Internal Resistance
Imagine you go to a shop to buy a 9V cell. In the shop you check the terminal p.d. with a voltmeter and, sure enough, it reads 9.0V
You have a bulb at home which apparently needs 9.0V to work. You connect your cell across it. The bulb lights up but you suspect that it is not quite as bright as you thought. A voltmeter across the bulb reads 8.7V. What's going on?
We appear to be losing 0.3V but only when a current is being drawn from the cell.
The more current is drawn from the cell the more voltage we lose. Why?
The cell is giving 9 J to every coulomb that flows through it but some of this is being transferred getting through the cell itself, after all it has resistance too. We can think of the cell as being another resistive component in the circuit, the 9V is being shared between it and the external circuit.
In the example at the top of the page:
E = 9V, V =
8.7V, v = 9 - 8.7 = 0.3 V, I =
0.4 A, R = V / I = 8.7 / 0.4 = 21.8
, r = v / I = 0.3 / 0.4 = 0.75
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Which of these 9V batteries could start your car? Your
starter motor needs a current of about 100A, which of these could deliver
it?
The first battery has a very small internal resistance so it can provide a large current at a relatively high p.d. ( it will drop to about 6V in use). The second battery has a much larger internal resistance. It can only provide very small currents. |
This graph shows how the p.d. across a cell varies with the current drawn from it.
In most questions you have to do internal resistance will not be a factor and you can assume that the voltage of the power supply is fixed. It should be obvious when you do need to consider it.
