Source voltage amplitude

1. Apr 27, 2017

superslow991

1. The problem statement, all variables and given/known data

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.

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

3. 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 ive 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.

2. Apr 27, 2017

cnh1995

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?

3. Apr 27, 2017

superslow991

Hmm isn't the power factor R/Z? Also after reading the question more wouldn't I used the average power equation?

4. Apr 27, 2017

cnh1995

Yes.
So what is the expression for active power now that you know the apparent power and the power factor?

5. Apr 27, 2017

superslow991

Have no clue

P=Z*R? That wouldn't make sense I think

6. Apr 27, 2017

cnh1995

7. Apr 28, 2017

superslow991

I read it but I'm not seeing anything related to the active power. Only the reactive power Q=IV.

Also are you saying when I solved for the impedance that was solving for the apparent power?

8. Apr 28, 2017

cnh1995

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.

9. Apr 28, 2017

superslow991

P=V^2/Z x cos(theta)?

10. Apr 28, 2017

cnh1995

Right.

11. Apr 28, 2017

superslow991

So I tried 529=v^2/458 * 163/458 but I don't think that's right am I missing something?

12. Apr 28, 2017

cnh1995

Why?

13. Apr 28, 2017

superslow991

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. Apr 28, 2017

ehild

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.

15. Apr 29, 2017

cnh1995

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.

16. Apr 29, 2017

superslow991

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. Apr 29, 2017

cnh1995

Voltage does not dissipate. Active power dissipates in the resistance only and not in the reactance.

18. Apr 29, 2017

superslow991

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. Apr 29, 2017

cnh1995

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.

Last edited: May 7, 2017
20. Apr 29, 2017

ehild

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 $P=\frac{U^2}{Z}\cos(\theta)$.

Last edited: Apr 29, 2017