Explanation

(c) The equivalent circuit can be redrawn as
we have, $\frac{P}{Q} = \frac{R}{S}$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{4}{4} = \frac{4}{4}$
So, the given circuit is a balanced Wheatstone’s bridge.
Hence, the equivalent resistance
$R_{AB} = \frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)}$
$\ \ \ \ \ \ \ \ = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\Omega$