A galvanometer has a resistance of 50 and it allows maximum current of 5 mA. It can be converted into voltmeter to measure up to 100 V by connecting in series a resistor of resistance :
- 19500
- 5975
- 20050
- 19950
Answer
(4) 19950
Solution
at full-scale current Ig the whole voltage V falls across (G+R):
Ig(G+R)=V R= VIg-G
= 1005 10-3-50=20000-50
R=19950
Answer: (4)