A series circuit consists of an ac source of variable frequency, a 125 resistor,

ID: 1450410 • Letter: A

Question

A series circuit consists of an ac source of variable frequency, a 125 resistor, a 1.10 F capacitor, and a 4.96 mH inductor.

A.) Find the impedance of this circuit when the angular frequency of the ac source is adjusted to the resonance angular frequency.

B.)Find the impedance of this circuit when the angular frequency of the ac source is adjusted to twice the resonance angular frequency.

C.)Find the impedance of this circuit when the angular frequency of the ac source is adjusted to half the resonance angular frequency.

here,

resistance, R = 125 ohms
Capacitor, C = 1.10*10^-6 F
inductor, I = 4.96 = 0.00496 H

Impedance in RLC circuit is given as:
Z = Sqrt(R^2 + (Xl - Xc)^2 )
Z = Sqrt(R^2 + (2*pi*f*L - 1/2pifC)^2 ) ------------------------(1)

Resonace Frequency in RLC circuit is given by expression, f
f = 1/2pi*sqrt(LC)
f = 1/(2pi*sqrt(0.00496*1.10*10^-6))
f = 2154.681 Hz

Part A:
Since the angular frequency of the ac source is adjusted to the resonance angular frequency. so Xl = Xc

Therefore, Z = sqrt(R^2)
Z = R = 125 ohms

Part B:
if f = (2*2154.681) Hz = 4309.362 Hz

Z = Sqrt(125^2 + ( (2*pi*4309.362*0.00496) - 1/(2pi*4309.362*1.10*10^-6) )^2 )
Z = 160.532 ohms

Part C:
if f = 2154.681/2 = 1077.341 Hz

Z = Sqrt(125^2 + ( (2*pi*1077.341*0.00496) - 1/(2pi(1077.341)1.10*10^-6) )^2 )
Z = 160.532 ohms

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