Q. 20

# A series LCR circ

(a) Given: L = 0.12 H

C = 480 nF

Converting into Farad, we get 480 × 10-9F

R = 23Ω

Voltage = 230 V

Peak voltage can be calculated as follows:

V0 = √2V

Substituting values we get

V0 = 230√2

V0 = 325.22 V

Current flowing through the circuit is given by the following:

I0 =

At resonating frequency, ωR L - 1/ωRC = 0

ωR = 1/√LC

= 1/ √0.12 (H) × 480 × 10-9 (F) = 4166.67 rad/s

Resonant frequency can be calculated using the formula:

vR = ωR/2π

vR = 4166.67(Hz)/2 × 3.14

vR = 663.48 Hz

Maximum current is calculated as follows:

(I0)max = V0/R = 325.22(Hz)/23 = 14.14 A

(b) The maximum power absorbed by the circuit can be calculated as follows:

Pav = 1/2(I0)2maxR

Substituting the values, we get

Pav = 1/2 × (14.14)2(A) × 23 (Ω)

Pav = 2299.33 W

Therefore, the resonating frequency is 663.48 Hz.

(c) The power transferred is equal to the half of the power at the resonating frequency.

Frequency at which power is half = ωR �∆ω

Or = 2π (vR �∆v)

Where ∆ω = R/2L

Substituting the values we get

∆ω = 23/2(Ω) × 0.12(H)

Therefore, change in frequency is written as follows:

∆v = 1/2π × ∆ω

Substituting the values, we get

∆v = 95.83(rad/s)/2π = 15.26 Hz

vR + ∆v is calculated as follows:

663.48(Hz) + 15.26(Hz) = 678.74 Hz

vR-∆v is calculated as follows:

663.48(Hz) – 15.26(Hz) = 648.22 Hz

The current amplitude is calculated as follows:

I’ = 1/√2 × (I0)max

I’ = 14.14(A)/ √2

I’ = 10 A

(d) Q factor can be calculated as follows:

Q = ωRL/R

Q = (4166.67 rad/s) × 0.12H/23Ω

Q = 21.74

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