State Kirchhoff's law for an electrical network. Using these laws deduce the condition for balance in a Wheatstone bridge.

Kirchhoff's two rules are as under: (a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction. (b) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero. The circuit diagram for Wheatstone bridge is as under: This bridge has four resistors R 1 , R 2 , R 3 and R 4 . Across one pair of diagonally opposite points, that is points A and C , a source is connected. This is called the battery arm. Between the other two vertices, B and D , a galvanometer G is connected. This is called the galvanometer arm. There will be currents flowing across all the resistors as well as a current I g through G . In the case of a balanced bridge, the resistors are such that I g = 0 . In this case, Kirchhoff's junction rule is applied to junctions D and B . Now, we apply Kirchhoff's loop rule to closed loops A D B A and C B D C . The loop A D B A gives, - I 1 R 1 + 0 + I 2 R 2 = 0 ( ∵ I g = 0 ) ∴ I 1 R 1 = I 2 R 2 or I 1 I 2 = R 2 R 1 ...(a) And the loop C B D C gives, I 4 R 4 + 0 - I 3 R 3 = 0 ( ∵ I g = 0 ) ∴ I 3 R 3 = I 4 R 4 or I 3 I 4 = R 4 R 3 ....(b) Now, let the current entering A be I then it divides into I 1 and I 2 such that, I 2 = I - I 1 . The current I 2 further divides into I g and I 4 such that I 4 = I 2 - I g . In the case of a balanced bridge, I g = 0 therefore, I 4 = I 2 . Similarly, we get I 3 = I 1 . Substituting I 3 = I 1 and I 4 = I 2 in the equation (b) we get, I 1 I 2 = R 4 R 3 ...(c) From the equations (a) and (c), we get, I 1 I 2 = R 2 R 1 = R 4 R 3 Hence, we obtain the condition, R 2 R 1 = R 4 R 3 which is the equation relating the four resistors called the balance condition for the galvanometer to give zero or null deflection.