Question

A galvanometer has a resistance of 49 $\Omega$ . If 2% of the main current is to be passed through the meter, then the value of the shunt will be

• A. 2$\Omega$
• B. 1$\Omega$
• C. 1/2$\Omega$
• D. 1/4$\Omega$

Answer 1

The correct option is B 1$\Omega$

We know that $R_S=\frac{I_G}{I-I_G}{R}_G$

Thus we get

$\frac{R_S}{R_G}=\frac{I_G}{I_S}$

or

$\frac{R_S}{R_G}=\frac{\frac{2}{100}I}{\frac{98}{100}I}=\frac{2}{98}$

Substituting $R_G=49$.

Thus we get $R_S=1ohm$