Question

In Wheatstone bridge, three resistors $P,Q,R$ are connected in three arms in order and 4th 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 $\frac{R(s_1+s_2)}{s_1{s}_2}$

Balanced condition for Wheatstone bridge
$\frac{P}{Q}=\frac{R}{S}$ .......(i)
where $\frac{1}{s}=\frac{1}{s_1}+\frac{1}{s_2}$
$\Rightarrow s=\frac{s_1{s}_2}{s_1+s_2}$ ........(ii)
$\therefore$ From equations (i) and (ii), we get
$\frac{P}{Q}=\frac{R(s_1+s_2)}{s_1{S}_2}$