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Question

A coil of resistance 300 Ω and inductance 1.0 H is connected across an alternating voltage of peak value 300 V of frequency 3002π Hz. Calculate the peak potential difference across inductor.

A
1502 V
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B
150 V
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C
1002 V
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D
100 V
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Solution

The correct option is A 1502 V
The given LR is represented in the below diagram.


The individual phasor diagrams for the resistor and inductor can be drawn as


V0R=i0R and V0L=i0XL

Combining the two phasor diagrams, we get


V0=V20R+V20L
V0=i0R2+X2L=i0Z

where Z (impedance of circuit) =R2+X2L
=R2+ω2L2
=R2+(2πf)2L2
=3002+(2π×(3002π))2×12
=3002+3002=3002 Ω

Peak current i0=V0Z=3003002
=12 A

Therefore, peak voltage across inductor =XL×i0
=12×300=3002 V
=1502 V

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