Find Resonant Freq, Q, Power Diss pwr Half Pwr Freq in RLC Circuit

[FrogPad]

Question:
A parallel RLC circuit, which is driven by a variable frequency 10-A source, has the following parameters:
$$R=500\Omega$$
$$L=0.5mH$$
$$C=20\muF$$

Find the resonant frequency, the $Q$, the average power dissipated at resonant frequency, the $BW$, and the average power dissipated at the half-power frequencies.

Answer:
$Q$, $\omega_0$, and $BW$ are all straightfoward calculations.

[tex] Q = 100 [/itex]
$$BW = 100 \frac{rad}{sec}$$
$$\omega_0 = 10000\frac{rad}{sec}$$

The half power frequencies are:
$$\omega_{hi} = 1005.01$$
[tex] \omega_{lo} = 995.01 [/itex]

I don't understand how to calculate the average power dissapated at resonant frequency, OR at the half power frequencies. If someone could give me a push in the right direction, that would be swell

 

[FrogPad]

nm, figured it out. That was embarrassingly easy.

 

[piercas]

Hello,

To calculate the average power dissipated at resonant frequency, we can use the formula P = I^2R, where I is the current through the circuit and R is the resistance of the circuit. In this case, the resistance is equal to R = 500Ω.

To find the current at resonant frequency, we can use the formula I = V/R, where V is the voltage across the circuit. The voltage across the circuit can be found using Ohm's Law, V = IZ, where Z is the impedance of the circuit. In a parallel RLC circuit, the impedance is given by Z = R/(1 + j(Q/Q0)), where Q0 = 1/(RCω0).

Thus, the current at resonant frequency is:
I = V/R = (10 A)/(500Ω) = 0.02 A

The average power dissipated at resonant frequency is then:
P = I^2R = (0.02 A)^2(500Ω) = 0.02 W

To calculate the average power dissipated at the half-power frequencies, we can use the same formula P = I^2R, but we need to find the current at these frequencies. The current through a parallel RLC circuit is given by I = V/Z, where Z is the impedance of the circuit. At the half-power frequencies, the impedance is equal to Z = R/√2.

Thus, the current at the half-power frequencies is:
I = V/Z = (10 A)/(500Ω/√2) = 0.028 A

The average power dissipated at the half-power frequencies is then:
P = I^2R = (0.028 A)^2(500Ω) = 0.039 W

I hope this helps! Let me know if you have any further questions.