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

I have made this thread for the June Unit 2 Physics exam for anyone interested in starting the revising process early and so that we can help eachother with what we're learning as we go. I started Unit 2 last week (an hour after my last exam, slightly brain dead).

I'm glad to hear some people thinking it would be a good idea, I haven't encountered all of this content but luckily I did a Lvl 3 VRQ in Electrical Engineering last year so hopefully I can help us quite a bit ^.^ Once I finished compiling my class notes I'll pop 'em up on this thread!

Edit: Would have finished typing up my notes tonight but got a bit distracted having my birthday, sorry will try to get everything done before the weekend's over.

(edited 11 years ago)

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Reply 1

Shinusuke_Akki

PART 1/5

Electrons drift slowly through a circuit carrying an electric current. As the electrons move they create a shift in the electrons pushing the electrons in front of them down the circuit; therefore despite the slow speed the moment an electrical circuit is switched on each of the electrical components (e.g. an LED or a resistor) receives current.

Electric Current is defined as the flow of charge through a conductive material. There are two types of electrical current:
Conventional Current is where current is considered to be the flow of positive charge from positive (e.g. 5V) to negative (0V) due to the negative electrons traveling in the opposite direction.
Electron Flow is the movement of electrons carrying a negative charge from negative to positive.

When there is no current electrons move randomly about the material causing no net drift.
When current is present (despite still moving randomly) the general drift of electrons is in a specific direction (towards negative).

Amperes (Amps) - Current is measured in Amperes (A) measuring the amount of electric current through a point per unit of time. 1A=1Cs^-1

Electric Charge is the total amount of current that is supplied over a certain period of time. It can be defined with the following equation:
Q=It (charge = current x time)

Coulombs - Charge is measured in coulombs (C) - 1C if the total charge supplied by a current of 1A in a time of 1s.

Kirchhoff's Law: "The sum of the currents entering any junction is always equal to the sum of the currents leaving the junction."
https://twitter.com/shinusuke_akki/status/293076910924263425/photo/1
Note: Charge will be conserved throughout the circuit.

Ammeters are used to measure electric current. A digital multimeter is more often used - they're a multifunctional piece of equipment that allows you to measure current, voltage and resistance. Theoretically an ammeter should have negligible resistance. In a circuit they need to be connected in series, not in parallel, otherwise the electrical current will pass through the ammeter, limiting or stopping the current from passing through the component it is connected in parallel to (should that component present any form of resistance).
https://twitter.com/shinusuke_akki/status/293085533599592448/photo/1

Mean Drift Velocity is the average velocity that an electron attains when an electric current flows. Mean drift velocity is directly proportional to current. It is given by the equation:
I=nAve
Where: I = current
n = number density of electrons (electrons per unit volume)
A = cross-sectional area
v = mean drift velocity
e= electron charge (1.6x10^-19C)*

*1.6x10^-19C is the negative charge of an electron; this is the number you need to know - like remembering 9.81ms^-2 was the magnitude of acceleration due to gravity in the last unit

Material Categories
Conductors contain a high number density of conducting electrons e.g. a copper wire.
Insulators contain very few or no conducting electrons e.g. rubber.
Semiconductors have properties that lie between those of conductors and insulators e.g. silicon. The conductivity of a semiconductor increases with increasing temperature, behavior opposite to that of a metal.

THE END OF PART 1/5

(edited 11 years ago)

Original post by Gotzz

Thanks for this, it's really helpful :smile:

Thank-you I'm glad that it is, will try to get the next one up when I've compiled a good enough note collection. If you have any questions or feel I've missed something, please do say ^^

Reply 4

Shinusuke_Akki

I'm about ready to start my write ups for part 2 of 5, I should be finishing up my topic on resistance and will write up notes and post them hopefully by this weekend if not sooner (sorry for the delay I've had some personal issues which have delayed my work temporarily but everything should be okie-dokie-lokie now) :3

