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

To find out the internal resistance and EMF of a given power supply

Extracts from this document...



Internal Resistance and EMF (ELC)

Physics Portfolio By Clement Ng 12.6

Aim: To find out the internal resistance and EMF of a given power supply.


  • Arrange apparatus as shown in the below circuit diagram
  • Start off by recording the corresponding voltage (terminal pd) and current by figuring out the values available in the voltmeter and ammeter.
  • Repeat previous procedures while the external load is varied, so that you obtain a set of voltage and current readings.
  • Calculate averages and plot a graph of current against voltage. Use the graph to figure out the internal resistance and the EMF of the power supply.
...read more.




















Reading #3:

Voltage (v) (± 0.01V)

Current (A) (± 0.01A)

Resistance (Ω)

(Provided on resistor)



















Observations: As different external loads, or resistors were plugged into the circuit, several voltage and current readings were recorded. It was also noticed that the readings on both meters kept on jumping, therefore uncertainties of ± 0.01 was deduced for both the current and voltage values.

Skill 4 Data Processing

Arranged in increasing voltage and resistance values:

...read more.


Improvement 3:

To improve on graph plotting skills, we could use a computer to help us. Many computer software’s nowadays can help us plot the results and calculate the gradients directly. After the experimental results are obtained, simply copy them into the program and plot graphs of V/I. These graphs would be much more accurate then hand plotted ones, and internal resistance values would be calculated to the highest degree of accuracy.

Unfortunately, allowing the computer to do the job for you does not show any skills in data processing. Thus another improvement can be done by using calculus. Calculus is a good tool in mathematics to calculate the gradient of a known equation. The final results maybe even more accurate then the computer values. However, this would be inappropriate in this experiment, since knowing the equation would already give as the internal resistance and EMF values. This improvement maybe effective on other investigations.

By Clement Ng 12.6

Thursday, July 4 2002 05:12AM


...read more.

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

    / (0.24 - 0.00) = 6.38 ?m = [(8.21-7) + (6.38-7)] /2 = 0.30 The e.m.f. of the dry cell: 1.24 V � 0.04 V Total internal resistance of the dry cell: 0.7? � 0.3 ? The e.m.f. of one cell = 1.29 V� 0.04V Since the resistance of the resistor was assumed to be 5.6 ?

  2. The potato - a source of EMF

    Internal Resistance: The internal resistance when the electrodes are 4cm apart is approximately half of what the internal resistance is at 8cm apart. However, the internal resistance does not decrease as much from when the electrodes are then moved to 2cm apart, therefore the distance between electrodes is not proportional

  1. Measuring the e.m.f. And Internal Resistance of a Cell

    * ? = V + I r Rearrange to get: V = -r I + ? This equation is now in the form y = m x + C This means that ; y is the potential difference m is the negative internal resistance x is the current C is the e.m.f.

  2. Sources of e.m.f. – internal resistance.

    Consider a source of e.m.f. E with an internal resistance r, delivering a current I into a load resistance R. The current I is given by: I=E/R + r The power P dissipated in the load is given by: P=I2R=(E/R + r)2R For a given internal resistance r, there will

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

    This is a very large internal resistance as the voltage produced was very high compared to the current, and means that, of the energy produced by the battery, only a small amount is put into the circuit, and a lot of energy is taken out and converted to waste heat energy.

  2. The Resolving Power Of The Eye

    For these reasons certain changes were made, firstly instead of one measurement being taken three sets of six will be taken for improved accuracy, a green LED will replace the white bulb as the LED has a known wavelength. The whole experiment will be conducted in complete darkness and pinpricks

  1. To find the factors that affect the amount of E.M.F. being produced. The amount ...

    I plug the circuit wire to it and turn it on. Then after I it on I change the frequency of the current from 0Hz to 10Hz and I record down the width and the height of the electro-motive-force wave.

  2. To find the factors that affect the amount of E.M.F. being produced. The amount ...

    I record down the result and then I turn off the power supply. I change the circuit wires now to 0V and 4V. Then I do the same thing as I did before except that I change the voltage for every experiment from 0 and 4 to 0 and 6, 0 and 8 and 0 and 12.

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