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Question

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


Solution

We apply Kirchoff's current law in the shown circuit.
At junction B, 
$$i_1=i_g+i_3$$
At junction D,
$$i_2+i_g=i_4$$
If current through the galvanometer is zero,
$$i_g=0$$
thus $$i_1=i_3$$
and $$i_2=i_4$$
Applying Kirchoff's voltage law for loop ABDA,
$$i_1P+i_gG=i_2R$$
Applying Kirchoff's voltage law for loop BCDB,
$$i_3Q=i_4S+i_gG$$
When $$i_g=0$$,
$$i_1P=i_2R$$
and $$i_3Q=i_4S$$
But $$i_1=i_3$$ and $$i_2=i_4$$,
Therefore $$\dfrac{P}{Q}=\dfrac{R}{S}$$

504736_472031_ans_cf42742c781f4ecfb784bfd3fab2a275.png

Physics

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