Describe Kelvin’s method to determine the resistance of the galvanometer by using a meter bridge.

#### Solution

**Kelvin's method:**

**Circuit:** The meter bridge circuit for Kelvin's method of determination of the resistance of a galvanometer is shown in the following figure. The galvanometer whose resistance G is to be determined is connected in one gap of the meter bridge. A resistance box providing a variable known resistance R is connected to the other gap.

**Kelvin's meter bridge circuit for the measurement of galvanometer resistance**

where, G: Galvanometer, R: Resistance box, AC: Uniform resistance wire, D: Balance point, E: Cell, K: Plug key, Rh: Rheostat

The junction B of the galvanometer and the resistance box is connected directly to a pencil jockey. A cell of emf E, a key (K) and a rheostat (Rh) are connected across AC.

**Working:** Keeping a suitable resistance R in the resistance box and maximum resistance in the rheostat, key K is closed to pass the current. The rheostat resistance is slowly reduced such that the galvanometer shows about 2/3rd of the full-scale deflection.

On tapping the jockey at end-points A and C, the galvanometer deflection should change to opposite sides of the initial deflection. Only then will there be a point D on the wire which is equipotential with point B. The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows no change in deflection. Point D is now called the balance point and Kelvin's method is thus an equal deflection method. At this balanced condition,

`"G"/"R" = "resistance of the wire of length l"_"G"/"- resistance of the wire of length l"_"g"`

where l_{G }≡ the length of the wire opposite to the galvanometer, I_{R} ≡ the length of the wire opposite to the resistance box.

If λ ≡ the resistance per unit length of the wire,

`"G"/"R" = (λ"l"_"G")/(λ"l"_"R") = "l"_"G"/"l"_"R"`

∴ G = `"R" "l"_"G"/"l"_"R"`

The quantities on the right hand side are known, so that G can be calculated.