Q117 / 160 · JIPMER 2013 · Physics
The equivalent resistance across $A$ and $B$ is
- A$2 \Omega$
- B$3 \Omega$
- Correct answerC$4 \Omega$
- D$5 \Omega$
Solution
The equivalent circuit can be redrawn as
We have, $\frac{P}{Q}=\frac{R}{S}$
i.e, $\quad \frac{4}{4}=\frac{4}{4}$
So, the given circuit is a balanced Wheatstone's bridge.
Hence, the equivalent resistance
$\begin{aligned} R_{A B} & =\frac{(4+4) \times(4+4)}{(4+4)+(4+4)}=\frac{8 \times 8}{8+8} \\ & =\frac{64}{16}=4 \Omega\end{aligned}$
We have, $\frac{P}{Q}=\frac{R}{S}$
i.e, $\quad \frac{4}{4}=\frac{4}{4}$
So, the given circuit is a balanced Wheatstone's bridge.
Hence, the equivalent resistance
$\begin{aligned} R_{A B} & =\frac{(4+4) \times(4+4)}{(4+4)+(4+4)}=\frac{8 \times 8}{8+8} \\ & =\frac{64}{16}=4 \Omega\end{aligned}$Hence, the correct answer is option: 'C'.