# Homework Help.

Discussion in 'Electronic Basics' started by Daniel Pitts, Jan 28, 2013.

1. ### Daniel PittsGuest

Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

..--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer. Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction, which
cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for the
..7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
out on the first attempt (as it does far too often), but I'm not always
sure.

2. ### Daniel PittsGuest

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a sort
of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4 when
the diode is in the circuit. This would lead to a Pavg = 15v^2/220ohm ~=
1watt.

Is that the proper approach?

Thanks,
Daniel.

3. ### Daniel PittsGuest

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

4. ### Daniel PittsGuest

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.
I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.
Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?

5. ### George HeroldGuest

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

George H.

Hide quoted text -

6. ### Daniel PittsGuest

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?

7. ### Daniel PittsGuest

Ah, okay. so I think I get it now. That makes sense.

8. ### Daniel PittsGuest

I see that now. Thanks.
I think I've got it now. Thanks everyone for your help and suggestions!

9. ### Mark StorkampGuest

Now that he's worked out the answer, it would be interesting to know
what his instructor expects the answer to be. I once took a community ed
night class at a local college. During the day the room was used for a
vocational electronics program. They had this same basic problems worked
out on the blackboard, but had the OP's original wrong answer at the end.

10. ### Tom BiasiGuest

Just because the instructor displayed a workout doesn't mean he or she
said it was correct.

11. ### Daniel PittsGuest

Some of my classmates were comparing answers, and my answer was the "odd
man out". Others had calculated the average voltage. When I explained
how I got my answer (both the "intuitive" way and by integration), they
power and average voltage. I'm not sure the instructor was paying full
attention but they said yes that would work.

Mathematically I can see it not working, so I'm sticking with my answer
and current understanding (Thanks Phil Hobbs), even if the instructor
does say it is wrong when he finally corrects the homework.

Anyway, thanks everyone for all your help. I appreciate it very much,
and I feel like I have a better grasp on this concept than many of my
classmates.

Ironic since most of them have taken an "AC" course, and I skipped that
one.

12. ### amdxGuest

Mikek
And don't wait for the next edition of, Art of Electronics. I've been
waiting 6 or 7 years, maybe 10!
Mikek

PS. I found this,

"UPDATE ( 01 / 24 / 2012 ): Sadly, the schedule for the 3rd edition of
The Art of Electronics has slipped. This post quotes a mail by Winfield
Hill in which he explains the current state of the book and says that it
"should be out by late 2012 or early 2013”. Considering the history of
delays of the 3rd edition, I think that early 2013 is more probable, and
I would not be surprised at all if it will be much later. It seems
obvious that both authors would rather delay the 3rd edition even
further than to rush their work and deliver a book of lower quality.
Anyway, as soon as I learn of a new schedule I will add it here."

Here:
http://www.wisewarthog.com/electronics/horowitz-hill-the-art-of-electronics.html

And just for giggles, this from 2004.

I have found in a german group, that the 3rd editon of the book
"Horowitz, Hill - art of electronics" shall be released in march 2004.
Paul

urban myth

Thanks,
- Win

<Bugs Bunny mode ON>

<Bugs Bunny mode OFF>
Ralph

13. ### Daniel PittsGuest

Yep, so the instructor has considered this approach and answer wrong...
His approach was to find the average voltage and use that in v^2/r. Of
course that isn't the answer I came up with... I used wolframalpha.com
to calculate my result:

http://www.wolframalpha.com/input/?...(integrate+(v*sinx)^2/r+from+0+to+(pi))/(2pi)

{r == 220, v == 60 Sqrt[2], Integrate[(v Sin[x])^2/r, {x, 0, Pi}]/(2 Pi)}

Unless I don't understand the definition of "Average" (sum over count,
or integral over range) or "Power" (current times voltage).

14. ### Guest

Doesn't require calculus at all, only a counterexample. The average
of a sine wave is zero. The RMS value is not. QED.