Question

An AC voltage source is connected in series to an inductor, a capacitor, and a resistor of 5 Ohms. At the frequency when the phase angle is zero, capacitive reactance is 6 Ohms. What is the total impedance of the of the circuit at a frequency which is a factor of 2.4 times less than this frequency? Answer in Ohms.

Answer 1

Given that phase Angle is zero, which means

arctan ((XL - XC)/R) = 0

XL - Xc = 0

XL = Xc

Suppose resonance frequency is f0, then

f0 = 1/(2*pi*sqrt (LC))

Given that Xc = 6 ohm at f = f0, which means

XL = 6 ohm at f = f0

Now when frequency us reduced by a factor of 2.4 times less than f0, than

f1 = f0/(2.4)

then

XL1 = 2*pi*f1*L = 2*pi*f0*L/(2.4) = XL/2.4

XL1 = 6/2.4 = 2.5 ohm

Xc1 = 1/(2*pi*f1*C) = 1/(2*pi*f0*C/2.4) = 2.4*(1/(2*pi*f0*C))

Xc1 = 2.4*Xc

Xc1 = 2.4*6 = 14.4 ohm

So at f = f1, impedance will be

Z = sqrt (R^2 + (XL1 - Xc1)^2)

R = 5 ohm (given)

Z = sqrt (5^2 + (2.5 - 14.4)^2)

Z = 12.91 ohm

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