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

(a) inductance, L = 80mH = 80 × 10^-3 H

frequency , f = 50Hz

capacitance of capacitor, C = 60μF = 60 × 10^-6 F

we know, inductive reactance, $X_L=2\pi fL$

= 2π(50) × 80 × 10^-3 = 25.12Ω

capacitive reactance, $X_C=\frac{1}{2\pi fC}$

= 1/{2π(50) × 60 × 10^-6} = 53.05Ω

impedance , Z = $X_C-X_L$

= 53.05 - 25.12 = 27.93Ω

rms value of current, $I=\frac{E}{Z}$

= 230/27.93 = 8.235A

peak value of current = √2I = (1.414) × (8.235) = 11.644A

(b) potential drop across L, $V_L=IX_L$

= 8.235 × 25.12 = 206.68 V

potential across C, $V_C=IX_C$

= 8.235 × 53.05 = 436.87V

(c) average power transferred to inductor is zero, because of phase difference π/2.

$P=EIcos\phi\\\\\phi=\pi/2,\therefore\: P=0$

(d) average power transferred to capacitor is also zero because phase difference is π/2.

$P=EIcos\phi\\\\\phi=\pi/2,\therefore\: P=0$

(e) Total pwrrr absorbed by the circuit, $P_{total}=P_L+P_C=0+0=0$

Answer 2

Answer:

L=80mH=80×10^-3H

C=60uf=60×10^-6F

for (a) and (b) check the attachment

(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.

Explanation:

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