What is the value of x when the Wheatstone’s network is balanced? P = 500 Ω, Q = 800 Ω, R = x + 400, S = 1000 Ω

225 Ω 250 Ω 350 Ω 400Ω

225 Ω

For balancing, the condition on resistances is P/Q = R/S $\Rightarrow \frac{500}{800} = \frac{x+400}{1000}$ $\Rightarrow \frac{5}{8} = \frac{x+400}{1000}\Rightarrow x+400 = 625$ $\Rightarrow x = 225\Omega$