Answer :

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:

V_{0} = V/√ 2

V_{0} = 230√2 V

Substituting the values we get

On calculating, we get

I_{0} = -11.63 A.

The amplitude of maximum current is given by the following relation:

I_{0} = 11.63 A.

The root mean square value of the current I = I_{0}/√ 2

⇒ I = -11.63(A)/√ 2

⇒ I = - 8.22 A

(b) The potential difference across the inductor can be calculated as follows:

V_{L} = 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:

V_{c} = I × 1/ωC

Substituting values we get

V_{c} = 8.22(A) × 1/ 100 π(Hz) × 60 × 10^{-6}(F)

On calculating, we get

V_{c} = 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.

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