HSC Physics Question Thread

[lazydreamer]

Hey Castform:

Here is your solution to the third question. The trick with this part is remembering to use v = u +at. This formula is not limited only to projectile motions, it can be applied versatilely across all physics calculations. In this case, because initially the rocket doesnt have any velocity, hence u=0 m/s. IF this question wanted us to consider the relativistic effects as well, it would be harder because relativistic effects of length contraction and time dilation would be different at each stage of our voyage, since velocity is gradually increasing and this has an impact upon the lorentz factor.

(Image removed from quote.)

Best Regards
Happy Physics Land

Hey HPL, was looking at your solution and in the section of 'Implementing v=u+at', i've noticed you used acceleration due to gravity as 'a' instead of the one calculated from the g-force.
Just making sure haha, your solutions are otherwise great, esp. that handwriting, so pretty

[Happy Physics Land]

Hey HPL, was looking at your solution and in the section of 'Implementing v=u+at', i've noticed you used acceleration due to gravity as 'a' instead of the one calculated from the g-force.
Just making sure haha, your solutions are otherwise great, esp. that handwriting, so pretty

Hey Lazydreamer!

Thanks for pointing out that fatally careless error! Yes it was indeed 19.6 not 9.8. This is an exemplar of what happens when you go on late night practise hahahaha. Thanks for your kind words regarding my handwriting as well, appreciate it!

[Neutron]

Hey guys!

In induction cooktops, you know how the resistance in the base of the saucepan produces the heat required to heat the food? I was wondering which equation this was from? Is it the P=I²R? So in this case, would the induced eddy currents also assist in producing heat? If this was the case, wouldn't having low resistance be better since it produces higher eddy currents and since the heat produced is proportional to current squared, it would have more effect?

Also, in transformers, in the power equation P=IV, current and voltage are essentially inversely proportional however in ohm's law V=IR, they are directly proportional so when do we use which relationship? Thanks

Neutron

[Happy Physics Land]

Hey guys!

In induction cooktops, you know how the resistance in the base of the saucepan produces the heat required to heat the food? I was wondering which equation this was from? Is it the P=I²R? So in this case, would the induced eddy currents also assist in producing heat? If this was the case, wouldn't having low resistance be better since it produces higher eddy currents and since the heat produced is proportional to current squared, it would have more effect?

Also, in transformers, in the power equation P=IV, current and voltage are essentially inversely proportional however in ohm's law V=IR, they are directly proportional so when do we use which relationship? Thanks

Neutron

Hey neutron!

Okay great questions, I especially like the second one because I was stuck on that question for a period of time as well. In regards to the induction cooktop, heat is generated when you have a current going through a high resistance conductor. Just like you stated, P = I²R, if we just have a look at the operation of induction cooktop quantatively, then we can see that Power (in the form of heat) increases as current increases and as resistance increases. If we think about this qualitatively, it makes sense as well. If we have higher current (i.e. more charges) travelling through a highly resistant conductor, then there is a higher possibility that the charges will collide with the conductor's crystal lattice, and this microscopic collision would produce heat as a result of friction experienced by these charges as they travel through the conductor. Remember however, when I say high resistance, I mean a conductor that has high internal resistance, relative to the insulators, they would still have low resistance. If a material has TOO high of an internal resistance, then eddy current wouldnt be able to pass through and we get no heat ... of course!

In regards to ohm's law and P=VI, they do seem contradictory. But if we think about it, in an ideal transformer, the POWER is the same in both the primary and secondary coil, not the RESISTANCE! Similarly in electricity transmission, we are dealing with POWER loss, and hence we use P=VI to determine that V and I are inversely proportional. If the situation doesnt deal with power, but with resistance, then ofc according to ohm's law I would be directly proportional to V. Tricky stuff there!

Thanks for asking that question buddy, if you have any further problems dont hesitate to ask!

Best Regards
Happy Physics Land

[jamonwindeyer]

Hey guys!

In induction cooktops, you know how the resistance in the base of the saucepan produces the heat required to heat the food? I was wondering which equation this was from? Is it the P=I²R? So in this case, would the induced eddy currents also assist in producing heat? If this was the case, wouldn't having low resistance be better since it produces higher eddy currents and since the heat produced is proportional to current squared, it would have more effect?

Also, in transformers, in the power equation P=IV, current and voltage are essentially inversely proportional however in ohm's law V=IR, they are directly proportional so when do we use which relationship? Thanks

Neutron

Hey Neutron! HPL has you covered well for both of those questions, we had a similar question on that high vs low resistance argument earlier in the thread. My thinking (copied from earlier):

The effect we are discussing here is called Joule Heating (or Resistive Heating), and essentially, the formula for the heating of an object is:



This heating is, as suggested by the formula, caused primarily by the passage of an electrical current through resistance. A phenomenon known as hysteresis also plays a role, but most engineers argue that it isn't as important as Joule heating.

So, this formula suggests that, given a constant current, that increasing the resistance will increase the heating effect, and this is true. However, it is important to remember that our resistance impacts the size of the induced eddy currents.

