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.

Explanation

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`