Two ideal batteries of emf V 1 and V 2 and three resistances R 1 , R 2 and R 3 are connected as shown in the figure. The current in resistance R 2 would be zero if

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TestHub · Accepted Answer

In order that the current is zero, the potential drop on $R_{2}$ due to $V_{1}$ must be nullified by that due to $V_{2}$ .

Accordingly,

$V_{1} = \frac{R_{1} (V_{1} + V_{2})}{R_{1} + R_{3}} \Rightarrow V_{1} R_{3} = V_{2} R_{1}$
$V_{2} = \frac{R_{3} (V_{1} + V_{2})}{R_{1} + R_{3}} \Rightarrow V_{2} R_{1} = V_{2} R_{3}$

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