Industrial electronics. Question with coils and diode.

[Femme_physics]

Homework Statement

http://img198.imageshack.us/img198/1421/vintm.jpg

Given that k = 2

A) Draw in a qualitative manner the following letters

Vin, VP, Vs, Uxy, UR1, UR2, UD

B) Calculate the power in R2

C) Calculate the average current through R1

The Attempt at a Solution

I'll start with a little theory. The way I see it there are 2 currents, one marked in blue and one marked in green. The current marked in red actually gets stuck, so it appeared no current hits the second circuit. Could it be?

http://img8.imageshack.us/img8/6656/mkkkmkkkk.jpg

 

[gneill]

Consider that voltage on the primary of the transformer is a sinusoid, so it has positive and negative excursions. Similarly, the voltage across the secondary will also have positive and negative excursions and follows a sinusoidal function.

In the secondary circuit, as you surmised, the diode will block current flow when the secondary's voltage causes the diode to be reverse biased. So, perhaps you could sketch the anticipated current function for the secondary circuit for a full cycle. How would you represent this function mathematically? (HINT: you might want to break the function into more than one time interval).

 

[I like Serena]

Morning Fp!

What do you mean by: "Given that k = 2"?

I suspect that when they asked to draw the letters in a qualitive manner, they meant as a graph, with the amplitude on the vertical axis, and the time on the horizontal axis.
Do you know how to draw for instance Vin in this manner (this looks like a sinusoid)?

I like your drawing with the blue and green arrows, which is exactly what happens. :)
And yes, in the right circuit the current following the red arrow is blocked, but as gneill already suggested, in the reverse direction it will still flow.

 

[Femme_physics]

Hi gneil! Morning ILS

What do you mean by: "Given that k = 2"?

Oh, I thought it's understood that k is "transmission relation of the transformer", which is basically i1/i2 or u1/u2 or n1/n1 (I'm still not sure what's N1/N2)...

Consider that voltage on the primary of the transformer is a sinusoid, so it has positive and negative excursions. Similarly, the voltage across the secondary will also have positive and negative excursions and follows a sinusoidal function.

To my understand coil Vp only transforms the current in one direction, no? The direction I drew in red. I do understand there are two directions for current flow, that's why I drew the one in green and blue

In the secondary circuit, as you surmised, the diode will block current flow when the secondary's voltage causes the diode to be reverse biased. So, perhaps you could sketch the anticipated current function for the secondary circuit for a full cycle. How would you represent this function mathematically? (HINT: you might want to break the function into more than one time interval).

I'm not sure. Can the current go like this?

http://img855.imageshack.us/img855/3387/vpvs.jpg

I suspect that when they asked to draw the letters in a qualitive manner, they meant as a graph, with the amplitude on the vertical axis, and the time on the horizontal axis.
Do you know how to draw for instance Vin in this manner (this looks like a sinusoid)?

I like your drawing with the blue and green arrows, which is exactly what happens. :)
And yes, in the right circuit the current following the red arrow is blocked, but as gneill already suggested, in the reverse direction it will still flow.

Ah, thanks for clearing it up

 

[I like Serena]

Oh, I thought it's understood that k is "transmission relation of the transformer", which is basically i1/i2 or u1/u2 or n1/n1 (I'm still not sure what's N1/N2)...

I suspected as much, although I'm still not clear which coil has the most windings.
Do you know?

As you can see the letters P and S are used.
Those stand for "primary" coil and "secondary" coil.
n1 and n2, or N1 and N2 are the number of windings in each coil.
Although in this case I'd prefer Np and Ns to eliminate the ambiguity.

So is k = Ns / Np?
Or is k = Np / Ns?

To my understand coil Vp only transforms the current in one direction, no? The direction I drew in red. I do understand there are two directions for current flow, that's why I drew the one in green and blue

Actually, the coils do not so much transform current, but they transform voltage (in both directions).
And due to the voltage the current will flow (if it can).
In the primary circuit the current will flow in both directions, while in the secondary circuit it will only flow in one direction due to the diode.

I'm not sure. Can the current go like this?

Yes, it can. :)

 

[Femme_physics]

So is k = Ns / Np?
Or is k = Np / Ns?

Good question! Can anyone help with it? I wish I knew!

But your preference of using Np and Ns is better IMO

I suspected as much, although I'm still not clear which coil has the most windings.
Do you know?

