Q. The resistance of a galvanometer is $2.5 \, \Omega$ and it requires $50 \, mA$ for full scale deflection. The value of shunt resistance required to convert it into an ammeter of range $0$ to $5 \, A$ is
Solution:
Here,
Resistance of the galvanometer, $G = 2.5 \, \Omega$
Full scale deflection current, $I_8 = 50 \, mA$
$ = 50 \times 10^{-3} \, A = 0.05 \, A$
To convert the galvanometer into an ammeter of range $0$ to $5 \, A \, i.e. \, I = 5 \, A, a$ shunt $S$ is connected in parallel with it such that,
$(I - I_g)S = I_g G$
$S = \frac{I_g G}{I - I_g}$
$ = \frac{ 0.05 \times 2.5 }{5 -0.05} = 0.025 \, \Omega = 2.5 \times 10^{-2} \Omega$
