# Homework Help.

D

#### Daniel Pitts

Jan 1, 1970
0
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.

D

#### Daniel Pitts

Jan 1, 1970
0
RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is just
the same as it would be if the diode weren't there, and therefore the
instantaneous power is also unchanged. There's no power dissipated on
the negative half cycle, so the average power is half of what it would
be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes down
by a factor of.....?

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.

D

#### Daniel Pitts

Jan 1, 1970
0
Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate
the two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root, which
is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

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?

D

#### Daniel Pitts

Jan 1, 1970
0
It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor. Even if the instructor
I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.
If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.
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.
In this case, RMS would be defined as the square root of the time average
of the square of the voltage. Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.
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?

G

#### George Herold

Jan 1, 1970
0
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?-

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 -

D

#### Daniel Pitts

Jan 1, 1970
0
That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

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

So, who is correct?

D

#### Daniel Pitts

Jan 1, 1970
0
The diode doesn't (as you originally posted) cut VRMS in half, it cuts the power
dissipated by the resistor in half. I think that's the difference. I think you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

D

#### Daniel Pitts

Jan 1, 1970
0
John and Phil. John's is _not_ the analysis that you did that Phil
didn't like. His method will get you the right answer, if you mind your
P's and Q's.
I see that now. Thanks.
(I ain't gonna say more until you go back and look at what John said
_really closely_, then revisit what RMS means, then come back. Sorry --
but you did say it's homework).

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

M

#### Mark Storkamp

Jan 1, 1970
0
Daniel Pitts said:
Ah, okay. so I think I get it now. That makes sense.

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.

T

#### Tom Biasi

Jan 1, 1970
0
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.
Just because the instructor displayed a workout doesn't mean he or she
said it was correct.

D

#### Daniel Pitts

Jan 1, 1970
0
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.

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.

A

#### amdx

Jan 1, 1970
0
The diode doesn't (as you originally posted) cut VRMS in half, it cuts the power
dissipated by the resistor in half. I think that's the difference. I think you
corrected the VRMS thing in a later post.

Never bet against Phil on stuff like this.

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by H+H.

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

D

#### Daniel Pitts

Jan 1, 1970
0
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.

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).

K

#### [email protected]

Jan 1, 1970
0
On 1/29/13 9:40 AM, Mark Storkamp wrote:
On 1/28/13 8:25 PM, John Larkin wrote:
On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold

On Jan 28, 4:06 pm, Daniel Pitts
On 1/28/13 12:03 PM, Tim Wescott wrote:

On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
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.

RMS is a time average based on the power dissipated.An
ideal diode
in
the secondary side conducts on the positive half-cycle and
not on
the
negative half-cycle. On the positive, the instantaneous
current is
just the same as it would be if the diode weren't there,
and therefore the instantaneous power is also unchanged.
There's no
power
dissipated on the negative half cycle, so the average power
is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS
voltage goes
down by a factor of.....?

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.

Power dissipated is RMS voltage squared divided by the
resistance.
That's the whole point of RMS--you can use AC or DC or some
nasty
switching waveform, but the RMS value will tell you how much
power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous
voltage)^2)

(note the order of the operations--square first, then
average).

Then to get back to the right way to express the voltage,
equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous
voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square
root,
which is where the name comes from: root( mean( square
voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the
same, the
RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

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?

It's probably close enough.

I would strongly suggest that you slap some graph paper down
and sketch
the voltage waveform seen by the resistor. Even if the
understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that
concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and
if you've
taken calculus, and if you have the time, you may want to take
your
sketch of the waveform at the resistor and calculate the RMS
voltage.

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.

In this case, RMS would be defined as the square root of the
time average
of the square of the voltage. Finding the time average of the
square
of
the voltage can either be done intuitively (you know what the
time average is for a whole cycle, so you can easily do it for
a half cycle),
but grinding through the exercise of integrating the
voltage-squared
over
one cycle, then dividing by the cycle time, will be good for

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?-

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?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and
drive them from diodes in opposite directions. They conduct on
opposite half
cycles. The transformer still sees R, so the power delivered from
the transformer is the same, but by conservation of energy and by
symmetry
each resistor burns half of the power P. Now back-calculate the
RMS voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

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

So, who is correct?

The diode doesn't (as you originally posted) cut VRMS in half, it
cuts the
power
dissipated by the resistor in half. I think that's the difference. I
think
you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

Never bet against Phil on stuff like this.
Good to know ;-)

Incidentally, his book on electro-optics is worth having for anyone
serious
about electronics. The other necessary book is The Art Of
Electronics by
H+H.

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

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 seemed very confused. One of them asked the
instruct about average 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.
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/?i=r=220,+v=60+sqrt(2),+% 28integrate+%28v*sinx%29%5E2%2Fr+from+0+to+%28pi%29%29%2F%282pi%29

{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).

Sigh.

Average voltage does _not_ equal RMS voltage, except for DC.

But I'm almost certain that any proof of that would end up requiring
calculus one way or another (for instance: appeals to AC power requires
use of the Fourier Transform, which requires use of calculus). So it
wouldn't be straightforward.

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

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