Page 75 - Physics_Form_2

Current electricity

If the current of 1 A flows through a Example 2.13
resistor of 1 Ω in 1 second, then the A resistor of 100 Ω is connected across
heat generated in water is 1 J. Thus, a battery of 12 V. How much heat is
k = 1. Hence, dissipated across the resistor in 5 s?

(Ignore the internal resistance of the
2
H = I Rt battery).
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This equation gives the relationship
between the resistance of a conductor, Solution
current passing through a conductor, Given: voltage, V = 12 V, resistance,
time a current flows and electrical R  100  , time t = 5s
energy generated. The electrical
energy generated is measured in From,
Joules. Thus, this equation is referred V 2
to as Joule’s law of heating. The law H = R t
states that, “When an electric current
is passed through a conductor, the 12  V 2
2
heat generated in a given time is  100   5 s 7.2 J
directly proportional to the resistance
of the conductor in ohms, the square The heat dissipated across the resistor
of current in amperes and the time in is 7.2 J.
seconds for which the current flows”.
Since the unit of electrical energy is Example 2.14
joule, then the Joule is defined as the
work done when a charge of 1 coulomb A bulb draws a current of 0.5 A from
flows through a conductor with a p.d a 240 V source. Calculate the energy
of 1 volt across it in 1 second. dissipated in 10 minutes.
Solution
By Ohm’s law, V = IR . Then, Given:
V 2
H = t I = 0.5 A, V = 240 V,
R t  10  60 s  600 s

Also,
Using the equation
V
R = H = IVt
I
Thus, 0.5 A  240 V 600 s

H = IVt 72000 J=
Therefore, the energy dissipated is
These are forms of stating up Joule’s 72000 J or 72 kJ.
law.

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Physics Form 2 Final.indd 69 25/10/2025 10:26