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I need to automate a power to ground test for a device which outputs 12V / 30A.

There is a .4 ohm resistive load connected to the device which it is driving, however if I remove that load and go straight to ground, the current will be much larger.

I = V/0 = Infinity

Of course the resistance of the conductor will not be exactly zero but I'm assuming the current will be very high, probably high enough to ruin my multimeter! I'm trying to rate a relay to switch the load.

Do I just measure the resistance through the relay and the associated wires in order to gauge the inrush current requirement or is there a better way to do it?
(The problem with the former method is that I need to measure a relay I don't have.)

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  • \$\begingroup\$ Does "Power to Ground Current" mean "Short Circuit Current"? \$\endgroup\$ – brhans Jul 6 '17 at 11:36
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    \$\begingroup\$ I will not be infinity as V will drop to zero due to the output impedance of your power supply or its current limiting circuitry will reduce the output current. (1) What type of PSU is it? (2) What is the output impedance? \$\endgroup\$ – Transistor Jul 6 '17 at 11:37
  • \$\begingroup\$ @brhans yes power to ground = short circuit \$\endgroup\$ – SeanJ Jul 6 '17 at 12:59
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Recognizing that your driver will never have a zero ohm source impedance, there will be an expected upper limit that it can reach in a true short circuit condition. You should be able to estimate this value. Then by applying a low resistance current shunt in series with the "short", you can measure the near short circuit condition. For example, a shunt of 0.01 ohms would allow you to measure up to 1200 amps.

You will need to include all resistances associated with the short circuit apparatus in your final measurement results. This includes connector resistances, wire resistance, relay contact resistances, etc. In order to reduce the relay contact resistance, you may wish to consider an industrial control "contactor" which is essentially a heavy duty relay designed for high current applications.

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