Help with this question please I'm at a crossroads. (1 Viewer)

[BoardofBoards1]

I just came across this question from CSSA I believe, and I was stuck between the answers A and B.

Most people are saying it's B, however, according this website (2nd image) of the graph relating the primary applied voltage to the inductor/secondary coil's current, and the fact that Ohm's law or V=IR (only for simple resistor circuits) cannot be applied to induction circuits, means that the cosine-like (I) curve on the website cannot be linearly translated into a secondary voltage curve, as shown by option B.

I also found out that for induction circuits as such transformers, in an ideal system, the magnetic flux is proportional to current, and by V or EMF = -nd(phi)/dt, V is proportional to d(current)/dt. So either I have developed some misconception about this, or there are some other factors at play here? It would be really helpful if someone could clarify this, thanks a bunch.

 

[qweeosh]

I just came across this question from CSSA I believe, and I was stuck between the answers A and B.

Most people are saying it's B, however, according this website (2nd image) of the graph relating the primary applied voltage to the inductor/secondary coil's current, and the fact that Ohm's law or V=IR (only for simple resistor circuits) cannot be applied to induction circuits, means that the cosine-like (I) curve on the website cannot be linearly translated into a secondary voltage curve, as shown by option B.

I also found out that for induction circuits as such transformers, in an ideal system, the magnetic flux is proportional to current, and by V or EMF = -nd(phi)/dt, V is proportional to d(current)/dt. So either I have developed some misconception about this, or there are some other factors at play here? It would be really helpful if someone could clarify this, thanks a bunch.
View attachment 50354View attachment 50355

Even using your textbook, isn't the answer still B?

 

[coolcat6778]

You need calculus and differentiation for this
Start by finding n from T = 2pi/a
this is a good novel question that requires calculus
I'll show the working out just wait

 

[cheesynooby]

im confused why you don't think it's B? the answers have voltage on the y axis, not current, and B seems to follow what the website says?

 

[C2H6O]

You need calculus and differentiation for this
Start by finding n from T = 2pi/a
this is a good novel question that requires calculus
I'll show the working out just wait

agree but “calculus” as in looking at where V is changing (sloped) and where it’s not changing (at the peaks), and emf only when it’s changing. Calculation would be overkill for this question, and obv just get amplitude using the ratio of turns 4:1 so magnitude of emf is a quarter the max V so B

 

[BoardofBoards1]

I'm just still a little confused because b) on the website says inductor current/milliamps in the y-axis and I just thought the relationship between Vp and Vs was linear so the output voltage would look something like this

 

[coolcat6778]

It's A right?
This is because
v secondary = 0.25 V
this occurs at t = 0.25

If it's A this question makes sense

 

[Trial&Error]

When the alternating current in the primary coil changes, it produces a changing magnetic flux in the iron core. That changing flux induces a voltage in the secondary coil according to Faraday’s Law:

V=−N delta(Φ)/delta(t)

That negative tells us the induced voltage opposes the change in flux (Lenz’s Law).

So, when the primary voltage increases positively (increasing current and flux), the secondary voltage is induced in the opposite direction.

Basically look at the gradient at point 0 on the original graph. It’s positive. But when you put it in the formula it becomes negative due to the -N. If you did that infinitely many times for each point on the original curve, you’d get a curve like option B. And the reason why the amplitude is different is because of Vp/Vs = Np/Ns and the transforming is reducing voltage by 1/4.

So each point on the curve is 1) reduced by 1/4 and 2) graphed as the derivative of the original graph.

I hope this makes sense.

 

[coolcat6778]

When the alternating current in the primary coil changes, it produces a changing magnetic flux in the iron core. That changing flux induces a voltage in the secondary coil according to Faraday’s Law:

V=−N delta(Φ)/delta(t)

That negative tells us the induced voltage opposes the change in flux (Lenz’s Law).

So, when the primary voltage increases positively (increasing current and flux), the secondary voltage is induced in the opposite direction.

Basically look at the gradient at point 0 on the original graph. It’s positive. But when you put it in the formula it becomes negative due to the -N. If you did that infinitely many times for each point on the original curve, you’d get a curve like option B. And the reason why the amplitude is different is because of Vp/Vs = Np/Ns and the transforming is reducing voltage by 1/4.

