1
\$\begingroup\$

I have a Milwaukee M18 Fuel 18v LI ion cordless impact wrench. It came with a 18v 2.0 ah 27 Wh battery pack. They also sell a 18v 5.0 ah battery pack among others. When reading online some people say the 5.0 ah battery will provide more power to the tool from the start while others say it will only provide the same power for a longer period of time. I tend to think the latter is true, but I am not an electrical engineer. There was a Youtube video showing that when the guy swapped out the smaller battery for the larger one he was able to break a lug nut loose where the smaller battery was not able to do so. Was that due to other factors like maybe the age of the batteries, or is there something to the larger battery pack providing more power at the motor from the start as well as for a longer duration?

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

A higher capacity battery probably has a lower internal resistance. That means its voltage will drop less than a smaller battery at the same current drain. This effect applies even when both batteries are fully charged.

In addition, the higher capacity battery will last longer at the same current drain.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer... I wonder if the resulting increase in torque would be minor i.e less than 1%, or would it be more significant? For me anything less than a 5% increase in torque would not be worth doubling the price of the tool by buying the higher capacity battery. If I do buy the higher capacity battery I will test both batteries max torque against a manual torque wrench and post the results. \$\endgroup\$ May 2, 2016 at 21:30
  • \$\begingroup\$ @68ch: Motor torque is proportional to current. For a stalled motor, this is also proportional to voltage. If one battery pack puts out 18V when the motor is stalled, but the other can only manage 16V, then you get 2V/18V = 11% less torque. This effect will lessen as the motor speeds up. A spinning motor will take less current. The first pack may still put out 18V, but the second now 17V due to the lower current. \$\endgroup\$ May 2, 2016 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.