Edexcel Physics Unit 2 &quot;Physics at work&quot; June 2013

Original post by CoolRunner

Edexcels gonna **** us over again. Gosh, how i hate edexcel physics.

I know. They may tear us apart. :frown:

Reply 201

Original post by muonz

Hm? P is a fixed resistor, when the lamp is in parallel with the fixed resistor, the resistance of the parallel combination is smaller than the resistance of P. Therefore now that the lamp is removed you have a potential divider, the ratio of the resistance between P and R changes so that P has a higher proportion of the voltage (as the resistance of P > the resistance of P and the lamp in parallel), hence the voltmeter reading increases.

But but...
I thought like this...where did I go wrong? :redface:

The 12 V is shared between P (and all its parallels) and Q(Q nah it was? Can't see the pic in this app!).
So v of P= 12 - v of Q

Now the total resistance of the circuit increases. So the current decreases. Resistance of Q is constant so its V decreases. So V of P increases!!
Sorry for the mixing up edits but as I posted firstly it increases

I don't seem to follow you. :confused:

EDIT: take Q as the other resistor not the bulb. The pic does not show up in the app. So Q in my above explanation is not the bulb
EDIT: yes I get it increases. Lol I just got it now and I assumed I typed in it increases. Not a typo a genuine mistake...lol now I'm not confused.

Thanks for pointing out :tongue:

(edited 10 years ago)

Reply 202

Guess this paper can't be as hard as the Unit 1 paper.

Reply 203

Original post by StUdEnTIGCSE

Guess this paper can't be as hard as the Unit 1 paper.

Who knows. Edexcel may throw us a real bomb as a surprise :frown:

Reply 204

popo

Its possible that it could be really hard. In one of the past papers i saw it was 45/80 for an A and 55/80 for 120 UMS. LOL Its kind of good having a hard paper cuz you can loose like 30 marks and get an A or B

Reply 205

Original post by popo

Ah yes...
But a hard paper always gives that suspension feeling that lasts untill results day. :redface:

Reply 206

DanielDaniels

Original post by CoolRunner

Edexcels gonna **** us over again. Gosh, how i hate edexcel physics.

how much u got in jan paper?

Reply 207

lucilleJR

did anyone do unit 1 in june 13? and if not does anyone know what 120 ums was in jan? both papers were hard so I'm thinking it was quite low

Reply 208

Jaydude

Original post by lucilleJR

did anyone do unit 1 in june 13? and if not does anyone know what 120 ums was in jan? both papers were hard so I'm thinking it was quite low

52=A

64= 120ums (or 62 marks, cant remember!)

Reply 209

CoolRunner

Original post by Daniel Atieh

how much u got in jan paper?

a D lol.

Reply 210

CoolRunner

Has anyone got markscheme notes for physics unit 2? similar to biology?

Reply 211

Zoeyyy

Just wanted to confirm why the answer to this was C and not D :smile:

image.jpg52.1KB

Reply 212

Original post by Zoeyyy

Just wanted to confirm why the answer to this was C and not D :smile:

Top part resistance=8 ohms
Bottom part resistance=8 ohms
1/R=1/8+1/8
1/R=1/4
R=4 ohms

Reply 213

Original post by Zoeyyy

Just wanted to confirm why the answer to this was C and not D :smile:

Well the total resistance of resistors depend on whether they are in parallel or series.
In series the total resistance is the sum of each individual resistances.
In parallel it follows this

\dfrac{1}{R}= \dfrac{1}{r_{1}} + \dfrac{1}{r_{2}} + \dfrac{1}{r_{3}} + \dfrac{1}{r_{3}} .....

Where R is the total resistance and r is the individual resistances.

Now before you solve for parallel you should find the resistance for each branch. The two 4ohm resistors are in series in each branch so adding it each branch has resistance of 8ohm. Now adding as in parallel gives 4ohms.

An easier way to find the resistance of two branches in parallel is

R= \dfrac{r_{1} \times r_{2}}{r_{1} + r_{2}}

And when there are n branches each with the same r resistance its easier. The total resistance us simple r/n

EDIT: I corrected one thing...typo :tongue:

(edited 10 years ago)

Reply 214

Jaydude

Does anyone have access to the 2013 jan paper? Thanks!

Reply 215

http://www.thestudentroom.co.uk/showthread.php?t=2262586

Original post by Jaydude

Does anyone have access to the 2013 jan paper? Thanks!

Reply 216

Branny101

Expectations for our Wednesday exam, guys?

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Reply 217

Zoeyyy

Original post by krisshP

Top part resistance=8 ohms
Bottom part resistance=8 ohms
1/R=1/8+1/8
1/R=1/4
R=4 ohms

Ahh, thanks a bunch! :biggrin:

Reply 218

Zoeyyy

Original post by StUdEnTIGCSE

Well the total resistance of resistors depend on whether they are in parallel or series.
In series the total resistance is the sum of each individual resistances.
In parallel it follows this

\dfrac{1}{R}= \dfrac{1}{r_{1}} + \dfrac{1}{r_{2}} + \dfrac{1}{r_{3}} + \dfrac{1}{r_{3}} .....

R= \dfrac{r_{1} \times r_{2}}{r_{1} + r_{2}}

And when there are n branches each with the same r resistance its easier. The total resistance us simple r/n

EDIT: I corrected one thing...typo :tongue: