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 $$\dfrac{P}{Q}=\dfrac{R}{S}$$Physics

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