Intro to Power Systems: Calculating Power

  • #1
SumDood_
30
6
Homework Statement
Calculate the power in the given circuit
Relevant Equations
P=IV
Help me understand what I am missing. So this is the circuit:
1696847734105.png

Given:
##V_{RMS} = 120V##
##I_{RMS}= 10A##
##v=Re \{Ve^{jwt}\}##
##i=Re \{Ie^{j(wt-\psi)}\}##
(a) Calculate and sketch real and reactive power P and Q as a function of the angle ψ
(b) Calculate and sketch the instantaneous power

What I've done so far:
##V = 120\sqrt{2}cos(\omega t)##
##I = 10\sqrt{2}cos(\omega t-\psi)##
##
\begin{align*}
P = iv &= 120\sqrt{2}cos(\omega t) \cdot 10\sqrt{2}cos(\omega t-\psi) \\
&= 120\sqrt{2}\frac{1}{2}(e^{j\omega t}+e^{-j\omega t}) \cdot 10\sqrt{2}\frac{1}{2}(e^{j(\omega t-\psi)}+e^{-j(\omega t-\psi)}) \\
&= 600(e^{j\psi} + e^{-j\psi} + e^{j(2\omega t - \psi)} + e^{-j(2\omega t - \psi)}) \\
&= 2400(cos(2\omega t - \psi) + cos(\psi))
\end{align*}
##

I am not sure if the work I've done is correct. If it is, I don't understand how to differentiate P and Q from the solution I have worked out.
Any help is appreciated!
 
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  • #2
First, which device is delivering power and which is receiving?
Then I would approach this using phsor notation.
 
  • #3
My knowledge of EE is a bit basic but shouldn't your circuit contain a load (R and/or L and/or C)? Does it make sense to consider a circuit consisting of only an ideal current source and an ideal voltage source? Or are the sources non-ideal and you have [edit - that should be 'haven't'] supplied all the data?
 
Last edited:
  • #4
scottdave said:
First, which device is delivering power and which is receiving?
Then I would approach this using phsor notation.
What I have provided is what is mentioned in the question. Let's assume that the current source is delivering power.
 
  • #5
Steve4Physics said:
My knowledge of EE is a bit basic but shouldn't your circuit contain a load (R and/or L and/or C)? Does it make sense to consider a circuit consisting of only an ideal current source and an ideal voltage source? Or are the sources non-ideal and you have suppled all the data?
You are right that this kind of circuit won't exist in the real world, but the question I've described is as stated in the book.
 
  • #6
With the current direction shown, the voltage source (I assume) shown as DC, and the text giving a voltage in RMS (which implies AC), just how much more confusing could it get?
 
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  • #7
It seems that everyone is finding this question to be weird. So, I'll post the solution that is at the end of the book:
##P = \frac{1}{2}VIcos(\psi)##
##Q = \frac{1}{2}VIsin(\psi)##
Instantaneous power
##p=2Pcos^{2}(\omega t)+Qsin(2\omega t)##
 
  • #8
SumDood_ said:
It seems that everyone is finding this question to be weird. So, I'll post the solution that is at the end of the book:
##P = \frac{1}{2}VIcos(\psi)##
##Q = \frac{1}{2}VIsin(\psi)##
Instantaneous power
##p=2Pcos^{2}(\omega t)+Qsin(2\omega t)##
I suppose you can just treat the question as a maths exercise and ignore the (unphysical!) physics.

Working backwards from the expected answers gives some clues.

The complex power, ##S##, is ##S= P+jQ## where ##P## is the required real power and ##Q## is the required reactive power. So you need to find a (complex) expression for ##S##.

According to Wikipedia (https://en.wikipedia.org/wiki/AC_power) the complex power, ##S##, is given by ##S = \hat V \hat I^*## where (in this question) ##\hat V = V_{rms}e^{j \omega t}## and ##\hat I = I_{rms}e^{j(\omega t- \phi)}##. Note the use of the complex conjugate.

To get the instantaneous power I guess one way is to evaluate ##p(t) = v(i)i(t)## with ##v(t) = V\cos (\omega t)## and ##i(t) = I \cos(\omega t - \phi)## and use some trig’ identities, and the expressions for ##P## and##Q##, to get the answer in the required format.
 
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Related to Intro to Power Systems: Calculating Power

What is electrical power and how is it calculated?

Electrical power is the rate at which electrical energy is transferred by an electric circuit. The basic formula to calculate electrical power is P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes.

What are the different types of power in AC circuits?

In AC circuits, there are three types of power: real power (P), reactive power (Q), and apparent power (S). Real power is measured in watts (W) and represents the actual power consumed by the circuit. Reactive power is measured in volt-amperes reactive (VAR) and represents the power stored and released by inductors and capacitors. Apparent power is measured in volt-amperes (VA) and is the product of the RMS voltage and RMS current.

How do you calculate power in a single-phase AC circuit?

In a single-phase AC circuit, the power can be calculated using the formula P = VI cos(φ), where P is the real power in watts, V is the RMS voltage, I is the RMS current, and φ is the phase angle between the voltage and current.

What is the power factor and why is it important?

The power factor is the ratio of real power to apparent power and is a measure of how effectively the electrical power is being used. It is important because a low power factor indicates poor utilization of electrical power, resulting in higher energy losses and increased costs. The power factor is calculated as PF = cos(φ), where φ is the phase angle between the voltage and current.

How do you calculate power in a three-phase AC circuit?

In a three-phase AC circuit, the power can be calculated using the formula P = √3 VL IL cos(φ) for a balanced load, where P is the real power in watts, VL is the line-to-line voltage, IL is the line current, and φ is the phase angle between the voltage and current. For an unbalanced load, the power must be calculated for each phase and then summed up.

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