is it possible to make an ammeter read more amps than the scale reads? For example, could a ten amp meter read 15 amps without destroying it?
No. Unless the meter has overload protection built it, you would probably destroy it.
If you have an idea of the current you expect, and the normal load on the circuit you could add a low value resistor, of appropriate wattage to the circuit. Measure the voltage across the resistor, and calculate the current. You need the value of the resistor to be low enough that the voltage drop across it is not ‘significant’ relative to the drop across the real load for this to be reasonably accurate. Of course, if you have a fixed resistive load, and you know the voltage, you already know the current, and don’t need to measure it.
No. An ammeter is basically a short circuit. Of which there are several types.
The old mechanical meters could be moving iron (AC), which was designed for a specific current. Another technique was a 5A AC movement connected to a transformer.
Or for DC it used a shunt and a small DC voltmeter across the shunt. The shunt was typically inside the case. Larger current versions had the shunt externally mounted. Which could be changed if needed.
The safest and easiest way to read higher currents is to buy a meter that can handle them.
The proper way to do what you ask is to put a resistor ACROSS the ammeter. This resistor would bypass some of the current. In your example, the shunt resistor would carry 5 amps while leaving ten amps for the ammeter. So, when the circuit is drawing 15 amps, the meter would only indicate its ten amps. The calculation of the shunt requires only that the DC resistance of tha ammeter be known. (I assume an analog, moving coil meter). Then ohms law will find the value of the shunt. You will also need to draw up a new scale for the meter with a new full scale reading of 15. For digital meters, I don’t think it be practical as they have a constant impedance of around 10 megohms. (I think)
MOST METERS have a built-in (or external) resistor that they may just sample the current passing through. Exceeding this will cause the meter to go offscale or fry the resistor - making accurate readings impossible.
15 AMPS is what most 110-120 volt fused Husehold Circuits provide.
If I’m monitoring Household power, 15 mps (and beow) is what I want. 10 amps (and below) a 10 amp meter is fine, but if I’M RUNNING MORE THAN 10, Better to find a 20 amp meter which will read 15 amps @ 3/4 scale.
Absolutely, but you need an attachment for your DMM. A current clamp will allow you to measure any circuit without actually breaking and becoming part of the circuit itself.
Instead of being limited by the internal circuit protection of the DMM itself, typically 10-20A you have a range of 0-400A. Some even go higher.
Probably overkill for a MRR but if you like to do home or automotive electrical work it is worth it.
Jim
Certainly - but it will require a bit of experimentation. Blind Bruce had it absolutely correct.
You can make a shunt for an analog meter from a piece of wire coathanger. It has low but non-zero resistance. Solder a wire to one end and another somewhere along the wire. Closer together means less resistance and less meter swing. We used to do this often in the lab.
Digital meters also use a shunt for the 10 amp range. The input impedance (resistance) should be in the spec sheet (we can at least be hopeful).
It would really help if you knew the series resistance of the meter. Then you can use a simple current divider equation to fugure out the shunt resistance needed. All in all though it’s more fun to experiment. Just begin with safe currents. IE: Pass five amps through the meter and adjust the shunt untill the meter reads 2/3 of five amps. Make a new scale and you’re done.
Good luck,
Karl
You could put a low-value resistor in series with the load, and then measure the voltage accross the resistor. The current through the resistor (and the load) is calculated as the voltage measured accross it divided by its resistance. The resistor value needs to be small - less than 1 Ohm, and it needs to be capable of dissipating the resulting power. It also needs to have as accurate as possible resistance value. It will affect the behavior of the circuit, making the load draw less current. You may be able to calculate the current without the resistor as a function of the measured current. You’ll need some background in the application of “Ohm’s Law” which states that E=IR, or the Voltage equals the product of the current and the resistance. The power a device is dissipating is the product of the voltage accross it and the current through it (P=IE).
Or, get a bigger meter 
Jim
Well it is railroad related in a convoluted way but it is for my riding lawnmower that just had its fortieth birthday. So I found an OEM ammeter for it on E Bay that is 10 amp but the charging system is 15 amp. however since it is 40 years old I am hoping it doesn’t quite reach full output. The railroad connection? It pulls the railroad car I built for my grandchildren to ride in when they visit.
Ya know, when all is said and done, that ammeter has no need to be super accurate. It is only there to confirm that the battery is charging or not. I would put a piece of small diameter wire across it and see what happens. #20 or even #22 wire should be large enough. About 6 or 8 inches long.
agreed!