Does the large hole fit over a battery terminal (the lead post kind)?
The bar underneath is screwed to the two posts. If you remove the screws, is the shorter post insulated from the body?
It looks like it is missing something that attaches to the rectangle with a bottom bump.
It doesn't show up on an image search. If you don't get an answer here, you can search on reddit r/whatisthisthing
The brass piece is reminiscent of a shunt. If it was associated with "battery testing" as the OP suggests, a shunt would be useful to read the battery current using nothing more than a voltmeter.
Check to see if the 'shunt' is mounted to the metal frame in such a way that the battery current could be made to flow through the shunt and possibly emerge from one of the knoblike structures on the top of the frame.
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Regards, Marv
Failure is just success in progress
That looks about right - Mediocrates
Thanks for the inputs. The shunt is isolated from the aluminum body as is the vertical stud and threaded knob. The large hole goes on the bat. post and the vertical post accepts the cable I think. Current can flow through the body and if the knob is tightened pass through the shunt to the cable. Must be some sort of accessory for troubleshooting with a meter. It would be great if I could use it to test current but would need some direction.
Dick
If you have an ohm meter that would have low enough resolution, you could measure the shunt. First measure the leads touching and then measure the shunt and take the difference. My guess is that you will find it nearly zero. Most people don't have low resistance measurement capability.
If you have a constant current source, then you run the curent through the shunt and measure the voltage across the shunt.
What if you don't have a constant current source, then what you do is find a low resistance (and stable) device and load a power supply with this low resistance device in series with the shunt.
What kind of device might you have? An incandecent bulb. Find a low wattage bulb and make a circuit with a power source, the bulb, and the shunt. Don't get shocked - be careful. What you have created is a voltage divider with two resistors. Now you need to measure the total voltage across the two loads and across each of the resistors. The current through the bulb and the shunt is the same. The resistance across the bulb should be much greater than the shunt, so you can estimate the resistance of the bulb and thus calculate the current and knowing the voltage and current through the shunt, you now have it's resistance.
Here is an article with schematics and formulas to calculate this voltage divider:
https://en.wikipedia.org/wiki/Voltage_divider
Last edited by BuffaloJohn; Feb 19, 2024 at 11:23 PM.
If BuffaloJohn is correct, and I think he is, in use the VM used to read the current would need to be attached between the frame (which is attached to the shunt by the left hand screw in the photo) and the wire to the load which is attached to the thumbscrew. The tab sticking out of the left side of the frame would be a convenient place to make the VM attachment to the frame.
If the OP knows of any 'hams' (amateur radio operators) near him, they may have equipment suitable for measuring the resistance of the shunt.
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Regards, Marv
Failure is just success in progress
That looks about right - Mediocrates
BuffaloJohn (Feb 20, 2024), jimfols (Sep 21, 2024)
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