Li-ion has 99 percent charge efficiency, and the discharge loss is small. If you are trying to charge a capacitor via a resistor from a DC source then the energy you lose (to heat) in the resistor is always the same irrespective of how large or how small that resistor is. The charge time (as dictated by RC) is irrelevant to the problem.

You can assume that there is no radiation or other losses happen.

Only heat losses from the resistor can be enough for calculation. Imene Mansour

Thanks for you answer!

I am going to consider only the heat losses from the battery resistance.

But the battery resistance is divided to:

- internal resistances (current collectors and conductive additives around the active material) .

- the polarization, due to diffusion (mass transport, with creation or relaxation of concentration gradients).

I have the internal resistance, my question is how to identify the heat generated due to the polarization.

Based on ECM models, you can calculate the heat generated due to polarization by Q = R0*i^2, where R0 is the internal resistance. Or, you can also calculate the heat generated using the following relationship Q = i(OCV-V).