NEET-XII-Physics
39: Alternating Current
- #5The 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
(a) 20 Hz
(b) 50 Hz
(c) 200 Hz
(d) 500 HzdigAnsr: bAns : (b) 50 Hz
The magnetic field energy in an inductor is given by,
`` E=\frac{1}{2}L{i}^{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.
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
`` \Rightarrow T=20\,\mathrm{\,ms\,}``
`` \therefore \,\mathrm{\,Frequency\,},\nu =\frac{1}{T}=\frac{1}{20\times {10}^{-3}}=50\,\mathrm{\,Hz\,}``
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