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. At a frequency ω1, the reactance of a certain capacitor equals that of a certain inductor.

a. If the frequency is changed to ω2 = 2 ω1, what is the ratio of the reactance of the inductor to the reactance of the capacitor? Which reactance is larger?

b. If the frequency is changed to ω3 = ω1/3, what is the ratio of the reactance of the inductor to the reactance of the capacitor? Which reactance is larger?

Respuesta :

The answer to the set of questions are as follows:

a. 2 ( 2/w2C) = 4xc2 = Xl2 / Xc2 = 4

The inductors reactance is greater than the capacitor

b. (1/ 3w1C ) = (1/ 9w3C ) = 1/9 Xc3 = X l2/ Xc2 = 1/9

The capacitors reactance is greater than inductor

I hope my answer has come to your help. God bless and have a nice day ahead!

Answer:

Part a)

ratio = 4

reactance of inductor will be more

Part b)

ratio = 1/9

reactance of capacitor will be more

Explanation:

As we know that at angular frequency ω1 the reactance of capacitor is equal to the reactance of inductor

so here we can say that this angular frequency must be resonance frequency

so we will have

[tex]x_L = x_C[/tex]

now we will have

part a)

if we have another frequency

[tex]\omega_2 = 2\omega_1[/tex]

so here in this case

[tex]x_L = 2\omega_1 L[/tex]

[tex]x_C = \frac{1}{2\omega_1C}[/tex]

now the ratio of the new reactance is given as

[tex]ratio = \frac{2\omega_1L}{\frac{1}{2\omega_1C}}[/tex]

[tex]ratio = 4[/tex]

so reactance of inductor will be larger in this case

part a)

if we have another frequency

[tex]\omega_3 = \frac{\omega_1}{3}[/tex]

so here in this case

[tex]x_L = \frac{\omega_1}{3} L[/tex]

[tex]x_C = \frac{3}{\omega_1C}[/tex]

now the ratio of the new reactance is given as

[tex]ratio = \frac{\frac{\omega_1}{3}L}{\frac{3}{\omega_1C}}[/tex]

[tex]ratio = \frac{1}{9}[/tex]

so reactance of capacitor will be larger in this case

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