Question

A galvanometer has a resistance of 50$\Omega$. If a resistance of 1$\Omega$ is connected across its terminals, the total current flow through the galvanometer is [$I_g$ represents the maximum current that can be passed through the galvanometer]

• A. 42$I_g$
• B. 53$I_g$
• C. 46$I_g$
• D. 51$I_g$

Answer 1

Answer: (D)

51$I_g$

Answer 2

In the galvanometer, $I_g$ = max. current through galvanometer, S = shunt resistance, G = galvanometer resistance then
$I_g(galvanometer)=1\bigg(\frac{S}{G+S}\bigg)$
$\Rightarrow I=I_g \bigg(\frac{G+S}{S}\bigg)=\bigg(\frac{50+1}{1}\bigg )I_g=51I_g$