In Wheatstone bridge, resistance P is formed by a parallel combination of $p_{1}$ and $p_{2}$ . The Q resistance is formed by a parallel combination of $q_{1}$ and $q_{2}$ . The remaining two resistances are R and S respectively, and in the balanced condition, $\frac{R}{S}$ is

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In the given Wheatstone bridge, the resistance of the top-left arm is formed by the parallel combination of $p_{1}$ and $p_{2}$ . Let this equivalent resistance be $P$ .

$P = \frac{p _{1} p _{2}}{p _{1} + p _{2}}$

Similarly, the resistance of the top-right arm is formed by the parallel combination of $q_{1}$ and $q_{2}$ . Let this equivalent resistance be $Q$ .

$Q = \frac{q _{1} q _{2}}{q _{1} + q _{2}}$

The resistances of the bottom-left and bottom-right arms are $R$ and $S$ respectively.

For a balanced Wheatstone bridge, the ratio of resistances in the adjacent arms is equal:

$\frac{P}{Q} = \frac{R}{S}$

Substituting the values of $P$ and $Q$ :

$\frac{R}{S} = \frac{\frac{p _{1} p _{2}}{p _{1} + p _{2}}}{\frac{q _{1} q _{2}}{q _{1} + q _{2}}}$

$\frac{R}{S} = \frac{p _{1} p _{2} ( q _{1} + q _{2} )}{q _{1} q _{2} ( p _{1} + p _{2} )}$

Answer: $\frac{( p _{1} p _{2} ) ( q _{1} + q _{2} )}{( q _{1} q _{2} ) ( p _{1} + p _{2} )}$

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