Question

To send 10% of the main current through a moving coil galvanometer of resistance 99 the shunt required is-

• A. 11
• B. 9
• C. 100
• D. 19

Answer 1

The correct option is A 11

The current through galvanometer is $\frac{I}{10}.$

Now, current through galvanometer is

$\frac{I}{10}=\frac{R}{R+R_g}I$

$⟹\frac{1}{10}=\frac{R}{R+99}$

$⟹10R=R+99$

$⟹9R=99\Omega$

$⟹R=11\Omega$

Answer-(A)