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Question -

A circuit containing a 80 mHinductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply.The resistance of the circuit is negligible.

(a) Obtainthe current amplitude and rms values.

(b) Obtainthe rms values of potential drops across each element.

(c) What isthe average power transferred to the inductor?

(d) What isthe average power transferred to the capacitor?

(e) What isthe total average power absorbed by the circuit? [‘Average’ implies ‘averagedover one cycle’.]



Answer -

Inductance, L =80 mH = 80 × 10−3 H

Capacitance, C =60 μF = 60 × 10−6 F

Supply voltage, V =230 V

Frequency, ν =50 Hz

Angular frequency, ω =2πν= 100 π rad/s

Peakvoltage, V0 =

(a) Maximumcurrent is given as:

The negative sign appears because 

Amplitude of maximum current, 

Hence, rms value of current, 

(b) Potentialdifference across the inductor,

VL= I ×ωL

= 8.22 × 100 π × 80 × 10−3

= 206.61 V

Potential difference across thecapacitor,

(c) Averagepower consumed by the inductor is zero as actual voltage leads the current by

(d) Averagepower consumed by the capacitor is zero as voltage lags current by

(e) Thetotal power absorbed (averaged over one cycle) is zero.

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