In a balanced Wheatstone’s network, the resistances in the arms $Q$ and $S$ are interchanged. As a result of this
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We know that in In a balanced Wheatstone's network,
$\frac{ P }{ Q }=\frac{ R }{ S }$
If $Q$ is replaced by $S$ then ratio becomes $\frac{P}{S}=\frac{R}{Q}$ which is not equal to the previous ratio. Hence, network is not balanced.
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