# Energy loss due to internal resistance of a battery?

Say I'm using a battery to power some process, and the internal resistance of the battery is given. The resistance of the process is not given, but assume it is minimal so the circuit runs at maximal current. I want all the energy to go to the process, but is some inherently lost in the battery itself? How would I calculate that?

• -1. No research effort. Did you try googling your title? The #2 hit (#1 is your question) is farside.ph.utexas.edu/teaching/302l/lectures/node62.html – sammy gerbil Oct 19 '17 at 10:27
• @sammygerbil I did actually research the question and I did find that page. My continued confusion was because, in the problem I am considering, the external resistance is not given, but the problem still asks for me to find how much total energy is output by the system assuming it runs at the maximum current possible. Thinking about it now, I guess maximal current implies no external resistance, and using the equation for power loss all the power would be lost in the battery. I was very tired when considering this so that answer did not seem satisfactory to me, but now it makes sense. – Miles Johnson Oct 20 '17 at 2:00

Yes. Note that all the current flowing through your process also flows through the battery. This means that if the internal resistance of the battery is R(i) and the current you measure flowing through your process is I(p), then the power loss in the battery is equal to (I(p))^2 x R(i). In practical terms this means you want the resistance of your process (the "load resistance") to be larger than the internal resistance of the battery, otherwise more power will be lost internally in the battery than will be delivered to the load... and the battery will get hot.

Yes, energy will be lost to the battery.

In electric batteries, electrochemical reactions which take place inside the battery, in addition to other factors such as material resistivity and temperature, will cause the battery to possess an internal resistance.

Just as you might calculate the power loss due to another type of resistance, you can calculate the power loss due to internal resistance in the same manner.

Assuming there is an emf of $\cal E$, internal resistance $r$, and external resistance $R$, the power consumed by the internal resistance is:

$P_r = I^2\,r = \frac{ {\cal E}^2\,r}{(r+R)^2}$

where

$I = \frac{ {\cal E}}{r + R}$

from

${\cal E} = I(R+r)$

• it is an interesting and useful exercise to derive the optimum value for the load resistance (i.e., that resistance value for which the power dissipated in it is maximized) when the internal resistance of the source is given. This is generically referred to as "impedance matching". – niels nielsen Oct 19 '17 at 5:13