Let Voc = open circuit voltage, Vm = measured voltage under load, RL = load resistance, RI = internal resistance. We have the formula (using voltage division) : RI = ((Voc/Vm) - 1)*RL. If we let IL = current under load this becomes (substituting for RL): RI = (Vm/Voc)*(Voc-Vm)/IL. As you can see from the last formula the lower IL (current under load) is the higher the predicted RI (internal resistance). So why is that the case for measurements using the DMM? That's because it is not accurate in detecting small resistances due the additional resistance of the leads. So, in that case you are measuring the lead resistance of the DMM as well as the internal resistance of the battery. This results in a lower current measurement and thus higher predicted RI. When using the kelvin clamps the lead resistance is not included resulting in a higher measured load current and thus lower calculated internal resistance. If you have a benchtop multimeter with kelvin leads try repeating this experiment and I predict that the differences will not be nearly as great as with your dedicated device.
You use the tester wrong. You need to open the clips so they are not touching the others. The reading is wrong if the clips touch directly together.
Let Voc = open circuit voltage, Vm = measured voltage under load, RL = load resistance, RI = internal resistance. We have the formula (using voltage division) : RI = ((Voc/Vm) - 1)*RL. If we let IL = current under load this becomes (substituting for RL):
RI = (Vm/Voc)*(Voc-Vm)/IL. As you can see from the last formula the lower IL (current under load) is the higher the predicted RI (internal resistance). So why is that the case for measurements using the DMM? That's because it is not accurate in detecting small resistances due the additional resistance of the leads. So, in that case you are measuring the lead resistance of the DMM as well as the internal resistance of the battery. This results in a lower current measurement and thus higher predicted RI. When using the kelvin clamps the lead resistance is not included resulting in a higher measured load current and thus lower calculated internal resistance. If you have a benchtop multimeter with kelvin leads try repeating this experiment and I predict that the differences will not be nearly as great as with your dedicated device.
Measuring resistance on direct current and on alternating current, as the device does, are completely different methods that give different results.
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Dear sir, why are you saying the 'resistance is milliamps'???
because AI is stupid.