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(D) Wheatstone bridge is balanced when the ratio of resistances in the adjacent arms is equal,i.e.,$P/Q = R/S$. In the given circuit,the resistances a

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$18\,\Omega$ and $6\,\Omega$

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(D) Wheatstone bridge is balanced when the ratio of resistances in the adjacent arms is equal,i.e.,$P/Q = R/S$. In the given circuit,the resistances are $P = 4\,\Omega$,$Q = 12\,\Omega$,$R = 18\,\Omega$,and $S = 6\,\Omega$. The current ratio is $P/R = 4/18 = 2/9$ and $Q/S = 12/6 = 2$. Since $P/R 
eq Q/S$,the bridge is currently unbalanced. For the bridge to be balanced,we need $P/Q = R/S$. Let us check option $(B)$: If we interchange $18\,\Omega$ and $12\,\Omega$,the new resistances become $P = 4\,\Omega$,$Q = 18\,\Omega$,$R = 12\,\Omega$,and $S = 6\,\Omega$. Now,the ratio $P/Q = 4/18 = 2/9$ and $R/S = 12/6 = 2$. This is still not balanced. Let us check option $(D)$: If we interchange $18\,\Omega$ and $6\,\Omega$,the new resistances become $P = 4\,\Omega$,$Q = 12\,\Omega$,$R = 6\,\Omega$,and $S = 18\,\Omega$. Now,the ratio $P/Q = 4/12 = 1/3$ and $R/S = 6/18 = 1/3$. Since $P/Q = R/S$,the bridge is balanced. Therefore,the correct option is $(D)$.

In the following diagram,the Wheatstone bridge is balanced when we interchange the resistances of:

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