OCR Physics Unit 2 - G482 - (June Exams Preparation) Watch

sophiekutie
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#881
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#881
(Original post by Gotzz)
Omg, why don't they?
Tbh I think it's because they like to make us think we've done bad. To inspire us to work harder for our other exams :lol:
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leahcee
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(Original post by Hamster96)
No. It was definitely 1000 and 1500

there was 0.02 amps of current going towards the 750 ohm resistor, so that works out to be 15v. In parallel, voltage is equal so R2 will have 15v in it too. From the diagram you knew 0.01 amps of current went to R2, 15/0.01=1500 resistance in R2 is 1500 ohms

to work out resistance in parallel:
1/750 + 1/1500 = 1/R, solve to find R and you get the combined resistance of the two in parallel= 500. Now lets call total resistance in the circuit 500+x, where x is R1,

current was 0.03 and total voltage was 45, so 45/0.03=total resistance=1500
1500=x+500, x is 1000 ohms, and x is R1
hope that's enough evidence for you guys.
This gives me hope of possible ONE method mark thanks
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charlotteg96
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#883
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#883
can someone please tell me how you did the resistivity one?? also what did everyone else get for the path difference??
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sophiekutie
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(Original post by Hamster96)
No. It was definitely 1000 and 1500

there was 0.02 amps of current going towards the 750 ohm resistor, so that works out to be 15v. In parallel, voltage is equal so R2 will have 15v in it too. From the diagram you knew 0.01 amps of current went to R2, 15/0.01=1500 resistance in R2 is 1500 ohms

to work out resistance in parallel:
1/750 + 1/1500 = 1/R, solve to find R and you get the combined resistance of the two in parallel= 500. Now lets call total resistance in the circuit 500+x, where x is R1,

current was 0.03 and total voltage was 45, so 45/0.03=total resistance=1500
1500=x+500, x is 1000 ohms, and x is R1
hope that's enough evidence for you guys.
Ooooooh dear!! I forgot what voltage was EQUAL in parallel!!! *doh*
Oh well, it was only 2 marks
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Gcayte
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#885
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#885
Can anyone remember the questions? Ill try to make an unofficial mark scheme if people can?
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Room4student
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#886
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#886
(Original post by leahcee)
I would be over the MOON with a C I got an A in my coursework and a U in the January exam, even though I did every single past paper to death and did so much revision for it

But that's only for AS, A2 is suppose to be much harder so it's likely to even go down to a D :/
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Gotzz
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#887
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#887
(Original post by Boy_wonder_95)
Fair enough, I saw you had Bio 2, Chem 2, and Physics 2 all within the space of a few days! How did you manage?
It was horrible. I focused mainly on biology because it had the most content and chemistry I already knew quite well. Hence why I didn't have much time for physics. But hopefully it'll pay off
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OllieGCSEs
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#888
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#888
Here is my reasoning with why R1 is 1000 and R2 is 1500:

We know that the p.d across R1 is 30V (45-15). So we know the expected p.d result (Vout) from a potential divider equation should be 30V.

The combination of the other two resistances in parallel will therefore need to make 500 when calculated, so that when you use the potential divider equation on R1, you get 30V, shown here:

Vout=R1/(R1+R2) *Vin

Note that R1 here does represent R1 in the circuit but R2 represents the resistance of the combo (R2 and the 750ohm Resistor)

So, we get: 30=1000/(1000+500) *45, which works.

Now that we know what the combination resistance needs to equal (500), we know that 1/500= 1/750 + 1/R2

Therefore 1/R2 = 1/1500 and R2 = 1500ohms

Does this clear things up for people? If there is another way to prove a different answer then please post it


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Gcayte
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#889
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(Original post by Room4student)
But that's only for AS, A2 is suppose to be much harder so it's likely to even go down to a D :/
I personally found Unit 4 to be alot nicer than Unit 2.
I think the content of Unit 5 is alot more interesting, no idea what the exam is going to be like but overall A2 has been better.
Plus, after getting the results from AS I have worked so much harder!
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OllieGCSEs
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#890
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#890
(Original post by Hamster96)
No. It was definitely 1000 and 1500

there was 0.02 amps of current going towards the 750 ohm resistor, so that works out to be 15v. In parallel, voltage is equal so R2 will have 15v in it too. From the diagram you knew 0.01 amps of current went to R2, 15/0.01=1500 resistance in R2 is 1500 ohms

to work out resistance in parallel:
1/750 + 1/1500 = 1/R, solve to find R and you get the combined resistance of the two in parallel= 500. Now lets call total resistance in the circuit 500+x, where x is R1,

current was 0.03 and total voltage was 45, so 45/0.03=total resistance=1500
1500=x+500, x is 1000 ohms, and x is R1
hope that's enough evidence for you guys.
Sigh, I thought really hard about the question and then you go ahead and beat me to it but thanks


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Gotzz
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(Original post by hannahsm)
was the current not 0.030, 0.020 ect? so you would get a voltage of 15V for the fourth question?
Ohhh, I'm so silly. I just got what you said. Yes, I made a mistake, it was 0.03, etc.
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Hamster96
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#892
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#892
(Original post by OllieGCSEs)
Sigh, I thought really hard about the question and then you go ahead and beat me to it but thanks


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You snooze you lose
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charlotteg96
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#893
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#893
(Original post by Gotzz)
Yeah we all got 0.3
what did you do to get 0.3? feeling stupid now haha also what did you get for the path difference one??
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Loiks94
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#894
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#894
1000 nanometeres for one on last page?
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Room4student
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#895
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#895
(Original post by Gcayte)
I personally found Unit 4 to be alot nicer than Unit 2.
I think the content of Unit 5 is alot more interesting, no idea what the exam is going to be like but overall A2 has been better.
Plus, after getting the results from AS I have worked so much harder!

That's why i would like to carry on with physics, i heard it involves solar system in A2 which i would find interesting but exams in physics are in my opinion the hardest and Uni's require A's and B's so a C in AS does not sound like enough to carry on but we'll see, good luck in your remaining exams man!
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Gotzz
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#896
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#896
(Original post by Loiks94)
1000 nanometeres for one on last page?
I got 1060?
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redtortoise
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#897
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#897
What was the percentage for the battery question?
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Loiks94
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#898
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#898
(Original post by Gotzz)
I got 1060?
It was 2*the wavelength then divide by 1*10^-9
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Gotzz
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#899
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#899
(Original post by charlotteg96)
what did you do to get 0.3? feeling stupid now haha also what did you get for the path difference one??
Read the length off the diagram for L, read off the width and height, multiply them to get A, and then just use the resistivity formula.
Which path difference one, sorry?
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Gotzz
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#900
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#900
(Original post by Loiks94)
It was 2*the wavelength then divide by 1*10^-9
Did you get exactly a thousand?
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