If I have connected an ammeter, with some specific shunt resistance Rₛₕ, to a DC source, through ideal, of zero resistance, conductors, then whatever the value of the shunt resistance the ammeter will all the time give the same reading, so in such a situation the use of an ammeter to measure the current is useless, right?
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\$\begingroup\$ Yes, it's of no use. Why did you ask? \$\endgroup\$– datenheimCommented Dec 27, 2022 at 10:16
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2\$\begingroup\$ The downvote is inappropriate because the question is meaningful and makes us think. \$\endgroup\$– Circuit fantasistCommented Dec 27, 2022 at 11:29
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1\$\begingroup\$ What kind of source? What load? (None, it seems?) If there is no load, why do you care what the reading is? \$\endgroup\$– Tim WilliamsCommented Dec 27, 2022 at 14:35
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4 Answers
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An ammeter is actually a voltmeter that measures the voltage drop across the shunt.
Voltage V = I x RSH
The voltmeter, graduated in amperes, displays the load current 'I'.
In practice, shunts are designed to drop 50, 60, 75, 100 or 150 mV when operating at their full rated current.
Matching voltmeters are also designed to have full scale deflection of 50, 60, 75, 100 or 150 mV.
The circuit in question is as shown below.
With the voltage across the shunt being as high as the source voltage, instead of being limited to the range of mV drops listed above, the following would result:
Pinning of the pointer to full scale and subsequent damage, in the case of a moving coil meter.
'Out-of-range' being displayed, in the case of a digital meter.
Hence the stated application of current measurement is not feasible.
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1\$\begingroup\$ You have answered a question that I have never asked. In my question I assumed no load is anywhere in the circuit in addition to connecting everything through ideal conductors. Given all that, the voltage drop across the ammeter is always constant whatever the value of the shunt resistor and this in turns forces the same current through the ammeter. \$\endgroup\$– JackCommented Dec 27, 2022 at 10:01
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\$\begingroup\$ Thank you for pointing it out, Jack. My answer has been duly edited. \$\endgroup\$– vu2nanCommented Dec 27, 2022 at 13:01
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\$\begingroup\$ But if the voltage is the full scale deflection voltage:-)? \$\endgroup\$ Commented Dec 27, 2022 at 15:34
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1\$\begingroup\$ @Circuit fantasist - Hi Cyril, I presumed that it's a standard shunt with a matching voltmeter (FSD in mV) connected across a standard power source ( V ). But I do agree that use of a power source having its voltage same as or lesser than voltmeter FSD will result in the same reading irrespective of the shunt. \$\endgroup\$– vu2nanCommented Dec 27, 2022 at 16:55
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You ask is it useless.. not really, at least in an ideal situation.
Assuming nothing else is in the circuit* the current will be what is indicated on the scale.
The value of the ideal DC voltage source will be the shunt resistance times the indicated current (assuming the meter resistance itself is much higher). So if you have a 100mV shunt with a meter that reads 50A at full scale (resistance 2mΩ) then if you apply a voltage such that the current reading is 25A you know the voltage must be 50mV.
Of course ammeters are typically designed to drop a very low voltage so they don't unduly affect the circuit being measured, and if you put a much higher voltage on the meter you may destroy the shunt and/or the meter, or at least blow a fuse*. Now, if there is a fuse in the circuit there is significant additional resistance added, and the first paragraph does not really apply. Even a real Kelvin-connected shunt without a fuse has some series resistance in the high current connections that will muck up that calculation. You can, even in a real situation, however place an upper bound on the applied voltage.
answered Dec 27, 2022 at 11:12
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My initial answer
You have connected a real ammeter to an ideal voltage source.
The problem of this arrangement is that the voltmeter inside the ammeter actually measures the entire source voltage... and this voltage does not depend on the shunt resistance Rsh (here we ignore the fact that the current will be significant). Thus your ammeter will act as a voltmeter.
The problem with this 22nd century meter is that it is not a "true" ammeter such as a 19th century "coil ammeter" which directly measures current through the magnetic field it creates. The modern one measures current indirectly by voltage across a resistance and that is why it is "mislead" in this situation.
Edit 1:
We can observe two cases:
1. Known resistance. If the meter "knows" what the resistance Rsh is and uses it to calculate the current (this is the situation when we switch ranges and Rsh changes stepwise), the reading is true (in the sense that this is the current flowing through the shunt). But actually this value is very wrong because it is entirely determined by Rsh (the ammeter). It turns out that by switching the ranges, we (the ammeter) set the current... and it should be exactly the opposite - the current should be set by the external circuit.
2. Unknown resistance. However, if the meter does not "know" what the resistance is (for example, we change it "invisibly" for it :-), the situation gets even worse. Now the ammeter not only changes the current in the circuit, but also reads it incorrectly.
Edit 2:
Above we considered the case where a real ammeter made by a voltmeter in parallel to a shunt resistor is connected in parallel to an ideal voltage source.
It would be interesting to consider the dual case where a real voltmeter made by an ammmeter in series with a ballast resistor is connected in series to an ideal current source.
Now the ballast resistor cannot change the current and the ammeter will show the current set by the ideal current source.
answered Dec 27, 2022 at 10:18
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1\$\begingroup\$ Hi, People are not allowed to write comments on other people's answers, within an answer of their own. (a) That is using the site like a forum - not allowed. (b) Those authors don't get notified. (c) It can cause big, big problems (comment wars) if someone referenced in that way tries to respond. (d) Your post becomes obsolete if those answers are changed. || Instead, the policy is: If you want to comment on someone else's answer, you write a comment (which must comply with policy) below their answer. || Use your answer to answer the question only. \$\endgroup\$– SamGibson ♦Commented Dec 28, 2022 at 14:25
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1\$\begingroup\$ If you are online, please edit your answer to remove that commentary on other answers. If you have gone offline, we will make the edit for you and you can (if you want to) write comments (which comply with the comment policy) below those answers later instead. Thanks. \$\endgroup\$– SamGibson ♦Commented Dec 28, 2022 at 14:26
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\$\begingroup\$ @SamGibson, I see... Sorry for the trouble. My desire was to be fair to these authors by quoting the texts from their answers related to my explanations. BTW what do you think about my new insight described in Edit 2? \$\endgroup\$ Commented Dec 28, 2022 at 14:59
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If I have connected an ammeter, with some specific shunt resistance Rₛₕ, to a DC source, through ideal, of zero resistance, conductors, then whatever the value of the shunt resistance the ammeter will all the time give the same reading
That depends what "same reading" means.
As other answers have pointed out, an ammeter is a voltmeter across a shunt resistance. If the shunt resistance is the only thing in the circuit with an ideal voltage source, the voltage across it will always be equal to that of the voltage source, regardless of the value of R_sh.
But does that mean that the "reading of the ammeter" will be independent? Not quite, because R_sh is the value used to convert volts (which are measured directly) to amps (which the ammeter presents to the outside world). A typical milliamp meter with a 1Ω resistance has a conversion factor of 1A/V, while a higher-current meter with a 33mΩ shunt has a conversion factor of 30 A/V. The latter will read 30 times as many amps in your scenario... which is fair, because it will be passing 30 times as many amps, with no other load in-circuit.
