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

Options

A. `2 A`
B. `1 A`
C. `0.5 A`
D. `0.25 A`

Answer

C

Explanation

The circuit is shown in the figure.
Resistance of the ammeter is
`R_(A)=((480Omega)(20Omega))/((480Omega+20Omega))=19.2Omega`
(As `480Omega and 20Omega` are in parallel)
As ammeter is in series with `40.8Omega`,
`therefore` Total resistance of the circuit is
`R=40.8Omega+R_(A)=40.8Omega+19.2Omega=60Omega`
By Ohm.s law,
Current in the circuit is
`I=(V)/(R)=(30V)/(60Omega)=(1)/(2)A=0.5A`
Thus teh reading in the ammeter will be 0.5 A.