A battery with e.m.f. E and internal resistance r is connected in series with a variable external resistor.

Question 23

A battery with e.m.f. E and internal resistance r is connected in series with a variable external resistor.

The value of the external resistance R is slowly increased from zero.

Which statement is correct? (Ignore any temperature effects.)

A The potential difference across the external resistance decreases.

B The potential difference across the internal resistance increases.

C The power dissipated in r increases and then decreases.

D The power dissipated in R increases and then decreases.

Reference: Past Exam Paper – June 2015 Paper 11 Q36

Solution:

Answer: D.

Ohm’s law: V = IR

In general, as the resistance of a component increases the p.d. across it increases.

So, as the value of external resistance R is increased, the p.d. across it increases while the p.d. across the internal resistance decreases (since the sum of p.d. in a circuit is equal to the e.m.f. in the circuit). [A and B are incorrect]

The external resistance R and the internal resistance r forms a potential divider.

Power dissipated in external resistance R = I²R

Initially, the value of R is zero, so the power dissipated across it is also zero.

Similarly, if the value of R is infinite, the power dissipated is also zero (as the current I will then be zero).

So, the power dissipated must increase and then decrease as R increases from zero. As the value of R increases, the current I decreases, the p.d. across the external resistance R increases. The p.d. across the internal resistance r decreases and the power dissipated in r (given by I²r) decreases.

The maximum power dissipated across R occurs when R = r.