Question

A galvanometer has a current range of $15 \, mA$ and a voltage range of $750 \, mV$ . To convert this galvanometer into an ammeter of range $25 \, A$ , 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 $I=25 \, A$
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 $