Well, the question doesn't say so

In the primary circuit the current will flow in both directions, while in the secondary circuit it will only flow in one direction due to the diode.

Makes sense

Yes, it can. :)

So what's the point of the dots in the coil? u see? one Vp dot and one Vs dot? Do you see the dots I'm talking about?

 

[Femme_physics]

This is my idea of Vin after reading the replies

http://img31.imageshack.us/img31/302/vinnnnnnn111.jpg

 

[I like Serena]

Good question! Can anyone help with it? I wish I knew!

But your preference of using Np and Ns is better IMO

Let's assume k=Np/Ns, because usually a transformer transforms the voltage from high to low.
(Like in the adapter for your laptop, where it is transformed from 230 V to 12 V.)

So what's the point of the dots in the coil? u see? one Vp dot and one Vs dot? Do you see the dots I'm talking about?

I believe the dots are only markers to indicate the points where they would like you to make a qualitative drawing of the voltage (in the form of a graph with an x-axis and an y-axis).

 

[I like Serena]

This is my idea of Vin after reading the replies

I'm almost seeing a smile.

I'm afraid the voltage of Vin does not make jumps.
It's just a fluent sine curve, it has to be since it's supposed to be a perfect voltage source.

 

[Femme_physics]

Let's assume k=Np/Ns, because usually a transformer transforms the voltage from high to low.
(Like in the adapter for your laptop, where it is transformed from 230 V to 12 V.)

Makes sense

I believe the dots are only markers to indicate the points where they would like you to make a qualitative drawing of the voltage (in the form of a graph with an x-axis and an y-axis).

I was thinking it had more significance than that, but I guess I had it wrong. Oh well.

I'm almost seeing a smile.

I'm afraid the voltage of Vin does not make jumps.
It's just a fluent sine curve, it has to be since it's supposed to be a perfect voltage source.

Doesn't it depends at what loop we're looking at? But if not, I guess it should be: (ignore lack of symmetry!)

http://img812.imageshack.us/img812/1227/secondtryy.jpg

 

[I like Serena]

Ignoring lack of symmetry... yep! :)

Btw, do you know how high the tops are?

 

[Femme_physics]

Well, there's

U(av)
and
U(rms)
and
U(max)

I think what's given to me is U(max) so I'd venture to say the tops are U(max)

i.e. 230V

but they ask to draw it in a qualitative manner, so that means without numbers no?
As far as Vp and Vs...

http://img42.imageshack.us/img42/2459/booooooz.jpg

once again ignoring symmetry heh...

 

[I like Serena]

Well, there's

U(av)
and
U(rms)
and
U(max)

I think what's given to me is U(max) so I'd venture to say the tops are U(max)

Hmm, I don't know U(av). What is that?

I can tell you that U(rms) is given - I believe it usually is.

but they ask to draw it in a qualitative manner, so that means without numbers no?

I'm not sure, what "qualitative" means exactly.
It think it means that you should not do calculations, but I think it does (or should) include which numbers go where.

Anyway, after the graphs they are asking for numbers, so I think it's a good thing if you're already aware of how the numbers apply to the graphs.

As far as Vp and Vs...

once again ignoring symmetry heh...

I see you've taken the smile-version of the sliced sine graph. :)

However, I'm afraid you're too early slicing it off.
The induced voltage in the secondary coil is still the full voltage, positive and negative.
I'm afraid that's how induced voltage (also known as electromotive force or emf) works.
If you want I can give a longer explanation, based on the related magnetic field.

You should take into consideration that Vs is on the side of the coil that has fewer windings, which has an impact on the amplitude.

 

[Femme_physics]

Hmm, I don't know U(av). What is that?

I can tell you that U(rms) is given - I believe it usually is.

http://img808.imageshack.us/img808/7753/mizzz.jpg

I'm not sure, what "qualitative" means exactly.
It think it means that you should not do calculations, but I think it does (or should) include which numbers go where.

Anyway, after the graphs they are asking for numbers, so I think it's a good thing if you're already aware of how the numbers apply to the graphs.

Fair enough makes sense.

I see you've taken the smile-version of the sliced sine graph. :)

oh heh, that's how it's called?^^

However, I'm afraid you're too early slicing it off.
The induced voltage in the secondary coil is still the full voltage, positive and negative.
I'm afraid that's how induced voltage (also known as electromotive force or emf) works.
If you want I can give a longer explanation, based on the related magnetic field.