Let me explain. The size of the induced EMF/voltage in the pot base is proportional to the rate of change of magnetic flux. This is Faraday's Law. Let's assume we have a constant voltage. We know that:



So, given our voltage is constant, if we want a higher current, we need a smaller resistance. Now, going back to the heating formula above, doubling the current will cause the heating effect to be quadrupled (since the current is squared), whereas doubling the resistance only results in the heating effect to be doubled.

Based on this interpretation, it is definitely more effective to have a lower resistance, and thus, larger eddy currents in the pot base, to achieve the maximum heating effect. This becomes more obvious if we use ohms law to re-write the heat dissipation law:



So, in fact, having a lower resistance means more current will flow, and thus, more heat will be generated than it would be by having a high resistance. This is why a low valued resistor, plugged into a huge voltage, blows up (for lack of a better term).

This is a very simplified model and by no means totally accurate, but it is an interesting way to look at the problem.

Now, this does not get rid of the idea that high resistance = more heat. It does. But we need a sizeable voltage drop across the said resistance for it to matter. This is quite a complex topic, definitely not worth investigating in depth. In actuality, there are many things at play here.

Just a different way of looking at the question. At its core, I am saying the same as Happy Physics Land. If the resistance goes too high, no current flows, and so heating does not occur. This is just a more detailed/mathematical explanation of that intuition 

[polpark]

hello !!!
do i asssume all circuts in the hsc are ohmic??

[polpark]

also how would you answer compare questions?
for example compare the torque created between motor 1 and motor 2?
would you show a ratio?

[Happy Physics Land]

hello !!!
do i asssume all circuts in the hsc are ohmic??

Hey Polpark!

Yes indeed! In HSC, if they ask you questions about electric circuits, they must all obey ohm's law, or equations that are derived from Ohm's law (E.g. P=IV, but according to ohm's law we can also express it in the form of P = I²R). After all, ohm's law is all that we have been taught!

Best Regards
Happy Physics Land

[RuiAce]

Hey Polpark!

Yes indeed! In HSC, if they ask you questions about electric circuits, they must all obey ohm's law, or equations that are derived from Ohm's law (E.g. P=IV, but according to ohm's law we can also express it in the form of P = I²R). After all, ohm's law is all that we have been taught!

Best Regards
Happy Physics Land

Just be careful there. P=RI² is specifically for power loss.

hello !!!
do i asssume all circuts in the hsc are ohmic??

Now. In terms of long distance power generation we assume that power is constant. In P=VI, we effectively assume that voltage and current are what vary. (Naturally, this is the whole fundamental basis of the AC current transistor)

But keep in mind that this MAY mean resistance does change as well. So I wouldn't say all conductors are, ohmic conductors. Ohmic conductors assumes that the resistance is constant whilst we change the voltage, whereas non-ohmic assumes that it is variable.

Typically, HSC questions that require V=IR are all related to power generation now. For these questions, generally you simply cannot be wrong when you assume POWER is the constant.

also how would you answer compare questions?
for example compare the torque created between motor 1 and motor 2?
would you show a ratio?

This is completely ambiguous. Present a past question for us to comment on it.

[jamonwindeyer]

This is completely ambiguous. Present a past question for us to comment on it.

I think what Rui meant to say is that we're not 100% sure what you mean, but if you could give us some more details we're super eager to help you out as quick as we can! 

[Neutron]

Hey guys!

Okay okay this is a sort of long winded question but I feel like I haven't grasped transmission properly.. I was wondering whether you could explain what the answer is for each one and why it is? D;

Which generator (AC or DC) is:
a) More reliable
b) Suffers less wear on critical parts
c) Can be transmitted over long distances with less loss of energy
d) Is easier to connect to houses
e) Is more likely to have shorting
f) Can have cables with less insulation
g) Produces less electrical interference
h) Is likely to experience a higher impedance from coiled cables


Thank you so much omg you guys are lifesavers xx

Neutron

[Happy Physics Land]

Hey guys!

Okay okay this is a sort of long winded question but I feel like I haven't grasped transmission properly.. I was wondering whether you could explain what the answer is for each one and why it is? D;

Which generator (AC or DC) is:

a) More reliable
b) Suffers less wear on critical parts
c) Can be transmitted over long distances with less loss of energy
d) Is easier to connect to houses
e) Is more likely to have shorting
f) Can have cables with less insulation
g) Produces less electrical interference
h) Is likely to experience a higher impedance from coiled cable


Thank you so much omg you guys are lifesavers xx

Neutron

Hey Neutron!

