Consider a series LRC-circuit in which C 120.0uF. When driven at a frequency w 200.0 rads the complex impedance is given by Z (100.0-j10.0) 2. (a) Calculate (i) the reactances X, and Xc of the circuit at w200.0 rads-1, (ii) the inductance of the circuit. (b) Calculate the resonant angular frequency of the circuit. Hence, calculate the driving frequencies for which the average power in the circuit is half the reso- nant power. [Hint: the solutions to the quadratic formula: ax2 + bx + c = 0 are given by: bt - 4ac)/2a

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klefenz klefenz · Answer 1 · 2019-09-08T19:49:44+02:00

Answer:

X = -10 ohms

Xc = 41.7 ohms

L = 160 mH

w resonant = 228 rad/s

Explanation:

The reactance is the imaginary component of the impedance:

X = -10 ohms

The reactance is composed of the capacitive and inductive reactances:

X = Xl - Xc

The capacitive reactance depends on the capacitance and frequency

[tex]Xc = \frac{1}{w * C} = \frac{1}{200 * 120e-6} = 41.7 ohms[/tex]

And the indcutive reactance is:

Xl = X + Xc = -10 + 41.7 = 31.7 ohms

Since inductive reactance depends on frequency and inductance

Xl = w * L

We can rearrange:

[tex]L = \frac{Xl}{w} = \frac{31.7}{200} = 0.16 Henry = 160 mH[/tex]

For the circuit to resonate X must be 0, so Xl = Xc

0 = Xl - Xc

[tex]1 = w * L - \frac{1}{w * C}[/tex]

[tex]0 = w^2 * L - \frac{1}{C}[/tex]

[tex]w^2 * L = \frac{1}{C}[/tex]

[tex]w^2 = \frac{1}{L * C}[/tex]

[tex]w = \frac{1}{\sqrt{L * C}} = \frac{1}{\sqrt{0.16 * 0.00012}} = 228 rad/s[/tex]