A galvanometer having 30 divisions has current sensitivity of 20 µA/div. It has a resistance of 25 ohm. How will you convert it to an ammeter measuring upto 1 ampere? How will you now convert this ammeter into a voltmeter reading upto 1 volt?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(0.015 Ω Ω in parallel ; R = 0.985 Ω Ω in series.)
Sol.
I g = 20 × 10 –6 × 30 = 0.6 × 10 –3 A
As we know
I g R g = S(I – I g )
25 × 0.6 × 10 –3 = S × (1 – 0.6 × 10 –3 )
S = 
0.015 Ω Ω
For voltammeter
V = (R A + R) i
Resistance of ammeter is R A = 
R A ~ S = 0.015
V = (R A + R) i
1 = (0.015 + R) × 1
R = 0.985 Ω Ω
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