Question

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

(a) Obtain the current amplitude and rms values.

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

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

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

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

Answer 1

Inductance, L = 80 mH = 80 × 10⁻³ H

Capacitance, C = 60 μF = 60 × 10⁻⁶ F

Supply voltage, V = 230 V

Frequency, v = 50 Hz

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

Peak voltage, V₀ = `"V" sqrt2 = 230 sqrt2 "V"`

(a) Maximum current is given as:

I₀ = `"V"_0/((ω"L" - 1/(ω"C")))`

= `(230 sqrt3)/((100π xx 80 xx 10^-3 - 1/(100π xx 60 xx 10^-6)))`

= `(230 sqrt2)/((8π - 1000/(6π)))`

= −11.63 A

The negative sign appears because `ω"L" < 1/(ω"C")`.

Amplitude of maximum current, |I₀| = 11.63 A

Hence, rms value of current, I = `"I"_0/sqrt2 = (-11.63)/sqrt2` = −8.22 A

(b) Potential difference across the inductor,

V_L = I × ωL

= 8.22 × 100 π × 80 × 10⁻³

= 206.61 V

Potential difference across the capacitor,

V_e = `"I" xx 1/(ω"C")`

= `8.22 xx 1/(100π xx 60 xx 10^-6)`

= 436.3 V

(c) Average power consumed by the inductor is zero as actual voltage leads the current by `π/2`.

(d) Average power consumed by the capacitor is zero as voltage lags current by `π/2`.

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