Q. A galvanometer has a resistance 50$\Omega$. A resistance of 5$\Omega$ is connected parallel to it. Fraction of the total current flowing through galvanometer is
A) $\frac{1}{10}$
B) $\frac{1}{11}$
C) $\frac{1}{50}$
D) $\frac{2}{15}$
Explanation
(b) The galvanometer G and shunt S is connected in parallel, hence potential difference is the same.
$\therefore i_{g} \times G = \left( i - i_{g} \right) \times S$
$\Longrightarrow \ \frac{i_{g}}{i - i_{g}} = \frac{S}{G} = \frac{5}{50} = \frac{1}{10}$
$\Longrightarrow 10i_{g} = i - i_{g}$
$\Longrightarrow 11i_{g} = i$
$\Longrightarrow \frac{i_{g}}{i} = \frac{1}{11}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.