RLC Circuit Homework: Find L, Z, I, and P

AI Thread Summary
The discussion revolves around solving a series RLC circuit homework problem with given values for resistance, capacitance, and resonance frequency. The calculations for inductance, impedance, rms current, and average power consumption were presented, with the inductance calculated as approximately 5.07e-7 H and impedance as 75 ohms. The rms current at resonance was found to be 1.6 A, leading to an average power consumption of 192 W, confirming the calculations were correct despite a minor typo. The importance of the power factor at resonance, which is 1, was also highlighted, ensuring accurate power calculations. Overall, the participants clarified and validated the calculations while addressing typographical errors.
SilverAu
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Homework Statement


A series RLC circuit has R=75 ohms, C=20 mF and a resonance frequency of 5.0 kHz. (i) What is the value of inductance? (ii)What is the impedance of the circuit at residence? (iii)What is the rms current at resonance if the rms voltage of the power supply is 120 V? (iv) How much average power is consumed in the circuit at resonance?

Homework Equations


f=1/(2π√LC), z= √(XL-XC)^2 +R^2, I=V/Z and P=VI [/B]

The Attempt at a Solution


(i) What is the value of inductance?
f=1/(2π√LC)
L=1/4π^2 * 1/f^2C
1/(4π^2) * 1/[(50e3)^2(20^-6)]
L=5.07 e-7 H
(ii)What is the impedance of the circuit at residence?
z= √(XL-XC)^2 +R^2
z= √WL-1/WC)^2 +R^2
z= √[(314e3 * 5.07e-7)-(1/314e3 * 20e-6)]^2 +5^2
z= 75 Ohms
(iii)What is the rms current at resonance if the rms voltage of the power supply is 120 V?
I=V/Z
I=120/75
I=1.6 A
(iv) How much average power is consumed in the circuit at resonance?
P=VI
P=120* 1.8
P=192 W[/B]
 
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What is your question about the problem? :)

Have you thougth about the power factor in (iv)?
 
Alettix said:
What is your question about the problem? :)

Have you thougth about the power factor in (iv)?
I just wanted to know if I was correct. Whether my process of thinking was correct and the numbers backed it up. The power factor is just p=1/2 VI cosϕ, correct? so is it then 1/2 [120*1.8 cos (0)]= 96 W.
 
SilverAu said:
A series RLC circuit has R=75 ohms, C=20 mF and a resonance frequency of 5.0 kHz. (i) What is the value of inductance? (ii)What is the impedance of the circuit at residence? (iii)What is the rms current at resonance if the rms voltage of the power supply is 120 V? (iv) How much average power is consumed in the circuit at resonance?
Are the capacitor units mF or μF? There's a 1000 times difference...
 
gneill said:
Are the capacitor units mF or μF? There's a 1000 times difference...
sorry microfarad (µF)
 
SilverAu said:
.The power factor is just p=1/2 VI cosϕ, correct? so is it then 1/2 [120*1.8 cos (0)]= 96 W.

Because there is resonance, the power factor is cos(0) = 1.
The average power is given by: P = Irms * Vrms * cos(θ)
so in your case: P = Irms * Vrms
If your calculated values for Vrms and Irms are right the power should be: P = 120 * 1,6 = 192 W as you have written.
(So I guess it was a typo to write P = 120 * 1,8 = 192 W :) )

I haven't checked the rest of the calculation in detail, but this part appears right to me.
 
Alettix said:
Because there is resonance, the power factor is cos(0) = 1.
The average power is given by: P = Irms * Vrms * cos(θ)
so in your case: P = Irms * Vrms
If your calculated values for Vrms and Irms are right the power should be: P = 120 * 1,6 = 192 W as you have written.
(So I guess it was a typo to write P = 120 * 1,8 = 192 W :) )

I haven't checked the rest of the calculation in detail, but this part appears right to me.
Yes, apologies it was a typo it is 1.6*120=192 W. If you could please check the rest of my work it would be great because this problem has been a pain in my side for a couple of days (:
 
SilverAu said:
z= √WL-1/WC)^2 +R^2
z= √[(314e3 * 5.07e-7)-(1/314e3 * 20e-6)]^2 +5^2

One more typo here (R=75Ω, and not R=5Ω), but otherwise I think it's right.
 
Alettix said:
One more typo here (R=75Ω, and not R=5Ω), but otherwise I think it's right.
Yes another typo lol...Thank you so much, all your help is greatly appreciated!
 
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