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
Answer & explanation
Correct answer: option 1
For balancing, the condition on resistances is P/Q = R/S
$\Rightarrow \frac{500}{800} = \frac{x+400}{1000}$
$\Rightarrow \frac{5}{8} = \frac{x+400}{1000}\Rightarrow x+400 = 625$
$\Rightarrow x = 225\Omega$