alleryA cell of emf $\varepsilon$ and internal resistance r is connected to an external resistance R. The current in the circuit is
A$I = \dfrac{\varepsilon}{R + r}$
B$I = \dfrac{\varepsilon}{R - r}$
C$I = \dfrac{\varepsilon R}{r}$
D$I = \varepsilon (R + r)$
Answer & Solution
Correct answer: A. $I = \dfrac{\varepsilon}{R + r}$
1. Terminal voltage equals the drop across R: $V = IR$ and $V = \varepsilon - Ir$.
2. Equate: $IR = \varepsilon - Ir$, so $I(R + r) = \varepsilon$.
3. Therefore $I = \varepsilon/(R + r)$.
4. The internal resistance adds in series, so $R - r$ (B) is wrong.
_Source: syllabus Physics Ch 3 "Current Electricity", p.14_
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