Find the real power consumed using MATLAB Simulink

12 replies · 4K views · last reply
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
  • Text problem asking for the real power consumed by an electric component, with v(t) = 10√2 sin(20πt) and i(t) = (5/√2) sin(20πt).
    Problem Statement.PNG
    10.3 KB · Views: 445
  • Simulink model with two blocks: top chain multiplies Voltage and Current sine sources, integrates, subtracts a delayed copy, gains by 10 to a display of 25; bottom chain squares the product, integrates, and takes a square root to a display of 30.62.
    1.PNG
    11.3 KB · Views: 770
  • Three Simulink block parameter dialogs side by side: Voltage sine wave source with amplitude 10*sqrt(2), frequency 10*2*pi, bias 0, phase 0; Current sine wave source with amplitude 5/sqrt(2), frequency 10*2*pi, bias 0, phase 0; Transport Delay block with time delay 1/10 and initial buffer size 1024.
    2.PNG
    25.6 KB · Views: 448
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: Calculate the real power consumed by an electric component where voltage and current are v(t) = 10√2 sin(20πt) and i(t) = (5/√2) sin(20πt).
    Problem Statement.PNG
    10.3 KB · Views: 359
  • Simulink model with two sine wave sources labelled Voltage and Current feeding a multiplier, then an integrator 1/s, a sum block, gain 10, and a display showing 25, with a second chain using u^2, sqrt(u) and a display showing 30.62.
    1.PNG
    11.8 KB · Views: 495
  • Three Simulink block parameter dialogs side by side: Voltage sine wave source with amplitude 10*sqrt(2), frequency 10*2*pi; Current sine wave source with amplitude 5/sqrt(2), frequency 10*2*pi; Transport Delay block with time delay 1/10.
    2.PNG
    25.6 KB · Views: 380
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.
 
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.
 
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.
 
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 ?
 
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