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

The circuit shown in the figure.Resistance of the ammeter isRA=(480Ω)(20Ω)(480Ω+20Ω)=19.2Ω (As 480Ω and 20Ω are in parallel)  As ammeter is in series with 40.8Ω , ∴   Total resistance of the circuit is R=40.8Ω+RA=40.8Ω+19.2Ω=60ΩBy Ohm's law, Current in the circuit is I=VR=30 V60Ω=12 A=0.5 AThus the reading in the ammeter will be  0.5  A.