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 [2015]

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Solution

Q.
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 [2015]

1 2 A

2 1 A

3 0.5 A

4 0.25 A

Ans.
(3)

The circuit is 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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