Q. A galvanometer coil has a resistance of 15$\ \Omega$ and gives full scale deflection for a current of 4mA. To convert it to an ammeter of range 0 to 6A
A) 10 m$\ \Omega$ resistance is to be connected in parallel to the galvanometer
B) 10$\ m\ \Omega$ resistance is to be connected in series with the galvanometer
C) 0.1$\ \Omega$ resistance is to be connected in parallel to the galvanometer
D) 0.1$\ \Omega$ resistance is to be connected in series with the galvanometer
Explanation
(a) To convert a galvanometer into an ammeter, a shunt resistance in parallel is connected to galvanometer.
Since, galvanometer G and shunt S are in parallel, hence
$i_{g}G = \left( i - i_{g} \right)$
$\Longrightarrow \ \ \ S = \frac{i_{g}G}{i - i_{g}} = \frac{4 \times 10^{- 3} \times 15}{6 - 4 \times 10^{- 3}}$
$= \frac{60 \times 10^{- 3}}{5.996} = 10 \times 10^{- 3}\Omega = 10m\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.