A galvanometer of resistance $40 \Omega$ gives a deflection of 10 divisions per $\mathrm{mA}$. There are 50…
A galvanometer of resistance $40 \Omega$ gives a deflection of 10 divisions per $\mathrm{mA}$. There are 50 divisions on the scale. Maximum current that can pass through it when a shunt resistance of $2 \Omega$ is connected is
$105 \mathrm{~mA}$
$155 \mathrm{~mA}$
$210 \mathrm{~mA}$
$75 \mathrm{~mA}$
Solution
Given, galvanometer resistance, $R_G=40 \Omega$
Shunt resistance, $R_\lambda=2 \Omega$
Reading $=10 \mathrm{div} / \mathrm{mA}$
and number of divisions, $n=50$
$\therefore$ Galvanometer current, $I_G=\frac{50}{10}=5 \mathrm{~mA}$
Let shunt current be $I$.
$
\begin{array}{ll}
\text { Since, } \frac{I}{R_G+R_\lambda} & =\frac{I_G}{R_\lambda} \\
\therefore \quad I & =\frac{5 \times(40+2)}{2} \\
& =105 \mathrm{~mA}
\end{array}
$