Find the real power consumed using MATLAB Simulink

Click For Summary

Discussion Overview

The discussion revolves around calculating the real power consumed by an electric component using MATLAB Simulink, focusing on the provided voltage and current waveforms. Participants explore the definitions and calculations related to real power in AC circuits, including RMS values and phase relationships.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about the absence of a clear answer or relevant equations for real power in the initial posts.
  • One participant emphasizes the need to calculate the real power using the definitions and relationships between voltage and current.
  • Another participant mentions that instantaneous power does not have real and imaginary parts, suggesting that real power should be calculated as an average over a complete cycle.
  • There is a discussion about the relevance of a transport delay in the context of the problem, with some participants questioning its necessity.
  • One participant clarifies that real power is produced when current and voltage are in phase, leading to a discussion on calculating RMS values for both current and voltage.
  • Another participant points out that the provided expressions for voltage and current indicate that the phase angle θ is zero, which affects the calculation of true power.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to calculate real power, with multiple viewpoints on the definitions and necessary calculations remaining unresolved.

Contextual Notes

There are limitations regarding the clarity of the problem statement and the assumptions made about the definitions of real power and the relevance of additional components like capacitors.

Fatima Hasan
Messages
315
Reaction score
14
Homework Statement
The question is attached below.
Relevant Equations
Output a sine wave:

O(t) = Amp*Sin(Freq*t+Phase) + Bias
T = 1/Freq
My work is attached below. I want to confirm my answer.
Any help would be greatly appreciated !
 

Attachments

  • Problem Statement.PNG
    Problem Statement.PNG
    10.3 KB · Views: 420
  • 1.PNG
    1.PNG
    11.3 KB · Views: 749
  • 2.PNG
    2.PNG
    25.6 KB · Views: 423
Physics news on Phys.org
Problem statement
244907


Answer
244908
244909
 
Still don't see your answer
Nor a relevant equation for the 'real power'

Is there a miscommunication here ?
 
BvU said:
Still don't see your answer
Nor a relevant equation for the 'real power'

Is there a miscommunication here ?

##v(t)=10√2 sin (20πt)##
##i(t)=\frac{5}{√2} sin(20πt)##
The DC and ac values that I got are :
P(dc)=25 W
P(ac)=30.62 W

I reattached the problem statement and my answer below.
 

Attachments

  • Problem Statement.PNG
    Problem Statement.PNG
    10.3 KB · Views: 330
  • 1.PNG
    1.PNG
    11.8 KB · Views: 466
  • 2.PNG
    2.PNG
    25.6 KB · Views: 352
Last edited:
There reeally is no need to attach them for the fourth time. I've seen all three in the first post already. The exercise is quite clear: you are supposed to
Calculate the real power consumed by an electric component where voltage and current are $$\begin {align*}
v(t) & = 10\sqrt 2 \sin (20\pi t) \\ \mathstrut \\ i(t) & = {5\over \sqrt 2} \sin (20\pi t)
\end{align*} $$
So, based on the first picture alone, I should think you need to find a definition of 'real power' and work out what it means in terms of the given ##v## and ##i## ...

I'm not familiar with your software package but I can guess what some of the blocks are supposed to do. (but about the function of the part in the middle

244918


I have no idea).
##\ ##
 
BvU said:
the function of the part in the middle
It's a transport delay which is equal to T (time=0.1 s).
 
Nothing to do with the question in this exercise, then !
 
Fatima,

This is confusing us.

"Find the real power consumed "

only needs algebra to derive the equation.

Output a sine wave:

O(t) = Amp*Sin(Freq*t+Phase) + Bias
T = 1/Freq

That has nothing to do with finding the real power.

Instantaneous power P(t)=V(t)*I(t) has no real and imaginary parts. Real and Imaginary powers apply only to the average over a complete cycle. Calculate the RMS (root mean square) over a cycle.
 
  • #10
There's no capacitor in the original problem statement, so why are you even talking about a capacitor? It seems to me that you're just throwing random things together to come up with an answer and hoping it's right. In the process, you're making this problem way too complicated.

I think you should heed @BvU's advice and look up the definition of real power. I'm not sure why you seem so reluctant to do so. It seems silly to start doing calculations if you don't know what you're trying to calculate.
 
  • #11
vela said:
There's no capacitor in the original problem statement, so why are you even talking about a capacitor?
Sorry ,I got confused . That was my solution for another question , I will delete it.
 
  • #12
vela said:
look up the definition of real power.
Real power is produced when the current and the voltage are in phase with each other .
P in ac circuit = current(rms) * voltage (rms) * cos θ
P = current (rms) * voltage (rms) , (θ=0)
So, I have to find the rms values for current and voltage separately and multiply them together , right ?
 
  • #13
You are given the expressions for ##v(t)## and ##i(t)##. What do they tell you about your ##\theta## ?
##\theta = 0## as I think you indicate.
That means that (according to e.g. https://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/ ) true power
##P## is ##I^2R = V^2/R = V\,I ##.

With sine square averaging to 1/2 your Prms is indeed Vrms * Irms
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 4 ·
Replies
4
Views
22K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K