(i) The elements of laundry irons are made  
of strip instead of wire. Why?  
Ans: So that they are as flat as possible  
and also present large surface for  
conducting away heat into the sole of the  
iron.  
When the switch is closed, electric current  
starts to flow. The electric energy is  
converted to heat against the resistance  
offered by the coil. The heating effect in the  
coil causes the coil to get warmer.  
(ii) The resistance of the conductor result  
heat energy  
JOULES’ LAW  
It  
tells  
us  
the  
relationship  
between  
Factors affecting heat quantity  
It depends on the following factors  
(i) Resistance of a conductor  
resistance, current and heat generated.  
States that” When electric current is passed  
through a conductor heat evolved in a given  
time is directly proportional to the resistance  
of the conductor in ohms multiplied by the  
square of the current in amperes”  
(ii) Magnitude of electric current  
(iii) Time taken for the current to pass  
Resistance of a conductor  
The higher the resistance the higher the heat  
퐻 ∝ 푅  
퐇 ∝ 퐈ퟐ퐑퐭  
The magnitude of the current  
Large current produces more heat. The  
higher the current the higher the heat  
produced.  
Hence  
퐇 = 퐤퐈ퟐ퐑퐭, but k = 1  
퐇 = 퐈ퟐ퐑퐭  
H ∝ I2  
Joule is the work done when a charge of one  
coulomb flows through a conductor with a  
potential difference (p.d) of 1 volt a cross it  
in one second.  
Time taken by the current to pass  
The more the time the current takes to pass  
through the conductor, the higher the heat  
produced in a conductor.  
퐻 ∝ 푡  
From  
2
Example 01  
( )  
H = I Rt … … … … … . i  
V2  
Study carefully the diagram below and use it  
to answer the next questions  
H =  
t … … … … … . (ii)  
R
H = ItV … … … … … … (iii)  
ELECTRICAL POWER  
Electrical power (P):is the rate at which  
electrical energy is dissipated by the  
appliance.  
The coil feels warmer after closing the  
switch, explain why this happens?  
Answers  
electrical energy  
Power =  
time taken  
ItV  
t
t푇 = t퐴 + t퐵  
t푇 = 10 min + 20 min  
t푇 = 30 min  
Power =  
Power = IV ……… (i)  
Power = I2R …….. (ii)  
2
V
Power =  
………. (iii)  
R
Example 02  
An electric kettle containing two heating  
coils A and B is used to boil water. If it  
takes 10 minutes for coil A to boil water and  
20 minutes for coil B to boil the same  
amount of water, how long does it take for  
the water to boil when the two coils are  
joined in parallel?  
Points to note  
If the electric energy is used to heat water,  
we assume that no energy is lost, therefore  
the heat energy gained by water and  
calorimeter is equal to the heat supplied by  
electricity.  
Solution  
Electric energy = heat gained by water +  
vessel  
t1t2  
tT =  
t1 + t2  
If the vessel is assumed to have no mass, or  
the heat capacity, or specific heat capacity is  
neglected then.  
10 × 20  
tT =  
10 + 20  
I2Rt = mc∆θ, for specific heat capacity  
tT = 6.67minutes  
Itv = c∆θ, for heat capacity  
It will take 6.67 minutes for water to boil.  
Example 01  
Example 03  
An electric kettle containing two heating  
coils A and B is used to boil water. If it  
takes 10 minutes for coil A to boil water and  
20 minutes for coil B to boil the same  
amount of water, how long does it take for  
the water to boil when the two coils are  
joined in series?  
A resistor of 100Ω is connected across a  
battery of 12V. How much heat is dissipated  
across the resistor in 5 seconds? (Ignore the  
internal resistance of the battery)  
Solution  
Solution  
V2  
The heat energy due to electric current is  
given by  
H =  
t
R
(12)2  
H = ItV  
H =  
× 5  
If the two heater coils are joined in series,  
the total het energy is given by:  
100  
H = 7.2J  
HT = HA + HB  
Example 04  
IVt푇 = IVt퐴 + IVt퐵  
A bulb draws a current of 0.5A from a 240V  
source. Calculate the energy dissipated in 10  
minutes.  
lamp uses to transfer electrical energy into  
hat and light energy.  
