Edits to “Statement”, “Answer”
en/11.4.20.md
+7 −2
| @@ -1,7 +1,12 @@ | |||
| ### Statement | |||
| − | $11.4.20.$ | ||
| + | $11.4.20.$ | ||
| + | The initial voltage across the capacitance capacitor C0is V0, and the capaci- | ||
| + | tance capacitor C is not charged. How long after the key K is closed will the | ||
| + | capacitor of capacitance C break through, if its breakdown occurs at voltage | ||
| + | V ? | ||
| + | |||
| ### Solution | |||
| Let q be the charge that flows from the positive plate of$ C_0$ toward C. Then the charge on$ C_0$ is $Q_0 - q $(with $ Q_0 = C_0 V_0$) and the charge on C is q. The voltages are | |||
| $V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$ | |||
| The circuit equation is obtained from the mesh: | |||
| $V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$ | |||
| Substituting the above expressions, | |||
| $\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$ | |||
| Rearranging, we obtain the differential equation of the forced harmonic oscillator: | |||
| $L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$ | |||
| The natural angular frequency of the circuit is | |||
| $\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$ | |||
| where | |||
| $ C_{\text{eq}} = \dfrac{C\,C_0}{C+C_0}$ is the equivalent series capacitance. | |||
| The general solution with the initial conditions $q(0)=0 $(capacitor C starts uncharged) and$ I(0)=\dot{q}(0)=0$ (the inductor prevents abrupt changes in current) is | |||
| $q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$ | |||
| Therefore, the voltage on capacitor C evolves as | |||
| $V_C(t) = \frac{q(t)}{C} | |||
| = \frac{C_{\text{eq}}}{C}\,V_0 \bigl(1 - \cos\omega t\bigr) | |||
| = \frac{V_0}{1 + \dfrac{C}{C_0}} \bigl(1 - \cos\omega t\bigr)$ | |||
| The maximum value that V_C can reach is | |||
| $V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$ | |||
| If the breakdown voltage V is greater than this maximum, capacitor C never breaks down. | |||
| Otherwise $(V < V_{C,\max})$ breakdown occurs at the instant $\tau $when $V_C(\tau) = V$ | |||
| Solving for the cosine and then for time: | |||
| $\cos\omega\tau = 1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}$ | |||
| $\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$ | |||
| @@ -52,4 +57,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$ | ||
| @@ -1,7 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $11.4.20.$ |
$11.4.20.$ | ||
| The initial voltage across the capacitance capacitor C0is V0, and the capaci- | |||
| tance capacitor C is not charged. How long after the key K is closed will the | |||
| capacitor of capacitance C break through, if its breakdown occurs at voltage | |||
| V ? | |||
| ### Solution | ### Solution | ||
| Let q be the charge that flows from the positive plate of$ C_0$ toward C. Then the charge on$ C_0$ is $Q_0 - q $(with $ Q_0 = C_0 V_0$) and the charge on C is q. The voltages are | Let q be the charge that flows from the positive plate of$ C_0$ toward C. Then the charge on$ C_0$ is $Q_0 - q $(with $ Q_0 = C_0 V_0$) and the charge on C is q. The voltages are | ||
| $V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$ | $V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$ | ||
| The circuit equation is obtained from the mesh: | The circuit equation is obtained from the mesh: | ||
| $V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$ | $V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$ | ||
| Substituting the above expressions, | Substituting the above expressions, | ||
| $\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$ | $\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$ | ||
| Rearranging, we obtain the differential equation of the forced harmonic oscillator: | Rearranging, we obtain the differential equation of the forced harmonic oscillator: | ||
| $L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$ | $L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$ | ||
| The natural angular frequency of the circuit is | The natural angular frequency of the circuit is | ||
| $\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$ | $\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$ | ||
| where | where | ||
| $ C_{\text{eq}} = \dfrac{C\,C_0}{C+C_0}$ is the equivalent series capacitance. | $ C_{\text{eq}} = \dfrac{C\,C_0}{C+C_0}$ is the equivalent series capacitance. | ||
| The general solution with the initial conditions $q(0)=0 $(capacitor C starts uncharged) and$ I(0)=\dot{q}(0)=0$ (the inductor prevents abrupt changes in current) is | The general solution with the initial conditions $q(0)=0 $(capacitor C starts uncharged) and$ I(0)=\dot{q}(0)=0$ (the inductor prevents abrupt changes in current) is | ||
| $q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$ | $q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$ | ||
| Therefore, the voltage on capacitor C evolves as | Therefore, the voltage on capacitor C evolves as | ||
| $V_C(t) = \frac{q(t)}{C} | $V_C(t) = \frac{q(t)}{C} | ||
| = \frac{C_{\text{eq}}}{C}\,V_0 \bigl(1 - \cos\omega t\bigr) | = \frac{C_{\text{eq}}}{C}\,V_0 \bigl(1 - \cos\omega t\bigr) | ||
| = \frac{V_0}{1 + \dfrac{C}{C_0}} \bigl(1 - \cos\omega t\bigr)$ | = \frac{V_0}{1 + \dfrac{C}{C_0}} \bigl(1 - \cos\omega t\bigr)$ | ||
| The maximum value that V_C can reach is | The maximum value that V_C can reach is | ||
| $V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$ | $V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$ | ||
| If the breakdown voltage V is greater than this maximum, capacitor C never breaks down. | If the breakdown voltage V is greater than this maximum, capacitor C never breaks down. | ||
| Otherwise $(V < V_{C,\max})$ breakdown occurs at the instant $\tau $when $V_C(\tau) = V$ | Otherwise $(V < V_{C,\max})$ breakdown occurs at the instant $\tau $when $V_C(\tau) = V$ | ||
| Solving for the cosine and then for time: | Solving for the cosine and then for time: | ||
| $\cos\omega\tau = 1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}$ | $\cos\omega\tau = 1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}$ | ||
| $\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$ | $\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$ | ||
| @@ -52,4 +57,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$ | ||