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. $210\, mA$
• B. $155\, mA$
• C. $420\, mA$
• D. $75\, mA$

Answer 1

Answer: (A)

$I_{G}=\frac{50}{5}=10\,mA; R_{G}=40\, \Omega, R_{s}=2\,\Omega $
Maximum current,
$I=\frac{R_{G}+R_{S}}{R_{G}}\times I_{G}$
$=\frac{\left(40+2\right)\times10}{2}$
$=210\, mA$