A galvanometer has an internal resistance of 50 Ω and current required for full scale deflection is 1 mA. Find the series resistances required (as shown in Fig.) to use it as a voltmeter with different ranges, as indicated in Fig.
Text Solution
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Sol. For range 1 volt, galvanometer and R, are in series
I g = 
or 10 –3 = 
or 50 + R 1 = 1000
R 1 = 1000 – 50 = 950 Ω
For range 10 volt, galvanometer and R 2 , R 3 are in series.
10 –3 = 
∴ G + R 1 + R 2 = 
= 10 × 10 3 or R 2 = 10000 – (50 + 950) = 9000
= 9 k Ω
and for range 100 V, galvanometer, R 1 , R 2 and R 3 are in series.
10
–3 = 
∴ G + R 1 + R 2 + R 3 = 
= 100 × 10 3 or R 3 = 100 × 10
3 – (G + R 1 + R 2 )
= 100 × 10 3 – 10 × 10 –3
= 90 × 10 3 = 90 k Ω
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