You should take into consideration that Vs is on the side of the coil that has fewer windings, which has an impact on the amplitude.

Oh, you're right, I was thinking "circuit", I wasn't thinking "coils". So Vs is the same as Vp (as I've drawn Vp)

 

[I like Serena]

http://img808.imageshack.us/img808/7753/mizzz.jpg

Aaah, "av" stands for "average".
Oh, well, that one is not usually used.

Anyway, I know those formulas slightly differently:
U_{av} = \frac {2U_{max}} {\pi}
U_{rms} = \frac {U_{max}} {\sqrt 2}
so your U(max) is not right yet.

Where did you get your formulas?

Oh, you're right, I was thinking "circuit", I wasn't thinking "coils". So Vs is the same as Vp (as I've drawn Vp)

Almost, but not quite.
Vs is on the coil with fewer windings...

 

[Femme_physics]

Where did you get your formulas?

From my textbook! I swear it says it... so it's wrong?

Almost, but not quite.
Vs is on the coil with fewer windings...

A bigger hint, maybe?^^

 

[Femme_physics]

Where did you get your formulas?

From my textbook! I swear it says it... so it's wrong?

Almost, but not quite.
Vs is on the coil with fewer windings...

A bigger hint, maybe?^^

 

[I like Serena]

From my textbook! I swear it says it... so it's wrong?

Errr... yes, it is wrong assuming U(rms) and U(max) are voltages.

There is an alternate formula:
P_{rms} = \frac {P_{max}} {2}
This is about power, instead of voltage.
Perhaps that one was intended?

Btw, the symbol U is also used for energy, although that seems to be misplaced here.
But that might almost be applicable.

A bigger hint, maybe?^^

You wrote before:

Oh, I thought it's understood that k is "transmission relation of the transformer", which is basically i1/i2 or u1/u2 or n1/n1 (I'm still not sure what's N1/N2)...

So I assumed you knew...
\frac {Vs} {Vp} = \frac {Ns} {Np}

 

[Femme_physics]

Well my textbook definitely says U and not P. But I'll ask my lecturer, for sure.

So I assumed you knew...
\frac {Vs} {Vp} = \frac {Ns} {Np}

[/QUOTE]

Well, I'm now reading the comments section of the question and it says

"If both the points are in the same place, the transformer keeps the phase. If both points are in a different location, the transformer reverses the phase. I don't know what it means by "reverses the phase"..hopefully I'm translating it correctly. "


It also tells me that: "The directing of the winding determines the current's direction. If it's the same winding directon, it'll be the same direction of current. Although it doesn't say here the direction of the winding. Do I read it from the sketch?

 

[I like Serena]

It also tells me that: "The directing of the winding determines the current's direction. If it's the same winding directon, it'll be the same direction of current. Although it doesn't say here the direction of the winding. Do I read it from the sketch?

It does not appear to be given in the sketch or in the problem statement.
My choice would be to assume the winding direction is the same, meaning the direction of the current is the same.
But since they do not specify both choices would be correct.

"If both the points are in the same place, the transformer keeps the phase. If both points are in a different location, the transformer reverses the phase. I don't know what it means by "reverses the phase"..hopefully I'm translating it correctly. "

As always you have a "voltage difference".
On the top of the coil the voltage goes up and then down (relative to the bottom or some zero in the middle).
But at the bottom of the coil the voltage goes down and then up (relative to the top or some zero in the middle).
This is what they call "reversed phase".

(Note that with reverse winding, this would also reverse the phase.)

 

[I like Serena]

Btw, since this information about winding direction and points is given in the comments section, that makes me reconsider the meaning of the dots.

Now I think the dots symbolize that the phase is reversed.
I have to admit that I was not aware of such a convention.

That means that when Vp goes up, Vs goes down.
And when Vp goes down, Vs goes up.

 

[Femme_physics]

Errr... yes, it is wrong assuming U(rms) and U(max) are voltages.

I was told it's true for "where AC current passes through a single diode creates a half wave rectified sine"


but I guess since what we have is a more complicated circuit your formulas apply? I just don't see them on the manual.

