VITEEE 2016 Paper

For no current through the galvanometer, the wheatstone bridge should be balanced. For this, we must have $\frac{P}{Q}$ = $\frac{S}{R} \Rightarrow \frac{10}{5} = \frac{4}{4}$ 2 = 1 is not balanced so we need to add 5 to $Q$ then the circuit will be balanced. This condition is satisfied with only option (a). When a 5Ω resistor is connected in series with Q, the equivalent resistance in the P-arm becomes 1 Ω ∴ $\frac{P}{Q}$. = $\frac{10}{10}$ = 1 and $\frac{S}{R}$ = $\frac{4}{4}$ = 1 ⇒ $\frac{P}{Q}$ = $\frac{S}{R}$