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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 $S\,?$ ............... $\Omega$
$3$
$6$
$1$
$2$
Solution
Let $X$ be the equivalent resistance between $S$ and $6\, \Omega$.
$\therefore \quad \frac{1}{X}=\frac{1}{S}+\frac{1}{6}$ …….$(i)$
Therefore, the equivalent circuit diagram drawn below.
For a balanced Wheatstone bridge, we get
$\frac{P}{Q}=\frac{R}{X} \quad \text { or } \quad \frac{2}{2}=\frac{2}{X} \quad \Rightarrow \quad X=2 \,\Omega$
From eqn. $(i)$, we get
$\frac{1}{2}=\frac{1}{S}+\frac{1}{6} \quad$ or, $\quad \frac{1}{S}=\frac{2}{6} \quad$ or, $\quad S=3\, \Omega$
