VITEEE 2012 Paper

Balanced wheatstone bridge condition $\frac{P}{Q}$ = $\frac{R}{S}$ No, current flows through galvanometer Now, P and R are in series, so Resistance $R_1$ = P + R = 10 + 15 = 25 Ω Similarly, Q and S are in series, so Resistance $R_2$ = R + S = 20 + 30 = 50Ω Net resistance of the network as R, and $R_2$ are in parallel $\frac{1}{R}$ = $\frac{1}{R_1}+\frac{1}{R_2}$ ∴ R = $\frac{25\times 50}{25+50}$ = $\frac{50}{3}$ Ω Hence, current, I = $\frac{V}{R}$ = $\frac{6}{\frac{50}{3}}$ = 0.36 A