Q. A galvanometer whose resistance is $120\Omega$ gives full scale deflection with a current of $0.005\ A$ so that it can read a maximum current of $10\ A.$ A shunt resistance is added in parallel with it. The resistance of the ammeter so formed is
A) $0.06\ \Omega$
B) $0.006\ \Omega$
C) $0.6\ \Omega$
D) $6\ \Omega$
Explanation
(c) Resistance of shunted ammeter $= \frac{GS}{G + S}$
Also $\frac{i}{i_{g}} = 1 + \frac{G}{S} \Rightarrow \frac{GS}{G + S} = \frac{i_{g}.G}{i}$
$\Rightarrow \frac{GS}{G + S} = \frac{0.05 \times 120}{10} = 0.6\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.