The Wheatstone bridge is in a balanced condit | vedclass

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(D) For a balanced Wheatstone bridge, the ratio of resistances in opposite arms must be equal, i.e.,$\frac{P}{Q} = \frac{R}{S}$.From the given figure,

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(D) For a balanced Wheatstone bridge, the ratio of resistances in opposite arms must be equal, i.e.,$\frac{P}{Q} = \frac{R}{S}$.From the given figure, the resistances are $P = 2\Omega$, $Q = 2\Omega$, $R = 6\Omega$, and the arm $S$ consists of a resistor $X$ in parallel with a $6\Omega$ resistor.The equivalent resistance of the parallel combination in arm $S$ is $S = \frac{X \cdot 6}{X + 6}$.Applying the balanced bridge condition: $\frac{2}{2} = \frac{6}{\left(\frac{6X}{X + 6}\right)}$.This simplifies to $1 = \frac{6(X + 6)}{6X}$.$1 = \frac{X + 6}{X}$.$X = X + 6$ is not possible, let us re-examine the circuit. The arm $CD$ has $X$ and $6\Omega$ in parallel. The condition is $\frac{AB}{BC} = \frac{AD}{CD_{eq}}$.Here $AB=2\Omega, BC=2\Omega, AD=2\Omega$. So, $CD_{eq} = 2\Omega$.Since $CD_{eq} = \frac{X \cdot 6}{X + 6} = 2$, we have $6X = 2(X + 6)$.$6X = 2X + 12$.$4X = 12$, which gives $X = 3\Omega$.

The Wheatstone bridge is in a balanced condition in the given figure, then $X = (in $Omega$)?$

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