Question

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

• 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
$l = \frac{V}{R}=\frac{30\, V}{60\, \Omega}=\frac{1}{2}= A = 0.5\, A$
Thus the reading in the ammeter will be $0.5\, A.$