electrical circuit problem

a i) The phase angle between the voltage and current is 45 since cos 45 = 0.707, therefore:

[ωL - 1/(ωC)] = 40

since when the impedance has a phase angle of 45 (leading), the reactive and resistive components should be equal.

With ω = 100π, we have:

50π - 1/(100πC) = 40

C = 27.19 μF

ii) For resonance to occur, reactive component should be zero, i.e.

ωL = 1/(ωC)

C = 1/(ω2L)

= 20.27 μF

b i) For the 3-dB bandwidth, with varying frequency, we have:

At resonance, impedance = 40 Ω

At boundary of bandwidth, impedance = 80 Ω

So at that time, the reactive component should be 40√3 Ω, i.e.

ωL - 1/(ωC) = 40√3

0.5fπ - 49348/(2fπ) = 40√3

(fπ)2 - 24674 = 80√3(fπ)

(fπ)2 - 80√3(fπ) - 24674 = 0

Solving for the positive root, fπ = 196, giving f = 62.2 Hz

Hence bandwidth = 2 x (62.2 - 50) = 24.4 Hz

(ii) For RLC series circuit, Q = (1/R)√(L/C) = 3.93

(iii) Since the circuit is in resonance, apparent power is the power delivered by the source = 2202/40 = 1210 W

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