Solution  
Solution  
Power = IV  
퐄 = 퐈퐭퐕  
Power = 240V × 0.25A  
Power = 60W  
E = 0.5A × 240V × 600 s  
E = 72000J  
The heat energy dissipated is 72000J or 72kJ  
The lamp transfers 60 J of electrical energy  
into heat and light each second.  
Example 05  
Three resistors with resistance 9Ω, 12Ω and  
15Ω are connected in series across a 12V  
battery. Calculate the energy dissipated by  
the 12Ω resistor in 10 s. Internal resistance  
of the battery is negligible.  
Example 07  
Television set is rated as 240V, 200W. What  
does this information mean?  
Answers  
Solution  
The operating voltage for the television set  
is 240Volts. When it is operating normally,  
the electrical power output is 200Joules per  
second (200 Watts).  
First step: To find the total resistance  
푅 = 9Ω + 12Ω + 15Ω  
푅 = 36Ω  
Second step: to find the current flowing in  
the circuit  
From: P = IV,  
푃
I =  
푉
푉표푙푡푎푔푒  
퐶푢푟푟푒푛푡, 퐼 =  
200푊  
I =  
푅푒푠푖푠푡푎푛푐푒  
240푉  
12푉  
퐼 =  
36Ω  
I = 0.83ꢀ  
퐼 = 0.33ꢀ  
The television set can allow a current of  
0.83A to flow through it when is  
connected to a 240V main power supply.  
Third  
step:  
To  
calculate  
the  
energy  
dissipated  
퐇 = 퐈ퟐ퐑퐭  
Example 08  
2
(
)
H = (0.33A) × 12Ω × 10 s  
An electric bulb is labeled 100W, 240V.  
Calculate  
H = 13.07J  
(a) The current through the filament lamp  
when the bulb works normally  
The energy dissipated by the 12Ω resistor in  
10s is 13.07J  
Example 06  
If the lamp on a 240V supply draws a  
current of 0.25A, calculate the power the  
Solution  
P = IV,  
푃
푉
A 2.5kW immersion heater is used to heat  
water. Calculate  
I =  
(a) The operating voltage of the heater is  
its resistance is 24Ω.  
100푊  
I =  
240푉  
Solution  
I = 0.417ꢀ  
2
V
Electrical power , P =  
R
The current in the filament lamp is  
0.417A  
푉2 = 푃 × 푅  
푉 = √푃 × 푅  
(b) The resistance of the filament used in  
the bulb  
Solution  
V = IR  
√
푉 =  
2.5 × 1000 × 24  
푉 = 245푉  
The operating voltage is 245V  
푉
R =  
퐼
240  
R =  
0.417  
(b) The electrical energy converted into  
heat energy in 2 hours.  
R = 575.5Ω  
R ≈ 576Ω  
Thus, the resistance of the filament lamp is  
576Ω  
Solution  
Electrical  
energy,  
E = (IV)t  
E = P × t  
Example 09  
An electric bulb is labeled 60W, 240V.  
Calculate the resistance of the filament used  
in the bulb.  
E = 2500 × 2 × 60 × 60  
E = 18 × 106J  
Solution  
E = 18MJ  
Power rating of the bulb is 60W, 240V  
V2  
P =  
R
The heat energy is 18MJ  
V2  
R =  
P
Example 11  
A radiation room heater consumes 28.8MJ  
of energy in 4 hours when connected to the  
main power supply of 240V. Calculate the  
current through the filament heater.  
Solution  
(240)2  
R =  
60  
R = 960Ω  
The resistance of the filament is 960Ω  
Energy supplied by electric current = energy  
consumed  
Example 10  
VIt = 28.8MJ