Page 194 - NCERT Science Class 10 English Medium

Applying Ohm’s law [Eq. (11.5)], we get
2
H = I Rt (11.21)
This is known as Joule’s law of heating. The
law implies that heat produced in a resistor is
(i) directly proportional to the square of current
for a given resistance, (ii) directly proportional to
resistance for a given current, and (iii) directly
proportional to the time for which the current flows
through the resistor. In practical situations, when
an electric appliance is connected to a known
voltage source, Eq. (11.21) is used after
calculating the current through it, using the Figure 11.13
Figure 11.1311.13
Figure
Figure
11.13
Figure 11.13
relation I = V/R. A steady current in a purely resistive electric circuit
Example 11.10
An electric iron consumes energy at a rate of 840 W when heating is
at the maximum rate and 360 W when the heating is at the minimum.
The voltage is 220 V. What are the current and the resistance in each
case?

Solution
From Eq. (11.19), we know that the power input is
P = V I
Thus the current I = P/V
(a) When heating is at the maximum rate,
I = 840 W/220 V = 3.82 A;
and the resistance of the electric iron is
R = V/I = 220 V/3.82 A = 57.60 Ω.
(b) When heating is at the minimum rate,
I = 360 W/220 V = 1.64 A;
and the resistance of the electric iron is
R = V/I = 220 V/1.64 A = 134.15 Ω.

Example 11.11
100 J of heat is produced each second in a 4 Ω resistance. Find the
potential difference across the resistor.

Solution
H = 100 J, R = 4 Ω, t = 1 s, V = ?
From Eq. (11.21) we have the current through the resistor as
I = √(H/Rt)
= √[100 J/(4 Ω × 1 s)]
= 5 A
Thus the potential difference across the resistor, V [from Eq. (11.5)] is
V = IR
= 5 A × 4 Ω
= 20 V.

Electricity 189

2024-25