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Two cells of emfs $\varepsilon_1$, $\varepsilon_2$ and internal resistances $r_1$, $r_2$ are connected in parallel (positives joined, negatives joined). The equivalent internal resistance is
A$r_1 + r_2$
B$\dfrac{r_1 r_2}{r_1 + r_2}$
C$\dfrac{r_1 + r_2}{2}$
D$\dfrac{r_1 r_2}{r_1 - r_2}$
Answer & Solution
Correct answer: B. $\dfrac{r_1 r_2}{r_1 + r_2}$
1. For cells in parallel, the internal resistances combine like resistors in parallel.
2. The rule is $1/r_{eq} = 1/r_1 + 1/r_2$.
3. Solving gives $r_{eq} = r_1 r_2/(r_1 + r_2)$.
4. The sum $r_1 + r_2$ (A) is the series result, not parallel.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.16_
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