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Find the maximum value of current when an inductance of 2H is connected to 150 V, 50 cycle supply. |
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Answer» Solution :Here `L = 2H , E_(rms) = 150 V, v = 50 HZ` `X_(L) = L omega = L xx 2pi v` `2 xx 2 xx 3.14 xx 50 = 628` ohm RMS value of current through the INDUCTOR, `I_(rms) = (E_(rms))/(X_(L)) = (150)/(628) = 0.24` Maximum value (or PEAK value) of current is given by `I_(rms) = (I_(0))/(SQRT(2))` or `I_(0) = sqrt(2)I_(rms)` `= 1.414 xx 0.24= 0.339 A` |
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