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Question

Use Kirchhoff's rule to obtain conditions for the balance condition in a Wheatstone bridge.

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Solution

We apply Kirchoff's current law in the shown circuit.
At junction B,
i1=ig+i3
At junction D,
i2+ig=i4
If current through the galvanometer is zero,
ig=0
thus i1=i3
and i2=i4
Applying Kirchoff's voltage law for loop ABDA,
i1P+igG=i2R
Applying Kirchoff's voltage law for loop BCDB,
i3Q=i4S+igG
When ig=0,
i1P=i2R
and i3Q=i4S
But i1=i3 and i2=i4,
Therefore PQ=RS

504736_472031_ans_cf42742c781f4ecfb784bfd3fab2a275.png

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