Question

The energy that can be extracted from a storage battery is always less than the energy that goes into it while it is being charged. Why?

Short answer

The battery has its own internal resistance, there is always energy dissipation during charging due to its internal resistance.

Step-by-step solution

Power and energy of an electrical circuit

Electrical energy is the product of power multiplied by the length of time it was consumed.

$$\begin{gathered} \mathbf{Energy}\mathbf{}\left(E\right)\mathbf{=}\mathbf{power}\mathbf{}\left(P\right)\mathbf{xtime}\mathbf{}\left(t\right) \\ \mathbf{Power}\mathbf{}\left(P\right)\mathbf{}\mathbf{=}\mathbf{voltage}\left(V\right)\mathbf{xcurrent}\mathbf{}\left(I\right) \end{gathered}$$

Determine the power dissipation and energy

The batteryhas internal resistance.

Let r be the internal resistance of the battery and I be the current flowing during charging.

Therefore, the power dissipation to Its internal resistance is

$\mathrm{P}={\mathrm{I}}^2\mathrm{r}$

And the total energy dissipation during time role="math" localid="1655782126418" $\mathrm{t}$is

$\mathrm{E}={\mathrm{I}}^2\mathrm{rt}$

Therefore, there is always energy dissipation during charging due to its internal resistance.