The electron drift speed in metals is small 10 - 4 m s - 1 and the charge of the electron is also very small 1 . 6 × 10 - 19 C , but we can still obtain a large amount of current in a metal. Why?

The electron drift speed in metals is small 10 - 4 m s - 1 and the charge of the electron is also very small 1 . 6 × 10 - 19 C , but we can still obtain a large amount of current in a metal because the electron number density in the metal is very large, (of the order of 10 29 m - 3 ). For one coulomb of charge, it just takes 6 . 25 × 10 18 electrons. And when these many electrons flow through the cross-section of a conductor made of metal in one second, they constitute one ampere of current. Thus, due to high electron number density in the metal as compared to number of electrons constituting one coulomb of charge, we get a large amount of current in a metal.

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.