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Edexcel Physics Unit 2 "Physics at work" June 2013

Reply 400

CWE

1

Original post by LegendX

Anyone who can please explain potential dividers? :smile:

They are quite simple really. Voltage around a circuit is split in proportion to the current. Imagine you have a series circuit with 2 resistors and a a battery. . The resistance of each resistor determines how much of this the EMF they get.

The proportion of total voltage a resistor uses is found by

Resistance of the resistor/ Total resistance in the ciruit.

This proportion is the same as the proportion of the voltage it will get, so the answer multiplied by the voltage in the circuit will equal how much of the voltage it gets.

Reply 401

B-Stacks

13

Original post by stealth_writer

Can someone please explain to me how to do Q 7 and 14 ( in depth tutorial please)

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202012%20-%20QP/6PH02_01_que_20120307.pdf

Both have been answered several times on this thread. Have a quick look a few pages back.

Reply 402

CWE

1

Original post by stealth_writer

Can someone please explain to me how to do Q 7 and 14 ( in depth tutorial please)

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202012%20-%20QP/6PH02_01_que_20120307.pdf

Not sure about question seven, but four I can.

The resistance of the first Voltmeter is 10,000 Volts. The resistance across the 2nd resistor combination of Voltmeter and resistor can be found by using the formula for resistors in parallel.
1/Rt = 1/R1+1/R2, you will find this gives you a very small resistance reading. Now the voltage is split in proportion to the resistance of each component. The second combination of resistor and voltmeter can be dealt with as one because they are parallel and so the voltage through both will be equal. What you will find is that the proportion of total resistance that the second voltmeter and resistor combination form is negligible. This tiny number* the voltage will only give you a very small reading, and to the accuracy used in this question will be 0. The other resistor is an extremely large proportion of the resistance (effectively all of it) and so it will receive the same proportion of the voltage, which will be 6V to the accuracy of this question.

Reply 403

stealth_writer

4

Original post by B-Stacks

Both have been answered several times on this thread. Have a quick look a few pages back.

where????

Reply 404

Swathi Kumar

Original post by Mimi85

Hey, unit 1 phy was tge worst exam ever!!! Dont knw abt unit two, im repeating unit 1 and ill be doing units 2&3 for the first time in june.. Hope they turn out to be easy... Didnt really start much with phy, and im dying with electricity.. Its soo hard, i really need help with it

Posted from TSR Mobile

I agree i didn't have enough time to finish the paper! I barely got a B

Reply 405

justinawe

19

Original post by stealth_writer

where????

I answered Q7 here:

Original post by justinawe

Think I explained this to someone else earlier... anyway,

I = \dfrac{V}{R}

We're given that V is constant, so the variables are R and I. So the graph of I against R would be a reciprocal graph.

And part b of Q14 here:

Original post by justinawe

First, you need to find the current.

The p.d. across V₁ is 4 V, and the resistance is 10 megaohms.

So,

V = IR

4 = I \times (10 \times 10^6)

I = 4 \times 10^{-7} \ \mathrm{A}

We know that current is the same throughout the circuit, and the overall voltage is 6 V, so the total resistance would be:

V = IR

6 = (4 \times 10^{-7}) \times R

R = 1.5 \times 10^7 \ \Omega = 15 \ \mathrm{M} \Omega

The total resistance of the circuit would be found by adding up the resistance of V₁ and the other voltmeter and the resistor that are in parallel.

So,

15 \ \mathrm{M} \Omega = 10 \ \mathrm{M} \Omega + \left( \dfrac{1}{10 \ \mathrm{M} \Omega } + \dfrac{1}{R} \right)^{-1}

Solve for R.

Reply 406

CWE

1

Can anyone explain question 18 of may 2012 - have no idea what principles the car battery question relies on?

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202012%20-%20QP/6PH02_01_que_20120525.pdf

Reply 407

lipgloss9

What are the practicals we should be familiar with?

Reply 408

saladesilva.13

Does anyone have any good notes for unit 2??

Reply 409

saifulahmed

1

Original post by CWE

Can anyone explain question 18 of may 2012 - have no idea what principles the car battery question relies on?

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202012%20-%20QP/6PH02_01_que_20120525.pdf

a(i)as negative terminal of the car battery connected to negative, and positive to the positive terminal of the charger, resultant e.m.f=voltage of charger-voltage of battery.
(ii)total resistance (in series)=0.05+0.65
(iii)initial charging current= resultant e.m.f/total resistant. (I=V/R)
b(i)p.d. across the resistance of charger=current in the circuit*resistance of the charger.
so p.d. across charger=15-p.d. across the resistance of the charger.
(ii)rate of energy supply=power input by battery= IV=4.3*15
(iii)wasted power (by the internal resistance)=(I^2)R
answer in 2 (power supply by battery)-waster power=useful power.
efficiency=(useful power/input power)*100%

Reply 410

sarah.102

5

Does anyone know from the top of their head what formulas we actually need to remember that aren't given on the formula sheet?

Reply 411

stealth_writer

4

Original post by justinawe

Think I explained this to someone else earlier... anyway,

I = \dfrac{V}{R}

We're given that V is constant, so the variables are R and I. So the graph of I against R would be a reciprocal graph.

why? Can you explain that further please

Reply 412

B-Stacks

13

Original post by sarah.102

Does anyone know from the top of their head what formulas we actually need to remember that aren't given on the formula sheet?

Potential divider.
Lambda=2l/n
Pir^2
Intensity=P/A

There's not that many.

Reply 413

sarah.102

5

Original post by B-Stacks

Potential divider.
Lambda=2l/n
Pir^2
Intensity=P/A

There's not that many.

That's what I thought, thanks.

Reply 414

justinawe

19

Original post by stealth_writer

why? Can you explain that further please

Do you know what a graph of

y = \dfrac{k}{x}

(where k is a constant) looks like?

Reply 415

Water_fall

Hey every one! So not looking forward to this exam tommorow :frown:

I dont even know why Im bothering to revise, not like Edexcel are going to aak us questions from the syllabus :/

Reply 416

sarah.102

5

Original post by Water_fall

Hey every one! So not looking forward to this exam tommorow :frown:

I dont even know why Im bothering to revise, not like Edexcel are going to aak us questions from the syllabus :/

Hahaha, so true, so true!

Reply 417

CharlieTT

4

Original post by lipgloss9

What are the practicals we should be familiar with?

Stationary Waves on a stretched string.
Stationary Waves using sound waves and microwaves
Measuring refractive index
Single slit diffraction
Double Slit Diffraction
Investigation of light polarisation
Measuring Resistivity of a metal wire
Investigating Power Dissipation
Calibrating a Thermistor

The most common ones are in bold.

Reply 418

Branny101

Original post by sarah.102

That's what I thought, thanks.

What's that second equation for?

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Reply 419

lipgloss9

Original post by CharlieTT

Stationary Waves on a stretched string.
Stationary Waves using sound waves and microwaves
Measuring refractive index
Single slit diffraction
Double Slit Diffraction
Investigation of light polarisation
Measuring Resistivity of a metal wire
Investigating Power Dissipation
Calibrating a Thermistor

The most common ones are in bold.

Hey thanx alot

Edexcel Physics Unit 2 "Physics at work" June 2013

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