In Wheatstone bridge, resistance P is formed by a parallel...

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 $P$.

$P = \dfrac{p_1 p_2}{p_1 + p_2}$

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 $Q$.

$Q = \dfrac{q_1 q_2}{q_1 + q_2}$

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:

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

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

$\dfrac{R}{S} = \dfrac{\dfrac{p_1 p_2}{p_1 + p_2}}{\dfrac{q_1 q_2}{q_1 + q_2}}$

$\dfrac{R}{S} = \dfrac{p_1 p_2 (q_1 + q_2)}{q_1 q_2 (p_1 + p_2)}$

Answer: $\dfrac{(p_1 p_2)(q_1 + q_2)}{(q_1 q_2)(p_1 + p_2)}$