So it really should be:

http://img600.imageshack.us/img600/4651/umaxxy.jpg

It does not appear to be given in the sketch or in the problem statement.
My choice would be to assume the winding direction is the same, meaning the direction of the current is the same.
But since they do not specify both choices would be correct.

I guess I'll add it to my notes then

Now I think the dots symbolize that the phase is reversed.
I have to admit that I was not aware of such a convention.

That means that when Vp goes up, Vs goes down.
And when Vp goes down, Vs goes up.

Aha! So in that case, this is wrong

http://img855.imageshack.us/img855/3387/vpvs.jpg


Should be


http://img171.imageshack.us/img171/136/reallyshould.jpg

 

[I like Serena]

I was told it's true for "where AC current passes through a single diode creates a half wave rectified sine"

Not quite.

Edited:
When an AC current passes through a diode you would have:
V_rms(after) = V_rms(before) / √2
since effectively only part of the voltage (half of the power) comes through.

The V_max however is unchanged.

Note that the given relation between V_max and V_rms is only true when we're talking about a sine wave.

but I guess since what we have is a more complicated circuit your formulas apply? I just don't see them on the manual.

No, it's not about a more complicated circuit.
It's about the relation between "max" and "rms" when we're talking about a sine wave.

So it really should be:

Yes.

I guess I'll add it to my notes then

Always a good thing. :)
In this case we can conclude that the construction is such that the phase reverses.
This would for instance happen if the coils are inside each other with a reverse winding.

Aha! So in that case, this is wrong

Should be

Yes, but this is only when the current in the primary circuit is flowing as you indicated.
It will still also flow the other way.

 

[ehild]

When an AC current passes through a diode you would have:
V_rms(after) = V_rms(before) / 2
since efffectively only half of the voltage comes through.

rms voltage means root-mean square of the time-dependent periodic voltage. It is connected to the power: The average power of the AC voltage is the same as that of a DC voltage, equal to the rms value of AC. The power is proportional to the square of the voltage: That means that the time-average of the square of voltage for a period is

V_{rms}^2=\frac{1}{T}\int_0^T{V(t)^2dt}

If it is a full wave, V=V_maxsin(wt)
so

V_{rms}^2=\frac{1}{T}\int_0^T{V_{max}^2s in^2(w t)dt}

As sin² (wt)=1/2(1-cos(2wt), the integral of cos (2wt) cancels, and

V_{rms}^2=\frac{1}{2}V_{max}^2 ,

V_rms=V_max/√2

If the voltage is zero during half period, the integral is half than before (and the power is half, too):

V_{rms}^2=\frac{1}{T}\int_0^{T/2}{V_{max}^2 sin^2(w t)dt}=\frac{V_{max}^2}{4}

Vrms²(half-wave)=Vrms²(full wave)/2. The half wave produces half the power of the full wave, so the V_rms(half wave)=V_rms(full wave)/√2

ehild

 

[I like Serena]

My bad. Thanks ehild.
I edited my previous statement.

 

[I like Serena]

@Fp: Can you now draw the graph for Vs?
It should have a lower amplitude than Vp, and it should have a reversed phase.

 

[gneill]

A couple of comments.

Conventionally the turns ratio of a transformer is defined to be the ratio of secondary turns to primary turns, n = Ns/Np. Whether or not FP's k value follows this convention is something that she should check with her text or class notes.

Also conventionally, for AC circuits a voltage source is usually taken to be given in RMS volts unless otherwise specified. Again, FP should check to see what convention is being used in her class.

 

[Femme_physics]

I'm sorry I'm still quite confused with respect to these formula:

http://img7.imageshack.us/img7/2083/umaxyman.jpg

I'm really not sure which formulas to use (the one you provided or the ones provided to me) until I figure that little tidbit out. You seem to have edited something but I'm not sure what!

Conventionally the turns ratio of a transformer is defined to be the ratio of secondary turns to primary turns, n = Ns/Np. Whether or not FP's k value follows this convention is something that she should check with her text or class notes.

I'll check, thanks sent my lecturer email just now.

Also conventionally, for AC circuits a voltage source is usually taken to be given in RMS volts unless otherwise specified. Again, FP should check to see what convention is being used in her class.

Yes the value is given to us in rms. that's what my notes say.

 

[I like Serena]

You seem to have edited something but I'm not sure what!

You can see my original text in the quote that is contained in ehild's post.
I only changed 2 into √2 to make it true, and I adjusted the text to match.

