What is the attenuation factor?

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Discussion Overview

The discussion revolves around understanding the concept of an attenuation factor in the context of a transformer circuit. Participants explore how to calculate the high voltage (HV) side voltage after applying an attenuation factor of 0.8, while also addressing potential errors in the problem statement and calculations. The scope includes theoretical reasoning, mathematical calculations, and practical implications of the transformer and attenuator setup.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes the transformer setup and attempts to calculate the HV side voltage using the turns ratio and attenuation factor.
  • Another participant suggests that the attenuation factor is typically expressed in decibels for audio applications, indicating a potential misunderstanding of its application here.
  • There is a correction regarding the turns ratio and the input voltage to the transformer, which some participants argue affects the calculations.
  • Participants discuss the implications of the attenuation factor, with one asserting that it should simply reduce the input voltage without amplification.
  • There is confusion about how current behaves in the circuit, with some participants questioning the role of current in the transformer operation when no load is present.
  • One participant expresses uncertainty about the correct answer from the textbook, leading to further debate about the validity of the provided solution.
  • Some participants propose that viewing the attenuator as a simple resistive divider may clarify the calculations.
  • There is a suggestion to seek a second opinion from another participant to resolve the disagreement regarding the correct answer.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct answer to the problem, with multiple competing views on the calculations and the role of the attenuation factor. The discussion remains unresolved regarding the validity of the textbook answer.

Contextual Notes

There are limitations in the problem statement, including potential errors in the turns ratio and input voltage values. The discussion also highlights the dependence on definitions of attenuation and the assumptions made about circuit behavior.

Who May Find This Useful

Readers interested in transformer theory, electrical engineering concepts, and practical applications of attenuation in circuits may find this discussion beneficial.

jaus tail
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Homework Statement


the question has a transformer, the hv side of which is connector to a pot that is said to be an attenuator in the question with attenuation factor = .8?
the turns ratio of transformer is 1:1.125.
I have to find HV side voltage after the attenuation...i know attenuator reduces voltage but how does the factor come into picture?

Homework Equations


transformer HV Voltage = n1/n2*LV Voltage = 1.125/1*230 = 258.75

The Attempt at a Solution


I considered attenuator as resistor with .8 value of total value. like a pot and took output voltage as V=V(HV)*.8/(.8+.2)=V *.8 = 258.75*.8 = 207V

I don't think i used the attenuatior factor rightly. Google suggested a log formula. Like 10 raised to A/20. where A is factor.
I'm not sure how to proceed.

Any help?
 
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Hi,

Not an expert, but I agree with what you did - provided you sketched the whole problem (in particular: no load)

The attenuation you quote is for purposes like audio where they express attenuation in dB (1 Bel = factor 10 in power so 20 dB is a factor 10 in voltage)
 
Thanks for the reply. Here is the print screen of the homework question paper...
Homework%20Question_zpskkd05azf.png
 
Ok, so that is a different problem than the one you calculated. First of all your turns ratio is not 1:1.125 but 1:1.25 .
On what you call (als a bit confusing) the LV side the voltage isn't 230 but 1000 Volt. (Although that isn't really important because the exercise asks for ratios :smile:)
Re-doing your calculation with the 1.25 should give you ##\displaystyle {V_{YZ1}\over V_{WX1}}## which already narrows it down.
Now the other way around for [edit: wx/yz, not yz/wx] ##\displaystyle {V_{WX2}\over V_{YZ2}} ## . What do you propose ?

And yes, the attenuation factor 0.8 is simply as you assumed.
 
How do you calculate VYZ1? What formula for using attenuation do you suggest?
 
Sorry about the errors posted in the problem...I thought I had mugged up the question, but I was wrong...sorry again.
 
No need to apologize. You learn and you have the right to make mistakes.
 
The answer in the book is C. I don;t know how that comes. When Vwx1 is 100 V. V of HV side is 100*1.25 = 125V.
So input to attenuator is 125V. How do i calcuate Vyz1 based on this? Attenuation factor is .8
 
See, I make mistakes too: 100 V instead of 1000 V :smile:

If you attenuate 125 V by a factor of 0.8 you get 125 V x 0.8 = 100 V !

( :rolleyes: the 100/100 is a likely result for this first part -- any of the other answers makes it unnecessary to solve the second part. I sort of gave that away by saying "narrows it down" instead of "gives the answer").
 
  • #10
Hmm...I had to read your posts a few times and go back at the question. I guess you're right. Attenuation simply works as o/p = i/p * attenuation factor.
The whole log and decibels stuff doesn't apply here.
 
  • #11
Right. Now, how do we choose between (B) and (C) ?
 
  • #12
I did that. The Vyz2 is 100 V. So o/p of attenuator is 100V. So i/p of attenuator is 100V divided by attenuation factor = 100/0.8 = 125V.
So HV of transformer is 125V.
LV of transformer is 125 * N1/N2 = 125*1/1.25 = 100V. So Vwx2 is 100V.
So ratio is 100/100 V.
So correct answer is C.
 
  • #13
Ha, you fell for it. An attenuator attenuates. It does not amplify.
 
  • #14
Darn...this is so confusing. So how is C the answer? What is the formula for attenuator? the formula for transformer is Voltage ratio = turns ratio. is there a simple way to solve this problem?:frown:
 
  • #15
You're almost there. Your transformer is ideal, meaning V x I is the same on both sides. So what is the input current on the righthand side ?
 
  • #16
Well as per KCl, current input at right side of attenuator = current input to transformer + current through attenuator.
so I (input to attenuator) = 100V /(attenuator resistance) + I transformer input current...
Sorry i know this is the wrong track. How can you find current without load or B-H curve or resistance?
 
  • #17
jaus tail said:
How does current play a role here?
Look at it from the other side (the lefthand side): it is open, so there is no current. V x I = 0 on that side. Ideal transformers have V x I the same on both sides. So there is no current going into the transformer on the righthand side. Therefore there is no current going "upward" from Y to the transformer.
That means the voltage drop over that top part of the "pot" is ... ?
So the righthand side of the transformer sees how many Volts ?
 
  • #18
But if there is no current in the transformer HV side, then transformer action wouldn't take place. No current, so no flux, so no linkage and no emf?
This is then an open circuit test, i guess.
So the right side of transformer would see input voltage from right side that is 100V, as the attenuator isn't really doing anything if no current is flowing...
 
  • #19
then the answer comes to 80V/100V that is B.
 
  • #20
I agree. I think it was a nice exercise (looks innocent but has some substance), but you might have a different opinion :smile: .
All clear now ? Then on to the next challenge :wink:
 
Last edited:
  • #21
No not clear. The answer in the book is C. How does that happen...?
 
  • #22
It doesn't happen. Book answer can't be right !
 
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  • #23
Thanks for this. You've been very helpful. Thanks. I learned that attenuator doesn't amplify. Lol...:)
 
  • #24
"Resistors resist" is another way of putting it :-p .
 
  • #26
Pi and T ( sounds like a tea party) (*) ? Way too advanced for my pay grade -- and most likely for your textbook level too. Looking at it as a simple potentiometer voltage divider (notice the term 'passive' there :smile: !) seems much more appropriate.

Still we have the nagging issue that the book answer seems wrong. Let's ask a second opinion from e.g. @NascentOxygen !
Ah, and we have @Merlin3189 approving of our bold claim too !

(*) [edit] I now learned here that our simple circuit is an L pad !
 
  • #27
Okie Professorie. Thou have helped moi quite a lot. Thank you, professorie :D
 
  • #28
With the attenuator being a resistive divider (as implied), then (b) would be the correct answer.
 

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