a)More reliable: DC, because commutator is in contact with carbon brushes, ensuring a more consistent flow of electricity
b) Suffers less wear on critical parts: AC, because slip rings are not directly in contact with carbon brushes, meaning that they dont suffer as much wear as commutators which will eventually wear out due to physical contact with brushes
c) Can be transmitted over long distances with less loss of energy: AC, because AC generators produce AC which are able to be stepped-up or stepped-down. Through stepping up AC to a very high voltage (550000V transmission voltage in Australia), the amount of currents being delivered has been reduced (Voltage and current are inversely proportional), hence less power loss because less heat dissipation.
d) Is easier to connect to houses: AC, because many appliances at homes require the standard 240V AC to operate
e) Is more likely to have shorting: AC (Im not too sure of the reason, perhaps when others come around they can provide you with the explanation?)
f) Can have cables with less insulation: DC, AC is a lot of dangerous than DC because it has the ability to cause fibrillation and can fatally propagate through human body. Hence theres more necessity to insulate AC than DC.
g) Produces less electrical interference: DC, because AC travels in a sinusoidal wave, hence its rising and falling propagation pattern makes it more likely to interfere with other waves. In contrast, DC currents only travel in straight lines, hence it is unlikely for it to come in contact and interfere with other waves.
h) Is likely to experience a higher impedance from coiled cables: DC, this is because DC travels only in a straight line, so if the cable is in the shape of a coil, DC would encounter a lot more resistance. Imagine yourself trying to run around in a circular maze, if you run in straight lines, you would frequently collide into walls and then you would turn and run straight in another direction, but again you would collide into another wall. Similarly, this is why DC would experience a lot more impedance as it travels through the coil. AC travels in a sine curve that is easier to propagate through the coil and because it has higher frequencies, it would experience less impedance

I hope these explanations all make sense to you, the technicality in my reasoning may not be very strong but Im sure the answers are correct.

Best Regards
Happy Physics Land

[Meckenza]

Hii,

I was looking at a multiple choice question that plotted Current (y-axis) vs speed (x-axis) of an electric motor, but the graph appeared to have a curved shape that approached zero as the speed increased. I understand that A) would be the most correct because as speed increases so does back emf which causes a decrease in current, but would someone be able to explain the shape of the curve please?

[Happy Physics Land]

Hii,

I was looking at a multiple choice question that plotted Current (y-axis) vs speed (x-axis) of an electric motor, but the graph appeared to have a curved shape that approached zero as the speed increased. I understand that A) would be the most correct because as speed increases so does back emf which causes a decrease in current, but would someone be able to explain the shape of the curve please?

Hey Meckenza!

Very nice question there! I completely agree with your reasoning and you should also be satisfied that you understand the main concern of this question, well done!

Like you have correctly stated, as speed increases, back emf increases and this opposes the supply current, hence causing the current to decrease. Now let's analyse what happens physically. Initially, at speed = 0, the DC motor is first started. At this time, the coil is stationary so the back emf is zero. Therefore the current flowing through the coil is high and equal to supply current. Once the armature rotates faster, the back EMF increases (This explains the relatively sharp drop in current at the start) and the difference between constant supply EMF and back EMF becomes smaller. However, because back EMF can never exceed the value of supply emf, there has to be a point where supply current pretty much stops decreasing because the impact of back emf upon it becomes smaller and smaller. This explains that the current is decreasing at a decreasing rate. Hence you have that curve that looks like as if its reaching a horizontal asymptote (you can think about it this way if you are maths-orientated).

Hope that helped! If you have further questions please dont hesitate to ask!

Best Regards
Happy Physics Land

[jamonwindeyer]

Hii,

I was looking at a multiple choice question that plotted Current (y-axis) vs speed (x-axis) of an electric motor, but the graph appeared to have a curved shape that approached zero as the speed increased. I understand that A) would be the most correct because as speed increases so does back emf which causes a decrease in current, but would someone be able to explain the shape of the curve please?

Hey Meckenza! Ah, Back EMF, this concept drove me nuts in the HSC aha 

So you've correctly said that A makes the most sense. I'll try to explain the curved shape as best I can, though it actually isn't something I know about for sure!

Back EMF is caused by the change in magnetic flux experienced by the coil in a motor as it spins. As it rotates, the coil moves through different parts of the field, and this changing flux induces a current. By Lenz's Law, this current acts against the supply. It sounds like you understand this pretty well, but let me know if you need a little more!

To answer your question on the shape of the curve, it probably makes sense to first consider the alternative. What if we had a straight line relationship? That would imply that we could push the speed of a motor high enough that no current flows. As speed increased, eventually the current would drop to zero. That can't be possible, since that would mean the motor would not be spinning! So a straight line relationship definitely doesn't work.

In actuality, Back EMF only increases until it reaches a point very similar to the supply EMF. This is the horizontal part of the graph you are seeing!

What we have here is an inversely proportional relationship. As speed increases, current decreases at a proportional rate, since Back EMF is proportional to speed. This creates a hyperbolic graph, roughly represented by:



If you graph a hyperbola, it has a shape similar to what you are showing in that question.

Now I am sure it is more complex than this, but this explanation gives you the idea of why the line is curved. I hope it helps! If you wanted a more concrete, detailed explanation I can definitely do some research for you, I am sure I have some more detailed explanations in my Electrical Engineering textbooks