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#### FlorianK

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« on: March 30, 2013, 08:10:52 pm »
+5

If you have general questions about the VCE Physics course or how to improve in certain areas, this is the place to ask! 👌

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Original post.
« Last Edit: February 16, 2018, 03:28:07 pm by Joseph41 »

#### ~T

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##### Re: VCE Physics Question Thread!
« Reply #1 on: April 01, 2013, 12:08:12 am »
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Because I don't "immerse" myself or believe that "beauty" is an appropriate term regarding the subject, can I still post here?

Seriously though, seconding a question that someone started as a thread, how much do we need to know about diodes? I'm confident with their applications and circuit questions, but what about the mechanics? No idea whether to spend time learning the workings of a diode or not.
Similarly, the Heinemann textbook is all like "Transistors aren't covered in this course" and now it's showing me a 'physics in action' segment that is telling me (very poorly) how a transistor works. I think I already understand them, but do I need to for the course?

And lastly, question 4 of chapter 4.1. I swear it can't actually happen. Unless I'm missing something... but it seems that when there is a grounded point on the circuit, there are two different currents at different points on the circuit. The answers say so too. Now as I'm typing I realise I'm not even sure about the point/function of a ground point. I know it's a reference point for the measurement of voltage, but does it affect the flow of current, or the circuit, in any way?

Many questions, many thanks in advance
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#### b^3

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##### Re: VCE Physics Question Thread!
« Reply #2 on: April 01, 2013, 01:10:03 am »
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Currents will always flow towards the lowest potential (i.e. in the context of VCE physics, the lowest 'voltage'), which in most cases is the ground, since it's 0 V.  Here though we have a voltage that is lower than the ground, that is -3 V, as in this case our ground is just a reference. For a, there is 5 V dropped across the first resistor, so I=V/R=5/5K=1 mA. For the second resistor there is 3 V dropped over the 2K resistor, so I=3/2k=1.5 mA.

As to the two different currents part, the second current is greater than the first current, that is because there will be some current leaking from the ground, as again, the current will flow towards the lower voltage. So from 0 V to -3 V, i.e. that is where your extra current is coming from.

For b, if we wanted to find the voltage dropped over each resistor, we would use a voltage divider, but we want the current through each, which is the same through each. So 8 V is dropped over 7 K worth of resistance, that is I=8/7K =1.14 mA.

As for the direction, the will all be flowing in the direction P, from higher potential to lower potential.

Hope that helps.
« Last Edit: April 01, 2013, 01:17:54 am by b^3 »
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#### ~T

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##### Re: VCE Physics Question Thread!
« Reply #3 on: April 01, 2013, 11:40:49 am »
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Thank you both! I think I'm fine with diodes then, if that's all we need.

As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
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#### EspoirTron

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##### Re: VCE Physics Question Thread!
« Reply #4 on: April 01, 2013, 11:56:25 am »
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Can someone please explain to me how an LDR works and how to do calculations involving LDR's. Also, can someone explain it using a ratio method, instead of 'regular formula' way. Anyone's help would be greatly appreciated!
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#### b^3

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##### Re: VCE Physics Question Thread!
« Reply #5 on: April 01, 2013, 03:07:42 pm »
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Thank you both! I think I'm fine with diodes then, if that's all we need.

As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
Normally the ground is the lowest reference, which is why we can do what we normally do with other circuits. But since there is a potential that is lower than this reference in this case, we have to look at the situation differently, as we could find the two currents in the two resistors. But as the second was greater than the first, the current had to come from somewhere (again current flows from higher potential to lower potential, so that's why some current was drawn from there). In the normal case, where the lowest potential is the ground, the currents are flowing towards this direction, and so we can look at the problem normally.

It's a little confusing, VCE physics doesn't explain it well (there was a fair bit of further electronics that I didn't fully understand (but was able to do the problems) until I got to uni and did an electrical engineering unit). Hope that makes sense.
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #6 on: April 08, 2013, 08:53:48 pm »
+1
could someone help me with the first two
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#### FlorianK

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##### Re: VCE Physics Question Thread!
« Reply #7 on: April 08, 2013, 09:34:24 pm »
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could someone help me with the first two
let IB1 be the voltage drawn from the battery 1 and IB2 be the voltage drawn from battery 2

IR1 = IB1
IR2 = IR3 = 0.5 * IB1
IR4 = 2/3 * IB2
IR5 = IR6 = 1/3 * IB2

Ok, since both circuits will have different resistances IB1=/= IB2

The total resistance of circuit 1 is: [Rx is the resistance of any resistor in the circuit]

Rtotal = Rx + 1/2 * Rx     [When 2 resistors of the same value are in parallel the resitance of this parallel circuit is half the resitance of each of them]

--> Rtotal = 1.5 * Rx

Ok the other one is a parallel circuit. The equation you should use for a parallel circuit, with 2 components is:
$\frac{R1*R2}{R1+R2}$

So here R1 is R4 so Rx and R2 is R5+R6 so 2*Rx
--> R1=Rx R2=2*Rx
we get:
$\frac{Rx*2Rx}{Rx+2Rx}$ = 2/3 * Rx

So now to the current

IB1=V/(3/2*Rx) = 2/3 V/Rx
IB2=V/(2/3*Rx) = 3/2 V/Rx

So now putting 'values' into the equations from the top

IR1 = IB1 = 2/3 * V/Rx
IR2 = IR3 = 0.5 * IB1 = 1/3 V/Rx
IR4 = 2/3 * IB2 = V/Rx
IR5 = IR6 = 1/3 * IB2 = 1/2 V/Rx

So we see we have the highest Current in R4 and the lowest current in R5 and R6.

Hope I didn't make some stupid mistake :p

#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #8 on: April 08, 2013, 09:48:51 pm »
+1
oooo, i see! thanks a bunch FlorianK
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#### LazyZombie

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##### Re: VCE Physics Question Thread!
« Reply #9 on: April 10, 2013, 09:42:47 pm »
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as a rollercoaster goes through a dip, which of the two cars, front or back, will be the fastest or will they have the same speed? Explain your answer.
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #10 on: April 10, 2013, 09:55:04 pm »
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im just guessing the second one since the first has to pull the second down, slowing itself down a lil?
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#### LazyZombie

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##### Re: VCE Physics Question Thread!
« Reply #11 on: April 11, 2013, 09:22:10 pm »
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Another:
Explain what happens to your body on corners using at least one of Newtons Laws.
What points would I have to cover?
I wrote: There is no force acting on the body initially so it conforms to newtons first law, continuing to move in the original forward motion so there is a feeling of being 'pushed' outwards from the corner.
Plus a little birds eye diagram of a roller coaster rounding a corner and an arrow pointing tangent to the curve labelled velocity. (right thing to label?)
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #12 on: April 16, 2013, 07:28:29 pm »
+1
in a force distance graph for satellite, is the kinetic energy greater at a greater distance or is the kinetic energy greater at a smaller distance
« Last Edit: April 16, 2013, 07:36:02 pm by Homer »
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#### LazyZombie

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##### Re: VCE Physics Question Thread!
« Reply #13 on: April 16, 2013, 08:46:28 pm »
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in a force distance graph for satellite, is the kinetic energy greater at a greater distance or is the kinetic energy greater at a smaller distance
kinetic energy is greater at a smaller distance.
by finding the area of the force distance graph, you can find the work done and so find the kinetic energy
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#### Robert123

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##### Re: VCE Physics Question Thread!
« Reply #14 on: April 17, 2013, 04:05:36 pm »
0