Reply 5

Better

Why isn't there this for Unit 5 arghhhhhhhhhhhhh

Reply 6

Shinusuke_Akki

Original post by Better

Why isn't there this for Unit 5 arghhhhhhhhhhhhh

Well apparently there is a thread... Yours by the looks of it... http://www.thestudentroom.co.uk/showthread.php?t=2244676 :P

Reply 7

Shinusuke_Akki

Original post by Law-Hopeful

I'm really lost on emf and polarisation. Anyone care to help? :frown:

Will post about E.M.F by this weekend not sure about polarisation yet tho, sorry. If the E.M.F is urgent I can try to post my notes before I've finished them for you?

Reply 8

Boy_wonder_95

I love this Unit so much it is the best! Much better than Unit 1 :unimpressed:

. I sat this exam in Jan so if anyone is unsure of anything just ask me :smile:

(edited 11 years ago)

Reply 9

Boy_wonder_95

Original post by Law-Hopeful

I'm really lost on emf and polarisation. Anyone care to help? :frown:

Definition for E.m.f is the energy transfer per unit charge when other forms of energy is transferred into electrical energy denoted by the equation V = W/Q.

There is also another equation for Emf which combines Kirchoff's Second Law have you reached to that stage yet?

For polarization what you need to know is that when light passes through a polarizing filter it will become partially polarized meaning the polarizing filter will only let through light which is acting in the same plane, all other planes of light will be blocked up.

However when you pass partially polarized light through a second polarizing filter, the intensity of light will depend on the angle the second filter is rotated. It is denoted by equation I = Io + cos^2(pheta)
Where:
- Io is the initial intensity (after passing through the first filter)
- I is the final intensity (after passing through the second filter)
- Pheta is the angle of rotation of the second filter.

Here you would note that if 2nd filter isn't rotated, such that the angle is 0 degrees light would have full intensity. Since cos 0 = 1. 1 squared = 1.

If the 2nd filter is rotated perpendicular to the first filter, such that the angle is 90 degrees no light will pass through. Since cos 90 = 0. 0 square = 0.

Also if the 2nd filter is parallel to the first filter, such that the angle is 180 degrees the light would have full intensity. Since cos 180 = -1. And -1 squared = 1.

Hope this helps :smile:

Reply 10

motivatedshroom

Can anyone please help me?
I understand the currents but not the answers to the voltages.

ANSWERS:
a) 5.0V, 0A
b)0V, 1.1mA

Reply 11

gunner18

The resistance between PQ can be regarded as very high(infinite) when the switch S is open. So pd across PQ = 5 x infinite/(infinite + 4.7kohm) = 5V
When the switch is closed the resistance between PQ is zero. Hence pd across PQ = 5 x zero/(zero+ 4.7kohm) = 0V

Reply 12

motivatedshroom

Original post by gunner18

Got it! Thanks.

http://www.cyberphysics.co.uk/topics/electricity/higher_electricity/resistivityQ.htm

I found this pretty useful. :smile:

Reply 13

Loiks94

Anybody does the physics practical courseworks? Could do with some tips.

Reply 14

Gotzz

I had my first physics practical before the half term, it was was the one out of 20 marks. Not sure how I did, will find out tomorrow, I think :eek:

ETA: I got 19/20! SO happy! :biggrin:

(edited 11 years ago)

Reply 15

Loiks94

Original post by Gotzz

I had my first physics practical before the half term, it was was the one out of 20 marks. Not sure how I did, will find out tomorrow, I think :eek:

Was it to do with amps and resistivity?

Reply 16

Gotzz

Original post by Loiks94

Was it to do with amps and resistivity?

Nope

Reply 17

brawlerpit

Has anyone started doing past exam papers? How are you finding it relative to the work you were given in class? The questions considerably harder or about the same?

Reply 18

Dusky Mauve

Original post by brawlerpit

Has anyone started doing past exam papers? How are you finding it relative to the work you were given in class? The questions considerably harder or about the same?