How can I find the DC component of an Output Sinusoidal Voltage?

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Homework Help Overview

The discussion revolves around finding the DC component of an output sinusoidal voltage, specifically in the context of rectification and RMS values. Participants explore the relationship between RMS values and DC components in electrical signals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the RMS method for determining the DC component and question its appropriateness. There are attempts to clarify the difference between RMS values and DC equivalents, as well as the average value of AC signals.

Discussion Status

The discussion is active, with participants raising questions about the definitions and calculations involved. Some guidance has been offered regarding the interpretation of RMS values and their application to fluctuating signals, but there is no explicit consensus on the correct approach to finding the DC component.

Contextual Notes

Participants reference specific equations and concepts related to half-wave rectification and the average values of AC signals. There is mention of confusion regarding the terminology used in the context of RMS and DC components, as well as the implications of periodicity in the signals being discussed.

Rongeet Banerjee
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Homework Statement
How can I find the RMS value of a DC Current?
Relevant Equations
Vrms=Vpeak/(2)½
IMG_20200822_121952466.jpg

I had previously solved this using Root Mean Square method by integrating the value of voltage from t=0 to t=T/2 and then from t=T/2 to t=T.Answer was Vo/2½.Yesterday I found this question👇🏾
15980792790661347500990.jpg

if I followed the previous approach then:
15980800547871983487461.jpg

15980801446431097768317.jpg

5 volts is not even in the option. How can I find the DC component of an Output Sinusoidal Voltage?
 
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Rongeet Banerjee said:
Homework Statement:: How can I find the RMS value of a DC Current?
Relevant Equations:: Vrms=Vpeak/(2)½

I had previously solved this using Root Mean Square method
This gives you the dc "equivalent" of the time varying signal, not the dc "component" of it.
 
Coming back to question 42,how can we find RMS value of a non alternating current?
 
For DC, the actual value is the RMS value.
 
Rongeet Banerjee said:
Coming back to question 42,how can we find RMS value of a non alternating current?
What is question 42?
Edit: Ok I can see that in your first image. Haven't you already solved it correctly? It is a non-alternating waveform. What else do you want to know?
Rongeet Banerjee said:
Ok.But isn't the average value of AC 2Vpeak/π.
No. The question mentions "a half wave" rectifier. What will the output voltage look like?

Also, ac in electricity usually means a periodic signal with both positive and negative halves (mostly sinusoidal signals). So the average value of ac is 0.
 
cnh1995 said:
This gives you the dc "equivalent" of the time varying signal, not the dc "component" of it.
So basically when the question asked me to find RMS value they actually meant"DC equivalent".I got a bit confused because I had learned that only alternating currents have RMS values.So in reality is RMS value used for any fluctuating physical quantity?
 
DC Component =(average of V from t=0 to t=T/2 + average of V fom t=T/2 to t=T) /2 i.e. =(20/π + 0)/2 =10/π .Is that the answer?
 
  • #10
Rongeet Banerjee said:
So in reality is RMS value used for any fluctuating physical quantity?
Yes, if they are fluctuating periodically.
Same is true for average value.
You compute rms and average values of a periodic signal and derive conclusions from it.
If the average of a periodic signal is 0, it is symmetric about the time axis (i.e. equal positive and negative parts).
If the average is non zero, it has a dc "component" in it which is making the positive and negative parts unequal.

If you are passing an alternating current through a resistor, the heat produced by the alternating current in its one cycle time is equal to the heat produced by its rms current(dc) in the same time. This means the rms current is the dc "equivalent" of the alternating current.
 
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  • #11
Rongeet Banerjee said:
DC Component =(average of V from t=0 to t=T/2 + average of V fom t=T/2 to t=T) /2 i.e. =(20/π + 0)/2 =10/π .Is that the answer?
Yes that is the answer, but I am not sure if the formula you wrote is correct. Is that some "trick" or "shortcut" considering the MCQ nature of some exam?
I prefer the general formula:
Average= integral of function/time period.
You can then split the integral as per the function behavior between t=0 to t=T.

In this case, your function is sinusoidal from 0 to T/2 and 0 from T/2 to T. You can split the integral accordingly and get the average value.
 

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