• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Design and Carry Out an experiment to determine the EMF and Internal Resistance of a standard laboratory power pack.

Extracts from this document...


FINDING THE INTERNAL RESISTANCE OF A POWER PACK AIM Design and Carry Out an experiment to determine the EMF and Internal Resistance of a standard laboratory power pack. THEORY This information was taken from the Collins advanced science Physics textbook, the Cambridge Advance Science Physics 1 textbook and from notes taken in class. E.M.F or Electro Motive Force is the opposite of potential difference, in that; it is the situation where a voltage is gaining energy. This seems unlikely, but it is required in order to allow an electric circuit to function. An electric current is a flow of electric charge, the charge flows around the circuit, transferring some of its energy to areas of resistance along the way (resistors, filament lamps, buzzers). At some point along the way the energy must be initially supplied, otherwise the whole process couldn't function. At the beginning of the process the power supply provides the charge with energy to pass to the circuit. This is the electro-motive force. EMF is a type of voltage along with potential difference and these are defined as: A Voltage where the charge is losing energy is a potential difference, V. A Voltage where the charge is gaining energy is an electromotive force, E. A relationship exists between volts and joules. A 10v power supply, for example, will give 10J of energy to each coulomb that it pushes round the circuit. ...read more.


need be worn, however, it should be noted that the rheostat will become very hot after power is put through it, and must be left to cool down before handling it. FAIR TEST In order to make this experiment a fair test, several factors have been taken into consideration to make the experiment accurate. Firstly all of the same equipment will be used throughout the investigation, as this will affect the results from the power pack. The rheostat will be allowed to cool down in between experiment repeats, to make sure that the temperature is not affecting the results in any way, and finally, the same number of results will be taken from each repetition, as this will be used on the table to calculate the gradient and will be the basis of the results. DIAGRAM PREDICTION I predict that, based on my initial research, the resistance will become reduced as the current is increased; this should not affect the internal resistance, which (hopefully) will remain the same. The e.m.f should also be slightly higher than the stated voltage on the power pack, as the e.m.f is the highest value for energy. Method 1. Set up the apparatus up as shown in the above diagram. 2. Set the rheostat to it's minimum value 3. Make note of the results on both the ammeter and the voltmeter 4. Increase the current using the rheostat and take at least 5 more readings. ...read more.


This provides an evidential mark of consistency for this method of data collection. I noticed during the collection of results there was instability in the voltmeter and ammeter of ? 0.2v and ? 0.1v respectively. This, obviously lead to certain mistakes being made through a simple case of choosing the wrong result, therefore leading to a error range of 4 volts and 2 amps. I have included a graph of current against potential difference, which includes error bars; this shows how the fluctuation on the voltmeter can have an effect on the results accumulated. There was also an issue when trying to take reading at low temperatures; this was due to a reduced sensitivity on the rheostat. The result of this incident was that the accuracy of results at low voltages has a greater error range. This cannot be displayed on the graph as there is no direct indication as to the new error ranges. If I could get an opportunity to repeat this investigation in ideal conditions I would use a rheostat with a greater range of selections, this (hopefully) would reduce the error on lower voltages. There were many limitations that I encountered during the conduction of this experiment; mainly this involved the equipment and digital readouts. Replacing these with significantly more accurate pieces of equipment would reduce the error range for the gradient, therefore giving more accurate final results. I would also spend a lot more time on taking many more results, especially in the lower voltages, as this would help in reducing the amount of error from the rheostat. ?? (Footnote continued) Greg Rogers ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Electrical & Thermal Physics essays

  1. Peer reviewed

    To determine the internal resistance of a dry cell using an ammeter and a ...

    3 star(s)

    - Ir 0.9=?- 0.085 X 6.00 1.41 V =? The e.m.f. of cell Y is 1. 41 V. Question for Discussion : 1. What do you notice about the values of the internal resistance of the cells X and Y? Internal resistance of X(2 ohm) is smaller than that of Y(6 ohm)

  2. In this experiment, we will measure the e.m.f. and the internal resistance of a ...

    following: V = IR - Ir or V = E - Ir For graphical method, we can plot a graph of Voltage against Current. The graph should result in a straight line because the graph of V against I is proportional.

  1. Investigating the Emf and the internal resistance of a dry cell.

    The internal resistance = 0.57 - 0.05 = 0.07 r = 0.5 0.07 Emf The Emf of a cell is the y-intercept of the line on the graph. Therefore: The Emf of the line of best fit = 1.68v Due to the uncertainty of the results I will use the

  2. Investigating the E.m.f and Internal Resistance of 2 cells on different circuit Structures.

    + (0.5*12) E = (1 + 6) = 7 V I also expect the internal resistance to be double the internal resistance of a parallel circuit with two cells because in a parallel circuit the resistors are halved so therefore in series the internal resistance is (r).

  1. The potato - a source of EMF

    I also predict that the EMF of the potato will decrease as more and more of the chemicals inside of the potato are being used up over the duration of the experiment; by the time the 3rd set of readings have been taken the potato cell will have reacted with

  2. The aim of my investigation is to determine the specific heat capacity of aluminium.

    * I chose to record the temperature, voltage and current for 10 minutes starting at room temperature. * Switch the power pack off and continue recording the temperature, current and voltage for a further 10 minutes. * Remove the lagging from the aluminium and allow to cool * Once cooled

  1. The Resolving Power Of The Eye

    Mark this point on the floor for measuring later when in full light. Continue this process changing the holes used for the different readings on different separations. Repeat the whole experiment three times. Safety Safety is not enforced strongly as the experiment has few dangerous elements.

  2. Internal resistance investigation - I will conduct the following investigation with the aim to ...

    If, as hoped, my method has been successful and my results are fair and accurate then the gradient should be the internal resistance of my lemon battery. If there are any apparent anomalies in my data I will do my best to detect and explain them with regard to the relevant theory.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work