In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be
- 1499 G
- 499500 G
- 1500 G
- 500499 G
Answer
(3) 1500 G
Solution
Ig = 0.002 I ⇒ current through shunt = 0.998 I
Shunt and galvanometer are in parallel (same p.d.):
Ig G = (I - Ig) S
S = 0.0020.998 G = G499
Resistance of ammeter:
RA = GSG + S = G G499G + G499
= G500
Answer: (3)