# Why is the energy lost due to the internal resistance always equal to $(1/2)CV^{2}$ in a simple capacitor battery circuit? [duplicate]

Suppose we have simple circuit in which a battery of EMF $E$ is connected to a capacitor of capacitance $C$.

Work done by battery $W_{b}=CV\times V$ and energy stored in capacitor $=\frac{1}{2}CV^{2}$. Therefore the difference, which I'm assuming to be energy loss due to battery's internal resistance $=\frac{1}{2}CV^{2}$.

My question is why doesn't the energy loss by internal resistance depend on the chemical nature of the battery used? How do we define the quality of a battery if same energy is lost irrespective of their chemical composition in all of them ?

If my approach is wrong, what does this remaining $\frac{1}{2}CV^{2}$ signify?

• For time being lets keep discussions fixed to a simple capacitor battery circuit. (PS: Sorry for the noob post) Commented Nov 25, 2015 at 12:36
• BTW, a very clear and simple explanation is here: hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4 Commented Nov 25, 2015 at 12:50
• @CarlWitthoft yes i read that, and i cant find any clear justification of why it is independent of internal resistance. Commented Nov 25, 2015 at 13:04
• @CarlWitthoft hmm, leads me to another doubt, why do we have different types of battery then (high/low quality ones)? Commented Nov 25, 2015 at 13:08