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 \(40.8\, \Omega\),

\(\therefore \quad\) 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

\(I=\frac{V}{R}=\frac{30\, \mathrm{V}}{60\, \Omega}=\frac{1}{2} \mathrm{A}=0.5\, \mathrm{A}\)

Thus the reading in the ammeter will be \(0.5\, \mathrm{A}\).