Exercise 3135 A series circuit consists of an ac source of v

Exercise 31.35

A series circuit consists of an ac source of variable frequency, a 105 resistor, a 1.50 F capacitor, and a 5.00 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.

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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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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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Solution

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

= 1/ [sqrt(5*10^-3 * 1.50*10^-6)]   

  w = 11547 rad/s

impedence   

W = 2*w = 2*11547 = 23094

Xc = 1/ ( wC)

Xl = L* w

Xr = R

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

= sqrt [ {1/ ( 2*11547 * 1.50*10^-6) - 5*10^-3 *  2*11547}^2 + 105^2]

= 136.10 ohm

part 3

W = w /2 = 11547 / 2 = 5773.5 rad/s

then

Z =sqrt [ {1/ ( 5773.5 * 1.50*10^-6) - 5*10^-3 * 5773.5}^2 + 105^2]

= 136.10 ohm   

both coming same interesting isn\'t it.