# transformers (1 Viewer)

#### kractus

##### New Member

How would you even determine the graph of the emf v t just based on transformers?? The answer is C

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following information help?

When the current is turned on in X, at the start a magnetic field is induced to oppose the increase in voltage, and therefore generate a back EMF, however it will eventually reach a constant voltage as it is a DC circuit and there will eventually be no magnetic field. A similar field will be created to oppose the voltage change when the switch is turned off but it will be in the opposite direction. Because there is only a changing magnetic field when the switch is turned off and on and not while it is operating at full voltage, a current will only be induced in coil Y when this field is changing, so therefore only when the switch at X is turned on and off. The induced voltage in Y when the switch is turned on will be in the opposite direction as that induced when the switch is turned off, so it would be C.

#### edwxnsamuel

##### Member
The graphs for C can be explained as follows:

LHS graph: since it is a DC power supply, when the switch is closed, there will be a rise in voltage which plateaus at a maximum value until it returns to when the switch is reopened. The voltage would not oscillate around the 0V mark, since it is a DC power source (from the diagram) not an AC power source.

RHS graph: There are 2 momentary changes in flux that is experienced by the secondary coil; these instances being when the switched is closed then opened. When the switch is closed, there is a current flowing through the primary coil which produces its own magnetic field, thus there is a change in flux experienced by the secondary coil. However, as it is a DC power source, when the voltage plateaus, there is no longer a changing magnetic field produced by the primary coil, so there is no EMF or current induced in the secondary coil by Faraday's Law (as dflux = 0). When the switch is reopened, the magnetic field of the primary coil drops to 0, so that change in flux induces an EMF in the secondary coil. They will peak in opposite directions because the net change in flux is in 'opposite directions' so the opposition to this change in flux (Lenz's Law) will also be in opposite directions

#### wizzkids

##### Member
Geeze, I hate these questions that are written by supposedly experienced HSC teachers, who have no specialist knowledge of electricity.
Yes the best answer is (C) but the graphs for the primary emf and secondary emf are wrong.
The actual graphs for emf would look like this:

When the switch is closed, the data logger will be connected to the applied DC voltage, through conductors that have negligible series resistance, so the applied voltage will rise pretty much instantaneously to the maximum DC value. When the switch suddenly opens, the collapsing magnetic field will induce a large back e.m.f. in the primary winding and the secondary winding. Anyone who has ever done this in a Physics lab will have seen the spark that is induced at the switch when the voltage spike appears across the contacts of the switch. This transient is actually utilised in internal combustion engines. The Kettering ignition system uses a coil and switched DC current to create a very high voltage spark to ignite the fuel/air mixture.
I don't like unrealistic problems being given to students in Stage 6, if they don't have the necessary Physics and Mathematics to understand fully what is going on.
To properly understand the behaviour of an inductor that is subject to a step-change in applied emf, we need 2nd Year university AC Circuit Theory for current and e.m.f. in an inductor. You would also need more data, such as the series resistance R of the primary winding.
(end of rant)

