Question

The maximum current that can be measured by a galvanometer of resistance $40\Omega$ is $10$$mA$. It is converted into a voltmeter that can read upto $50V$. The resistance to be connected in series with the galvanometer(in ohms) is:

• A. $2010$
• B. $4050$
• C. $5040$
• D. $4960$

Answer 1

Answer: (D)

$4960$

Given : Galvanometer resistance $G=40\Omega$

Current flowing through a galvanometer $I_g=10mA$

Let a resistor of resistance $R$ is connected in series with galvanometer resistance to convert it into a voltmeter.

Using , $V=I_g(G+R)$

$\therefore$ $50=10\times {10}^{-3}(40+R)$

OR $5000=40+R$ $⟹R=4960\Omega$