Discussion
A battery of e.m.f. 1.5 V has a terminal potential difference of 1.25 V when an external resistance of 25 Ω is connected to it. The internal resistance of the battery will be
Answer: C
The internal resistance is 5 Ω.
The current through the 25 Ω external resistance follows from the terminal voltage, and the lost volts across the internal resistance from the difference between e.m.f. and terminal voltage:
\[ \begin{aligned} I &= \frac{1.25}{25} = 0.05\ \text{A} \\ r &= \frac{1.5 - 1.25}{0.05} = \frac{0.25}{0.05} = 5\ \Omega \end{aligned} \]
Equivalently, r = R(E/V - 1) = 25(1.5/1.25 - 1) = 5 Ω. Zero internal resistance would make the terminal voltage equal the e.m.f., and 25 Ω or 15 Ω would give much larger voltage drops.
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