Download Internal Resistance

NOTES
INTERNAL RESISTANCE
As a concept, this is slightly tricky to grasp, but not impossible.
Although we normally deal with resistors having resistance, and light-bulbs and a
whole host of other components having resistance too, it’s important to recognise that
the connecting wires in circuits also provide a little resistance of their own. By
thinking about potential dividers for a second, you should realise that so long as the
components in the circuit have quite large amounts of resistance, then the
comparitively small amount of resistance found in copper wires will be insignificant,
and need not worry us. In these circumstances, we can say the resistance of the wire is
negligible.
Do remember that this assumption is only for the sake of convienience, though. The
resistance of the wires, whilst being small, is still there.
Let’s now consider a transformer. Our power supplies invariably contain a transformer
to convert the dangerous 240V a.c domestic supply into a lower (and safer) p.d.
Transformers are really long lengths of wire wrapped around an iron block. So the
transformer has a little bit of resistance of its own, in the wire that it’s made up of.
This is what Internal Resistance is. It’s constant for any particular power source, and
it acts as if it was in series with the main circuit.
Now imagine that you are an electron, whizzing round a circuit powered by a
transformer. All the way round the circuit you will be meeting at least some
resistance... ...in the components, in the connecting wires, and of course in the
transformer itself. And just as resistances in the main circuit cause p.d to be dropped
across them, the transformer’s own resistance causes a loss of p.d too. So the upshot
of the transformer’s own resistance is that instead of getting the full p.d that the
transformer could supply (called the e.m.f), you actually get slightly less, because
some p.d is lost to this internal resistance. Through Ohms Law (p.d = I x R), you can
see that the amount of p.d lost this way is dependent on the current being drawn by the
main circuit. We call the main circuit the Load.
Example 1
The circuit opposite shows a 1000 resistor
as the load in a circuit, powered by a 10V
supply. To establish the effect of the internal
resistance of the power supply, we would be
better redrawing the circuit again to show the
internal resistance in some way. Because
internal resistance acts as if it was in series,
and is really quite small, we show it as a
series resistance or value ‘r’.
O
10V
O
1000
Current Electricity & Elastic Properties of Solids - AS Physics Module 3 (cjh-01)
o
We have simply replaced the ‘power supply’
symbol with a more detailed version.
o
r
The internal resistance is shown as a series
resistance (r), and the full p.d that could be
supplied is given as ‘‘, which is the true
10V.
In this example, let’s take the internal
resistance ‘r’ as being 5. The 1000
resistance is the load, ‘R’.
1000
If the load is 1000, then the total resistance of the whole circuit is r + R =1005.
With a 10V e.m.f, we can use I= V/ R to determine the current flowing in the circuit.
I = V/ R
= 10/ 1005
= 0.00995A or 9.95mA
With this current flowing, how much p.d will be lost to the internal resistance?
p.d = I r
= 9.95 x 10-3A x 5 = 0.05V
So with a load of 1000 the actual supplied voltage would only be 9.95V, not 10V
What happens if we reduce the load resistance, and cause more current to
flow....?
If the load is 100, then the total resistance of the whole circuit is r + R = 105.
With a 10V e.m.f, we can use I= V/ R to determine the current flowing in the circuit.
I = V/ R
= 10/ 105
= 0.095A or 95mA
With this current flowing, how much p.d will be lost to the internal resistance?
p.d = I r
-2
= 9.5 x 10 A x 5
= 0.475V
So with a load of 100 the actual supplied voltage would only be 9.525V, not 10V
If the load is 10, then the total resistance of the whole circuit is r + R = 15. With
a 10V e.m.f, we can use I= V/ R to determine the current flowing in the circuit.
I = V/ R
= 10/ 15
= 0.667A or 667mA
With this current flowing, how much p.d will be lost to the internal resistance?
p.d = I r
= 6.67 x 10-1A x 5 = 3.33V
So with a load of 10 the actual supplied voltage would only be 6.67V, not 10V
Current Electricity & Elastic Properties of Solids - AS Physics Module 3 (cjh-01)
Current Electricity & Elastic Properties of Solids - AS Physics Module 3 (cjh-01)