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2015 HSC Q18 Physics HSC ??? (1 Viewer)

DinShmiesal

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I was just doing past papers for the physics test on Monday and I am confused about the 18th multiple choice question for the 2015 physics test. According to the marking guidelines the solution is A. How did students doing the test know if the galvanometer is measuring current or voltage? Or is it implied in some way by the diagram?
If measuring current wouldn't it be B ???

Thanks for any replies, sorry if the question is stupid I may just be having a mental blank, my brain is cooked after the maths test yesterday.

Test is here
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
 

kashkow

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I think general consensus is that question was wrong. I don't really understand how that mistake made it through (unless it was a typographical error on the published mg) since it seems pretty obvious that it is B.

Anyway here's two links concerning your question: http://community.boredofstudies.org/264/motors-generators/350684/mc-question.html#post7161469

https://www.youtube.com/watch?v=KM1XFOCzKDE

The first is the general consensus from bored of studies. The second is an "Explanation" from a HSC teacher, but his explanation doesn't really make much sense - he may be just trying to find reasons for why it is A. I'll leave you with these links to decide for yourself... I'm sorry I can't really help because I don't know which is right, though if this was given to me in 2015 I'd probably pick B.
 

quarkquark

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See precisioninfo comment in relation to this video:

https://www.youtube.com/watch?v=KM1XFOCzKDE&feature=youtu.be

The comment reads as follows:

Responses to Question 18
In terms of the HSC theory and step-up / step-down the correct answer is (b) - as the narrator initially identifies. Anything beyond that seems beyond the scope of the HSC course. In reality, the secondary current is determined by the input power (voltage and current), the ratio of the windings and, importantly, the load on the transformer. Hence, with effectively no load, the point of the question seems moot. The “correct answer” is put to put a meter on the unloaded transformer and see with a given transformer.

In attempting to justify returning an answer of (A) the narrator experiences a number of difficulties and argues inconsistently with the requirements of an ideal transformer. The narrator is undoubtedly correct in that this is not the best question (nor answer) and consequently justifications are not readily available.

Nevertheless, the application of Ohm’s law to the input coil here is inadmissible. Ohm’s Law is a Law of Conduction in name only. It is not a law that is embedded in a mechanism or a deeper theory of guiding principles such as Newton’s Laws of motion.

Ohm’s law is a statement of observation of the behaviour of a thermal resistor over the range of (constant input) voltages for which resistance is constant. To achieve anything approximating Ohms law over a range of applied (DC) voltages it is typically necessary to control other variables including temperature (using water baths etc).

Ohm’s law does not apply in semi-conduction or super-conduction.

Ohm’s law does not apply in any cases where current is limited by some (external) factor(s).

Ohm’s law does not apply in high frequency AC applications such as operational motors, generators and transformers. In all of these cases Back EMF and or inductive loads come into play.

To claim, via Ohm’s law, that by removing coils from an input loop, effective input current increases in an “Ideal” transformer, misconstrues the facts that “ideal” transformers cannot have (thermal) resistance. Secondly, removing loops in the input wire does not require shortening it (as the video implies in claiming reduced resistance on decreasing the number of turns). Particularly, in the case where the transformer is stated to be ideal then the length of the wires should make no difference as thermal resistance would be zero.

The only way to develop transformers problems even close to properly in the HSC is to start from the premise that Faraday’s Law is written in terms of magnetic flux and voltage, use the argument that changes in magnet flux will induce a voltage from which current may then follow (depending on circuit resistance). Further, transformers have a range of operational input voltages and currents for which they are optimised. It is only at these voltages for which they are optimised that highly efficient voltage and current transformations will be achieved.

A transformer (such as a computer charger) can be left plugged in with no load of it for days and consume very little power (and nor will it get particularly hot). Nevertheless, the transformer will get quite warm when a battery is plugged in for charge. Ohm’s Law cannot explain this situation as it would predict massive currents in a loop regardless of whether or not there is a transformer core there.

In an unloaded transformer input AC current is limited by non-ohmic factors (principally back EMF). As the load on the non-input (output) side of the transformer increase the transformer will carry more current and will be observed to heat up since the loops are then carrying more current.

Similarly, using Ohms law, it is not possible to explain why a fast running motor draws little current; but if you slow it down (by putting a load on it) then current draw increases dramatically. Other major factors are at play.

Capture 2015Q18.jpg
 
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pikachu975

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I'm just curious if anyone did the 2015 HSC and bought their raw marks, and found out they put B (the correct answer) but the HSC marked it wrong with the right answer A.

Would be unfair to miss out on a band 6 by getting 89 because BOSTES got a question wrong.
 

quarkquark

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Various further feed-back has been received in relation to this question. All of the feedback supports the view that appropriate analysis of this question is beyond the scope of the HSC Course and further that the realm of correct analysis is not specifically within (HSC) transformer theory.

In the normal course of events transformers are supplied AC power - and while they will respond to interrupted DC (& students may know this) it is Induction Coils, not transformers, that are the primary users of interrupted DC power.

Advanced theory on AC and Inductance, is, as is stated in the Blog post, not within the scope of the HSC Course. It may lead to different interpretations and also inconsistencies with what students would know.

In transformers the issue of load is vital. To proceed toward answer A requires consideration of current induced in linked solenoids. Inductive loads were referred to in the original blog post and induced currents in linked solenoids are quantified by Ampere's Law and self-inductance.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.html

In particular, Inductance, arising from Faradays law, is the phenomenon whereby a coil of wire will resist a change in electric current flowing within.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/induct.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html#c2

Self inductance, L, limits the rate of change of the current. This in-turn determines the rate of change of the magnetic flux (at the secondary). The turns (ratio) determines the emf and current in the secondary. On this basis the scope of possible answers extends beyond B.

Quantification or application of both Ampere's Law and self-inductance in the manner required to approach this question is beyond the scope of the HSC course. It is for reasons including these that the original post referred to a range of voltages (& currents) for which transformers are optimised and also stated:

"Anything beyond that (B) seems beyond the scope of the HSC course. In reality, the secondary current is determined by the input power (voltage and current), the ratio of the windings and, importantly, the load on the transformer. Hence, with effectively no load, the point of the question seems moot. The “correct answer” is put to put a meter on the unloaded transformer and see with a given transformer." Testing in this way will yield device specific results for particular voltages and currents.
 

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