel out to give a zero result if they are identical. In such cases it is customary to take the average value over a half-cycle. Symbols used are Uav for voltage Iav for current.

time(ms)volts(V)

00

0.2545

0.572

0.7591

time(ms)volts(V)

1.0104

1.25118

1.5142

1.75185

time(ms)volts(V)

2.0240

2.25278

2.5295

2.75300

time(ms)volts(V)

3.25248

3.5195

3.7585

4.00

198

The average value of voltage will be the average length of lines (expressed in volts). To find this, we add the voltage represented by each line and divide by the number of lines.

)(15717

0851952482803002952782401851421181049172450VUav =

=

The effective or root mean square (RMS) value of the system is the square root of the average value of the squares of the instantaneous values. The symbols used for RMS values are U, I.

)(6.18617

0851952482803002952782401851421181049172450 22222222222222222

VU =

=

199

200

Sinusoidal waveforms

externalprime mover

Loop connected to slip rings rotates in magnetic field

Fleming‘s right hand rule

201

Induced EMF

tNBAte ww sin)( =

tEte wsin)( =

fpw 2=

Since the circumference of a circle is 2p×radius, there are 2p radians in 360, so 1 radian = 360/(2p) = 57.3 approximately.The total angular movement after t seconds of a wire loop rotating at f revolutions per second and giving an output of f cycles per second will be 2pft radians.

202

Peak, average and rms values of sinusoidal waves

A graph of Em sin φ, is referred to as a ‘sine wave‘, and is said to be ‗sinusoidal‘ in shape.

203

average value =2 × maximum value/p=0.637 × maximum value

Eavg = (2/p )Emax =0.637 x Emax

RMS value = maximum value/√2=0.707 × maximum value

E = Emax / √2 (V)

form factor =RMS value/average value=0.707Emax/0.637 Emax = 1.11

204

Example 1

Find the maximum and average values for a 230 V supply.

Emax = √2 E = 1.4142 * 230 = 325.27 V

Eavg = E /form factor = 230 / 1.11 = 207.21 V

or

Eavg = (2/p )Emax =0.637 x 325.27 = 207.2 V

205

Concept of capacitance

Whenever an electric voltage exists between two separated conductors, an electric field is present within the space between those conductors.

Capacitors are components designed to take advantage of this phenomenon by placing two conductive plates (usually metal) in close proximity with each other. There are many different styles of capacitor construction, each one suited for particular ratings and purposes.

206

When a voltage is applied across the two plates of a capacitor, a concentrated field flux is created between them, allowing a significant difference of free electrons (a charge) to develop between the two plates.

Because capacitors store the potential energy of accumulated electrons in the form of an electric field, they behave quite differently than resistors (which simply dissipate energy in the form of heat) in a circuit. Energy storage in a capacitor is a function of the voltage between the plates, as well as other factors.

207

Factors affecting capacitance

There are three basic factors of capacitor construction determining the amount of capacitance created. These factors all dictate capacitance by affecting how much electric field flux (relative difference of electrons between plates) will develop for a given amount of electric field force (voltage between the two plates).

1. Plate area2. Plate spacing3. Dielectric material

C = e A

d

C = Capacitance in Faradse = Permittivity of dielectric (absolute, not relative)A = Area of plate overlap in square metersd = Distance between plates in meters

(F)

208

‖Relative‖ permittivity means the permittivity of a material, relative to that of a pure vacuum.The greater the number, the greater the permittivity of the material. Glass, for instance, with a relative permittivity of 7, has seven times the permittivity of a pure vacuum, and consequently will allow for the establishment of an electric field flux seven times stronger than that of a vacuum, all other factors being equal.

209

Material Relative permittivity (dielectric constant)Vacuum ------------------------- 1.0000Air ---------------------------- 1.0006PTFE, FEP ("Teflon") ----------- 2.0Polypropylene ------------------ 2.20 to 2.28ABS resin ---------------------- 2.4 to 3.2Polystyrene -------------------- 2.45 to 4.0Waxed paper -------------------- 2.5Transformer oil ---------------- 2.5 to 4Hard Rubber -------------------- 2.5 to 4.80Wood (Oak) --------------------- 3.3Silicones ---------------------- 3.4 to 4.3Bakelite ----------------------- 3.5 to 6.0Quartz, fused ------------------ 3.8Wood (Maple) ------------------- 4.4Glass -------------------------- 4.9 to 7.5

210

"Ohm‘s Law" for a capacitor

Capacitors do not have a stable ‖resistance‖ as conductors do. However, there is a definite mathematical relationship between voltage and current for a capacitor, as follows:

i = Cdu

dt

i = instantaneous current through the capacitor

C = capacitance in Farads

du/dt = instantaneous rate of voltage change (volts per second)

211

The capacitor acts as a LOAD

to the rest ofthe circuit

increasingvoltage

Energy being absorbed by the capacitor from the rest of the circuit -charging

The capacitor acts as a SOURCE

to the rest ofthe circuit

decreasingvoltage

Energy being released by the capacitor to the rest of the circuit -discharging

212

Capacitors act somewhat like secondary-cell batteries when faced with a sudden change in applied voltage: they initially react by producing a high current which tapers off over time.• A fully discharged capacitor initially acts as a short circuit (current with no voltage drop) when faced with the sudden application of voltage. After charging fully to that level of voltage, it acts as an open circuit (voltage drop with no current).• In a resistor-capacitor charging circuit, capacitor voltage goes from nothing to full source voltage while current goes from maximum to zero, both variables changing most rapidly at first, approaching their final values slower and slower as time goes on.

