91影视

The energy stored in capacitor and it is given by${\mathbf{鈭�}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{CV}}^{\mathbf{2}}$Where${\mathit{U}}_{\mathbf{C}}$is the energy stored in capacitor. The self-inductance of the coil L is related to the energy stored in the coil by${\mathit{U}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}{{\mathit{i}}_{\mathbf{L}}}^{\mathbf{2}}$. Voltage across the capacitorrole="math" localid="1668256194446" $V_C=\sqrt{\frac{2U_C}{C}}$

The capacitor is connected to the battery with voltage${\mathrm{V}}_o=24V$and it is charged with C$6.40脳{10}^{-9}\mathrm{F}$. This is the initial stage before disconnecting. The energy stored in the capacitor when it is connected to the battery is given by ${\mathrm{U}}_0=\frac{1}{2}C{V}_0^2$, Substitute the values we have,

$\begin{array}{r}{U}_0=\frac{1}{2}\left(6.40脳{10}^{鈭�9}\mathrm{F}\right)(24\mathrm{V}{)}^2\\ =\left(1.84脳{10}^{鈭�6}\mathrm{J}\right)\end{array}$

After disconnecting from the battery, the capacitor is connected to an inductor L and discharges. After connecting to the coil, the charges on the plates of the capacitor are Q = 0.0800C. The energy stored in the capacitor is related to the charge Q in the form $U_C=\frac{Q^2}{2C}$now, plug the values for Q and C to get the energy stored in the capacitor after connecting to the coil.

localid="1668313696368" $\begin{array}{r}{U}_c=\frac{{\left(0.0800脳{10}^{鈭�6}\mathrm{C}\right)}^2}{2\left(6.40脳{10}^{鈭�9}\mathrm{F}\right)}\\ =0.50脳{10}^{鈭�6}\mathrm{J}\end{array}$

The initial energy stored in the capacitor [Jo decreased by$0.50脳{10}^{-6}J$. To obey the conservation law of energy, the remaining energy is stored in the inductor L, so the difference between$U_0$ and $U_C$represents the energy stored in the coil ${\mathrm{U}}_L={\mathrm{U}}_0-{\mathrm{U}}_C$.Substitute the values we have,

$\begin{array}{r}{U}_L=1.84脳{10}^{鈭�6}\mathrm{J}鈭�0.50脳{10}^{鈭�6}\\ =1.34脳{10}^{鈭�6}\mathrm{J}\end{array}$

The self-inductance of the coil L is related to the energy stored in the coil$U_L$ where$U_L$ is given by ${\mathrm{U}}_L=\frac{1}{2}L{{\mathrm{i}}_{\mathrm{L}}}^2$ Substitute the values we have

$\begin{array}{r}{\mathrm{I}}_{\mathrm{L}}=\sqrt{\frac{2U_{\mathrm{L}}}{\mathrm{L}}}\\ =\sqrt{\frac{2\left(1.34脳{10}^{鈭�6}\mathrm{J}\right)}{\left(66脳{10}^{鈭�3}\mathrm{mH}\right)}}\\ =6.37\mathrm{mA}\end{array}$

When the charges on the plates of the capacitor are 0.0800$\mathrm{渭}$ C, the energy stored in the capacitor is ${\mathrm{U}}_C$= $\left(0.50脳{10}^{-6}J\right)$.is related to the voltage across the capacitor $V_C$by$V_c=\sqrt{\frac{2U_C}{C}}$ Solve this equation we get ,

$\begin{array}{r}{V}_c=\sqrt{\frac{2\left(0.50脳{10}^{鈭�6}\mathrm{J}\right)}{6.40脳{10}^{鈭�9}\mathrm{F}}}\\ =12.5\mathrm{V}\end{array}$

The capacitor is connected to the inductor in series, so both devices have the same voltage$V_c=V_L=12.5V$

When the current i changes with time in any circuit causes a self-induced emf E. This inductance depends on the geometry of this circuit and the used material and it is given by equation (30.7) in the form$蔚=L\left|\frac{鈭�i}{鈭�t}\right|$

$\begin{array}{r}\left|\frac{\mathrm{螖颈}}{\mathrm{螖迟}}\right|=\frac{蔚}{\mathrm{L}}\\ \left|\frac{\mathrm{螖颈}}{\mathrm{螖迟}}\right|=\frac{12.5\mathrm{V}}{0.0660\mathrm{H}}=189\mathrm{A}/\mathrm{s}\end{array}$

Therefore, the current is 189 A/s