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Question

The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0ms, when connected to an AC source. The frequency of the source is

A
20 Hz
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B
50 Hz
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C
200 Hz
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D
500 Hz
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Solution

The correct option is B 50 Hz
Let inductor=L, instantaneous current =I(t)=Ipeaksin(ωt)
The magnetic field energy in an inductor is given by

12LI2(t)=12LI2peaksin2(ωt)
the energy goes from a maximum value ( at time t1) to the next minimum value (at time t2) when

ωt1=π2&ωt2=π

Or ωt1=3π2&ωt2=2π
In either case,
ω(t2t1)=π2
ω×5×103=π2
2f×5×103=12
f=1032×2×5
f=50Hz

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