9702/Oct Nov/13/2013/Q34

An electrical device of fixed resistance 20 Ω is connected in series with a variable resistor and a
battery of electromotive force (e.m.f.) 16 V and negligible internal resistance.

What is the resistance of the variable resistor when the power dissipated in the electrical device is 4.0W?

A 16 Ω 

B 36 Ω 

C 44 Ω 

D 60 Ω

Solution:

Answer: A

Power dissipated, P = I²R

For electrical device,

4.0 = I² (20)

Current I in circuit = 0.45A

Since this is a series circuit, the same current flows through the variable resistor.

Ohm’s law: V = IR

p.d. across electrical device = 0.45 (20) = 9.0V

Let the resistance of the variable resistor = R

From Kirchhoff’s second law, the sum of p.d. in a loop is equal to the e.m.f. in the circuit.

16 = 9.0 + 0.45R

Resistance R = 16Ω

Reference: PYQ - Oct/Nov 2013 Paper 13 Q34