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Shunt in DC ammeter circuit

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imranahmed

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I have DC ammeter and shunt for sending current.
Shunt is 30A@60mA and ammeter also have same rating @24VDC supply.

But I connected 24VDC line from 2 series batteries with shunt 1st main terminal and load connected to shunt 2nd main terminal.And Ammeter connected with 2 small screws in shunt but when I energised power ammeter was burned and on-off switch for ammeter also burnt.

I tested negative terminal of ammeter and most left terminal was showing continuity.

Please try to understand.
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I assume you mean 30A->60mV not mA. Shunts convert current to voltage.

Use a volt-meter to measure the shunt.
 
Unfortunately I don't know that the 'digital panel meter' does exactly.

The marking on it for 30A/60mV indicate it may be configured to scale the signal for your shunt but if that's true you probably wouldn't have blown it.

One thing to test: Before you connect a black box to the shunt measure the R of the two pins you're about to connect. They should be high R. If they're a short then current will bypass your shunt and flow into the back box, probably blowing it (unless it's a 30A meter in which case you don't need a shunt at all).
 

But I connected 24VDC line from 2 series batteries with shunt 1st main terminal and load connected to shunt 2nd main terminal.And Ammeter connected with 2 small screws in shunt

As asdf44 said, a voltmeter goes across the small terminals of the shunt.
Read the voltage and calculate the current using the resistance of the shunt.
Voltmeters have a very high input impedance so they do not carry any current, compared to whatever you're measuring

When you connected the ammeter across the small terminals of the shunt,
you connected a low impedance across the shunt, maybe less than the shunt,
the current went through the ammeter, not the shunt
since it was more current than the meter was designed for, it burned up.
 
Hi,

From the explanation it seems you connected the load on the terminals of the voltmeter....and not directly to the shunt.
This is not the same since the wires between shunt and voltmeter have resistance that causes additional voltage drop and thus measurement errors.
In worst case the resistance is much higher than the shunt resistance ( 60mV / 30A= 0.002 Ohms) causing too high voltage at the voltmeter inputs destroying the voltmeter.

Thus the correct wiring is shown in post#3 ... with galvanically isolated power supply.

Klaus
 

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