# Why does putting the leads of an ammeter in a receptacle short the outlet?

• An ammeter needs to be connected in series with a circuit to test current, which sticking the leads in an outlet does not do?
• Also, there is no amperage to measure on an outlet with nothing plugged into the outlet to draw current, correct?
• Why does attaching the leads to each hole of the outlet short the outlet and trip the breaker?

Any explanations would be appreciated! I'm trying to refresh myself on circuit theory, so feel free to be as technical as needed.

• your ammeter draws full current, because it presents itself as a near short circuit Apr 21, 2021 at 0:38
• An ammeter is like measuring how much water flows through a pipe. How do you do that? Well, one thing you can't do is block the flow of water when you connect it up. So when you connect an ammeter up to something, the ammeter is like an open free-flowing pipe, aka a piece of wire aka a short-circuit between its two terminals. Apr 21, 2021 at 0:39
• sticking the leads in an outlet ... definitely in series with current flow .... An ammeter needs to be connected in series with a load Apr 21, 2021 at 0:43
• If you like to think about it this way, you are connecting the ammeter in series with a short circuit. Apr 21, 2021 at 10:03
• "there is no amperage to measure on an outlet with nothing plugged into the outlet to draw current" - right, but now it's not an outlet with nothing plugged into it anymore Apr 21, 2021 at 12:24

An ammeter has a very low resistance, and is normally inserted in series with the circuit whose current is to be measured.

When you plug the ammeter leads into the Line and Neutral holes in an outlet, you are placing a short circuit across the outlet - the meter will draw a very large current and (hopefully) trip the circuit breaker feeding the outlet (and perhaps sacrifice itself in the process).

Normally, you would connect the ammeter in series with some load, and the load would limit the current. The ammeter would then display the current drawn by that load.

An ammeter needs to be connected in series with a circuit to test current, which sticking the leads in an outlet does not do?

It does. Your meter is in series with a circuit: power plant, a long cable, your meter, same length back to the power plant.

Also, there is no amperage to measure on an outlet with nothing plugged into the outlet to draw current, correct?

That would be correct, but:

You just plugged in the meter. That's a Short circuit. A huge current will flow.

Why does attaching the leads to each hole of the outlet short the outlet

Because an amperemeter is effectively a short. It's just a thick piece of wire with a sensor next to it.

and trip the breaker?

Because a huge current flows.

• There are also significant impedances in the circuit from transformers along the way, but the current through a short at a residential electric service is still on the order of 10,000A. Apr 21, 2021 at 14:46

Ammeters normally drop 50 mV full scale thus R=50mV/I-full scale so putting any voltage source across it will blow the shunt.

Exceptions are ones with very high currents or extremely low pico amps.

It is meant to go in series with a DC load not to exceed this.

The other answers are correct but may not quite 'gel'.

User253751's comment is correct.
An ammeter is in series with the load SO THAT the load current flows through it so that it can be measured. The meter resistance is typically under 1 Ohm so that minimum voltage drops across the meter and minimum power is dissipated in the meter.
eg if meter resistance is 0.1 Ohm, at 10A current V = IR = 10A x 0.1 Ohm = 1V.
Power dissipated in meter = I^2 x R = 10 x 10 x 0.1 = 10 Watt.

When you place an ammeter "across the mains" the load is the meter leads - some fraction of an Ohm. With just the leads and no meter the current would be VERY high. With the meter added it's only slightly less and almost always FAR more than the meter is designed to measure.
In a domestic situation a mains outlet may deliver well over 100A before the fuse blows.

All your questions actually lead to the simple question: "What is inside the ammeter?" or, in other words, "How do we make an ammeter by a voltmeter?" Here is my story which I often tell to my students...

In the 19th century, there were very good ammeters (galvanometers) but there were no voltmeters. So they had to build them, according to Ohm's law I = V/R, by connecting a resistor in series with an ammeter.

In the 21st century, there are perfect voltmeters (ADCs) but there are no ammeters. So we have to build them, according to Ohm's law V = I.R, by connecting a resistor in parallel to the voltmeter. So, today's ammeter is a voltmeter in parallel to a resistor. I have considered this arrangement in Class 2 of Basic Electronics course.

The problem is that this current-sensing resistor must have a low enough resistance for two reasons: first, not to change the current being measured and second, not to exceed the range of the voltmeter.

For example, if you are in the 10 A range, and the voltmeter has a maximum range of 10 V, the resistance should be less than 1 ohm. This means that in practice, as other answers also say, there is a short circuit (a piece of wire) between the probes:

Fig. 1. The current-sensing resistor (at the end of the photo) is a piece of thick wire.

Unfortunately, at small current ranges, the resistance becomes significant and the error can be large. Then a clever trick can be applied - compensating the voltage drop (resistance) by an equivalent voltage ("negative resistance"). It brings us to the famous circuit of the transimpedance amplifier. Here are some of my materials dedicated to this idea:

Reinventing the transimpedance amplifier

Op-amp inverting current-to-voltage converter

Voltage compensation

How do we create an almost ideal ammeter

How do we build an op-amp ammeter?

Electrical outlets at home are typically connected in parallel with each other. When you connect an ammeter directly to an outlet, you are connecting an infinitesimally small resistance in parallel with other connected devices.

Now consider the equivalent resistance of all of your connected devices combined, which is given by:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$$

From the above equation one can deduce that the equivalent resistance is lower than the lowest of the parallel resistances. Hence the equivalent resistance is lower than the ammeter's resistance and has a value close to zero. Ohm's law gives the current in the whole circuit:

$$I = \frac{V}{R}$$

Which implies that an enormously high current flows into the circuit, hence tripping the breaker. It is also very likely that this high current would damage the ammeter and perhaps melt some of its parts.