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A shunt, or more properly called a current shunt, is just a precision calibrated power resistor designed to generate a specific voltage drop when a specific current is passing through the shunt. One uses Ohm's law to calculate the passing current by the formula I = E/R where R is the fixed resistance value of the shunt. A typical resistance value for a high current shunt is .001 ohms. Such a low resistance is made by using a metal bar and filing away material until the resistance is an accurate .001 ohms. Special metals are usually used so the the heat being generated by the high current flowing through the metal does not change it's resistance. A current shunt can be used for measuring both AC and DC currents by measuring the corresponding voltage drop with a AC or DC volt meter. The .001 ohm value allows displayed millivolts to equal amps, that is 1 amp will measure 1 millivolt and 50 amps will measure 50 millivolts.
Here is an interesting E-bay auction selling a current shunt showing pictures of how it would be wired up to read the current being drawn by a 12 volt automobile head lamp.
**broken link removed**
More information here: go to wikipedia.org and enter.. Shunt (electrical)
"Is this mean that with using a shunt I can measure high and small value of current? "
Yes, to limit of resolution of the volt meter used to measure the voltage drop. If using a .001 ohm shunt, then if 100 ma is flowing through the shunt a voltmeter would display only 100 microvolts. Not all voltmeters read this small od a voltage value.
"Can shunts be used in voltmeters like in ampmeters?"
Yes, that is how most multimeters measure current, they use an internal shunt when measureing current and measure the voltage drop across it to display current value.
20 Ω is technically nearly correct. However, the question implies the world's worst ammeter.
1.5 A and 200 Ω is 300 V drop. That would dissipate 450 W, even if you could find a circuit that wasn't seriously affected by a voltage drop of 'just' 300V
Your 20 Ω shunt isn't quite right. The overall resistance should be 20 Ω so the shunt should be 22.2222 Ω and (15 - 1.5A) will go though the shunt at the stupidly high voltage drop of 300 V. Your shunt needs to dissipate 4050 W.
I think someone has missed a "milli" from the ohms.
That would give the original ammeter dropping 0.3 V and dissipating 0.45 W. With the shunt the total power dissipation is 4.5 W.
Thats where I get stuck, your explanation is indeed right, because thats what everyone else says, but it doesn't make sense to me (and perhaps others new to this).
So the correct shunt value vir a 1.5 A meter, to display 15 AMPS, is 22.2 milli Ohms?
How do you get to this? How did you figure it should be milli? I am asking out of curiousness to understand this.
I have done the math here over and over, and I keep getting at 20ohm.
Please help
Do the math for my original scenario for example?
We know that it is milliohms because of the voltage drop and power dissipation. 450 W is just far too much for a meter. That is the power of a big desktop computer and meters just never get that hot.
What you are forgetting is that if you want the meter to read 15A when it is a 1.5A meter, you want (15 - 1.5) = 13.5 A through the shunt.
If the resistance of the meter is 200 mΩ or 0.2 Ω then the voltage on the meter (and therefore the shunt) is 1.5 * 0.2 = 0.3 V
The current is 13.5 A so the shunt is 0.3 / 13.5 = 0.02222222 Ω
In your calculation, I don't see why you used mA, as in 15000/1500 rather than 15/1.5 but that isn't what you got wrong.
ah ok, I see.
But I see where the misunderstanding comes in, the meter Ive got here, chosen for this argument has an internal resistance of 200 ohms, NOT 0.2 ohms
(see MONACOR INTERNATIONAL:New products )
That means according to you equations (which are perfect )
1.5 * 200 = 300
Shunt being 300 / 13.5 = 22.2 Ohm
This is how I got to 22 ohm, and not 22 mOhm.
The technical data sheet has more realistic figures for all of the meters except the 30 V / 1.5 A
I think that the meter is 1 mA/200 Ω, just marked 30 V / 1.5 A
To make it read to 30 V, it needs 29800 Ω in series
To make it read to 1.5 A it needs 0.133422 Ω in parallel
If they are offering them as 30 V / 1.5 A, then they could be including both resistors with the meter. It could just be that the marking is 30 / 1.5 A and they haven't actually stated the meter current.
There is no way that they could have one resistance to do 30 V and 1.5 A. It would have to be 20 Ω resistance and it would then produce 45 W of heat at full scale, which would melt things in seconds in that case. It would certainly be a bloody awful ammeter, as it would drop 30V. Most of the other ones in their range drop 60 mV. Now in a typical DC circuit of 5 - 24 V, a drop of 60 mV isn't usually significant, but a 30 V drop is just silly.
I dont know where they came up with the 200. But I suppose you have a point.
I have this exact meter, and no resistances came with it.
I dug it out of a box now, and used my fluke73 ohm meter to measure it's terminals. 5 ohms. Not sure if this is the correct way, but 5ohms sounds better than 200. Blackbeard alone knows where they come up with the 200ohms in the documentation.
With a 5K resistor, it becomes a 0-30V voltmeter and for shunt, I would assume(for a 15AMP scale):
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