91Ó°ÊÓ

We need to find the total energy supplied by the battery as the capacitor is being charged.

The current in the circuit is given as:

$$\begin{gathered} I=\frac{dQ}{dt} \\ I=Q_{max}\left(-\frac{1}{\mathrm{Ï„}}\right)\left(-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Ï„}$}\right.}\right) \\ I=\frac{\mathrm{Î\mu }C}{RC}{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.} \\ I=\frac{\mathrm{Î\mu }}{R}{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.} \end{gathered}$$

Now, $P_{bat}$is:

$$\begin{gathered} P_b=I×\mathrm{Î\mu } \\ {P}_b=\left(\frac{\mathrm{Î\mu }}{R}{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.}\right)\mathrm{Î\mu } \\ {P}_b=\frac{{\mathrm{Î\mu }}^2}{R}{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.} \end{gathered}$$

We know that, role="math" localid="1650877098929" $P_b=\frac{dU}{dt}$.

$$\begin{gathered} P_b=\frac{dU}{dt} \\ dU=P_{b}dt \\ ∫dU=P_{b}dt \\ ∫dU=\frac{{\mathrm{Î\mu }}^2}{R}e\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right. \\ ∫dU=\frac{{\mathrm{Î\mu }}^2}{R}\left[-RC{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.}\right] \\ ∫dU={\mathrm{Î\mu }}^{2}C \end{gathered}$$

We need to find the total energy dissipated by the resistor as the capacitor is being charged.

Now we finding the power dissipation :

$$\begin{gathered} P_r=I^{2}R \\ {P}_r=\frac{{\mathrm{Î\mu }}^2}{R^2}{e}^{\raisebox{1ex}{$-2t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.} \end{gathered}$$

So, $P_r=\frac{dU}{dt}$.

$$\begin{gathered} P_r=\frac{dU}{dt} \\ dU=P_{r}dt \\ ∫dU=\frac{{\mathrm{Î\mu }}^2}{R}{e}^{\raisebox{1ex}{$-2t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.}dt \\ {U}_r=\frac{{\mathrm{Î\mu }}^2}{R}{\left[-\frac{RC}{2}{e}^{\raisebox{1ex}{$-2t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.}\right]}_0^{∞} \\ {U}_r=\frac{{\mathrm{Î\mu }}^{2}C}{2} \end{gathered}$$

We need to find the energy stored in the capacitor when it is fully charged. Your expressions will be in terms of $E,R$and $C$.

We finding the energy stored in the capacitor :

$$\begin{gathered} U_c=\frac{1}{2}×\frac{{Q^2}_{max}}{C} \\ {U}_c=\frac{1}{2}×\frac{C^2{\mathrm{Î\mu }}^2}{C} \\ {U}_c=\frac{1}{2}C{\mathrm{Î\mu }}^2 \end{gathered}$$

We need to find that energy is conserved for parts a to c.

The energy of the battery is dissipated into the resistor and the capacitor. Therefore, the energy is conserved through the discharge.