213

214

215

Series and parallel capacitors

When capacitors are connected in series, the total capacitance is less than any one of the series capacitors‘ individual capacitances.

CT =C1 x C2

C1 + C2

CT =1

1

C1+

1

C1+…..

1

Cn

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors‘ capacitances.

CT = C1 + C2

216

Capacitors, like all electrical components, have limitations which must be respected for the sake of reliability and proper circuit operation.

Working voltage: Since capacitors are nothing more than two conductors separated by an insulator (the dielectric), one has to pay attention to the maximum voltage allowed across it. If too much voltage is applied, the ‖breakdown‖ rating of the dielectric material may be exceeded, resulting in the capacitor internally short-circuiting.

Polarity: Some capacitors are manufactured so they can only tolerate applied voltage in one polarity but not the other. This is due to their construction: the dielectric is a microscopically thin layer of insulation deposited on one of the plates by a DC voltage during manufacture. These are called electrolytic capacitors, and their polarity is clearly marked.

217

Resistance, Capacitance and Inductance in AC circuits

Resistance in AC circuit

Because the resistor simply and directly resists the flow of electrons at all periods of time, the waveform for the voltage drop across the resistor is exactly in phase with the waveform for the current through it.

218

The power dissipated by the resistor

The power is never a negative value. This consistent ―polarity‖ of power tells us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy. Whether the current is positive or negative, a resistor still dissipates energy.

Average powerP = Urms x Irms (W)

219

Inductance in AC circuits

Inductors oppose changes in current through them, by dropping a voltage directly proportional to the rate of change of current. u=L (di /dt)In accordance with Lenz's Law, this induced voltage is always of such a polarity as to try to maintain current at its present value. That is, if current is increasing in magnitude, the induced voltage will ―push against‖ the electron flow; if current is decreasing, the polarity will reverse and ―push with‖ the electron flow to oppose the decrease. This opposition to current change is called reactance, rather than resistance.

220

The instantaneous power

Because instantaneous power is the product of the instantaneous voltage and the instantaneous current (p = i x e), the power equals zero whenever the instantaneous current or voltage is zero. Negativepower means that the inductor is releasing power back to the circuit, while a positive power means that it is absorbing power from the circuit. Since the positive and negative power cycles are equal in magnitude and duration over time, the inductor releases just as much power back to the circuit as it absorbs over the span of a complete cycle.

Average power = 0

221

The power that surges back and forth in thus manner is called reactive power (Q) to distinguish it from the active power (P). The unit of reactive power is VAr, while the unit for active power is W. Both powers function independently of each other and they can not be converted into the other and they have to be treated as separate quantities in electrical circuit.

222

This opposition to alternating current is similar to resistance, but different in that it always results in a phase shift between current and voltage, and it dissipates zero power.Because of the differences, it has a different name: reactance. Reactance to AC is expressed in ohms, just like resistance is, except that its mathematical symbol is X instead of R. To be specific, reactance associate with an inductor is usually symbolized by the capital letter X with a letter L as a subscript, like this: XL.

The exact formula for determining reactance is as follows:

XL = 2pf L (W)

223

Capacitance in AC circuits

Whereas resistors allow a flow of electrons through them directly proportional to the voltage drop, capacitors oppose changes in voltage by drawing or supplying current as they charge or discharge to the new voltage level. The flow of electrons ―through‖ a capacitor is directly proportional to the rate of change of voltage across the capacitor.

i = Cde

dt

224

The instantaneous power

Average power = 0

Capacitor does not dissipate power as it reacts against changes in voltage; it merely absorbs and releases power, alternately.

225

Since capacitors ―conduct‖ current in proportion to the rate of voltage change, they will pass more current for faster-changing voltages (as they charge and discharge to the same voltage peaks in less time), and less current for slower-changing voltages. What this means is that reactance in ohms for any capacitor is inversely proportional to the frequency of the alternating current.

XC = 1

2pf C

The relationship of capacitive reactance to frequency is exactly opposite from that of inductive reactance. Capacitive reactance (W) decreases with increasing AC frequency. Conversely, inductive reactance (W) increases with increasing AC frequency.

(W)

226

Inductors oppose faster changing currents by producing greater voltage drops; capacitors oppose faster changing voltage drops by allowing greater currents.

227

Concept of reactance and impendence

Resistance is essentially friction against the motion of electrons. It is present in all conductors to some extent (except superconductors!), most notably in resistors. When alternating current goes through a resistance, a voltage drop is produced that is in-phase with the current.Resistance is mathematically symbolized by the letter ―R‖ and is measured in the unit of ohms (W).

