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(D) The given circuit is a balanced Wheatstone bridge.In a Wheatstone bridge,if the ratio of resistances in opposite arms is equal,the central resisto

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$20$

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(D) The given circuit is a balanced Wheatstone bridge.In a Wheatstone bridge,if the ratio of resistances in opposite arms is equal,the central resistor carries no current.Here,all four outer resistors are $20 \ \Omega$,so the ratio is $20/20 = 20/20 = 1$.Since the bridge is balanced,the central $20 \ \Omega$ resistor can be removed.Now,the circuit consists of two parallel branches,each having two $20 \ \Omega$ resistors in series.Resistance of the upper branch = $20 \ \Omega + 20 \ \Omega = 40 \ \Omega$.Resistance of the lower branch = $20 \ \Omega + 20 \ \Omega = 40 \ \Omega$.These two branches are in parallel,so the equivalent resistance $R_{eq}$ is given by:$1/R_{eq} = 1/40 + 1/40 = 2/40 = 1/20$.Therefore,$R_{eq} = 20 \ \Omega$.

The equivalent resistance between $P$ and $Q$ in the given figure is ............... $\Omega$.

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