Apparent Power, Real Power and Reactive Power

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Discussion Overview

The discussion revolves around calculating apparent power, real power, and reactive power in an AC circuit with a given voltage source, resistance, inductance, and capacitance. Participants explore the relationships between these quantities and the methods for calculating them, including the use of RMS values and impedance in the context of AC circuits.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to start solving the problem without a specific time given for the voltage.
  • Another participant calculates the RMS voltage from the peak voltage and questions the origin of the sqrt(2) factor in the conversion.
  • A participant explains that the sqrt(2) is used for converting peak voltage to RMS voltage, which is commonly used in AC power calculations.
  • One participant outlines a method for calculating current and power by using impedance and vector addition, emphasizing the importance of understanding these concepts in AC circuits.
  • The same participant notes that the concept of time is not necessary for finding the answers to the power calculations.

Areas of Agreement / Disagreement

Participants generally agree on the methods for calculating power in AC circuits, but there is some confusion regarding the initial steps, particularly about the voltage and the use of RMS values. The discussion remains unresolved regarding the participant's initial question about starting the calculations.

Contextual Notes

Some participants may have varying levels of proficiency with vector analysis and AC circuit concepts, which could affect their understanding of the problem and the proposed solutions.

Who May Find This Useful

This discussion may be useful for students or individuals learning about AC circuit analysis, particularly those interested in understanding power calculations and the significance of RMS values in electrical engineering.

fatmoe
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A voltage source of v(t) = 169.7sin200t V is connected across a series combination of a resistance of 10 ohm, an inductance of 10mH, and a capacitance of 500 uF. Determine the current in the circuit, the apparent power, the real power, and the reactive power supplied by the source. What is the power factor of the circuit.


I am having problems with this questions, as I don't know how to get the voltage since no time is given, and without the voltage, none of the equations are going to be any good.. Can someone point me in the right direction on how to get start with this question?
 
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Well, I figured out how to calculate the voltage = v = 169.7/sqr(2).. But I don't understand where the sqrt(2) comes from in this formula?
 
fatmoe said:
Well, I figured out how to calculate the voltage = v = 169.7/sqr(2).. But I don't understand where the sqrt(2) comes from in this formula?

That's the conversion from peak to RMS for the magnitude of the voltage source. If you're dealing with power in an AC circuit you almost always want to be working with RMS values.
 
If you are proficient in vectors it should be doable for you.

impedance of cap: 1/jwC
impedance of inductor: jwL

w=200 in this case. (omega=200)

J simply equals 1<90

Add up your three impedances.
10 + 1/jwC + jwL
10+ (1/wC)<-90 + (wL)<90

This will give you one vector at 0 degrees, one at -90 degrees and one at 90 degrees.


Add them together and you will have one vector with an angle. Now divide your RMS voltage by this Vector with the angle. You now have the current...also a vector with an angle.

Now multiply your RMS voltage by your current. This will give you your apparent power...also with an angle.

Current and power should be complex conjugates.

Now find the x and y-axis of your apparent power from the angle. The x-axis is your real power and the y-axis is your reactive power.

To find power factor, divide real power by apparent power.

Notice how the "time" you mention above was not needed to find any of the answers.

It takes new electrical guys quite a while to grasp all these concepts together. So if you don't get it the first time, or even 10th time, understandable.
 
Last edited:

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