Two batteries with e.m.f 12 V and 13 V are connected (in same polarity) in parallel across a load resistor of 10Ω . The internal resistances of the two batteries are 1Ωand 2Ω respectively. The voltage across the load lies between

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11.7 &11.8VV

11.6 &11.7VV

11.5 &11.6VV

11.4 &11.5VV

answer is B.

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Detailed Solution

$\begin{array}{l} ε_{eq} = \frac{\frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}}} = \frac{37}{3} V \\ r_{eq} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{2}{3} Ω \\ i = \frac{ε_{eq}}{r_{eq} + R} = \frac{\frac{37}{3}}{10 + \frac{2}{3}} = \frac{37}{32} \end{array}$
$V_{10 Ω} = i \times 10 = \frac{37}{32} \times 10 = \frac{370}{32} = 11 .56 V$

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