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:
Resistance of arm AB $=P$
Resistance of arm BC $=Q$
Resistance of arm AD $=S$
Resistance of arm DC $=R$

The condition for the bridge to be balanced is:
$\frac{P}{Q}=\frac{S}{R}$

The resistance $P$ is formed by three resistors $P_1$, $P_2$, and $P_3$ connected in parallel. The equivalent resistance $P$ is given by:
$\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}$

$\frac{1}{P}=\frac{P_2{P}_3+P_1{P}_3+P_1{P}_2}{P_1{P}_2{P}_3}$

$P=\frac{P_1{P}_2{P}_3}{P_1{P}_2+P_2{P}_3+P_1{P}_3}$

Substituting the value of $P$ in the balance condition, we get:
$\frac{P_1{P}_2{P}_3/(P_1{P}_2+P_2{P}_3+P_1{P}_3)}{Q}=\frac{S}{R}$

Answer: $\frac{P_1{P}_2{P}_3/(P_2{P}_3+P_1{P}_3+P_1{P}_2)}{Q}=\frac{S}{R}$