A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the re…

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asked Jan 23, 2020 196k views

1 Answer

Saxoier · Answer 1 · 2020-01-26T17:56:13+0000

Answer:

E = 0.13 J

Step-by-step explanation:

At resonance condition we have

$\omega = \sqrt{(1)/(LC)}$

$\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}$

$\omega = 1217.2 rad/s$

now if the frequency is double that of resonance condition then we have

$x_L = 2\omega L$

x_L = 2(1217.2)(13.5* 10^(-3))

x_L = 32.86 ohm

now we have

x_c = (1)/(2(1217)(50* 10^(-6)))

x_c = 8.22 ohm

now average power is given as

P = i_(rms)V_(rms)* (R)/(z)

P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))

P_(avg) = 50.67 Watt

Now time period is given as

$T = (2\pi)/(\omega)$

so total energy consumed is given as

$E_(avg) = 50.67((2\pi)/(2(1217.2)))$

E = 0.13 J