So each point on the curve is 1) reduced by 1/4 and 2) graphed as the derivative of the original graph.

I hope this makes sense.

wtf you actually needed calc knowledge for this? at this point they should've just added calculus into physics.

everything we do in mod 5 and 6 requires calculus, indirectly or directly. indirectly is the work as another example

 

[Trial&Error]

wtf you actually needed calc knowledge for this?

Not necessarily, no. It just makes it easier to explain.

Otherwise you just look at the formula and say that between point (0,0) and (0.25,1) on the original graph delta(Φ) is positive so when you put it in the equation, the -N makes it negative. Do that for each interval. And know that the change is 0 at all turning points.

 

[BoardofBoards1]

When the alternating current in the primary coil changes, it produces a changing magnetic flux in the iron core. That changing flux induces a voltage in the secondary coil according to Faraday’s Law:

V=−N delta(Φ)/delta(t)

That negative tells us the induced voltage opposes the change in flux (Lenz’s Law).

So, when the primary voltage increases positively (increasing current and flux), the secondary voltage is induced in the opposite direction.

Basically look at the gradient at point 0 on the original graph. It’s positive. But when you put it in the formula it becomes negative due to the -N. If you did that infinitely many times for each point on the original curve, you’d get a curve like option B. And the reason why the amplitude is different is because of Vp/Vs = Np/Ns and the transforming is reducing voltage by 1/4.

So each point on the curve is 1) reduced by 1/4 and 2) graphed as the derivative of the original graph.

I hope this makes sense.

Really thank you for the reply, I understand the amplitude and the negation in the formula that all makes sense now, but I don't understand the significance of the derivative of the original voltage curve, I thought by this formula V=−N delta(Φ)/delta(t), one would only yield a derivative curve only if the original curve was a magnetic flux-time curve not primary voltage curve. And I also saw many sources show curves such as this. Thank you for the support eitherway.

 

[Trial&Error]

Really thank you for the reply, I understand the amplitude and the negation in the formula that all makes sense now, but I don't understand the significance of the derivative of the original voltage curve, I thought by this formula V=−N delta(Φ)/delta(t), one would only yield a derivative curve only if the original curve was a magnetic flux-time curve not primary voltage curve. And I also saw many sources show curves such as this. Thank you for the support eitherway.View attachment 50370

I think I see where you’re confused.

The important part is to first understand that this relationship is unique for transformers only since the magnetic flux passing through the core at every instant (not overall) links both coils, and thus, delta(Φ)/delta(t) for both coils is the same. The only thing which changes the voltage is N (the number of coils). This is the fundamental mechanics of a transformer system and is what allows it to follow the conservation of energy.

Thus:
Vp=−N delta(Φ)/delta(t)
Vs = -N delta(Φ)/delta(t)

where delta(Φ)/delta(t) is the same for both of them.

That’s how we derive the formula Vs/Vp = Ns/Np

So that’s why, even though the original relationship is about flux, we can use the ratio for voltages at any instant in time.

But this all is probably not required for hsc Phsyics, it just makes it better to understand.

 

[BoardofBoards1]

I think I see where you’re confused.

The important part is to first understand that this relationship is unique for transformers only since the magnetic flux passing through the core at every instant (not overall) links both coils, and thus, delta(Φ)/delta(t) for both coils is the same. The only thing which changes the voltage is N (the number of coils). This is the fundamental mechanics of a transformer system and is what allows it to follow the conservation of energy.

Thus:
Vp=−N delta(Φ)/delta(t)
Vs = -N delta(Φ)/delta(t)

where delta(Φ)/delta(t) is the same for both of them.

That’s how we derive the formula Vs/Vp = Ns/Np

So that’s why, even though the original relationship is about flux, we can use the ratio for voltages at any instant in time.

But this all is probably not required for hsc Phsyics, it just makes it better to understand.

Sorry, I don't mean to be argumentative. What you said about the instantaneous change in magnetic flux across both primary and secondary inductors is correct, however, doesn't this equivalence imply that Vs ∝Vp, and therefore, how is the secondary voltage the derivative of the primary voltage? And from the aforementioned colored image, this proportionality seems to hold graphically.

 

[C2H6O]

I just came across this question from CSSA I believe, and I was stuck between the answers A and B.

