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Question

A coil of negligible resistance is connected in series with a 90 Ω resistor across a 120 V, 60 Hz line. An ac voltmeter reads 90 V across the resistance, then the inductance of the coil is approximately

A
0.2 H
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B
0.3 H
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C
0.4 H
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D
0.7 H
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Solution

The correct option is A 0.2 H
Using Ohm's law
V=IR
where V is potential difference across resistance and I is current in it.
As both inductor and resistor are connected in series the current will be same in both the inductor and resistor, which is given as

I=120R2+L2ω2

From given data V=90 V and R=90 Ω

90=RR2+L2ω2120

1120=1902+L2(2π60)2

(120)2=(90)2+1202π2L2L2=12029021202π2

L=30×2101202π2

L=0.21H

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