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Find the maximum value of current when inductance of two henry is connected to 150 volt, 50 cycle supply. |
Answer» Here, `L = 2H, E_("rms") = 150 V, f - 50 Hz` `X_(L) = L omega = L xx 2 pi f = 2 xx 2 xx 3.14 xx 50 = 628` ohm `RHS` value of current of thorugh the inductor `I_("rms") = (E_("rms))/(X_(L)) = (150)/(628) = 0.24 A` 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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