Q. To convert a $800\ mV$ range $milli\ voltmeter$ of resistance $40\ \Omega$ into a galvanometer of $100\ mA$ range, the resistance to be connected as shunt is
A) $10\ \Omega$
B) $20\ \Omega$
C) $30\ \Omega$
D) $40\ \Omega$
Explanation
(a) $\frac{i}{i_{g}} = 1 + \frac{G}{S} \Rightarrow \frac{i.G}{V_{g}} = 1 + \frac{G}{S} \Rightarrow \frac{100 \times 10^{- 3} \times 40}{800 \times 10^{- 3}} = 1 + \frac{40}{S}$
$\Rightarrow S = 10\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.