I'm sorry I'm still quite confused with respect to these formula:

I'm really not sure which formulas to use (the one you provided or the ones provided to me) until I figure that little tidbit out.

I can imagine.
I can't make sense of these formulas either.
Can you provide some contextual information on what u(t) represents?
And what U(max) represents?

As far as I can tell, the only way these formulas can be true, is if these symbols are defined differently than what I'm expecting.

As I expect them, the top formula for U(av) should be zero, and the bottom formula for U(rms) is basically the same as the formulas ehild posted - just with a different result!

 

[I like Serena]

Hey Fp, goodmorning!

Do you need a better hint?

Since we have a k=2, which presumable transforms the primary voltage to a lower secondary voltage, the amplitude of Vs would be half of Vp.
Since Vs is reversed in phase with respect to Vp, Vs would be negative when Vp is positive and Vs would be positive when Vp is negative.

 

[Femme_physics]

Sorry it took me so long!


Well, I figured out when to use which formula, it all depends on the phase!

http://img560.imageshack.us/img560/5786/figuredit.jpg

And the Umax in the case of Vin

http://img812.imageshack.us/img812/1227/secondtryy.jpg [/QUOTE]

Should be

http://img641.imageshack.us/img641/8950/umaxxor.jpg

Voila! Got it. I'm on a roll!

Conventionally the turns ratio of a transformer is defined to be the ratio of secondary turns to primary turns, n = Ns/Np. Whether or not FP's k value follows this convention is something that she should check with her text or class notes.

Again, sorry it took me a while, but the answer I got was that it's the PRIMARY divided by the secondary. In fact, this is what "p" and "s" stand for (p = primary ; s = secondary)

Now, back to business

Yes, but this is only when the current in the primary circuit is flowing as you indicated.
It will still also flow the other way.

I don't see how it can flow the other way.

See?

http://img14.imageshack.us/img14/2126/bumpsc.jpg

 

[I like Serena]

Hi Fp!

Glad you got all the issues of this problem resolved.

So the max voltage is indeed 230 \sqrt 2.
It would be nice if you marked that in the graph for Vin, Vp and Vs.

And now we know that the voltage is indeed transformed down by a factor of 2.

I don't see how it can flow the other way.

See?

In the secondary circuit the current does indeed *bump* against the diode and won't flow.

However, the primary circuit does not have this problem, and the current will flow there.
And this will still induce a voltage in the secondary circuit, which will be counterbalanced by the diode, preventing the current to flow.

 

[Femme_physics]

Hi ILS! whew, refreshing to see you here again... <3 it's good to see engines are back to normal state, cap'n, even temporarily^^

It would be nice if you marked that in the graph for Vin, Vp and Vs.

Will do, as soon as I figure out how to draw them correctly

And now we know that the voltage is indeed transformed down by a factor of 2.

Yep!

In the secondary circuit the current does indeed *bump* against the diode and won't flow.

Once again I must

However, the primary circuit does not have this problem, and the current will flow there.
And this will still induce a voltage in the secondary circuit, which will be counterbalanced by the diode, preventing the current to flow.

Hold on, we're talking about Vs here. How does Vs plays a role? The way I see it, the current that flows in the first circuit, doesn't even touch the secondary coil, it just flows through the first coil. The only current that flows through Vs, is in the position where the diode allows it to flow. In reality, what I drew with the *bump* does not exist, since it's not a closed loop, so there is no current flowing and actually bumping the diode.

 

[I like Serena]

Hold on, we're talking about Vs here. How does Vs plays a role? The way I see it, the current that flows in the first circuit, doesn't even touch the secondary coil, it just flows through the first coil.

First things first, so you agree that the current flows in the primary circuit?



And no, the current in the primary circuit does not touch the secondary coil, it just flows through the primary coil.
However, the current does induce a magnetic field that does touch the secondary coil.
In turn in the secondary coil a voltage Vs is induced (across the secondary coil).

Here's a (schematical) picture:

The only current that flows through Vs, is in the position where the diode allows it to flow.

Current flowing through Vs?

Current does not flow through Vs.
Current flows through the secondary circuit.
Vs is the voltage in the secondary circuit.
A voltage difference across a resistor causes current to flow through the resistor.

In reality, what I drew with the *bump* does not exist, since it's not a closed loop, so there is no current flowing and actually bumping the diode.

