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 × 10⁻³ A

I = 0.4 A

V = 0.5 v

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

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

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

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

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

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

= `40/99` Ω

(b) R_s = `V/I_g - G`

= `0.5/(4 xx 10^-3) - 40`

= `5/4 xx 100 - 40`

= 125 − 40

= 85 Ω