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

• superslow991
In summary,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 EquationsXC = 1 / (2 × π × f × C)Z = √(R2 + X2)P=V^2/R
superslow991

## 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.

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

cnh1995 said:
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?

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)?

superslow991 said:
P=V^2/Z x cos(theta)?
Right.

cnh1995 said:
Right.
So I tried 529=v^2/458 * 163/458 but I don't think that's right am I missing something?

superslow991 said:
but I don't think that's right
Why?

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

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.

superslow991
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.

superslow991
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

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.

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?

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.

Last edited:
superslow991
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)##.

Last edited:
superslow991

## 1. What is source voltage amplitude?

Source voltage amplitude refers to the maximum voltage that is supplied by a power source, such as a battery or generator. It is measured in volts and represents the strength or intensity of the electrical signal.

## 2. How is source voltage amplitude different from voltage?

Voltage is a general term that refers to the potential difference between two points in an electrical circuit. Source voltage amplitude specifically refers to the maximum voltage output of a power source, whereas voltage can vary depending on the resistance and components in a circuit.

## 3. What factors can affect source voltage amplitude?

Source voltage amplitude can be affected by various factors, including the type of power source, the quality and condition of the power source, and the load or demand on the circuit. Other factors such as temperature and external interference can also impact the source voltage amplitude.

## 4. Why is source voltage amplitude important?

Source voltage amplitude is important because it determines the amount of power that can be supplied to a circuit. If the source voltage amplitude is too low, the circuit may not function properly, and if it is too high, it can damage the components in the circuit.

## 5. How is source voltage amplitude measured?

Source voltage amplitude can be measured using a voltmeter, which is a device that measures electrical potential difference. It is typically connected in parallel to the power source and provides a numerical value in volts. It is important to use a calibrated and accurate voltmeter to ensure an accurate measurement of source voltage amplitude.

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