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A pure inductor of 25mH is connnected to a source of 220V and 50 Hz. Find the inductive reactance, rms value of current and peak current in the circuit. |
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Answer» Solution :Here : `L=25x10^(-3)H,V_("RMS")=220V,V=50Hz,X_(L)=?I_("rms")=?I_(0)=?` INDUCTIVE reactance is `""X_(L)=omegaL=2pi vL` `=2xx3.14xx50xx25xx10^(-3)` `=7850xx10^(-3)` `X_(L)=7.850Omega` rms value of current in the CIRCUIT is given by `I_("rms")=(V_("rms"))/(X_(L))` `=(220)/(7.850)` `=28.025" A"` Peak current in the circuit is `I_(0)=sqrt(2)I_("rms")=1.414xx28.025` `=39.627" A"`. |
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