Hello. I am not sure I can wirte down my question here but let me try.
I am using a locktronics and when I install series/parrell circuit to measure ohm, voltage, and current, the calcuation V=IR is not matched. for example, I got 6V and 2 ohm and the current must be 3A but what I got is 0.3A which is very low. I assume there is some hidden resistance but couldn't find anything in anywere. I tried to make closed circuit without any resistance but I still get only 0.3A somehow(it must burn the ciurcuit!).
could you help me????
Thank you.
Josh
Regarding locktronics, measuring ampere, voltage and ohm
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Re: Regarding locktronics, measuring ampere, voltage and ohm
Hello Josh,
What current is the power supply rated to generate?
Also do you have a fuse in the chain somewhere, our DC/AC power supply carriers have poly fuses underneath which will kick in at around 1A and provide the 'hidden' resistance.
6V at 3A is 18 Watts, Can your 2 Ohm resistor cope with this much power? If not then it will get very hot and probably end up damaging itself either by going open circuit or by increasing resistance.
What current is the power supply rated to generate?
Also do you have a fuse in the chain somewhere, our DC/AC power supply carriers have poly fuses underneath which will kick in at around 1A and provide the 'hidden' resistance.
6V at 3A is 18 Watts, Can your 2 Ohm resistor cope with this much power? If not then it will get very hot and probably end up damaging itself either by going open circuit or by increasing resistance.
Regards Ben Rowland - MatrixTSL
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