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

Question

A series circuit consists of an ac source of variable frequency, a 125 resistor, a 1.40 F capacitor, and a 4.30 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.

125

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Correct

Part B

Find the impedance of this circuit when the angular frequency of the ac source is

adjusted to twice the resonance angular frequency.

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Incorrect; Try Again; 3 attempts remaining

Part C

Find the impedance of this circuit when the angular frequency of the ac source is

adjusted to half the resonance angular frequency.

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A series circuit consists of an ac source of variable frequency, a 125 resistor, a 1.40 F capacitor, and a 4.30 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.

Z =

125

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Correct

Part B

Find the impedance of this circuit when the angular frequency of the ac source is

adjusted to twice the resonance angular frequency.

Z =   

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Incorrect; Try Again; 3 attempts remaining

Part C

Find the impedance of this circuit when the angular frequency of the ac source is

adjusted to half the resonance angular frequency.

Z =   

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Explanation / Answer

resonance resosnce anguler frequency w = 1/[ 2*pi * sqrt( LC) ]

= 1/ [sqrt(4.3*10^-3 * 1.40*10^-6)]   

  w = 12888.5 rad/s

impedence   

W = 2*w = 2*12888.5

Xc = 1/ ( wC)

Xl = L* w

Xr = R

Z = sqrt [ ( Xc-Xl)^2 +Xr^2 ]

= sqrt [ {1/ ( 2*12888.5 * 1.40*10^-6) - 4.3*10^-3 *  2*12888.5}^2 + 125^2]

= 150.119 ohm

part 3

W = w /2 = 12888.5 / 2 = 6444.25 rad/s

then

Z =sqrt [ {1/ ( 6444.25 * 1.40*10^-6) - 4.3*10^-3 * 6444.25}^2 + 125^2]

= 150.119 ohm   

both coming same interesting isn't it.

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