Three resistance `P, Q, R` each of `2 Omega` and an unknown resistance `S` from the four amrs of a Wheatstone's bridge circuit. When a resistance of `6 Omega` is connected in parallel to `S` the bridge gets balanced. What is the value of `S` ?

Options

A. `3Omega`
B. `6 Omega`
C. `1 Omega`
D. `2 Omega`

Answer

A

Explanation

Let X be the equivalent resistance between S and `6Omega`
`therefore 1/X = 1/S + 1/6`

Therefore, the equivalent circuit is as shown in diagram (b).
`P/Q = R/X` or `2/2 = 2/X`
or `X = 2Omega`
From equation (i), we get
`1/2 = 1/S + 1/6` or `1/S = 2/6`
or `S= 3Omega`