91影视

The energy stored in the capacitor is given by,

${\mathbf{鈭�}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{C}}$

Consider the following circuit, where C = 2.00 pF, R = 6.00 k惟 and 詯 = 90.0 V.

The circuit is completed and switched on at t = 0.

Initially, the capacitor is uncharged, that is,

q(t = 0) = 0

but since q = CV, then the potential across the capacitor is zero, and the capacitor acts like a wire with no resistance.

So the circuit can be reduced to an emf connected to a resistor, the current through this resistor is,

$$\begin{gathered} \mathrm{I}=\frac{蔚}{\mathrm{R}} \\ =\frac{90.0\mathrm{V}}{6.00脳{10}^3\mathrm{惟}} \\ =0.015\mathrm{A} \end{gathered}$$

The rate at which electrical energy is dissipated in the resistor is, therefore,

$$\begin{gathered} P_R=I^{2}R \\ =(0.015\mathrm{A}{)}^2脳6.00脳{10}^3\mathrm{惟} \\ =1.35\mathrm{W} \end{gathered}$$

The rate at which electrical energy is dissipated in the capacitor is, therefore,

$$\begin{gathered} P_C=\frac{d{U}_C}{dt} \\ =\frac{q}{C}\frac{dq}{dt} \\ =\frac{ql}{C} \end{gathered}$$

We need to find the time at which electrical energy is dissipated in the resistor equals the rate at which electrical energy is being stored in the capacitor.

$$\begin{gathered} P_R=P_C \\ \frac{ql}{C}=I^{2}R \\ I=\frac{q}{RC} \end{gathered}$$ (1)

Let Q be the total charge at which the capacitor reaches this charge t tends to infinity.

In the RC circuit, the charge over the capacitor is given by,

$q=Q(1-exp(-t/RC))$

But the charge equals the capacitance multiplied by the potential of the emf Q= C詯, so we get

$q=蔚C(1-exp(-t/RC))$ (2)

The current is the rate of the charge, that is,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{dq}}{\mathrm{dt}} \\ =\frac{\mathrm{d}}{\mathrm{dt}}\left[蔚\mathrm{C}\right(1鈭�\exp (鈭�\mathrm{t}/\mathrm{RC})\left)\right] \\ =\frac{蔚}{\mathrm{R}}\exp 鈦�(鈭�\mathrm{t}/\mathrm{RC}) \end{gathered}$$ (3)

Substitute from (2) and (3) into (1) we get,

$$\begin{gathered} \frac{C蔚}{RC}(1鈭�\exp 鈦�(鈭�t/RC\left)\right)=\frac{蔚}{R}\exp 鈦�(鈭�t/RC) \\ \exp 鈦�(鈭�t/RC)=1鈭�\exp 鈦�(鈭�t/RC) \\ \exp 鈦�(t/RC)=2 \\ t=RC\ln 鈦�2 \end{gathered}$$

Substitute the given data in the above expression, and we get,

$$\begin{gathered} t=RC\ln 鈦�2 \\ =\left(6.00脳{10}^3\mathrm{惟}\right)脳\left(2.00脳{10}^{鈭�6}\mathrm{F}\right)脳\ln 鈦�2 \\ =8.31脳{10}^{鈭�3}\mathrm{s} \end{gathered}$$

Now we need to find the rate of the energy dissipated in the resistor at this time, using equation (3) we can write,

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

Substitute given values in the above expression, and we get,

$$\begin{gathered} P_R=\frac{(90.0\mathrm{V}{)}^2}{\left(6.00脳{10}^3\mathrm{惟}\right)}脳\exp 鈦�\left(\frac{鈭�2脳8.31脳{10}^{鈭�3}\mathrm{s}}{\left(6.00脳{10}^3\mathrm{惟}脳2.00脳{10}^{鈭�6}\mathrm{F}\right)}\right) \\ =0.338\mathrm{W} \end{gathered}$$

Hence the rate at which electrical energy is being dissipated in the resistor 0.338 W