What Is the Correct Source Voltage Amplitude for Desired Power in an AC Circuit?

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Homework Help Overview

The problem involves an AC series circuit with a voltage source, resistor, and capacitor, where the goal is to determine the correct source voltage amplitude for a specified average power consumed in the resistor. The circuit operates at a frequency of 60 Hz, and the parameters include a resistance of 163 ohms and a capacitance of 6.2x10^-6 F.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of reactance and impedance, questioning the relationship between apparent power and active power. There are discussions about the power factor and its role in determining active power.

Discussion Status

Participants are actively engaging with the problem, raising questions about the calculations and the definitions of power types in AC circuits. Some guidance has been offered regarding the relationship between voltage, impedance, and power factor, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a noted confusion regarding the distinction between rms voltage and amplitude, as well as the definitions of active and apparent power in the context of the circuit. Participants are also referencing external resources for clarification.

superslow991
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Homework Statement



An ac series circuit consists of a voltage source of frequency f = 60 Hz and voltage amplitude V, a resistor of resistance R = 163 ohms and a capacitor of capacitance C = 6.2x10^-6 F. What must the source voltage amplitude V be for the average electrical power consumed in the resistor to be 529 watts? There is no inductance in the circuit.

Homework Equations


XC = 1 / (2 × π × f × C)
Z = √(R2 + X2)
P=V^2/R

The Attempt at a Solution


So first i calculated the reactance and got 1/(2*π*60*6.2*10^-6)= 428.05
next impedance-√(163^2 + 428.05^2)= 458.0
last i used the power equation - 529=v^2/458.0 and solved for v and got 492.2 V

My only question is I've checked different answers online and saw some as V = 1165 or 1166

just wondering if i did the math right on this question. Any help is appreciated.
 
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superslow991 said:
for the average electrical power consumed in the resistor to be 529 watts
You are asked to find the active power in the circuit, which is dissipated in the resistor only.
You have calculated the apparent power, which is the sum of active and reactive powers.

What is the power factor of this circuit? How will you find the active power with the help of power factor?
 
cnh1995 said:
You are asked to find the active power in the circuit, which is dissipated in the resistor only.
You have calculated the apparent power, which is the sum of active and reactive powers.

What is the power factor of this circuit? How will you find the active power with the help of power factor?
Hmm isn't the power factor R/Z? Also after reading the question more wouldn't I used the average power equation?
 
superslow991 said:
Hmm isn't the power factor R/Z?
Yes.
So what is the expression for active power now that you know the apparent power and the power factor?
 
cnh1995 said:
Yes.
So what is the expression for active power now that you know the apparent power and the power factor?
Have no clue

P=Z*R? That wouldn't make sense I think
 
superslow991 said:
Only the reactive **apparent** power Q **S**=IV.
superslow991 said:
Also are you saying when I solved for the impedance that was solving for the apparent power?
Yes.
superslow991 said:
I'm not seeing anything related to the active power
There is a formula in that article which you can directly use to solve this problem. Read 'instantaneous power in ac circuits' again.
That formula describes the relation among active power P, voltage V, impedance Z and power factor cosθ.
 
cnh1995 said:
Yes.

There is a formula in that article which you can directly use to solve this problem. Read 'instantaneous power in ac circuits' again.
That formula describes the relation among active power P, voltage V, impedance Z and power factor cosθ.
P=V^2/Z x cos(theta)?
 
  • #10
superslow991 said:
P=V^2/Z x cos(theta)?
Right.
 
  • #11
cnh1995 said:
Right.
So I tried 529=v^2/458 * 163/458 but I don't think that's right am I missing something?
 
  • #12
superslow991 said:
but I don't think that's right
Why?
 
  • #13
cnh1995 said:
Why?
Idk maybe cause some of the answers I saw online but I'll just work with the answer you provided em. Thanks a lot
 
  • #14
superslow991 said:
So I tried 529=v^2/458 * 163/458 but I don't think that's right am I missing something?
The power is average power, calculated with the rms value of the AC voltage. P=Vrms2R/Z2. Vrms=amplitude/√2. You got the rms voltage of the source, while the problem asked the amplitude of the generator voltage.
 
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  • #15
superslow991 said:
Idk maybe cause some of the answers I saw online but I'll just work with the answer you provided em. Thanks a lot
They are asking for the amplitude of the voltage. You are calculating the rms voltage. As ehild said, multiply it by 1.4142 and verify your answer with the online one.
 
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  • #16
ehild said:
The power is average power, calculated with the rms value of the AC voltage. P=Vrms2R/Z2. Vrms=amplitude/√2. You got the rms voltage of the source, while the problem asked the amplitude of the generator voltage.
cnh1995 said:
They are asking for the amplitude of the voltage. You are calculating the rms voltage. As ehild said, multiply it by 1.4142 and verify your answer with the online one.
Yea got the answer that i saw online, 1166.
Thanks for the help
So to recap first i calculated for the reactance. Then i calculated for the impedence or the apparent power. Then i calculated for the rms of the voltage or the power dissipated through the impedance? Then i solved for the amplitude of the voltage using the equation v(rms)=amplitude/√2 and solved for amplitude
 
  • #17
superslow991 said:
rms of the voltage or the power dissipated through the impedance?
Voltage does not dissipate. Active power dissipates in the resistance only and not in the reactance.
 
  • #18
cnh1995 said:
Voltage does not dissipate. Active power dissipates in the resistance only and not in the reactance.
hmm ok so what exactly could you call the equation P=V^2/Z x cos(theta)? Apparent power dissipated through resistance? but what about the impedance?
 
  • #19
superslow991 said:
Apparent power dissipated through resistance? but what about the impedance?
No.
Apparent power has two components: Active power and reactive power. Active power is dissipated in the resistive elements only.
Reactive power is associated with the reactive elements (inductance and capacitance). It oscillates between the source and the load.
 
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  • #20
superslow991 said:
hmm ok so what exactly could you call the equation P=V^2/Z x cos(theta)? Apparent power dissipated through resistance? but what about the impedance?
Power is dissipated (transforms to heat) on the resistors only. The equivalent complex impedance seen by the source has a real and an imaginary part:: Z=R+iX. The magnitude of the impedance is ##Z=\sqrt{R^2+X^2}##. R=Zcos(theta), X=Zsin(theta), and the magnitude of the current is I=U/Z, where U is the voltage of the source. When calculating power, we use the rms value of both the voltage and current.
The power dissipated in the circuit is that dissipated on its equivalent resistance: the square of the rms current, multiplied by R:
##P=I^2 R=\left(\frac{U}{Z}\right)^2 R##. As R/Z=cos(θ), you can write ##P=\frac{U^2}{Z}\cos(\theta)##.
 
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