Question

A galvanometer has a current range of $15mA$ and a voltage range of $750mV$ . To convert this galvanometer into an ammeter of range $25A$ , the required shunt is

• A. $0.8\Omega$
• B. $0.93\Omega$
• C. $0.03\Omega$
• D. $2.0\Omega$

Answer 1

Answer: (C)

$V=750\times 10^{-3}V$
$I_g=15\times 10^{-3}A$
and $I=25A$
Using the relation, $R=\frac{V}{I_g}$ , resistance of galvanometer is
$=\frac{750\times 10^{-3}}{15\times 10^{-3}}$
$=50\Omega$
$I_g=\frac{S}{S+R}\times I$
$15\times 10^{-3}=\left(\frac{\mathrm{S}}{\mathrm{S}+50}\right)\times 25$
$S=\text{0.03}\Omega$