Find the real power consumed using MATLAB Simulink

Click For Summary
The discussion focuses on calculating the real power consumed by an electric component using MATLAB Simulink, with voltage and current given as sinusoidal functions. Participants express confusion over the lack of a clear answer or relevant equations for real power. The importance of understanding the definition of real power is emphasized, highlighting that it involves calculating RMS values of current and voltage when they are in phase. The correct formula for real power is reiterated, indicating that it can be derived from the RMS values of voltage and current. Ultimately, participants stress the need for clarity and proper methodology in solving the problem.
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: 408
  • 1.PNG
    1.PNG
    11.3 KB · Views: 736
  • 2.PNG
    2.PNG
    25.6 KB · Views: 408
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: 319
  • 1.PNG
    1.PNG
    11.8 KB · Views: 453
  • 2.PNG
    2.PNG
    25.6 KB · Views: 342
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
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K