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