Q. A galvanometer of $50\ ohm$ resistance has 25 divisions. A current of $4 \times 10^{- 4}\ ampere$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25\ volts,$ it should be connected with a resistance of
A) $2500\ \Omega$ as a shunt
B) $2450\ \Omega$ as a shunt
C) $2550\ \Omega$ in series
D) $2450\ \Omega$ in series
Explanation
(d) Full deflection current $i_{g} = 25 \times 4 \times 10^{- 4} = 100 \times 10^{- 4}A$
Using $R = \frac{V}{I_{g}} - G = \frac{25}{100 \times 10^{- 4}} - 50 = 2450\Omega$ in series
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.