Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistance $S$ form the four arms of a

Question

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 $S$ ?\

Professor PC Thomas Classes · Answer

(a):\ Let $X$ be the equivalent resistance between $S$ and $6 \Omega$.\ $	herefore \quad \frac{1}{X}=\frac{1}{S}+\frac{1}{6}$\ Therefore, the equivalent circuit diagram drawn here\ For a balanced Wheatstone's bridge, we get\ $\frac{P}{Q}=\frac{R}{X} \quad$ 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, $S=3 \Omega$.\