The condition such that the potential difference, $V_{A} - V_{B} = 0$ is.

\frac{R_{1}}{R_{2}} = \frac{R_{4}}{R_{3}}

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\frac{R_{1}}{R_{3}} = \frac{R_{4}}{R_{2}}

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\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}

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Solution

The correct option is C $\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$
In a simple setup of Wheatstone bridge as shown below, $\frac{R_{1}}{R} = \frac{R_{2}}{R_{s}}$

There is a simple trick so as to not get confused. Just take the product of resistors across the galvanometer, in an 'x' fashion, and equate. For example in the setup as shown, $R_{1} R_{s} = R_{2} R$

So accordingly for this setup, $R_{1} R_{4} = R_{2} R_{3}$
Which when rearranged, gives
$\frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}$ and also
$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$

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