228

Reactance is essentially inertia against the motion of electrons. It is present anywhere electric or magnetic fields are developed in proportion to applied voltage or current, respectively; but most notably in capacitors and inductors. When alternating current goes through a pure reactance, a voltage drop is produced that is 90o out of phase with the current. Reactance is mathematically symbolized by the letter ―X‖ and is measured in the unit of ohms (W).

229

Impedance is a comprehensive expression of any and all forms of opposition to electron flow, including both resistance and reactance. It is present in all circuits, and in all components.When alternating current goes through an impedance, a voltage drop is produced that is somewhere between 0o and 90o out of phase with the current. Impedance is mathematically symbolized by the letter ―Z‖ and is measured in the unit of ohms (W), in complex form.

IR

R (W) U

IL

jXL (W) U

IC

-jXC (W) U

IR =U

RIL = -j

U

wLIC = jUwC

230

AC series circuits

Series resistor-inductor circuits

Ztotal = (500) + (3.142900)

XL = 2pf L = 2 x p x 50 x 0.01 = 3.142W

5 W + j3.142 W

Ztotal = 52 + 3.1422 = 5.9W32.10

Ztotal = R + jXLj = arc tan

XL

R

j = arc tan3.142

5

j = 32.10

Ztotal =

231

XL= 3.142W; R = 5W; Z = 5.9W

I = ET

Z=

10

5.9= 1.695 A

EL = I x XL = 1.695 x 3.142

ER = I x R = 1.695 x 5 = 8.48 V

EL = 5.33 V

Active power

P = I2 R

P = 14.37 W

Reactive power

Q = I2 XL

Q = 9.03 VAr

Apparent power

S = 16.97 VA

S = P2 + Q2

232

j

S (VA)

P (W)

Q (VAr)

Q = P x tan (j)

Power triangle

cos j = P

S

S = P2 + Q2

The term cos j is referred to as the power factor. Power factor is equal to 0 for purely inductive load and equal to 1 for purely resistive load. In every other case 0 < pf < 1.

in our case p.f. = 0.847ind (cos 32.10)

Power factor

233

Series resistor-capacitor circuits

XC = 1/2pfC = 1/(2 x p x 50 x 100x10-6) = 31.83 W

Ztotal = (50o) + (31.83-900)

Ztotal = R - jXC

Ztotal =5 W - j3.142 W

Ztotal = 52 + 31.832 = 32.22 W 81.10

j = tan-1XC

R

j = tan-131.83

5

j = 81.10

234

XC= 31.83 W; R = 5 W;

Z = 32.22 W

I = ET

Z=

10

32.22= 0.31 A

EC = I x XC = 0.31 x 31.83

EC = 9.87 V

ER = I x R = 0.31 x 5 = 1.55 V

Active power

P = I2 R = 0.312 x 5

P = 0.481 W

Reactive power

Q = I2XC = 0.312 x 31.83

Q = 3.06 VAr

Apparent power

S = 3.1 VA

S = P2 + Q2

235

j

S (VA)

P (W)

Q (VAr)

Q = P x tan (j)

Power triangle

cos j = P

S

S = P2 + Q2

The term cos j is referred to as the power factor. Power factor is equal to 0 for purely capacitive load and equal to 1 for purely resistive load. In every other case 0 < p.f. < 1.

in our case p.f. = 0.155cap (cos 81.10)

Power factor

236236

Series inductor-capacitor circuits

XC = 1/2pfC = 1/(2 x p x 50 x 80x10-6) = 39.79 W

Ztotal = (31.4290o) + (39.79-900)

Ztotal = jXL - jXC

Ztotal =j31.42 W – j39.79 W

Ztotal = -j8.37 = 8.37 W -900

XL = 2pf L = 2 x p x 50 x 0.1 = 31.42 W

237

XL= 31.42 W; XC= 39.79 W;

Z = 8.37 W

I = ET

Z=

10

8.37= 1.195 A

EC = I x XC = 1.195 x 39.79

EC = 47.55 V

EL = I x XL = 1.195 x 31.42= 37.55 V

Active power

P = 0.00 W

Reactive power

Q = I2Z = 1.1952 x 8.37

Q = 11.95 VAr

Apparent power

S = 11.95 VA

S = P2 + Q2

238

Effect of frequency on inductive reactance

XL = 2pf L (W)

As the frequency is increased (reading to the right), the inductive reactance is shown to increase in direct proportion.

239

Effect of frequency on capacitive reactance

XC = 1/2pfC (W)

As the frequency is increased (reading to the right), the inductive reactance is shown to decrease in inverse proportion.

240

Resonance in series circuits

When a state of resonance is reached (capacitive and inductive reactance equal), the two impedances cancel each other out and the total impedance drops to zero.

f = 159.155 Hz

XL = 2pfL = j100 (W)

XC = 1/(2pfC) = -j100 (W)

I = V/Z = V/ R2+(jXL-jXC)2 = V/R = 1A

241

This series-resonant effect, with inductive and capacitive reactances equal and opposite, may be brought about in a number of ways:

1. Change in inductance, give a proportional change in inductive reactance (note that XL = 2πf L, so XL ∝L if f is constant).

