Question

In a Wheatstone's bridge, there resistances $P,O$ and $R$ connected in the three arms and the fourth arm is formed by two resistances $S_1$ and $S_2$ connected in parallel. The condition for bridge to be balanced will be

• A. $\frac{P}{O}=\frac{R}{S_1+S_2}$
• B. $\frac{P}{O}=\frac{2R}{S_1+S_2}$
• C. $\frac{P}{O}=\frac{R(S_1+S_2)}{S_1{S}_2}$
• D. $\frac{P}{O}=\frac{R(S_1+S_2)}{2S_1{S}_2}$

Answer 1

Answer: (C)

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

Equivalent resistance of the resistors $S_1$ and $S_2$ connected in parallel $\frac{1}{S}=\frac{1}{S_1}+\frac{1}{S_2}$

$⟹$ $S=\frac{S_1{S}_2}{S_1+S_2}$

For balanced bridge, $\frac{P}{O}=\frac{R}{S}$ $⟹\frac{P}{O}=\frac{R(S_1+S_2)}{S_1{S}_2}$