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OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

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they had also asked us about which law had been used in a current question. i misspeled kirchoffs. lol what a dumb name. i put in 2 c's
Original post by Nick_
only if the line would go through the origin if you continued it which it did not, hence resistance was changing and in this case it was decreasing

this is because R does not equal dV/dI, it just equals V/I


OK look i dont care if you understand it or not. but what you need to realize is that resistance IS dV/dI. now it might not be in the spec but i dont care. its the correct physics.
Original post by patterson
they had also asked us about which law had been used in a current question. i misspeled kirchoffs. lol what a dumb name. i put in 2 c's


its a shame but ive spelled Kirchhoff worng. i spelled it KIRSHHOF.:biggrin:
Original post by patterson
they had also asked us about which law had been used in a current question. i misspeled kirchoffs. lol what a dumb name. i put in 2 c's


I'm sure you had to explain the law as well?
On 1 line? Nope. They just asked what law is used to calculate the current in the resistor.
Original post by Sarabande
On 1 line? Nope. They just asked what law is used to calculate the current in the resistor.


It was two lines if I'm not mistaken. I put Kirchoff's first in brackets at the end anyway, hopefully will get me the mark D:
Reply 1266
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-Lychee
i have been waiting for the mark scheme for ages now =.=
Original post by pixelfrag
It was two lines if I'm not mistaken. I put Kirchoff's first in brackets at the end anyway, hopefully will get me the mark D:


It was the 2nd Law. You used V=IR to calculate it as you knew the voltage for both resistors was the same as they were in parallel. You didn't know the magnitude of the current entering the junction so it could not have been the 1st Law.

Why the neg? You have no idea what current entered the junction so it cannot, logically, be the 1st law.
(edited 11 years ago)
Reply 1268
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Nick_
Original post by mashmammad
OK look i dont care if you understand it or not. but what you need to realize is that resistance IS dV/dI. now it might not be in the spec but i dont care. its the correct physics.


Let me prove you wrong:
I make only one assumption in my proof and that is that V=IR
Let us imagine the example of a V/I graph with a curved line (non-constant resistant).
Now let us take a tangent from a point on the curve. The equation of this line will be V=mI+c , where m is the gradient of the line and c is the y-intercept.
As V=IR we can equate V to give: mI+c=IR
The divide both sides by I to give: R=m+c/I
As m is the gradient of this line,which is dV/dI, that means the R is only equal to dV/dI when c=0, which will often not be the case for a curved line.
Any questions?
Original post by Sarabande
It was the 2nd Law. You used V=IR to calculate it as you knew the voltage for both resistors was the same as they were in parallel. You didn't know the magnitude of the current entering the junction so it could not have been the 1st Law.


Sorry yeah I put that one down then since i know I put down an explanation and the explanation fit the scenario ^^
Reply 1270
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geditor
3
Original post by Nick_
Let me prove you wrong:
I make only one assumption in my proof and that is that V=IR
Let us imagine the example of a V/I graph with a curved line (non-constant resistant).
Now let us take a tangent from a point on the curve. The equation of this line will be V=mI+c , where m is the gradient of the line and c is the y-intercept.
As V=IR we can equate V to give: mI+c=IR
The divide both sides by I to give: R=m+c/I
As m is the gradient of this line,which is dV/dI, that means the R is only equal to dV/dI when c=0, which will often not be the case for a curved line.
Any questions?


You smell. But yeah, great proof :P It's such a simple question which I would have answered correctly under any conditions except for when in an exam. It's a shame really, I was really looking forward to high marks :'(
(edited 11 years ago)
Reply 1271
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HarryW95
Original post by mashmammad
OK look i dont care if you understand it or not. but what you need to realize is that resistance IS dV/dI. now it might not be in the spec but i dont care. its the correct physics.


