internal resistance

Last week, we learned about internal resistance of cells. Page 24 of your printed notes explains how to use a simple series circuit containing a cell, resistance box, ammeter and voltmeter to determine the internal resistance of the cell.  By plotting a graph of your dat, with current on the x-axis and voltage on the y-axis, you can find the internal resistance of the cell.

The video below shows the same type of experiment, but uses a potato and two different metals in place of a normal cell.  Watch the video and note the values of I and V each time the resistance is changed – remember to pause the video each time so you can write the results.  Just scroll back if you miss any.

Now plot a graph with current along the x-axis and TPD along the y-axis.  If you don’t have any sheets of graph paper handy, there is a sheet available to download using the button at the end of this post.  Alternatively, print a sheet from a graph paper site or use google sheetsto plot your results.

Draw a best-fit straight line for the points on your graph and find the gradient of the line.  When calculating gradient, remember to convert the current units from microamps (uA) to amps (A).

The gradient of your straight line will be a negative number. The gradient is equal to -r, where is the internal resistance of the potato cell used in the video.

You can obtain other important information from this graph;

  • Extend your best fit line so that it touches the y-axis.  The value of the TPD where the line touches the y-axis is equal to the EMF of the cell. (Explanation: on the y-axis, I is zero so TPD = EMF)
  • Now extend the best-fit line so that it touches the x-axis, the current at that point is the short-circuit current – this is the maximum current that the potato cell can provide when the variable resistor is removed from the circuit altogether and replaced with just a wire.

3 thoughts on “internal resistance”

  1. Hello,
    Is there any whole unit notes on CFE Higher Electricity available? I find Mr Noble’s notes extremely helpful. Thanks in advance.

    Reply

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