Question

A galvanometer with resistance $100\Omega$ gives full scale deflection with a current of $10mA$. The value of shunt, in order to convert it into an ammeter of 10 ampere range, will be

• A. $-10\Omega$
• B. $1\Omega$
• C. $0.1\Omega$
• D. $0.01\Omega$

Answer 1

Answer: (C)

$0.1\Omega$

$I=I_g+I_s$
$(I-I_g)R_s=I_g{R}_g$
Given,
$I=10A$
$I_g=10mA=0.010A$
$R_g=100\Omega$
Substituting the values,
$(10-0.010)R_s=0.010\times 100$
$\Rightarrow R_s=\frac{1}{9.99}=0.1$