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Let's say that I have two wires spliced together with a screw terminal: I don't know exactly how they are called in English, maybe Mammouth? These things here, where you have one wire "screwed" on one side and the other wire connected to the other side: Mammouth

Since the screws are in contact with the conductors inside, can I measure the current with a multimeter by probing the screws?

I feel like it's a bad idea: I'm picturing the connector as a node that connects the two wires and the probes together, which is wrong, since the multimeter should be in series with the circuit. Given that when used for measuring current it has a very low resistance, current should flow pretty much all through the multimeter, shorting out the connector, thus risking to blow a fuse inside the meter or even worse.

Am I correct, and should never try to measure current by probing the screws?

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No, you cannot measure the current using a multimeter without breaking the circuit.

You may be able to use a clamp-on meter to measure the current by clamping it over one of the wires. Older ones were all AC-only but many newer ones can measure AC or DC. Photo from Fluke.com.

enter image description here

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    \$\begingroup\$ If you connect it across the connector (shorted) you'll get some unpredictable reading depending on the resistance of the terminal strip. If you connect it between two non-shorted terminals with voltage across them you'll probably blow the fuse in the meter, and maybe much worse (especially with a cheap non-fused meter current range and mains voltage). \$\endgroup\$ Commented Apr 16, 2021 at 17:46
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    \$\begingroup\$ You could, however, treat the block of conductor that is inside the terminal block as a shunt and simply measure the voltage across it. Granted it's not calibrated but it is possible, in theory at least. \$\endgroup\$
    – jwh20
    Commented Apr 16, 2021 at 18:10
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    \$\begingroup\$ @RussellTeapot, "I'm shorting out the connector if I try that?" Two screw terminals on one terminal block are already a short circuit. You might as well just touch the meter probes together. It would be like measuring the height difference between your feet when both are on the same step on the stairs. \$\endgroup\$
    – Transistor
    Commented Apr 16, 2021 at 19:15
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    \$\begingroup\$ The resistance will be non-zero but very small. Just a quick back-of-the-envelope comparison: if the block is a similar effective resistance to 1 cm of 12 AWG wire, the resistance should be around 0.05 milliohms. Given that most shunts are around 50mV at rated current, you're going to have a hard trouble pulling any useful reading out of the noise. There are a lot of expensive non-contact Hall-effect or fluxgate current sensors but also a few relatively cheap (~$20) ones too. You might want to look into those. \$\endgroup\$
    – vir
    Commented Apr 16, 2021 at 19:52
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    \$\begingroup\$ I wasn't suggesting that it was simple to do, just that it's possible. An actual calibrated shunt might be a better choice. \$\endgroup\$
    – jwh20
    Commented Apr 16, 2021 at 20:24

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