Correct.

So the voltage difference across the resistors will be zero.

 

[Femme_physics]

First things first, so you agree that the current flows in the primary circuit?

Yes.

And no, the current in the primary circuit does not touch the secondary coil, it just flows through the primary coil.

I agree, that's what I thought!

However, the current does induce a magnetic field that does touch the secondary coil.
In turn in the secondary coil a voltage Vs is induced (across the secondary coil).

Duly noted, thanks for that!

Current flowing through Vs?

Current does not flow through Vs.
Current flows through the secondary circuit.
Vs is the voltage in the secondary circuit.

Oh. Ok, checked!

So, let me just take a look at the second "circuit".

http://img708.imageshack.us/img708/4882/vsss.jpg If this is correct then my graph for the wave should be correct, because when the current changes direction there is no flow. So there is indeed only half a phase. No?

So the voltage difference across the resistors will be zero.

No it won't. There is still current flowing in the other direction! So there are voltage differences across the resistors. Or "voltage drop" as we know it.

 

[I like Serena]

So, let me just take a look at the second "circuit".


If this is correct then my graph for the wave should be correct, because when the current changes direction there is no flow. So there is indeed only half a phase. No?

You seem to be mixing up voltage and current.
You're concentrating on the current and what you have drawn is indeed the graph for the current.
The graph for the voltage Vs across the coil is different.

Look at it this way.
Suppose we replace the secondary coil by a battery giving off a voltage.
If the plus pole is on the bottom, the current will *bump* against the diode and no current will flow.
The battery still gives off its voltage though.

No it won't. There is still current flowing in the other direction! So there are voltage differences across the resistors. Or "voltage drop" as we know it.

What I meant is that there are 2 states for the voltage Vs: it's positive or it's negative.
When it is positive there is indeed a voltage difference across the resistors and as you say current will flow.
When it is negative, no current will flow, so there won't be a voltage across the resistors at that time.

 

[I like Serena]

Here's an analogy with mechanics.

Suppose you have a sliding block on a slope.
When we jiggle the slope back an forth, the block will alternately slide left and right.
However, if you insert a stick in the slope on the right side of the block, it won't slide to the right, but there is still a slope.

What's the "current" in this analogy?
And what is the "voltage"?

 

[Femme_physics]

You seem to be mixing up voltage and current.
You're concentrating on the current and what you have drawn is indeed the graph for the current.
The graph for the voltage Vs across the coil is different.

Look at it this way.
Suppose we replace the secondary coil by a battery giving off a voltage.
If the plus pole is on the bottom, the current will *bump* against the diode and no current will flow.
The battery still gives off its voltage though.

Good point, so Vs is a sinus wave!

What's its Umax though...hmm... that has to be calculated.

Actually, I'd presume it's also 230V since the first circuit doesn't have any resistors. Could it be?

What I meant is that there are 2 states for the voltage Vs: it's positive or it's negative.
When it is positive there is indeed a voltage difference across the resistors and as you say current will flow.
When it is negative, no current will flow, so there won't be a voltage across the resistors at that time.

Oh, OK, agreed, makes sense

Here's an analogy with mechanics.

Suppose you have a sliding block on a slope.
When we jiggle the slope back an forth, the block will alternately slide left and right.
However, if you insert a stick in the slope on the right side of the block, it won't slide to the right, but there is still a slope.

What's the "current" in this analogy?
And what is the "voltage"?

voltage is the jiggle, block is current, I think. But I really understood what you mean with your post above it! Thanks. I'll work on the solution of the thing soon.

 

[I like Serena]

Good point, so Vs is a sinus wave!

What's its Umax though...hmm... that has to be calculated.

Actually, I'd presume it's also 230V since the first circuit doesn't have any resistors. Could it be?

Umax is not 230 V. You're talking about Urms here.

And you're forgetting that the secondary coil (where we have Vs) has less windings, causing the voltage to be lower.

voltage is the jiggle, block is current, I think. But I really understood what you mean with your post above it! Thanks. I'll work on the solution of the thing soon.

Good. :)

A little sharper:

The block is the electron.
The current is the speed of the block.

The voltage difference is the difference in height between two points on the slope (left and right of the block).

 

[Femme_physics]

Top of the morning ta ye ILS :)

Umax is not 230 V. You're talking about Urms here.

