Question

In Wheatstone's bridge, three resistors P, Q, R are connected in three arms in order and $4^{th}$ arm 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 the equivalent resistance of the resistors $s_1$ and $s_2$ connected in parallel be $S$.

Equivalent resistance in parallel combination $\frac{1}{S}=\frac{1}{s_1}+\frac{1}{s_2}$

$⟹$ $S=\frac{s_1{s}_1}{s_1+s_2}$

Wheatstone bridge is balanced if $\frac{P}{Q}=\frac{R}{S}$

Or $\frac{P}{Q}=\frac{R}{\frac{s_1{s}_2}{s_1+s_2}}$

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