Question

A galvanometer has a resistance of 40 Ω and a current of 4 mA is needed for full scale deflection. What is the resistance and how is it to be connected to convert the galvanometer

1. into an ammeter of 0.4 A range and
2. into a voltmeter of 5 V range?

Answer 1

Given: G = 40 Ω

I_g = 4 mA = 4 × 10⁻³ A

I = 0.4 A

V = 5 V

(a) Shunt resistance (S) = `(I_g G)/(I - I_g) `

= `(4 xx 10^-3 xx 40)/(0.4 - 4 xx 10^-3)`

= `(4 xx 10^-3 xx 40)/((4 xx 10^-1) - 4 xx 10^-3)`

= `(4 xx 10^-3 xx 40)/(4 xx 10^-1 (1 - 10^-2))`

= `(4 xx 10^-1)/(1 - 1/100)`

= `(4 xx 10^-1)/99 xx 100`

= `40/99` Ω

= 0.404 Ω

∴ Shunt resistance is connected in parallel with the galvanometer.

(b) Resistance in series (R_s) = `V/I_g - G`

= `5/(4 xx 10^-3) - 40`

= `5/0.004 - 40`

= 1250 − 40

= 1210 Ω

∴ The series resistance is connected in series with the galvanometer.