Obtain the balancing conditions in case of Wheatstone’s bridge.

Answer: At balance \(\dfrac{P}{Q} = \dfrac{R}{S}\).

[Figure: Wheatstone’s network — four resistances P (A to B), Q (B to C), R (A to D) and S (D to C) form a quadrilateral ABCD. A cell with key is connected between A and C; a galvanometer G is connected between B and D.]

Let current \(I\) enter at A and divide into \(I_1\) through P and \(I - I_1\) through R. Let \(I_g\) flow through the galvanometer (resistance \(G\)) from B to D, so the current in Q is \(I_1 - I_g\) and in S is \(I - I_1 + I_g\).

Applying Kirchhoff’s voltage law:

Loop ABDA: \(I_1P + I_gG - (I - I_1)R = 0\) … (1)

Loop BCDB: \((I_1 - I_g)Q - (I - I_1 + I_g)S - I_gG = 0\) … (2)

The bridge is balanced when the galvanometer shows no deflection, \(I_g = 0\). Then

from (1): \(I_1P = (I - I_1)R\) … (3)

from (2): \(I_1Q = (I - I_1)S\) … (4)

Dividing (3) by (4):

This is the balancing condition. At balance the points B and D are at the same potential. If three resistances are known, the fourth can be found.