Q. In order to pass 10% of main current through a moving coil galvanometer of $99\ ohm,$ the resistance of the required shunt is
A) $9.9\ \Omega$
B) $10\ \Omega$
C) $11\ \Omega$
D) $9\ \Omega$
Explanation
(c) Shunt resistances $S = \frac{i_{g}G}{(i - i_{g})} = \frac{10 \times 99}{(100 - 10)} = 11\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.