**1 Answer**

written 2.4 years ago by |

**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,

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

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.**