Most people are saying it's B, however, according this website (2nd image) of the graph relating the primary applied voltage to the inductor/secondary coil's current, and the fact that Ohm's law or V=IR (only for simple resistor circuits) cannot be applied to induction circuits, means that the cosine-like (I) curve on the website cannot be linearly translated into a secondary voltage curve, as shown by option B.

I also found out that for induction circuits as such transformers, in an ideal system, the magnetic flux is proportional to current, and by V or EMF = -nd(phi)/dt, V is proportional to d(current)/dt. So either I have developed some misconception about this, or there are some other factors at play here? It would be really helpful if someone could clarify this, thanks a bunch.
View attachment 50354View attachment 50355

Just flagging everyone here, after doing some research I have reasonable doubt on B, will dig further into this but something to do with self inductance (i think thats what the term is)

 

[C2H6O]

I just came across this question from CSSA I believe, and I was stuck between the answers A and B.

Most people are saying it's B, however, according this website (2nd image) of the graph relating the primary applied voltage to the inductor/secondary coil's current, and the fact that Ohm's law or V=IR (only for simple resistor circuits) cannot be applied to induction circuits, means that the cosine-like (I) curve on the website cannot be linearly translated into a secondary voltage curve, as shown by option B.

I also found out that for induction circuits as such transformers, in an ideal system, the magnetic flux is proportional to current, and by V or EMF = -nd(phi)/dt, V is proportional to d(current)/dt. So either I have developed some misconception about this, or there are some other factors at play here? It would be really helpful if someone could clarify this, thanks a bunch.
View attachment 50354View attachment 50355

I'm assuming you dont have answers? If you can confirm if it was CSSA what year?

 

[BoardofBoards1]

I'm assuming you dont have answers? If you can confirm if it was CSSA what year?

Yes thank you, I appreciate all the replies I recieved, however, the problem I still have is that the general statement "the secondary voltage is the first derivative of the primary voltage" I feel is kind of unheard of.

 

[BoardofBoards1]

I'm assuming you dont have answers? If you can confirm if it was CSSA what year?

The person who uploaded the question said "CSSA said it was B" and I'm not sure if I could completely take his word for it, but I could message him about the question that may require a bit of time.

 

[C2H6O]

The person who uploaded the question said "CSSA said it was B" and I'm not sure if I could completely take his word for it, but I could message him about the question that may require a bit of time.

I found solutions and they say B

 

[C2H6O]

Best explanation from my very limited understanding and 1 hour of chat gpt and googling (DISCLAIMER I AM MAKING THIS UP AS I GO BC I DONT KNOW ALL THIS ELECTRICAL ENGINEERING TERMINOLOGY):

WARNING: DO NOT READ IF YOU DO NOT WANT TO HAVE A DILEMMA IN HSC BETWEEN "IS THIS A TRICK QUESTION (A) OR DO I APPLY NORMAL IN SYLLABUS KNOWLEDGE (B)?", TRUST ME NOW IM FREAKING OUT

Spoiler: READ WITH CAUTION, THERE IS NO TURNING BACK, YOU CANNOT UNKNOW SOMETHING ONCE YOU KNOW IT

SELF INDUCTANCE: basically back emf applied by a solenoid TO ITSELF
Same thing as with normal back emf net voltage is supply voltage - emf
EMF is out of phase by 90 degrees
So the voltage thats actually driving primary is not the supply voltage but the voltage shifted 90 degrees by the self induced EMF
Then this out of phase voltage induces the emf in secondary which is out of phase by another 90 degrees
because direction doesn't matter too much cause it depends how you wire the voltmeter, this can be treated as no phase shift (just flip the graph across x axis)

Thus secondary voltage is IN PHASE with primary voltage

Based on the fact that solutions said B, I assume the effects of self inductance are OOS and should not be accounted for.

DISCLAIMER AGAIN THIS DOES NOT YET MAKE 100% SENSE TO ME AND I FORMED THIS CONCLUSION FROM BRIEF RESEARCH AND IT IS LIKELY AT LEAST SLIGHTLY INACCURATE

As a company such as CSSA has not accounted for this, the answer at an hsc level is B @BoardofBoards1

 

[C2H6O]

Ffs so now what do we put if this comes up
How do we know if its testing syllabus knowledge or an oos trick question? I HATE THIS OVERSIMPLIFIED SYLLABUS
I will censor explanation so yall read at your own risk