HEATING EFFECT OF AN ELECTRIC CURRENT
Energy Changes and Potential Difference
In electric circuits, electrical energy is supplied from a source such as a battery to an electrical
device, where it is converted into other forms of energy. For example:
(i) an electric motor transfers most of the electrical energy supplied to mechanical energy.
(ii) an electric fire (radiant heater) transforms most of the electrical energy supplied to heat energy in the fire element and some to light energy.
(iii) an electric lamp transform most of the electrical energy to heat energy with some produced
as light (luminous) energy.
EXPERIMENT 8.1: To investigate the effect of current on a coil (resistance wire)
Apparatus
A battery of four or more cells, connecting wires, switch, thick copper wire, coil of resistance
wire, ammeter, variable resistor, stopwatch
Procedure
- Assemble the apparatus as in figure 8.l.
- Feel with your fingers the temperature of the coil and other parts of the circuit when
circuit is open.
- Close the circuit.
- After sometime, say 2 minutes, feel the temperature of the coil and other parts of the
- Increase the amount of current flowing through the coil and feel the temperature of the
coil after sometime.
- Switch off the circuit.
Observations
The coil feels warmer after closing the switch. This shows that the electric current produces
heating effect in the coil. A higher current produces more heat.
Explanation
The e.m.f of the battery forces a flow of electrons round the circuit against the resistance
offered by the various components in the circuit. The work needed to keep the current flowing
through the high resistance wire (coil) is much greater than the work needed to keep the current
flowing through the low resistance copper wire. The coil therefore gets warmer than the parts
of the circuit.. .. . .
If the amount of heat produced is increased, the coil may get red hot. Further heating will
cause the atoms of the metal to break free from the crystal lattice, making the wire melt.
Factors Determining Heat Produced by Electric Current
In the heating effect, heater transforms electrical energy supplied to heat
energy and light energy. Electric device uses electric current to generate heat
energy.
The following factors affect heat energy produced by electric device.
(a) Time
The longer the time the heater is on, the more electric current is converted to heat. Heat generated is directly proportional to time.
Eαt
(b) Resistance of the heating filament.
Heat generated is directly proportional to resistance.
EαR
(c) Magnitude of current (I)
Heat generated is directly proportional to square of current.
E α I2
In summary:
H αI2Rt
From Ohms law V= IR
H αIvt
H=Ivt
Conclusion
The heating effect produced by an electric current I flowing in a conductor of resistance R in time t is given by H = Ivt.
This equation is known as Joule’s law of electrical heating, in honor of James Joule who first investigated the heating effect of an electric current.
Joule’s law of electrical heating states that the energy of developed in a wire is directly proportional to:
(a) the square of the current F (for a given resistance and time).
(b) the time t (for a given resistance and current) .
(c) the resistance R ofthe wire (for a given current and time).
Example 1
The potential difference across a lamp is 12 volts. How many joules of electrical energy are changed to heat and light when:
(a) a charge of 5 coulombs passes through it?
(b) a current of 2 A flows through the lamp for 10 seconds?
Solution
(a) W = QV
Energy changed to heat and light, W = 5 x 12
= 60J
(b) W = IVt
Energy changed to heat and light, W = 2 x 12 x 10 = 240J
.Example 2
, An iron box has a resistance coil of 30 Q and takes a current of 10 A. Calculate the heat in kJ
developed in 1 minute.
, Solution
| R |
= |
30 Q, I = 10 A, t = 60 s |
| H |
= |
PRt |
| |
= |
102 x 30 x 60 |
| |
= |
18 x 1()4 |
| |
= |
‘180 kJ |
Example 3
A heating coil providing 3 600 Jmin-1 is required when the p.d. across it is 24 V. Calculate, the length of the wire making the coil given that its cross section area is 1 x 10-7 m2 and resistivity 1 x 10-6Ωm.
