Question

A circuit contains an ammeter, a battery of $30V$ and a resistance $40.8ohm$ all connected in series.
If the ammeter has a coil of resistance $480ohm$ and a shunt of $20ohm$, the reading in the ammeter will be

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

Answer 1

Answer: (C)

The circuit is shown in the figure.

Resistance of the ammeter is
$R_A=\frac{(480\Omega )(20\Omega )}{(480\Omega +20\Omega )}=19.2\Omega$
(As $480\Omega$ and $20\Omega$ are in parallel)
As ammeter is in series with $40.8\Omega$,
$\therefore$ Total resistance of the circuit is
$R=40.8\Omega +R_A=40.8\Omega +19.2\Omega =60\Omega$
By Ohm's law,
Current in the circuit is
$l=\frac{V}{R}=\frac{30V}{60\Omega }=\frac{1}{2}=A=0.5A$
Thus the reading in the ammeter will be $0.5A.$