A series circuit consists of an ac source of variable frequency, a 150 resistor,

Question

A series circuit consists of an ac source of variable frequency, a 150 resistor, a 1.25 F capacitor, and a 4.44 mH inductor.

Part A

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

Part B

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

Part C

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

Explanation / Answer

A)
Impedance is given by,
Z = sqrt(R^2 + (Xl -Xc)^2)

At Resonance freq, Xl = Xc
So, XL - XC = 0
Z = R
Impedance of the circut, Z = 150

B)
Finding resonance angular freq, w = 1/sqrt(L*C)
w = 1/sqrt(1.25*10^-6*4.44*10^-3)
w = 13423.1

As source is adjusted to twice the resonance angular frequency, w = 2*13423.1 = 26846.2
Now,
Z = sqrt(R^2 + (Xl -Xc)^2)
Z = sqrt(150^2 + (26846.2* 4.44*10^-3 - 1/(26846.2*1.25*10^-6)))
Z = 150.3
Impedance of the circut, Z = 150.3

C)
As source is adjusted to twice the resonance angular frequency, w = 0.5*13423.1 = 6711.5
Now,
Z = sqrt(R^2 + (Xl -Xc)^2)
Z = sqrt(150^2 + (6711.5* 4.44*10^-3 - 1/(6711.5*1.25*10^-6)))
Z = 149.7
Impedance of the circut, Z = 149.7

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