Question

A galvanometer of resistance $40\Omega$ gives a deflection of $5$ divisions per $mA$. There are $50$ divisions on the scale. The maximum current that can pass through it when a shunt resistance of $2\Omega$ is connected is:

• A. $210mA$
• B. $155mA$
• C. $420mA$
• D. $75mA$

Answer 1

The correct option is A $210mA$

$I_g$=$\frac{50}{5}=10mA$;

$R_g$ = 40 $\Omega$, $R_s$= 2$\Omega$

Maximum current,

$I=\frac{R_g+R_s}{R_g}$ X $I_g$ = $\frac{(40+2)\times 10}{2}$ = $210mA$