I am trying to do some DIY capacity testing of my NiMH batteries. However when I connect, for example, a fully charged D cell in series with a 1 Ohm 5W power resistor I measure a current of 0.31A. With a 0.33Ω I get 0.38A and with 0.22Ω I get 0.41A.

According to Ohm's law I was expecting a current of 1.2A with the 1Ω given a 1.2V battery. Am I mistaken and I there any way I can draw currents of above 1A using resistors?

Thanks in advance.

 

Not using a D-cell. Look up internal resistance.

 

Sorry. Can't get there from here. If you want a whole amp, try lithium batteries. If you want to test ni-cads, do what MikeML said; First understand internal resistance, then work within their operating range.

 

Not using a D-cell. Look up internal resistance.

I've looked up internal resistance but I am still cofused. From what I've read NiMH has a lower internal resistance than Alkaline making them more suitable for high current applications.

Sorry. Can't get there from here. If you want a whole amp, try lithium batteries. If you want to test ni-cads, do what MikeML said; First understand internal resistance, then work within their operating range.

I wanted over an Amp because the batteries I have are rated at 4000mAh and 8000mAh respectively so will take much longer to discharge at low currents.

I'm pretty ignorant in this field but are you both saying that D cell NiMH are incapable of giving over an Amp or is it my use of resistors that is the issue?

Thanks for the replies btw.

 

Battery capacity is measured in mAh or Ah. A 4000mAh battery will deliver 1mA for 4000h, 10mA for 400h, 100mA for 40h, but not necessarily 1000mA for 4h, as you have already found out.

Different battery chemistries have different internal resistance.

What is it that you are trying to power?

 

Battery capacity is measured in mAh or Ah. A 4000mAh battery will deliver 1mA for 4000h, 10mA for 400h, 100mA for 40h, but not necessarily 1000mA for 4h, as you have already found out.

Different battery chemistries have different internal resistance.

What is it that you are trying to power?

I am just getting into rechargeable batteries and I'm trying to educate myself about the technology involved. The 8000mAh batteries I own have been tested by a reputable reviewer as being capable of 8779mAh at a 3A load. I was basically trying to replicate his tests to see what capacity I would get.

 

You need to ask "1. what is the no-load terminal voltage of the battery?" and then ask : "2. what is the battery terminal voltage when I draw 3A from this battery?".

The voltage drop from 1 to 2 is due to the battery's internal resistance which can be figured thus:
{V(1) - V(2)}/3.

If voltage 1 is 1.54V and voltage 2 is 0.9V, the internal resistance would be (1.54-0.9)/3 = 0.213 Ohms. Knowing that, you can figure the voltage drop at other load currents.

 

You need to ask "1. what is the no-load terminal voltage of the battery?" and then ask : "2. what is the battery terminal voltage when I draw 3A from this battery?".

The voltage drop from 1 to 2 is due to the battery's internal resistance which can be figured thus:
{V(1) - V(2)}/3.

If voltage 1 is 1.54V and voltage 2 is 0.9V, the internal resistance would be (1.54-0.9)/3 = 0.213 Ohms. Knowing that, you can figure the voltage drop at other load currents.

Thanks for the explanation.