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Derive the condition for balancing the Wheatstone's network.
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Solution:

  • Wheatstone bridge is a very accurate and precise method of determining an unknown resistance and it is an important application of Kirchoff's laws.

  • Since it is a null method, the measurement of unknown resistance is not affected by the internal resistance of the battery used.

  • Wheatstone bridge or network consists of four resistances P,Q,R and S connected in the form of a quadrilateral ABCD.

  • A galvanometer of resistance G is connected between the opposite end points B and D. A cell of emf E and a key are connected between the opposite end points A and C.

  • When the circuit is closed different currents flow through various branches of the network as shown in the diagram.

  • According to Kirchoff's current law Ig is the current passing through the galvanometer. Considering the network ABDA, applying Kirchoff's second law,

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$$ \begin{array}{ll} \mathrm{I}_{1} \mathrm{P}+\mathrm{I}_{\mathrm{g}} \mathrm{G}-\left(\mathrm{I}-\mathrm{I}_{1}\right) \mathrm{R}=0 \\\\ \text { or } \quad \mathrm{I}_{1} \mathrm{P}+\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{1}\right) \mathrm{R} \\\\ \end{array} $$

Considering the network $\mathrm{BCDB}$, applying Kirchoff's second law

$$ \begin{array}{ll} & \left(I_{1}-I_{g}\right) Q-\left(I-I_{1}+I_{g}\right) S-I_{g} G=0 \\\\ \text { or } \quad & \left(I_{1}-I_{g}\right) Q=\left(I-I_{1}+I_{g}\right) S+I_{g} G \\\\ \end{array} $$

The four resistances $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ are adjusted so that there is no deflection in the galvanometer. i.e. $\operatorname{Ig}=0$. Now the bridge is said to be balanced.

Now, equation (1) becomes,

$$ \mathrm{I}_{1} \mathrm{P}=\left(\mathrm{I}-\mathrm{I}_{1}\right) \mathrm{R} \\ $$

and equation (2) becomes,

$$ \mathrm{I}_{1} \mathrm{Q}=\left(\mathrm{I}-\mathrm{I}_{1}\right) \mathrm{S} \\ $$

dividing (3) by (4) we get,

$$ \begin{aligned} &\frac{\mathrm{I}_{1} \mathrm{P}}{\mathrm{I}_{1} \mathrm{Q}}=\frac{\left(\mathrm{I}-\mathrm{I}_{1}\right)}{\left(\mathrm{I}-\mathrm{I}_{1}\right)} \frac{\mathrm{R}}{\mathrm{S}} \\\\\ &\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}} \\ \end{aligned} $$

This is the balancing condition of Wheatstone's bridge network.

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