Question

On connecting a shunt of $12\Omega$, the deflection of galvanometer reduces from $50$ to $20$ divisions. Calculate resistance of galvanometer.

Answer 1

Shunt resistance $S=12\Omega$
Let the galvanometer resistance be $R_G$.
Current flowing through $R_G$, $I_G=20$ units
Current flowing through the ammeter $I=50$ units
Using $I=I_G(1+\frac{R_G}{S})$
$\therefore$ $50=20(1+\frac{R_G}{12})$
Or $2.5=1+\frac{R_G}{12}$
Or $1.5=\frac{R_G}{12}$
$⟹R_G=1.5\times 12=18\Omega$