Q 12-04-217JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
A galvanometer having a resistance of $20\ \Omega$ and $30$ divisions on both sides has figure of merit $0.005$ ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter up to $15$ volt, is:
Answer: (C) $80\ \Omega$
Full-scale current: $I_g = 30\times0.005 = 0.15$ A.
$$R = \frac{V}{I_g} - G = \frac{15}{0.15} - 20 = 80\ \Omega$$
Solution by Sreeraj P, M.Sc Physics