Consider the figure shown. The condition for which the potential difference between $A$ and $B$ is zero is :

C_{1} C_{2} = C_{3} C_{4}

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C_{1} C_{4} = C_{2} C_{3}

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C_{1} C_{3} = C_{2} C_{4}

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none of these

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Solution

The correct option is C $C_{1} C_{4} = C_{2} C_{3}$
Let the charge be as shown. (Capacitors in series have the same $t$ charge)
Taking loop containing $C_{1}, C_{2}$ and $E$ we get
$\frac{q}{C_{1}} + \frac{q}{C_{2}} - E = 0 o r q = E [\frac{C_{1} C_{2}}{C_{1} + C_{2}}]$
Similarly, from loop containing $C_{3}, C_{4}, E$ we get
$\frac{q^{'}}{C_{3}} + \frac{q^{'}}{C_{4}} - E = 0 o r q^{'} = E [\frac{C_{3} C_{4}}{C_{3} + C_{4}}]$
Now,
$V_{A} - V_{B} = \frac{q}{C_{2}} - \frac{q^{'}}{C_{4}}$
$= E [\frac{C_{1}}{C_{1} + C_{2}} - \frac{C_{3}}{C_{3} + C_{4}}]$
$= E [\frac{C_{1} C_{4} - C_{3} C_{2}}{(C_{1} + C_{2}) (C_{3} + C_{4})}]$
Now, $V_{A} - V_{B} = E \Rightarrow C_{1} C_{4} - C_{3} C_{2} = 0$
$\Rightarrow \frac{C_{1}}{C_{2}} = \frac{C_{3}}{C_{4}}$

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