Physics - Current Electricity Question with Solution | TestHub

For a balanced Wheatstone bridge, the ratio of resistances in the adjacent arms must be equal.

From the given circuit diagram, the resistances in the four arms are $P$, $Q$, $R$, and the equivalent resistance of the parallel combination of $S_1$ and $S_2$.

Let the equivalent resistance of the fourth arm be $S_{eq}$. Since $S_1$ and $S_2$ are connected in parallel, we have:

$\frac{1}{S_{eq}}=\frac{1}{S_1}+\frac{1}{S_2}\Rightarrow S_{eq}=\frac{S_1{S}_2}{S_1+S_2}$

The balancing condition for the Wheatstone bridge is:

$\frac{P}{Q}=\frac{R}{S_{eq}}$

Substituting the value of $S_{eq}$:

$\frac{P}{Q}=\frac{R}{\left(\frac{S_1{S}_2}{S_1+S_2}\right)}$

$\frac{P}{Q}=\frac{R(S_1+S_2)}{S_1{S}_2}$

Answer: $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left({\mathrm{S}}_1+{\mathrm{S}}_2\right)}{{\mathrm{S}}_1{\mathrm{S}}_2}$