Maximum power from a battery with source resistance

AI Thread Summary
The discussion revolves around calculating the maximum power that can be extracted from a 5V battery with a source resistance of 10 Ohms. Participants clarify that the maximum power transfer theorem applies, indicating that the load resistance should equal the source resistance for optimal power extraction. The internal resistance of the battery is set at 10 Ohms, leading to a calculated power output of 625mW, which matches one of the provided multiple-choice answers. There is some confusion regarding the significance of the battery's 1800mAh rating, with consensus that it does not directly impact the maximum power calculation. Ultimately, the focus remains on maximizing output power rather than energy delivery.
JordanHood
Messages
9
Reaction score
0

Homework Statement


A 5V battery rated at 1800mAh has a soucre resistance of 10 Ohms. What is the maximum power that can be extracted from the battery?

Homework Equations

The Attempt at a Solution


I understand that had the battery not had any source resistance then the power out would have been equal to Energy x Time giving us P=VIT, but i do no understand what to do when there is a source resistance. Do I calculate another current that is passing through the battery? using V=IR to get a current of 0.5A? If so then what does the 1800mAh mean? I have then tried putting this current into P=I^2R but this means I am totally ignoring one of the values in the question which means I can't be right

EDIT
Apologies I didnt realize this was a multiple choice question, the possible answers are
a) 625mW
b) 1.25mW
c) 2.5 mJ
d) 11.664 MJ
 
Last edited:
Physics news on Phys.org
JordanHood said:

Homework Statement


A 5V battery rated at 1800mAh has a soucre resistance of 10 Ohms. What is the maximum power that can be extracted from the battery?

Homework Equations

The Attempt at a Solution


I understand that had the battery not had any source resistance then the power out would have been equal to Energy x Time giving us P=VIT, but i do no understand what to do when there is a source resistance. Do I calculate another current that is passing through the battery? using V=IR to get a current of 0.5A? If so then what does the 1800mAh mean? I have then tried putting this current into P=I^2R but this means I am totally ignoring one of the values in the question which means I can't be right
I don't see the significance of 1800mAh in this problem. Do you know the maximum power transfer theorem in electrical circuits?
 
cnh1995 said:
I don't see the significance of 1800mAh in this problem. Do you know the maximum power transfer theorem in electrical circuits?
Is this when the source resistance and load resistance are equal?
 
JordanHood said:
Is this when the source resistance and load resistance are equal?
Yes. But I just realized that the theorem will give the value of load resistance for maximum power dissipation in the load, while the question asks about the maximum power in the entire circuit.
 
I believe 1800mAh means the battery can supply 1800mA current for 1hr.
Edit: I think the proper meaning of this is that the battery can supply 1mA current for 1800 hrs.
 
Last edited:
cnh1995 said:
I believe 1800mAh means the battery can supply 1800mA current for 1hr.
So do I need to calculate the current per second?
 
JordanHood said:
So do I need to calculate the current per second?
No. Seeing the options, I think this problem does belong to the maximum power transfer theorem. After all, it's the load that extracts power from the battery. So when we say 'power extracted from the battery', it means the power consumed by the load from the battery.
 
Last edited:
cnh1995 said:
No. Seeing the options, I think this problem does belong to the maximum power transfer theorem. After all, it's the load that extracts power from the battery. So when we say 'power extracted from the battery', it means the power consumed by the load from the battery.
Ok so I set the internal resistance of the battery to 10 Ohms and work from there?
 
JordanHood said:
Ok so I set the internal resistance of the battery to 10 Ohms and work from there?
I think so.
 
  • #10
cnh1995 said:
I think so.
That gives me 625mW, which is one of the answers, so assuming that is right. Bit concerned about not using the 1800mAh at all though, possibly there to confuse you?
 
  • #11
JordanHood said:
power out would have been equal to Energy x Time giving us P=VIT, but i do no understand
No. Energy delivered = power x time

But you are not asked about energy, the question involves maximizing output power.
 
  • Like
Likes JordanHood
  • #12
JordanHood said:
Bit concerned about not using the 1800mAh at all though, possibly there to confuse you?
I think so. I haven't seen mAh rating being used for calculating maximum power so far.
 
  • Like
Likes JordanHood
Back
Top