Solution
H = Pxt
P = H
t
= 60W
Electrical Energy and Power
The work done in pushing a charge round an electrical circuit is given by;
W=VIt
So, W =VI
t
But work/time is power
Therefore power p=IV
Hence, electrical power is given by, P = VI (1)
From Ohm’s law;
V = IR …………..(2)
Substituting (2) in (1);
P = I2V………………………………………………….. (3)
Substituting (4) in (3);
Current I=V/R
P=V2/R
In summary, electrical power consumed by an electrical appliance is given by;
P= VI, or,
P= I2R, or,
P=V2
R
The unit for power is the watt (W) and is equal to the energy change rate of 1 joule per second,
i.e., 1 W = 1 Js-I. Thus, an electrical lamp with a power rating of 100 W converts 100 J of
electrical energy into heat and light every second.
A larger unit of power is the kilowatt (kW).
1 kW= 1000W
Example 4
How much electric energy in joules does a 150 watt lamp convert to heat and light in:
(a) 1 second?
(b) 5 seconds?
(c) 1 minute?
Solution
1 W = 1 JS-1
So, energy changed in:
(a) 1 second is 150 x 1 = 150 J
(b) 5 seconds is 150 x 5 = 750 J
(c) 1 minute is 150 x 60 = 9 ()()() J
ExampleS
How much current does a bulb rated at 100 W and designed for a mains supply of 250 V draw
when operating normally?
Solution
P=VI
When the bulb is operating normally;
P= 100, V = 250
100= I x 250
:. I = ~~
= 0.40A
Example 6
What is the maximum number of 100 W bulbs which can be safely run from a 240 V source supplying a current of 5 A?
Solution
Let the maximum number of bulbs be n. Maximum energy developed in the circuit per second equals total energy converted by the bulbs per second.
Thus, 240 x 5 = 100 n
240 x 5
So, n = 100
= 12 bulbs
Example 7
What is the operating resistance of an electric lamp rated by the manufacturer at 60 W, 240 V?
Solution
From P = VI, current I flowing in the lamp when used normally is given by;
60=lx240
– 60
– 240
=0.25A
Resistance R = i= 5i~
= 960.Q
Example 8
An electric light bulb has a filament of resistance 470 O. The leads connecting the bulb to the 240V mains have a total resistance of 10 O. Find the power dissipated in the bulb and in the leads.
Solution
R total = 470 + 10
=4800
Therefore, I = ~~
=0.5A
For the bulb alone;
R=4700andI= 0.5A
:. Power dissipated = FR
= (0.5)2 X 470
= 117.5 W
For the leads alone, R = 10 0 and I = 0.5 A.
:. Power dissipated = (0.5)2 X 10
= 2.5W
Example 9
An electric iron of resistance 50 0 and an electric indicator of 6 000 0 are connected in parallel to a 240 V mains supply. Find the dissipated in the electric iron and in the indicator.
Solution
P= V2
R
For the iron alone;
V = 240 V, R = 50 0
Power = 2402
50
= 1152 W
For the indicator alone;
V=240V,R=60000
Power= 2402
6000
= 9.6W
Example 10
A house has twenty 60 W bulbs, two 1 000 W heater and two 500 W security lights. If the appliances are running on 230 V, calculate:
(a) the total power in kW used when all are switched on.
(b) the total current drawn from the mains supply.
Solution
(a) PT= (20 x 60) + (2 xl 000) + (2 x 500)
=4200W
=4.2kW
(b) T – 4200
‘T – 230
= 18.3A
Example 11
A hoist motor powered by a 240 V mains supply requires a current of 30 A to lift a load of mass 3 tonnes at the rate of 5 m per minute. Calculate:
(a) the power input.
(b) the power output.
(c) the overall efficiency.
Solution
(a) Power input = N
= 30×240
= 7200W
(b) Power output = force x velocity
= 3 x 1000 x 10 x go
= 2500W
(c) Efficiency = 2500 x 100
7200
= 34.72
Example 12
Two heaters A and B are connected in parallel across a 10 volts supply. Heater A produces 1 000 J of heat in one hour while B produces 200 J in half an hour. Calculate:
(a) the ratio
(b) RA, if RB = 100 Q.
(c) the amount of heat produced if the two heaters are connected in series across the same
voltage for 4 minutes.
