Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.27
Additional Exercise
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
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Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection, Ig = 3 mA = 3 x 10⁻3 A Range of the voltmeter is 0, which needs to be converted to 18 V.
Therefore ,V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
R = V/Ig – G
= [18 /(3 x 10⁻3) ]-12 = 6000 -12 = 5988 Ω
Hence, a resistor of resistance 5988 Ω is to be connected in series with the galvanometer.