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Question
The magnetic field energy in an inductor changes from maximum to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is
Options
20 Hz
50 Hz
200 Hz
500 Hz
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Solution
50 Hz
The magnetic field energy in an inductor is given by,
`E = 1/2 Li^2`
The magnetic energy will be maximum when the current will reach its peak value, i0, and it will be minimum when the current will become zero.
v
From the above graph of alternating current, we can see that current reduces from maximum to zero in T/4 time, where T is the time period.
So, in this case, T/4 = 5 ms
⇒ T = 20 ms
∴ Frequency , `V =1/T = 1/(20xx10^-3)=50Hz`
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