A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be
- 2 A
- 1 A
- 0.5 A
- 0.25 A
Answer
(3) 0.5 A
Solution
Resistance of the ammeter (coil shunt):
RA = 480 20480 + 20 = 19.2
Total resistance = 40.8 + 19.2 = 60
I = 3060 = 0.5 A
Answer: (3)