23 November Internal Resistance Questions

The topic of Internal Resistance builds on and uses many of the concepts and equations we've learned earlier – it’s a good summative topic.

All you really need for internal resistance questions are a good understanding of:

• Ohm's law V = IR
• The fact that potential differences add up in series
• To remember that p.d.s or voltages are measuring the energy of the charge, in joules per coulomb
• In addition, that the 'lost volts' is the p.d. 'across' the internal resistance.  Lost volts = Ir where r is the internal resistance
• It might also be useful in some questions to know that Power = VI = I2R
• Power dissipated in the cell or other source = I2r
• Also, power from the source is e.m.f. x current = Iε

By doing questions on internal resistance you will be revisiting these topics – this will consolidate your understanding and recall.

Remember that the internal resistance equation is in the formula book:

 V_load = ε – Ir

Also, remember that V_load = IR where R is the load resistance, Ir = the lost volts and ε is the e.m.f. of the source

Questions from the Internal Resistance booklet:  Internal resistance of power supplies.


Exam Questions Q 1.  Remember that power = VI =  I2R


Exam Questions Q 2.  Involves some algebra!


Exam Questions Q 3.  This is quite a challenging question requiring you to think about uncertainties carefully.

11 November Part 1 of Internal Resistance (AKA "the mystery of the lost volts")

Internal resistance - the mystery of the lost volts.

Extract from the revision guide (slightly edited) : 

The energy provided by a source (e.g. a cell or battery) is delivered to the components of the circuit by charge flowing round the circuit.  However for most sources some of the energy is dissipated inside the source due to the source's internal resistance. When a current is drawn from the source this internal resistance causes the potential difference across the terminals of the source to be less than the emf of the source. 

The lost p.d. in the source is the energy dissipated per unit charge (i.e. per coulomb) inside the source due to its internal resistance. The lost p.d. depends on the current and on the internal resistance of the source.    Lost p.d. = current x internal resistance
For a source of emf ε with internal resistance r connected to a load of resistance R, as shown in the circuit below  ε = I R + I r  where IR is the potential difference across the load resistance and Ir is the lost p.d. 

The external p.d. V = I R = ε – Ir. The graph below shows how the external p.d. V varies with the current drawn. This graph has a gradient of  –r and a y-intercept equal to ε.   The intercept on the x-axis is the short circuit current, with a terminal p.d. of zero volts.

Note that the p.d. V falls as the current increases. This is why the output potential difference of an electrical source of energy (including a laboratory power supply unit) falls if more current is drawn from the source. The headlights of a car often dim for a moment as you operate the starter motor. 

The videos

Here are a number of introductory videos - watch them and add to  your  notes.

Also make sure you go over the topic in the Chapter 2 revision guide and in the textbook chapter 2.


Useful introduction from www.Fizzics.org.uk



Further introduction and explanation of the graph of V against I -



Quite a good lecture on the theory



Demonstration of a potato cell.  Different metal plates are inserted into a potato to make a cell that produces about one volt.  However it has a high resistance and can only provide very small current - in micro amps.  The current is varied by switching a resistance box to different values.  No sound track - just watch the meters.




This final video shows some practical work - the image is bit fuzzy but the physics explanations are clear.  Note the results plotted on the graph of V against I

11 November Part 2 of Internal Resistance (AKA "the mystery of the lost volts")

These videos are from Steve4physics (not me) and give a really good, detailed explanation of internal resistance.

Part 1: Introduction - the basics



Part 2: Calculations - the basics



Part 3: Calculations - more complex



Part 4: Calculations - how to measure internal resistance N.B. the graphical method here plots Current on the y axis against terminal potential difference on the x axis.  Our exam board normally show the graphs the other way round, so the section after 5 minutes 19 seconds is less useful and might be confusing unless you are a very confident mathematician.

3 November, second section of revision question answers and explanations.

Make sure you've attempted the questions yourself first, and marked them using the mark scheme provided at the end of the booklet.

Then watch the videos below - even for questions you got completely correct, to reinforce your learning.

Questions on the test will be similar, and some will be the same as these!!

Questions 7 and 8



Questions 9 to 11 - note that there is a new Question 11 as the previous one was a duplicate of Q 2!  A copy of the new question is here

Questions 12 and 13

3 November. Revision Questions 1 to 6

Make sure you've attempted the questions yourself first, and marked them using the mark scheme provided at the end of the booklet.

Then watch the videos below - even for questions you got completely correct, to reinforce your learning.

Questions on the test will be similar, and some will be the same as these!!

This first video is just a reminder about the key equations that are in the formula booklet - I've extracted the key ones and put them onto one page here.


This video covers Q 1 to 3


This one explains Q 4 to 6

22 October Potential Dividers (also known as Voltage dividers in the USA) part 2

Part 2 of potential dividers
This video shows a potato used as a potentiometer - fun but don't try it at home! (If you ask nicely we might try it using a safe p.d. like 24 V and a high resistance voltmeter)  http://youtu.be/1l2kgi4DOOk




The website below might be useful, particularly for the potential divider calculator.  Also students aiming at high grades could make sure that they understand the approximations section - understand, not learn!!.   (Please note, if only one connection is shown for Vout, the other one is the negative of the supply, usually labelled 0 volts in electronics circuits.)  https://learn.sparkfun.com/tutorials/voltage-dividers/ideal-voltage-divider



Further work, more examples to practice on.   http://www.youtube.com/watch?v=rIEnMpgIaU
However this video makes the calculations look more complicated than they need to be as he leaves the current value as a calculation of V/R instead of working out the actual value.  For example in this screenshot 5V/(2+3)kilo_ohms is of course equal to 0.0010 A (1.0 mA) .  This approach can be useful as it shows directly how the ratio of the resistors is the key factor in working out the output p.d.  See further note and screenshot below the embedded video (at the bottom of this blog)




Note:  Instead of labelling V1, V2 as the voltages across the resistors he labels them V2kohm instead, as you can see in this screenshot.  The final answer for V2kohm is clearly 5V x (2/5) = 2 Volts!

22 October Potential Dividers (also known as Voltage dividers in the USA) part 1

On 25 October and in a later lesson we will be working on Potential Dividers. It is essential that you get a good understanding of these because:

• passive sensors are normally connected in a potential divider
• the resistance of the sensor changes in response to the environment (e.g. an LDR has a high resistance in the dark and a low resistance in the light)
• the change in resistance produces a change in the output voltage of the LDR

1)  The screenshot  below is from http://www.bbc.co.uk/schools/gcsebitesize/design/electronics/componentsrev4.shtml  which explains how to use a potential divider in an electronic circuit.  Please note that in electronics it's common practice not to show the power supply or battery - instead the diagram just shows a 'top rail' which would have the positive of the supply connected to it (V in) and a 'bottom rail' which is connected to 0 volts, the negative of the battery or supply.

2) The video below is a very good introduction to potential dividers - watch it and make notes for yourself.
http://www.youtube.com/watch?v=Mn2i3DPI-a4
 b



The video below includes some clear explanations - watch and add examples to your notes.  He mentions the concept of 'loading' the potential divider but the examples given only cover 'no load' situations - which are what you can expect to come across in AS physics ('loading' is putting a resistor across the output of the potential divider.   http://www.youtube.com/watch?v=u8pAGROJ5N4