Is it possible to divide the current in a dc circuit and measure only half of the current?

Thread Starter

injtsvetkov

Joined Dec 12, 2015
7
Hi everyone!
I need to measure direct current which is a bit higher than 10 amps, but my multimeter can measure maximum 10 amps. Is there a way to make current divider circuit (to split the current in 2) and measure the current on one branch, then derive the total current by multiplying the reading by 2?

Thanks for your answers!
 

crutschow

Joined Mar 14, 2008
29,466
You can put a resistance in parallel with, and equal to, the meter shunt resistance.

If you can't accurately measure that shunt resistance you could do the following:
Adjust the current limit on a power supply to its maximum (10A or less) and measure with your meter.
Then connect a resistance across it until the meter reads 1/2 that value.

I did a similar thing with a multimeter I had.
Since the meter shunt resistance is very low (perhaps 10mΩ if you meter has a minimum 100mV full scale setting), I used a steel paper clip for the added resistance.
I connected the paper clip wire between the two screw terminal connections in a dual banana plug (the two small openings below) and changed the wire length to adjust the resistance.
The banana jack is connected in parallel with the 10A current-measurement sockets on your multimeter.

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k1ng 1337

Joined Sep 11, 2020
417
You can put a resistance in parallel with, and equal to, the meter shunt resistance.

If you can't accurately measure that shunt resistance you could do the following:
Adjust the current limit on a power supply to its maximum (10A or less) and measure with your meter.
Then connect a resistance across it until the meter reads 1/2 that value.

I did a similar thing with a multimeter I had.
Since the meter shunt resistance is very low (perhaps 10mΩ if you meter has a minimum 100mV full scale setting), I used a steel paper clip for the added
I connected the paper clip wire between the two screw terminal connections in a dual banana plug (the two small openings below) and changed the wire length to adjust the resistance.
The banana jack is connected in parallel with the 10A current-measurement sockets on your multimeter.
I basically used your method to find the resistance of two 0.1ohm 10w resistors I bought since I don't have a precision ohm meter. Obviously the accuracy is only as good as my multimeter but is sufficient although I found an error between the shunt voltage and current through the meter. Is this error a direct variation (ohmic) between the two or does change at different voltage/currents?

I can rephrase my question if the above is confusing.
 

crutschow

Joined Mar 14, 2008
29,466
I found an error between the shunt voltage and current through the meter. Is this error a direct variation (ohmic) between the two or does change at different voltage/currents?
The error is likely due to the various tolerances of the measurements.
How exactly did you do the measurement?

The only reason it would vary at different voltage/currents is if the resistance heated from the current and changed value due to the temperature change.
 
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