Oops, you're right *smackey the foreheady*

And you're forgetting that the secondary coil (where we have Vs) has less windings, causing the voltage to be lower.

Oops, you're also right! So since k =2

I'd do

230 / 2 = 115

So Vs(max) = 115(square root of 2)

Aye, cap'n?

Good. :)

A little sharper:

The block is the electron.
The current is the speed of the block.

The voltage difference is the difference in height between two points on the slope (left and right of the block).

Woah. Ok, mmm, as long as mechanics analogy won't be in the test, we're safe


Ok, so now I figured

Vin, Vp an Vs. Now I need Vxy

Which according to my logic

Since Vs = 115V

Vyx also equals 115V, because if we treat the diode as an ideal diode and ignore the voltage drop there, using Kirchhoff law we know that the voltage drops in the resistors (RT) must equals the voltage source. Ergo,

Vs = Vyx


*does a suspicious victory dance waiting for approval*

 

[I like Serena]

Top of the morning ta ye ILS :)

Hi Fp!

So Vs(max) = 115(square root of 2)

Aye, cap'n?

Yep!

Ok, so now I figured

Vin, Vp an Vs. Now I need Vxy

Which according to my logic

Since Vs = 115V

Vyx also equals 115V, because if we treat the diode as an ideal diode and ignore the voltage drop there, using Kirchhoff law we know that the voltage drops in the resistors (RT) must equals the voltage source. Ergo,

Vs = Vyx


*does a suspicious victory dance waiting for approval*

Not quite.
You have to distinguish the two cases, where Vs is positive, and where Vs is negative.
When Vs is positive you are right.
When Vs is negative, it is different.

 

[Femme_physics]

Hi Fp!

Yep!

Did you edit it by any chance?

Not quite.
You have to distinguish the two cases, where Vs is positive, and where Vs is negative.
When Vs is positive you are right.
When Vs is negative, it is different.

Ah, well, that's why we have graphs

http://img683.imageshack.us/img683/1575/vxywt.jpg

Did I get it?

 

[I like Serena]

Yep to all!

 

[Femme_physics]

http://img194.imageshack.us/img194/7199/indiet.jpg

Are these graphs correct? (my dashed lines really mean no line)

 

[gneill]

Hi FP, you may wish to think about the relative phase of the primary and secondary waveforms; how do the transformer 'dots' affect the phase?

While this phase difference won't make any practical difference in the operation of this simple circuit, it's worthwhile making note of it since you're drawing Voltage vs Time graphs.

 

[Femme_physics]

Hi FP, you may wish to think about the relative phase of the primary and secondary waveforms; how do the transformer 'dots' affect the phase?

Well it reverses the phase on Vs. But I don't know how to show the difference in a graph though since Vp is both directions anyway, and Vs is one direction.

I hope I understood what you meant.

 

[gneill]

Well it reverses the phase on Vs. But I don't know how to show the difference in a graph though since Vp is both directions anyway, and Vs is one direction.

I hope I understood what you meant.

The difference shows up as a reversal of polarity in the secondary voltage waveform with respect to the primary's waveform. This is equivalent to a half wavelength shift in time for the secondary voltage waveform. It boils down to shifting your graph's peaks over by a half wavelength (either direction will do!).

Note that this is only necessary if you are showing voltages in both the primary and secondary circuits on the same time axes.

All your analysis conclusions still apply, only the time-wise locations of the peaks are shifted over.

 

[Femme_physics]

The difference shows up as a reversal of polarity in the secondary voltage waveform with respect to the primary's waveform. This is equivalent to a half wavelength shift in time for the secondary voltage waveform. It boils down to shifting your graph's peaks over by a half wavelength (either direction will do!).

Note that this is only necessary if you are showing voltages in both the primary and secondary circuits on the same time axes.


All your analysis conclusions still apply, only the time-wise locations of the peaks are shifted over.

Ah, makes perfect sense I get it now.

All your analysis conclusions still apply, only the time-wise locations of the peaks are shifted over.

I'll fix it then in my next scan this evening or tomorrow. Thanks

 

[I like Serena]

Morning Fp!

Looking good!
I see you have most graphs now. Nicely labeled with the voltage.
Still Vs is along the dotted line, while Vxy is without the dotted line.

Btw, you have to consider the 2 half periods separately.

In one half your current and power are correct.

But in the other half they are zero.