A simple circuit is formed by connecting a resistor of resistance R between the terminals of a battery of electromotive force (e.m.f.) 9.0 V and constant internal resistance r.

Question 21

A simple circuit is formed by connecting a resistor of resistance R between the terminals of a battery of electromotive force (e.m.f.) 9.0 V and constant internal resistance r.

A charge of 6.0 C flows through the resistor in a time of 2.0 minutes causing it to dissipate 48 J of thermal energy.

What is the internal resistance r of the battery?

A 0.17 Ω          B 0.33 Ω          C 20 Ω            D 160 Ω

Reference: Past Exam Paper – June 2015 Paper 13 Q34

Solution:

Answer: C.

A charge of 6.0 C flows through the resistor in a time of 2.0 minutes.

Current I = Q / t

Current through resistor = 6.0 / (2×60) = 0.05 A

Power dissipated in a resistor = I²R

Power = Energy / time

So, Energy dissipated = P × t = I²Rt = 48 J

(0.05)² × R × (2×60) = 48

Resistance R = 48 / [(0.05)² × (2×60)] = 160 Ω

The internal resistance of the battery and the resistor are in series. The same current flows through them.

So, the sum of p.d. across these 2 is equal to the e.m.f. in the circuit.

I × (R + r) = 9.0

R + r = 9.0 / 0.05 = 180

160 + r = 180

Internal resistance r = 180 – 160 = 20 Ω