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Find the maximum value of current when a coil of inductance 2H is connected to 150V, 50 cycles/sec supply. |
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Answer» SOLUTION :Here `L=2H, E_("rms")=150V, V=50Hz` `X_(L)=Lomega=Lxx2piv=2xx2xx3.14xx50="628 ohm"` RMS value of current through the INDUCTOR, `I_("rms")=(E_("rms"))/(X_(L))=(150)/(628)=0.24A` Maximum value *or PEAK value) of current is given by `I_("rms")=(I_(0))/(sqrt2)` or `I_(0)=sqrt2""I_("rms")=1.414xx0.24=0.339A` |
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