Edexcel Physics Unit 2 "Physics at work" June 2013

Well if you ever want to define a physical quantity you can form an equation and make the denominator equal to one. (If you want to define a unit then make both the numerator and denominator one)

$voltage = \dfrac{energy}{charge}$

So voltage is energy transferred per unit charge or per coulomb of charge. (And one volt is one joule transferred per coulomb charge).

This does not specify if it is inputting or outputting electrical energy. So we have emf.

Emf of a battery or cell is the total work done by the cell against both the internal and external circuits per coulomb of charge that passes through it. It is different to the terminal pd learnt at GCSE which does not recognise internal resistance so terminal PD is the work done by the cell in the external circuit only.

In other words emf is the total electrical energy provided to each coulomb of charge by a power supple. This energy is then used to do work in the internal circuit (against internal resistance) and in the external resistance. Lost volts us the work done in the internal circuit against internal resistance.

So emf=lost volt+terminal pd
Terminal PD= emf - lost volt

$V=E-Ir$


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Measuring wire resistivity

Bit confused by what actually is emf and potential difference and the differences between them

Well the syllabus statement is very important but takes a great deal of time to explain.
Hope this helps
http://www.s-cool.co.uk/a-level/physics/kirchoffs-laws-and-potential-dividers/revise-it/potential-dividers

The two things that you must remember are:

Firstly
$[br]Resistance =\dfrac{resistivity \times length} { cross sectional area}[br]$
When the resistivity/material is the same as well as the cross sectional area resistance is directly proportional to length.

Secondly

Using V=IR when current is the same voltage is directly proportional to the resistance, and voltage is the same as in parallel.

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must be a then

What about L I don't get it.

Can somebody please explain q5 in Jan 2011. I know why N is same, but not L.

N isn't the same. The answer is D.

I will have a go at it.
Basically if M goes off the resistance in the circuit would increase, because of the parallel combination. Linking it to I=V/R if R increases then I, current, would decrease in the circuit. Linking this to power, P=IV, for bulb L there would be an increase in power, meaning an increase in brightness because of the increase incurrent. For N I am not too sure how the brightness would decrease.
Not 100% on this, so If anyone knows better please add and correct me if I was wrong.

Are you doing this exam? I thought you fortunately switched exam boards?

De Broglie's wavelength?

Anyone know how I would work out the reading on the voltmeter?

Something like that, yeah!

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Exam time and date fellas?

P.d across branches in parallel with the power supply are always equal to the voltage across the power supply I think. So the answer would just be 15v?