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#### edwxnsamuel

##### Member
Geeze, I hate these questions that are written by supposedly experienced HSC teachers, who have no specialist knowledge of electricity.
Yes the best answer is (C) but the graphs for the primary emf and secondary emf are wrong.
The actual graphs for emf would look like this:
View attachment 37582
When the switch is closed, the data logger will be connected to the applied DC voltage, through conductors that have negligible series resistance, so the applied voltage will rise pretty much instantaneously to the maximum DC value. When the switch suddenly opens, the collapsing magnetic field will induce a large back e.m.f. in the primary winding. Anyone who has ever done this in a Physics lab will have seen the spark that is induced at the switch when the voltage spike appears in the primary winding. This transient is actually utilised in internal combustion engines. The Kettering ignition system uses a coil and switching DC voltage to create a very high voltage spark to ignite the fuel/air mixture.
I don't like unrealistic problems being given to students in Stage 6, if they don't have the necessary Physics and Mathematics to understand fully what is going on.
To properly understand the behaviour of an inductor that is subject to a step-change in applied emf, we need 2nd Year university AC Circuit Theory for current and e.m.f. in an inductor. You would also need more data, such as the series resistance R of the primary winding.
(end of rant)
The diagram you showed for the LHS v/t graph is pretty unrealistic for a Year 12 student to recognise because the ending portion of it will not be something that they are familiar with. The content given in the question contain assumptions that are taught to students (as assumptions which are clearly made known at times) within the syllabus itself and does not go out of the scope of the syllabus. Both graphs in the question are pretty stock-standard HSC level Module 6 questions because a lot of the mathematical theory behind Mod 6 formulae are taken for granted and students are just expected to memorise them for what they are. Surprisingly, I'd say the new syllabus is much more rigorous than the old one, especially with the removal of electives, and the module 6 content is definitely the most conceptually challenging in the syllabus. Yes, it's unfortunate that students need to take some things for how it is taught (and force to memory) but that's the reality of introducing a calculation heavy module in a course that isn't allowed to have a minimum restriction on the level of mathematics taken by candidates (i.e. integration, etc is not assumed knowledge so it cannot be used in any question or concept within HSC Physics).

At the end of all of this, don't get me the wrong way, I completely agree with where you're coming from actually; but it's the reality of the HSC syllabus without higher level math unfortunately.

#### wizzkids

##### Member
The diagram you showed for the LHS v/t graph is pretty unrealistic for a Year 12 student to recognise because the ending portion of it will not be something that they are familiar with. The content given in the question contain assumptions that are taught to students (as assumptions which are clearly made known at times) within the syllabus itself and does not go out of the scope of the syllabus. Both graphs in the question are pretty stock-standard HSC level Module 6 questions because a lot of the mathematical theory behind Mod 6 formulae are taken for granted and students are just expected to memorise them for what they are. Surprisingly, I'd say the new syllabus is much more rigorous than the old one, especially with the removal of electives, and the module 6 content is definitely the most conceptually challenging in the syllabus. Yes, it's unfortunate that students need to take some things for how it is taught (and force to memory) but that's the reality of introducing a calculation heavy module in a course that isn't allowed to have a minimum restriction on the level of mathematics taken by candidates (i.e. integration, etc is not assumed knowledge so it cannot be used in any question or concept within HSC Physics).

At the end of all of this, don't get me the wrong way, I completely agree with where you're coming from actually; but it's the reality of the HSC syllabus without higher level math unfortunately.
I take your points about taking much of the mathematical theory of electromagnetism for granted in the HSC Physics syllabus. However you can teach everything required by the syllabus in transformers using well-behaved continuous functions like sine or cosine. There is no need to introduce transient step-functions and other discontinuous functions that complicate the mathematics, in my opinion. The syllabus doesn't require it, and it should not be examinable at high school level. The other problem with erroneous marking schemes like the one above is that less experienced Science teachers will be misled into believing that's how self-induction actually works, which is very dangerous. That's my 2-cents worth.

#### edwxnsamuel

##### Member
I take your points about taking much of the mathematical theory of electromagnetism for granted in the HSC Physics syllabus. However you can teach everything required by the syllabus in transformers using well-behaved continuous functions like sine or cosine. There is no need to introduce transient step-functions and other discontinuous functions that complicate the mathematics, in my opinion. The syllabus doesn't require it, and it should not be examinable at high school level. The other problem with erroneous marking schemes like the one above is that less experienced Science teachers will be misled into believing that's how self-induction actually works, which is very dangerous. That's my 2-cents worth.
yep i agree with the end half, however, overall its a recipe for disaster since Standard Mathematics students are not aware of the cosine and sine function graphs as it is, so honestly, how simple can we really make Mod 6 to cater for this. but definitely i agree with everything you said, it's overall very problematic haha

#### kractus

##### New Member
Thank you! this rlly helped and now my brain isn't fried