Two batteries of different emf and internal resistances connected in series with each other and with an external load resistor. The current is 3.0 A. When the polarity of one battery is reversed, the current becomes 1.0 A. The ratio of the emf of the two batteries is

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2 : 1

3 : 2

2.5 : 1

1 : 1

answer is B.

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Detailed Solution

$\begin{array}{l} 3 = \frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} - - - - (1) \\ 1 = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} - - - - (2) \frac{(1)}{(2)} \end{array}$

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