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Question: In a Wheatstone’s network,P= 2\(T_{1} = T_{2}\),Q=2\(T_{1} = \frac{1}{T_{2}}\), R= 2 \(7.5 \times 10...

In a Wheatstone’s network,P= 2T1=T2T_{1} = T_{2},Q=2T1=1T2T_{1} = \frac{1}{T_{2}}, R= 2 7.5×10−4ms−17.5 \times 10^{- 4}ms^{- 1} and S = 3 3×10−10Vm−13 \times 10^{- 10}Vm^{- 1}. The resistance with which S is to be shunted in order that the bridge may be balanced is

A

16.5×10−6m2V−1s−16.5 \times 10^{- 6}m^{2}V^{- 1}s^{- 1}

B

22.5×106m2V−1S−12.5 \times 10^{6}m^{2}V^{- 1}S^{- 1}

C

42.5×104m2V−1S−12.5 \times 10^{4}m^{2}V^{- 1}S^{- 1}

D

66.5×104m2V−1s−16.5 \times 10^{4}m^{2}V^{- 1}s^{- 1}

Answer

66.5×104m2V−1s−16.5 \times 10^{4}m^{2}V^{- 1}s^{- 1}

Explanation

Solution

: Let x be the resistance shunted with S for the bridge to be balanced.

For a balance wheatstone’s bridge

PQ=RS′\frac{P}{Q} = \frac{R}{S'} or 22=2S′\frac{2}{2} = \frac{2}{S'} or S′=2ΩS' = 2\Omega

From figure.,

1S′=1S+1x\frac{1}{S'} = \frac{1}{S} + \frac{1}{x} or 12=13+1x\frac{1}{2} = \frac{1}{3} + \frac{1}{x}

or 1x=12−13=16\frac{1}{x} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} or x=6Ωx = 6\Omega