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Three resistances P, Q, R each of $2\,\Omega$ and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of $6\,\Omega$ is connected in parallel to S the bridge gets balanced. What is the value of S?
A
$3\,\Omega$
B
$6\,\Omega$
C
$1\,\Omega$
D
$2\,\Omega$
Detailed Solution
A balanced Wheatstone bridge requires $\dfrac{P}{Q} = \dfrac{R}{S'}$, where $S'$ is the resistance of the fourth arm.
$\dfrac{2}{2} = \dfrac{2}{S'}$
Therefore the effective resistance of the fourth arm should be $S' = 2\,\Omega$.
This effective resistance is the parallel combination of S and the $6\,\Omega$ resistance.
In a parallel combination: $\dfrac{1}{S'} = \dfrac{1}{6} + \dfrac{1}{S}$
$\dfrac{1}{2} = \dfrac{1}{6} + \dfrac{1}{S}$
$\dfrac{1}{S} = \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{3 - 1}{6} = \dfrac{1}{3}$
$\Rightarrow S = 3\,\Omega$
$\dfrac{2}{2} = \dfrac{2}{S'}$
Therefore the effective resistance of the fourth arm should be $S' = 2\,\Omega$.
This effective resistance is the parallel combination of S and the $6\,\Omega$ resistance.
In a parallel combination: $\dfrac{1}{S'} = \dfrac{1}{6} + \dfrac{1}{S}$
$\dfrac{1}{2} = \dfrac{1}{6} + \dfrac{1}{S}$
$\dfrac{1}{S} = \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{3 - 1}{6} = \dfrac{1}{3}$
$\Rightarrow S = 3\,\Omega$