Moving coil Galvanometer

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

Moderate

Solution

The circuit shown in the figure.

Resistance of the ammeter is

${R}_{A}=\frac{\left(480\Omega \right)\left(20\Omega \right)}{(480\Omega +20\Omega )}=19.2\Omega $

$\text{(As}480\Omega \text{and}20\Omega \text{are in parallel)}$

$\text{As ammeter is in series with}40.8\Omega \text{,}$

$\therefore \text{Total resistance of the circuit is}$

$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 A.

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