91影视

The electric power or the rate at which electric energy gets dissipated is calculated as:

$\mathit{P}\mathbf{=}\frac{{\mathbf{蔚}}^{\mathbf{2}}}{\mathbf{R}}\mathbf{,}\mathit{P}\mathbf{=}\frac{\mathbf{d}\mathbf{U}}{\mathbf{d}\mathbf{t}}\mathbf{,}\mathit{P}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R}\mathit{a}\mathit{n}\mathit{d}\mathit{P}\mathbf{=}\mathit{蔚}\mathit{I}$, whereis the electromotive force, I is current, R is the resistance, and U is the energy stored in the capacitor.

The charge over the capacitor during the charging process is given by$\mathit{q}\mathbf{=}{\mathit{q}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{R}\mathbf{C}}\mathbf{)}$

a) (i) The rate at which electrical energy is dissipated in the resistor just after the connection is made.

After the connection, the capacitor acts like a short circuit, so the power dissipated in the resistor is

$$\begin{gathered} P_R=\frac{{蔚}^2}{R} \\ =\frac{{(120V)}^2}{5.86惟} \\ =2460W \end{gathered}$$

(ii) The rate at which the electrical energy stored in the capacitor is (use $U_C=\frac{Q^2}{2C}$)

$$\begin{gathered} P_C=\frac{d{U}_C}{dt} \\ =\frac{1}{2C}\frac{d\left(q^2\right)}{dt} \\ =\frac{iq}{C} \\ =0W \end{gathered}$$

(iii) The rate at which electrical energy is being supplied by the source is

$$\begin{gathered} P_{蔚}=蔚I \\ =\frac{{蔚}^2}{R} \\ =\frac{{(120V)}^2}{5.86惟} \\ =2460W \end{gathered}$$

the power supplied by the source is the sum of the power dissipated in the resistor and the power stored in the capacitor.

At a long time after the connection is made, the capacitor acts like an open circuit, and thus the current is zero since all the power rate is proportional to the current.

Thus, all the rates are zero, that is,

$P_R=0,P_C=0,P_{蔚}=0$

c) (i) The charge over the capacitor during the charging process is given by

$$\begin{gathered} q=q_{\max }(1-e^{-t/RC}) \\ Putq=\raisebox{1ex}{$q_{max}$}\!\left/ \!\raisebox{-1ex}{$2Then$}\right. \\ {e}^{-t/RC}=\frac{1}{2}.................................\left(1\right) \end{gathered}$$

But , so $q_{\max }=C蔚$current in the circuit is

$$\begin{gathered} I=\frac{dq}{dt} \\ =\frac{蔚}{R}{e}^{-t/RC} \end{gathered}$$

The power dissipated in the resistor is thus

$$\begin{gathered} P_R=I^{2}R \\ =\frac{{蔚}^2}{R}{\left(e^{-t/RC}\right)}^2 \end{gathered}$$

Substitute from (1)

$$\begin{gathered} P_R=I^{2}R \\ =\frac{{蔚}^2}{4R} \\ =\frac{{(120V)}^2}{4(5.86惟)} \\ =614W \end{gathered}$$

(ii) The rate at which the electrical energy stored in the capacitor is

$$\begin{gathered} P_C=\frac{d{U}_C}{dt}=\frac{1}{2C}\frac{d\left(q^2\right)}{dt} \\ {P}_C=\frac{d}{dt}\left[\frac{{q^2}_{\max }}{2C}{\left(1-e^{-t/RC}\right)}^2\right] \\ {P}_C=\frac{{q^2}_{\max }}{R{C}^2}\left(1-e^{-t/RC}\right)e^{-t/RC} \end{gathered}$$

Substitute from (1),

${\text{P}}_{\text{C}}=\frac{{蔚}^2}{4R}$

Put the values,

$$\begin{gathered} P_C=\frac{{蔚}^2}{4R} \\ =\frac{{(120V)}^2}{4(5.86惟)} \\ =614W \end{gathered}$$

(iii) The rate at which electrical energy is being supplied by the source is

$$\begin{gathered} P_{蔚}=蔚I \\ =\frac{{蔚}^2}{R}\left(e^{-t/RC}\right) \end{gathered}$$

Substitute from (1),

$$\begin{gathered} P_{蔚}=\frac{{蔚}^2}{2R} \\ =\frac{{(120V)}^2}{2(5.86惟)} \\ =1230W \end{gathered}$$

the power supplied by the source is the sum of the power dissipated in the resistor and the power stored in the capacitor.