Q. A galvanometer, whose resistance is $50\, ohm$, has $25$ divisions in it. When a current of $4 \times 10^{-4}$ A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5\, V$, it should be connected to a resistance of:
Solution:
$I_g \, = \, 4 \, \times \, 10^{-4} \, \times \, 25 \, = \, 10^{-2} A$
$2.5 \, = \, (50+R) 10^{-2 } \, \, \therefore \, R \, = \, 200 \Omega$
