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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
1 A
B
0.5 A
C
0.25 A
D
2 A
Detailed Solution
Ammeter resistance: $\frac{480\times20}{480 + 20} = 19.2\ \Omega$
$R_{eff} = 40.8 + 19.2 = 60\ \Omega$
$I = \frac{30}{60} = 0.5$ A
$R_{eff} = 40.8 + 19.2 = 60\ \Omega$
$I = \frac{30}{60} = 0.5$ A