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Why ammeter voltage drops can be ignored The resistance of an ammeter is negligible.
This is because if you put a measurable voltage drop across an ammeter a very large current will flow through it and its circuitry will melt - it may even cause a fire! Because its resistance is very much smaller than the other components in the circuit strand, the voltage share that it gets is negligible also and generally we can ignore the voltage drop that an ammeter would get and assume that for all intents and purposes it is zero. This is not really the case - its resistance is not zero - but we can ignore it anyway. It does have a resistance and therefore does take some of the voltage drop but if your voltmeter can only be read to three or four significant figures (which is usually the case) it will not record that difference and read zero when placed in a circuit.
Suppose an ammeter has a resistance of 6.3 mΩ and it is positioned on a strand of a circuit with a 420 mΩ resistor and a potential difference of 9.0 V is provided by a power supply.
Total resistance of the strand is 420 mΩ plus 6.3 mΩ as the resistor and the ammeter are in series. Therefore the total resistance of the strand = 0.420 Ω + 0.000 006 3Ω = 0.420 006 3 .Ω The voltage is shared out across the strand according to the resistance values… each ohm gets the same voltage drop. So, 0.420 006 3 Ω share 9V therefore each ohm gets 9 / 0.420 006 3 V and the ammeter gets 0.000 006 3 x 9 / 0.420 006 3 V = 0.000 14 V This voltage value is so low it would not show up on the multimeters we use in class. The voltmeter would read zero and we can therefore ignore voltage drop across the ammeters in the circuit.
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That is why an ammeter should always be connected in series with a component
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