Question

A galvanometer has a current sensitivity of 10 div/mA and a voltage sensitivity of 2 div/mV. The instrument has 20 divisions. How can it be converted to
a) an ammeter reading upto 2 amperes, and
b) a voltmeter reading upto 20 volts?

Answer 1

$S_i=10div/mA=10\times {10}^{3}div/A$

$S_v=2div/mA=2\times {10}^{3}div/V$

$\theta =20division$

$I=2A$

$V=20V$

The resistnace of the galvanometer,

$G=\frac{S_i}{S_v}=\frac{10\times {10}^3}{2\times {10}^3}$

$G=5\Omega$

The current sensitivity is $S_i=\frac{\theta }{I_s}$

$I_s=\frac{\theta }{S_i}=\frac{20}{10\times {10}^3}$

$I_s=2\times10^{-3}$

The shunt resistance is,

$S=\frac{I_s\times G}{I-I_s}=\frac{G}{\frac{I}{I_s}-1}$

$S=\frac{5}{\frac{2}{2\times {10}^{-3}}-1}$

$S=\frac{5}{\frac{2\times {10}^3}{2}-1}$

$S=\frac{5}{\frac{2000-2}{2}}$

$S=\frac{10}{1998}$

$S=0.005\Omega$

The high resistance is,

$R=\frac{V}{I_s}-G$

$R=\frac{20\times {10}^3}{2}-5$

$R=\frac{20000-10}{2}$

$R=9995\Omega$