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Q.

A 6V cell with 0.5 $Ω$ internal resistance and 10 V cell with 1 $Ω$ internal resistance and a 12 $Ω$ external resistance are connected in parallel. The current in amp through the 10 V cell is

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a

0.6

b

2.27

c

5.14

d

2.87

answer is C.

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$\begin{array}{l} i = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2} + R (r_{1} + r_{2})} = \frac{(6 \times 1) + (10 \times 0 .5)}{(1 \times 0 .5) + 12 (1 .5)} = \frac{11}{18 .5} A \\ V_{2} = E_{2} - i_{2} r_{2} ∴ iR = E_{2} - i_{2} r_{2} \\ i_{2} = \frac{E_{2} - iR}{r_{2}} = \frac{10 - \frac{11}{18 .5} \times 12}{1} = 2 .87 A \end{array}$

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