I do not understand what happens in the following situation regarding internal resistance of an analog ammeter. The internal resistance of analog ammeter changes if the scale is changed.

For example if the ammeter has a scale such that the max current is 50 mA and then I switch to another scale, where the max current is 500 mA, the internal resistance will be different because changing scale is equivalent to add or remove shunt resistances.

Consider the circuit in picture, where the lamp is a tungsten filament lamp (the resistance value is at room temperature and rises with temperature). I represented ammeter and voltmeter and their internal resistance next to them. Consider the two scales 1. and 2. of the ammeter:

  1. The maximum current is i=50 mA and internal resistance is Ra=6 omh
  2. The maximum current is i=500 mA and internal resistance is Ra=1 omh


simulate this circuit – Schematic created using CircuitLab

Now suppose that I set AM1 to scale 1., I measure the voltage difference across R with VM1 and I find a certain value.

If now I switch AM1 to scale 2. what does happen to the current I and to the voltage V across R?

I tried this with a real circuit and I found that the

  • Voltage V across R increases
  • Current I decreases

How can this be explained? In particular why does the current decrease?

I tried to give me the following explanations: Ohm law states

V=I*R|| =R||*10/(Ra+R||) (where R|| is the parallel given by the lamp and the voltmeter which is almost equal to the resistance of lamp)

I see why V should increase if Ra decreases but, looking at the formula, the current I should increase too!

I think that the problem is to consider the rise in resistance of the lamp, hence R|| increases if V increases and this would explain why current I decreases.

But at the same time, following this reasoning, R|| should also increase when current I increases (since more heat is produced by the lamp and it should get hotter)..

I don't see how to get the explanation right and I do not understand how to "prove" that the current must decrease, just starting from the fact that the lamp has a resistance strongly dependent on temperature.

  • \$\begingroup\$ Please edit your question so that the "R"s used in the text match the "R"s in your schematic. That is: Ra & Rv, not simply R. Also, ammeters usually change range by adding/subtracting shunt (i.e. parallel) resistors, not a series resistor as you show in your diagram. \$\endgroup\$
    – FiddyOhm
    Jan 23 '17 at 18:04

To change the scale of an ammeter, you are changing a resistor in parallel with the meter not in series with it. The full scale voltage will be the same for all ranges. So for a 1 Amp scale on a 1V meter you would put a resistor that gives you 0.1V across the meter. For a 100mA scale, you would put a resistor that gives you 1v across the meter. Ra should be across the meter.


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