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 rms values of protential drops across each element.

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
Potential drop across L
`V_("rms")L=I_("rms")xx X_(L)=8.29xx2xx3.14xx50xx80xx10^(-3)`
`= 208.25 V`
Potential drop across C
`V_("rms")C=I_("rms")xx X_(C )`
`= 8.29xx(1)/(2xx3.14xx50xx60xx10^(-6))`
`= 440.02 V`
`therefore` Applied rms voltage `= V_(C )-V_(L)`
`= 440-208.25=231.75 V`