2. Change in capacitance, giving an inversely proportional change in capacitive reactance (note that XC = 1/2πf C so XC ∝ 1/C if f is constant).

3. Change in frequency. If L and C are constant, XL ∝f and XC ∝ 1/f , so an increase in frequency will

increase inductive reactance and decrease capacitive reactance.

fr =1

2π√LCAt some frequency these two values (inductive and capacitive reactance) wouldbe equal and series resonance would occur.

242

Variation of current with frequency in a

R-L-C series circuit

Variation of resistance,

reactance and impedance

with frequency in a R-L-C

series circuit

Extremely high voltages can be formed across the individual components of series LC circuits at resonance, due to high current flows and substantial individual component impedances.

243

Power in single phase circuits

Power is the rate of doing work, or of use up energy. The electrical unit of power is the watt, which represents a rate of expending energy of one joule each second. (W=J/s)

If a resistor of R ohms has a direct voltage of V volts applied to it, so that a direct current of I amperes flows, the power dissipated, P watts, will be given by

R

VRIVxIP

22 ===

for an AC circuit, P is the average power, while V and I and RMS voltage and RMS current, respectively. This power is dissipated in the resistor as heat.

244

Since both voltage and current values are continuously changing in the AC system, power will also fluctuate and the rate of dissipating energy is the instantaneous power, which is given by

vip =

Power in the resistive AC circuit

For a resistive AC circuit, current and voltage are in phase, and the power at any instant can be found by multiplying the voltage and current at that instant.

245

Voltage, current and power waves for resistive AC circuit

A 3 kW immersion heater is connected to a 230 V AC supply. Calculate the current.

Example 1

)(13230

3000A

V

PI ===

246

Power in the capacitive AC circuit

The current leading the voltage by 90, and the instantaneous power is: p = vi. In the first quarter-cycle of voltage, v and i are both positive, so their product, the power wave is also positive. In the second quarter-cycle of voltage, v is positive but i is negative, so the power wave goes negative.

247

During its first and third quarter-cycles, the voltage is increasing and the supply provides energy to charge the capacitor. During the second and fourth quarter-cycles of voltage, the reducing PD across the capacitor allows it to discharge, returning itsenergy to the supply.

The positive pulses represent energy supplied to the capacitor, while the negative pulses represent energy supplied by the capacitor as it discharges.

248

The interchange of energy dissipates no average power in a pure capacitor, so no heating occurs.Since we have voltage and current, but no average power, the expression P = VI is no longer true. The product of voltage and current in this case is called reactive power and is measured in reactive voltamperes (VAr). The current to a capacitor which does not contain resistance does not dissipate energy, and is called reactive current.

249

A 10 μF capacitor is connected to a 230 V, 50 Hz supply. Calculate the reactive current and reactive voltamperes.

Example 1

)(31810502

1

2

16

W===-xfC

Xcpp

)(723.0318

230A

Xc

VI ===

)(732.0230 VArxVxIQc ==

250

Power in the inductive AC circuit

The current lags the voltage by 90, and the instantaneous power is: p = vi. In the first quarter-cycle of voltage, v and i are both positive, so their product, the power wave is also positive. In the second quarter-cycle of voltage, v is positive but i is negative, so the power wave goes negative.

251

During its first and third quarter-cycles, the current is increasing and the supply provides energy to magnetic field of the inductor. During the second and fourth quarter-cycles of voltage, the reducing current across the inductor allows it to discharge, returning its energy to the supply.

The positive pulses represent energy supplied to the inductor, while the negative pulses represent energy supplied by the inductor as it discharges.

252

Power in resistive and capacitive AC circuits

In a circuit consisting of resistance and capacitive reactance in series, the voltage and current will have a relative phase angle between 0 and 90, depending on the ratio of resistance to reactance

The net energy drawn from the supply will be dissipated as heat in the resistive part of the circuit.

Although, there is still some energy returned to the supply, (negative pulses), the energy drawn from the supply (positive pulses), is greater.

253

The ratio of resistance to reactance in the circuit must have some bearing on the power dissipated, because power is expended in a resistive circuit, but not in a reactive circuit.

Example 2

A circuit connected to a 230 V AC supply consists of a resistance of 28.8 (W) in series with a capacitor of reactance 38.4(W). Calculate (a) the circuit current, (b) the circuit phase angle, and (c) the power dissipated.

W=== 484.388.28 2222 XcRZ

AZ

VI 79.4

48

230=== 6.0

48

8.28cos =====

Z

R

IZ

IR

V

VRf

)(6616.079.4230cos WxxVIP === f

)(6618.2879.4 22 WxRIP ===

254

Example 3

A 10 W resistor and a capacitor are connected in series to a 120 V, 60 Hz supply. If the power lost in the circuit is 360 W, calculate the capacitance.

FxfXc

CfC

Xc

RZXc

I

VZ

AR

PI

R

PI

R

PIRIP

mppp

1533.17602

1

2

1.....

2

1

)(3.171020

)(206

120

)(63610

380

.............

2222

22

====

W=-=-=

W===

====

===

255

Power in resistive and inductive AC circuits

When resistance and inductive reactance are in series, current lags supply voltage by an angle of φ, which will vary from almost 0 to nearly 90.

