Two batteries of different $e.m.f.'s$ and internal resistance 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 $e.m.f.'s$ of the two batteries is ............
Medium
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(b)
$3=\frac{E_1+E_2}{R+r_1+r_2}$
$1=\frac{E_1-E_2}{R+r_1+r_2}$
$3=\frac{E_1+E_2}{E_1-E_2}$
$2=\frac{E_1}{E_2}$
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