P, Q, R and S are four wires of resistances 3, 3, 3 and \( 4\,\Omega \) respectively. They are connected to form the four arms of a wheatstone bridge circuit. The resistance with which S must be shunted in order that the bridge may be balanced is ____________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 12\,\Omega \)
A Wheatstone bridge is balanced when the ratio of the arms is equal, \( P/Q = R/S \). Shunting S means connecting a resistance \( x \) in parallel with S, which lowers the resistance of that arm.
\( \displaystyle \frac{P}{Q} = \frac{R}{S_{\mathrm{eff}}} \;\Rightarrow\; \frac{3}{3} = \frac{3}{S_{\mathrm{eff}}} \;\Rightarrow\; S_{\mathrm{eff}} = 3\,\Omega \)
The parallel combination of S and \( x \) must therefore be \( 3\,\Omega \):
\( \displaystyle \frac{S\,x}{S + x} = 3 \;\Rightarrow\; \frac{4x}{4 + x} = 3 \)
\( 4x = 12 + 3x \)
\( x = 12\,\Omega \)
So S must be shunted with a \( 12\,\Omega \) resistance.