Energy is both taken from the supply and returned to it, that taken from the supply exceeding the energy returned.

The net energy drawn from the supply will be dissipated as heat in the resistive part of the circuit.

256

Example 4

A 4 W resistor and a pure inductive reactance of 3 Ware connected in series to a 200 V AC supply. Calculate (a) the current, (b) the circuit phase angle and (c) the power dissipated.

)(6400440

)(64008.040200cos

9.36..........8.05

4cos

405

200

534

22

0

2222

WxRIP

WxxVIP

Z

R

IZ

IR

V

V

AZ

VI

XcRZ

R

===

===

======

===

W===

f

ff

257

Example 5

A choke connected to a 130 V, 50 Hz supply has a resistance of 5 W and dissipates 500 W. Calculate its inductance.

)(2.38)(0382.0502

12

22

12513

1310

130

)(105

500

2222

2

mHHf

XLfLX

RZX

I

VZ

AR

PIRIP

LL

L

=====

W=-=-=

W===

====

ppp

258

Power in general

The dissipated power can then be calculated by any one of the three methods:

R

VPRIPVIP R

22 ............................cos === f

Where:P = power dissipated (W);V = supply voltage (V);I = circuit current (A);φ = circuit phase angle;R = circuit resistance (W); VR = PD across the resistive component (V).

259

Concept of power factor and its effect

It is possible for current to flow in a circuit and todissipate no power. In most practical cases this will not happen, but where the phase angle between current and voltage is large, the ‗in-phase‘ or ‗active‘ component of current will be smaller than the ‗reactive‘ component. In AC circuits, the product of voltage and current need not result in the power dissipated in watts, this product that gives volt-amperes, is called apparent power.

Power factor (often abbreviated to PF) is defined as

PF=active power

apparent power=

fcosVI

VI= fcos =

RV=

V

R

Z

260

In a predominantly inductive series circuit, where current lags voltage, the power factor is called a lagging power factor. Similarly, in a predominantlycapacitive series circuit, where current leads voltage, the power factor is called a leading power factor.The power factor can vary between definite limits, being 1 (unity) for purely resistive circuits, where the phase angle is 0 and P = VI; or 0 (zero) for purelyreactive (inductive or capacitive) circuits, where the phase angle is 900 and P = 0.

261

An AC single-phase motor takes 5 A at 0.7 power factor lagging when connected to a 230 V, 50 Hz supply. Calculate the power input to the motor. If the motor efficiency is 80% calculate the output.

Input power: P = VI cos φ = 230 × 5 × 0.7 = 805 W

Output power = input power × efficiency= 805 × 0.80 watts = 644 W

Example 1

262

Example 2

Instruments connected toa single-phase AC motor givethe following readings:wattmeter, 1800 W,voltmeter, 230 V,ammeter, 10 A. Calculate the operating power factor of the motor.

PF=active power

apparent power= =

1800

230x100.783 lagging

263

Components of power

Power diagram for resistive and inductive AC circuit

Power diagram for resistive and capacitive AC circuit

Active power is the in-phase component of apparent power, thus P = VI cos φ = apparent power × power factorReactive power is the quadrature component of apparent power, thus Q = VAr = VI sin φ

Since these three power relationships form the sides of a right-angled triangle,(VA)2 = (W)2 + (VAr)2

A single-phase load consists of:(1) 12 kW of lighting and heating at unity power factor(2) 8 kW of motors at 0.8 power factor lagging(3) 10 kVA of motors at 0.7 power factor lagging.Calculate:(a)the total kW,(b)the total kVAr,(c)the total kVA,(d)the overallpower factor andthe total supplycurrent at 230 V.

264

Example 1

12kW

10kVA

10kVA

j1j2

8kW7kW

PF=kW

kVA

6kVAr

7.14kVAr

9.030

27

)14.76()1287(

1287

22==

=

I=S/V=30000/230=130.43 A

265

Example 2

A single-phase 3.73 kW motor is 85% efficient at full load and is fed from a 230 V supply. Calculate its full-load current if it operates at a power factor of(a) unity, (b) 0.85 lag, and (c) 0.6 lag.

jhjh

coscos)(

V

PIVIWP ==

)(24.3385.06.0220

3730

)(47.2385.085.0220

3730

)(95.1985.01220

3730

Axx

I

Axx

I

Axx

I

c

b

a

==

==

==

266

All the disadvantages of a low power factor are due to the fact that a given load takes more current at a low power factor than it does at a high power factor. The most important disadvantages of operating a load at a low power factor are as follows.

(1) Larger cables, switchgear and transformers may be necessary both within an installation and in the supply mains feeding it.(2) Low-power-factor working causes operating difficulties on high-voltage transmission lines.(3) Because of the effects of items (1) and (2), electricity companies usually penalise the consumer whose load is at a poor power factor by charging more for the electrical energy used.(4) Larger cables may be needed within an installation to carry the extra current at low power factor. Alternatively, extra load can be connected to a cable if the power factor of the existing load it carries is improved.(5) Higher currents give rise to higher copper losses in cables and transformers.(6) Higher currents give larger voltage drop in cables, and a change in load gives a larger change in voltage drop if the power factor is low. This is called ‗poor voltage regulation‘.

