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A circuit containing a 80 mH inductor and a `60 mu F` capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values. |
Answer» Given Inductance `L = 80 mH = 80xx10^(-3)H` Capacitance of capacitor `C = 60 mu F = 60xx10^(-6)F` The rms value of voltage `V_("rms")=230 V` Frequency f = 50 Hz , Resistance R = 0 Impedance of circuit `Z = sqrt(R^(2)+(X_(L)-X_(C ))^(2))` `= sqrt(0+(2pi fL-(1)/(2pi fC))^(2))=(2pi fL-(1)/(2pi FC))` `= sqrt((2xx3.14xx50xx80xx10^(-3)-(1)/(2xx3.14xx50xx60xx10^(-6)))^(2))` `= 25.12-53.07=-27.95 Omega` As ZCO, means `X_(L)lt X_(C )`, emf lags by current by `90^(@)` The rms value of current `I_("rms")=(V_("rms"))/(z)=(230)/(27.95)=8.29 A`. The maximum value of current `I_(0)=sqrt(2)I_("rms")=1.414xx8.29=11.64 A` |
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