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A pure inductor of 25.0 mH is connected to a source of 220 V . Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. |
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Answer» Solution :Magnitude of inductive reactance is, `X_(L) = omega L ` `= 2 pi F L` `= ( 2) ( 3.14 ) ( 50) ( 25 xx 10^(-3))` ` = 7.85 Omega` Here, `V_(RMS) = I_(rms) X_(L)` `:. I_(rms) = ( V_(rms)/ X_(L))` `= ( 220)/( 7.85)` `= 28.025 A` |
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