267

Measurement of resistance by substitution

Measurement of resistance by substitution is based on the fact that two resistance must be equal, which, when substituted for each other in the same circuit, give the same current-strength.

The position (deflection) of the galvanometer‘s needle must be adjusted with rheostat, to suitable position.

R must excluded and rheostat adjusted in such a way to bring galvanometer‘s needle back in same position. The added rheostat resistance is equal to resistance R.

268

Example 1

R Rh 14W Rh 15W

deflection 4503 4709 4405

W=-

-= 76.14

5.449.47

3.459.4714R

The method of substitution is almost universally applicable if the resistances are not too small, and requires only galvanoscope to prove the equality of two circuits.

269

Measurement of resistance by direct methods

1. Connect the instruments using unknownresistance Rx. Pay attention about polarity.

2. Adjust Rd so that the voltage measured by voltmeter is in range 2-4V. Read the value of voltage V and current I and write down in table. Adjust Rd to another value so that V is in the range 2-4V.

270

3. Repeat points 1 and 2 with resistor Rx2

4. Connect both resistors Rx1 and Rx2 in the circuit in serial and perform measurements according points 1 and 2 to determine equivalent resistance Req.

5. Connect both resistors Rx1 and Rx2 in the circuit in parallel and perform measurements according points 1 and 2 to determine equivalent resistance Req.

6. Calculate the mean values of Rx1 and Rx2 and of equivalent resistance Req for serial and parallel combination.

7. Compare experimentally determined equivalent resistances Req for serial and parallel combination with formulas for series and parallel connections.

271

272

Wheatstone Bridge

The measurements with bridge circuits differ fromdirect measurements in that, the quantity being measured is compared with a known referencequantity. The balancing strategy avoids undesirable interaction effects and generally results inmore accurate measurement than the direct one.

By far the most common is the Wheatstone bridge designed for precise measurement of resistance.

273

In the basic circuit in which the measurement of an unknown resistance Rx is performed by balancing the variable resistances Ra and Rb until no current flows through meter A. Under this null condition,

RsRb

RaRx =

where Rs is the known standard resistance.

find: (a) R4, and (b) the current through R4.

Exercise 1

274

Measurement of insulation and conductor resistance of cables in series and parallel

Factors that affect cable insulation resistancemeasurements are length, type, temperature, and the equipment connected in the circuit. Each of these factors must be evaluated to reliably determine the condition of the cable from the measurements obtained.

Insulation resistance is defined as the resistance (in MW) offered by the insulation to an impressed direct voltage. The resulting current is called Insulation current.

OR

The resistance offered, by the insulation of a cable, in the path of leakage current, is called Insulation resistance of the cable

275

A Megger is an ohmmeter-type instrument by means of which the value of a resistance can be measured and directly indicated by the position of a pointer on a scale. The megger consists of two principal elements: a hand-driven magneto type direct current generator, which supplies the current for making the measurement; and the moving element with pointer, by means of which the value of the resistance under measurement is indicated.

276

Meggers are equipped with three connectionterminals, labeled Line, Earth, and Guard. To measure insulation resistance from a conductor to the outside of the cable, we need to connect the "Line" lead of the megger to one of the conductors and connect the "Earth" lead of the megger to a wire wrapped around the sheath of the cable:

Resistance is measured between the Line and Earth terminals, where current will travel through coil 1.

277

The "Guard" terminal is provided for special testing situations where one resistance must be isolated from another, for instance where the insulation resistance is to be tested in a two-wire cable:

Rather than just measure the resistance of the second conductor to the sheath (Rc2-s), what we'll actually measure is that resistance in parallel with the series combination of conductor-to-conductor resistance (Rc1-c2) and the first conductor to the sheath (Rc1-s).

278

If we want to measure only the resistance between the second conductor and the sheath (Rc2-s), then we need to use the megger's "Guard" terminal:

279

Connecting the "Guard" terminal to the first conductor places the two conductors at almost equal potential. With little or no voltage between them, the insulation resistance is nearly infinite, and thus there will be no current between the two conductors. Consequently, the megger's resistance indication will be based exclusively on the current through the second conductor's insulation, through the cable sheath, and to the wire wrapped around, not the current leaking through the first conductor's insulation.

280

Double wound and auto-transformers, principle of operation, application, precautions, advantages and

disadvantages

Transformers are very important in electrical engineering, because almost all of its many branches make use of them. The efficient transmission and distribution of electricity would be impossible without power transformers. Electronic equipment inindustry uses transformers in very large numbers. Communications systems, including television and telephony, rely on transformers for their operation. Although transformers differ in size and in application, all rely on the principle of mutual inductance for their operation.

281

If the two coils are arranged on a core of magnetic material, this will increase the amount of magnetic flux set up by one coil and will make sure that most of it links with the other coil. In this way mutual inductance is increased.

The arrangement is a simple double-wound transformer. The winding fed with current is called the ‗primary winding‘ and the other the ‗secondary winding‘.

