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

Given, coil resistance of ammeter = 480
Shunt resistance of ammeter = 20 Ω
Effective resistance of ammeter = $\frac{1}{\frac{1}{480}+\frac{1}{20}}=\frac{480\times 20}{500}=19.2\mathrm{\Omega }$
Since the ammeter is connected in series
with the 40.8 Ω resistance,
the effective resistance of the circuit is
$\begin{array}{l}{\mathrm{R}}_{\text{eff }}=4.08\mathrm{\Omega }+19.2\mathrm{\Omega }\\ =60\mathrm{\Omega }\end{array}$
From Ohm’s law,
v = IR
where V is the potential difference
I is the current
R is the resistance
$\begin{array}{l}\left(30\mathrm{v}\right)=I\times \left(60\mathrm{\Omega }\right)\\ \Rightarrow I=\frac{30}{60}=0.5\mathrm{A}.\end{array}$