Question

In Wheatstone's bridge, three resistors P, Q, R are connected in three arms in order and $4^{th}$ arm S is formed by two resistors $s_1$ and $s_2$ connected in parallel. The condition for bridge to be balanced is: $\frac{P}{Q}=$

• A. $\frac{R(s_1+s_2)}{s_1{s}_2}$
• B. $\frac{s_1{s}_2}{R(s_1+s_2)}$
• C. $\frac{R{s}_1{s}_2}{(s_1+s_2)}$
• D. $\frac{(s_1+s_2)}{R{s}_1{s}_2}$

Answer 1

The correct option is A $\frac{R(s_1+s_2)}{s_1{s}_2}$

Let $S$ denote the equivalent resistance of the two parallel resistors.

We then have:

$\frac{P}{Q}=\frac{R}{S}$ (for the bridge to be in the balanced state)

and $S=\frac{s_1{s}_2}{s_1+s_2}$

This gives, $\frac{P}{Q}=\frac{R(s_1+s_2)}{s_1{s}_2}$