282

Each winding must be made with insulated conductors to prevent short circuits within the winding itself, or to the magnetic circuit or core, which is usually earthed for safety on power transformers.Alternating current in the primary winding will set up an alternating magnetic flux in the core, the self-inductance of the winding inducing in it an EMF opposing the supply voltage. This EMF will be almost the same in value as the applied voltage, and for practical purposes the two may be assumed to be equal. If all of the changing magnetic flux set up by the first winding links with the second, the EMF induced in each turn will be the same regardless of whether it forms part of the primary winding or of the secondary winding.

283

Autotransformers

An autotransformer has only one tapped winding, which is both the primary and secondary of the machine.

step-downautotransformer

step-upautotransformer

284

The major advantage of the autotransformer which will be smaller, lighter and cheaper than its double-wound counterpart. The disadvantages of the autotransformer are as follows:

1 There is a direct metallic connection between the input and the output, whereas the coupling in a double-wound transformer is magnetic only, giving electrical isolation of the two windings.

2 In the event of an open-circuit fault in the common part of the winding, the input voltage of a step-down autotransformer would appear on the output terminals. Because of this danger, the IEE Wiring Regulations limit the use of autotransformers. However, they are used in high-voltage transmission systems, as starters for induction and synchronous motors, and for voltage control in some types of discharge lamp.

285

To indicate the danger of input voltage appearing at output terminals of a step-down autotransformer in the event of an open circuit in the common winding

286

Construction details. Simple calculations

The single-phase transformer will consist of primary and secondary windings mounted on a magnetic core.

287

Core materials

Since it is always be subjected to alternating magnetisation, the core material and construction must be chosen to reduce iron losses to a minimum, or the transformer will not be efficient.Most transformer cores are made from laminated silicon steel, the laminations reducing eddy currents and the silicon steel keeping hysteresis loss to a minimum. Laminations must be arranged so as to reduce the air gaps in the magnetic circuit.

The laminations must be tightly held together by clamping or by taping, or they are likely to vibrate and produce excessive noise. Some small high-frequency communications transformers have cores cast of solid ferroxcube, the eddy-current loss thus being kept to a reasonable level.

288

packs of laminations being laid up to form shaped core

289

Core arrangements

A core-type transformer, the windings being split, with part of each wound on each side of the magnetic circuit to reduce leakage flux.

Leakage is reduced still further by using the shell-type arrangement. Both windings are placed on the centre limb, the two outer limbs providing parallel return paths for the magnetic flux.

290

Windings

Windings are usually made of copper, although some experiments using aluminium as a conductor material have been carried out. Cylindrical or concentric windings, where the lower-voltage winding is completely surrounded by the higher-voltage turns, are used mainly for core-type circuits.

291

Sandwich or disc-type windings, where the two windings are split into alternately mounted sections, are used generally on shell-type circuits, except for very high voltage transformers which use the cylindrical type of winding.

High voltagewinding

Low voltagewinding

292

A transformer with an output voltage greater than its input is called a step-up transformer, whereas a step-down transformer has a lower output voltage than its input. If a voltage or turns ratio is quoted for a transformer, this is always put in the orderinput : output, which is primary : secondary.

The voltage per turn of the two windings are equal.

primary volts per turn =primary volts

primary turns =

V1

N1

secondary volts per turn =secondary volts

secondary turns =

V2

N2

V1

N1

V2

N2

=V1

V2

N1

N2

=

293

Example 1

A transformer with 1000 primary turns and 250 secondary turns is fed from a 230 V AC supply. Calculate the secondary voltage and the volts per turn.

V1/V2 =N1/N2 so V2 = V1 × N2/N1

V2 = 230 x 250/1000 = 57.5 V

A neon-sign transformer has an output of 4500 V and is fed at 230 V. If the secondary has 2000 turns, calculate the number of primary turns.

Example 2

V1/V2 =N1/N2 so N1 = N2 × V1/V2

N1 = 2000 × 230/4500 = 102.2 V

294

If we assume that our transformer is 100% efficient, then power input = power output,

V1 x I1 = V2 x I2 V1/V2 = I2/I1

Neither of the assumptions made is strictly true but, since the error involved is small, the resulting expression is a useful one.

Example 3

A 50 kVA transformer has a voltage ratio of 3300:400 V. Calculate the primary and secondary currents.

S =V1 x I1 so I1 = S / V1 = 50000/3300 = 15.2 A

S =V2 x I2 so I2 = S / V2 = 50000/400 = 125 A

295

Exercise 1

The single-phase transformer feeding a soil-warming system is supplied at 230 V, 50 Hz, and must provide a 20 V output. The full-load secondary current is 180 A, and the secondary has 45 turns. Calculate(a) the output kVA of the unit(b) the number of primary turns(c) the full load primary current(d) the volts per turn.

296

A 75 kVA transformer has step-down ratio of 12:1, 2400 primary turns and a primary voltage of 3.3 kV. Calculate(a) the number of secondary turns(b) the secondary voltage(c) the volts per turn(d) the full load primary and secondary currents.

