First slide

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

A

2 A
B

1 A
C

0.5 A
D

0.25 A

Solution

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

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