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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 150√2 VThe 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=i0√R2+X2L=i0Z where Z (impedance of circuit) =√R2+X2L =√R2+ω2L2 =√R2+(2πf)2L2 =√3002+(2π×(3002π))2×12 =√3002+3002=300√2 Ω Peak current i0=V0Z=300300√2 =1√2 A Therefore, peak voltage across inductor =XL×i0 =1√2×300=300√2 V =150√2 V

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