Question

ln, a Wheatstone network, P = Q = R = 8 `Omega` and S is 10 `Omega`. The required resistance to be connected to S so that network is balanced is ______.

Options

• 40 `Omega` in parallel

• 48 `Omega` in parallel

• 32 `Omega` in parallel

• 24 `Omega` in parallel

Answer 1

ln, a Wheatstone network, P = Q = R = 8 `Omega` and S is 10 `Omega`. The required resistance to be connected to S so that network is balanced is 40 `Omega` in parallel.

Explanation:

S = X `abs` 10 = 8 `Omega`

`therefore (10"X")/(10 + "X") = 8`

`therefore 10"X"` = 80 + 8X

`therefore 2"X" = 80`

`therefore "X" = 40  Omega`