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’.]
Given: L = 80 mH = 80 × 10-3 H
capacitor C = 60 Μf = 60 × 10-6 F
Voltage V = 230 V
Frequency v = 50Hz
(a) Peak voltage is given by the formula:
V0 = V/√ 2
V0 = 230√2 V
Substituting the values we get
On calculating, we get
I0 = -11.63 A.
The amplitude of maximum current is given by the following relation:
I0 = 11.63 A.
The root mean square value of the current I = I0/√ 2
⇒ I = -11.63(A)/√ 2
⇒ I = - 8.22 A
(b) The potential difference across the inductor can be calculated as follows:
VL = I × ω × L = 8.22(A) × 100 π(Hz) × 80 × 10-3 (F) = 206.61 V
Potential difference across the capacitor is given by the formula as follows:
Vc = I × 1/ωC
Substituting values we get
Vc = 8.22(A) × 1/ 100 π(Hz) × 60 × 10-6(F)
On calculating, we get
Vc = 436.3 V
(c) Since the actual voltage leads the current by π/2, so the average power consumed by the inductor is zero.
(d) Since voltage lags the current by π/2, so the average power consumed by the capacitor is zero.
(e) The total power over a complete cycle is zero.