Exercise 2

297

Transformer losses, efficiency and regulation

Losses actually occurring can be considered under two headings: iron (core) losses and copper (I2R) losses.

Iron losses occur in the magnetic core of the transformer, causing it to heat up. Iron losses can be divided into1) hysteresis losses2) eddy-current lossesIron losses depend on the frequency of the supply, and the maximum magnetic flux density in the transformer core. For power transformers, the supply frequency is almost always constant, and since the supply voltage is virtually constant,there is very little change in the core flux density.

298

Thus it is reasonable to assume that the iron losses in a transformer remain constant regardless of the load conditions – for example, the iron loss on no load will be the same as that on full load.

Magnetising current is quite small, so that the copper loss due to it may be ignored, and the total copper losses of a power transformer on no load may be assumed to be zero. Power loss in a resistive circuit is given by the expression P = I2R, and since winding resistances are largely constant, copper losses depend on the square of the load current.

Thus a transformer operating on half load will have only one quarter of the copper loss it has when providing full load.

load

loss

299

Efficiency

As well as providing for the output power, the input to a transformer must supply the transformer losses.

efficiency = (output power/input power) × 100%

efficiency =input power − power losses

input power× 100%

efficiency = × 100%output power + power losses

output power

300

Example 4

The full-load copper and iron losses for a large power transformer are 16 kW and 12 kW, respectively. If the full-load output of the transformer is 950 kW, calculate the losses and efficiency of the transformer(a) on full load(b) on 60% of full load(c) on half load.

Total loss = copper loss + iron loss =16 + 12=28kW

efficiency = × 100%output power + power losses

output power

a) efficiency =[950/(950 + 28)] × 100% = 97.1%

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b) At 60% full load, iron loss remains at 12 kW.

Copper loss = (60/100)2× 16 kW = 5.8 kW

Total loss = 5.8 + 12 = 17.8 kW

efficiency =[950x0.6/(950x0.6 + 17.8)]×100%= 97.0%

efficiency = × 100%output power x 0.6 + power losses

output power x 0.6

c) At half load, iron loss remains at 12 kW.

Copper loss = (50/100)2× 16 kW = 4.0 kW

Total loss = 4.0 + 12 = 16.0 kW

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efficiency =[950x0.5/(950x0.5 + 16.0)]×100%= 96.7%

efficiency = × 100%output power x 0.5 + power losses

output power x 0.5

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Regulation

On no load there will be no secondary currentand no voltage drop. On full load the output voltage will fall, and the difference between no-load voltage and full-load voltage, expressed as a percentage of no-load voltage, is called the voltage regulation.

regulation = × 100%no-load voltage

no-load voltage − full load voltage

A power transformer provides 400 V on no load and 390 V on full load. Calculate the voltage regulation.

Example 5

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regulation = × 100%no-load voltage

no-load voltage − full load voltage

(400−390)/400=(10/400)×100% = 2.5%regulation=

A transformer with a voltage regulation of 4% provides 220.8 V on full load.Calculate its no-load terminal voltage.

Exercise 3

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Resistance values for conductors at any temperature other than the standard temperature (usually specified at 20 deg Celsius) on the specific resistance table must be determined through yet another formula:

RT = Rref [1 + a(T - Tref)] (W)

RT = conductor resistance at temperature "T"Rref = conductor resistance at reference temperaturea = temperature coefficient of resistance for theconductor material (1 / oC).T = conductor temperature in degrees Celsius.Tref = reference temperature that a is specified atfor the conductor material. Tref, usually 20o C, but sometimes 0o C.

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Material ―a" per deg CNickel Element 0.005866Iron Element 0.005671Molybdenum Element 0.004579Tungsten Element 0.004403Aluminum Element 0.004308Copper Element 0.004041Silver Element 0.003819Platinum Element 0.003729Gold Element 0.003715Zinc Element 0.003847Steel* Alloy 0.003Nichrome Alloy 0.00017Nichrome V Alloy 0.00013Manganin Alloy +/- 0.000015Constantan Alloy 0.000074* Steel alloy at 99.5 % iron, 0.5 % carbon

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Moving Iron Vane Movement The moving iron vane movement can be used to measure both AC current and voltage. By changing the meter scale calibration, the movement can be used to measure DC current and voltage. The moving iron vane meter operates on the principle of magnetic repulsion between like poles. The measured current flows through a field coil which produces a magnetic field proportional to the magnitude of current. Suspended in this field are two iron vanes attached to a pointer. The two iron vanes consist of one fixed and one moveable vane. The magnetic field produced by the current flow magnetizes the two iron vanes with the same polarity regardless of the direction of current through the coil. Since like poles repel one another, the moving iron vane pulls away from the fixed vane and moves the meter pointer. This motion exerts a force against a spring. The distance the moving iron vane will travel against the spring depends on the strength of the magnetic field. The strength of the magnetic field depends on the magnitude of current flow. Figure 3 Moving Iron Vane Meter Movement As stated previously, this type of meter movement may also be used to measure voltage. When this type of movement is used to measure voltage, the field coil consists of many turns of fine wire used to generate a strong magnetic field with only a small current flow.