Question

A moving coil galvanometer of resistance $20\Omega$ produces full scale deflection for a current of $50$mA. How will you convert the galvanometer into:

(i) An ammeter of range 20A and

(ii) A voltmeter of range 120V.

Answer 1

Given data:

$G=20\Omega$ , $I_g=50\times {10}^{-3}A;I=20A$,

$S=?;V=120V,R=?$

Solution:

i) $\therefore S=G.\frac{I_g}{I-I_g}=\frac{20\times 50\times {10}^{-3}}{20-50\times {10}^{-3}}=\frac{1}{20-0.05}$

$S=0.05\Omega$

A shunt of $0.05\Omega$ should be connected in parallel.

ii) $R=\frac{V}{Ig}-G$

$=\frac{120}{50\times {10}^{-3}}-20=2400-20=2380\Omega$

$R=2380\Omega$.