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I need to measure the amperes of the current flowing through a cable; the power is consumed by an electric motor.

For that purpose, I am looking for an amperemeter that requires no splitting of the cables. (As I understand that cables generally transport both types of currents together, I need to be able to measure the amps of the system without actually disarming the cable).

Where can I find such device?

Any type of feedback is appreciated. Thanks.

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  • If you want to avoid cutting or disconnecting any wires, then you would use a clamp meter. The clamp is put around one of the wires of the circuit to measure the current through that wire. However, this requires being able to access a single wire rather than both wires in the circuit; if the clamp meter were put around both, the opposite currents would cancel each other out resulting in a reading of zero.

  • If you want to avoid separating the two wires making up the cable from each other, and you have access to exposed conductors at two ends of the cable, you can use the cable itself as a current shunt. First, shut off the power, then measure the resistance of one wire between its two ends. Then, with the motor running, measure the voltage between those same two points. (Since there is almost no current flowing through the voltmeter, the resistance of the test leads does not affect the results.) Then divide the voltage by the resistance to obtain the current (Ohm's law).

  • If what you want to measure is an AC line-powered plug-in device, then there are plug-in passthrough devices that can be used; either meters themselves (example) or which separate the conductors for use with a clamp meter (example) — this is breaking the circuit to insert the meter, but in a safe, packaged way.

But if none of the above apply and you cannot separate the wires or insert a device in the circuit (everything is insulated and there is a single cable with no connectors to unplug), then you cannot effectively measure the current.

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  • \$\begingroup\$ Thanks, I meant no need to separate the cables carrying different currents. \$\endgroup\$ – Andrés Segovia Oct 10 '15 at 23:10

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