Ohhh my, what ever, I can't be bothered to attempt to teach you anymore. Why do people bother taking AS Physics if they can't read figures off a graph. Stop being so stubborn and just accept the fact your wrong. Resistance was obviously decreasing..
So many more people would have got this if the line was clearly still curving above 1.8V, it seems unnecessarily hard for something which should be such a simple question!

Oh well. Hopefully it will drag the grade boundaries down a mark or 2.
Reply 1273
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101101
8
Original post by Sarabande
It was the 2nd Law. You used V=IR to calculate it as you knew the voltage for both resistors was the same as they were in parallel. You didn't know the magnitude of the current entering the junction so it could not have been the 1st Law.

Why the neg? You have no idea what current entered the junction so it cannot, logically, be the 1st law.


Surely it was the first law, it asked how you could calculate the current in the single series resistor, coming out of the two in parallel was 0.2 and 0.3, which is equal to 0.5? What am i doing wrong?
Original post by 101101
Surely it was the first law, it asked how you could calculate the current in the single series resistor, coming out of the two in parallel was 0.2 and 0.3, which is equal to 0.5? What am i doing wrong?


It was asking for the law used in the question before that when you had to SHOW that the current in one of the resistors is 0.2A. You didn't know the current going into the junction so it was definitely not Kirchhoff's 1st law. Moreover, you use V=IR so it definitely has to be his 2nd.
Original post by Sarabande
It was asking for the law used in the question before that when you had to SHOW that the current in one of the resistors is 0.2A. You didn't know the current going into the junction so it was definitely not Kirchhoff's 1st law. Moreover, you use V=IR so it definitely has to be his 2nd.


It was Kirchhoff's First law, you didnt know what was going in but you did know what was going out so its exactly the same principle. Current isnt just lost through the junction so that is exactly what they were looking for, the first law. I did the maths.

Besides Kirchhoff's Second law only applies to a closed loop. I.E in series. The resistors were parallel.
Original post by Dubai Nan Bread
It was Kirchhoff's First law, you didnt know what was going in but you did know what was going out so its exactly the same principle. Current isnt just lost through the junction so that is exactly what they were looking for, the first law. I did the maths.

Besides Kirchhoff's Second law only applies to a closed loop. I.E in series. The resistors were parallel.


The question was asking what law did you use in your calculation. You didn't use the 1st Law as you were trying to prove that the current of one of the resistors in the loop was 0.2A. YOU DIDN'T KNOW WHAT CURRENT ENTERED OR LEFT THE JUNCTION. You found that out in the question afterwards. You only knew the current of the other resistor. You applied the knowledge that components in parallel have the same voltage and used V=IR.
(edited 11 years ago)
Original post by Sarabande
The question was asking what law did you use in your calculation. You didn't use the 1st Law as you were trying to prove that the current of one of the resistors in the loop was 0.2A. YOU DIDN'T KNOW WHAT CURRENT ENTERED OR LEFT THE JUNCTION. You found that out in the question afterwards. You only knew the current of the other resistor. You applied the knowledge that components in parallel have the same voltage and used V=IR.


You did though it was 0.5A exiting. You already knew one was 0.3 or something like that so you thus had to prove it was 0.2
Original post by Dubai Nan Bread
You did though it was 0.5A exiting. You already knew one was 0.3 or something like that so you thus had to prove it was 0.2


You calculated that after the question stating which law it is. The 0.3 + 0.2 one.

I did the proof like this.

V=IR

0.3 * 4 = 1.2V

1.2/6 = 0.2A

No sign of Kirchhoff's 1st law here. Though there's no sign of the 1st law, admittedly, it's probably more likely that the 2nd law. I didn't use either law in my calculation.
(edited 11 years ago)
Reply 1279
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HarryW95
There was a question asking for the total current in the circuit after working out 0.20 ohms in one of the resistors. Because we had 0.30 and 0.20, it had to be 0.50. Maybe the question asking for the law was referring to that? I wouldn't have done it otherwise..

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