| ### Statement | | ### Statement |
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| $11.4.20.$ | | $11.4.20.$ |
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| The initial voltage across the capacitance capacitor C0is V0, and the capaci- | | 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 | | 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 | | capacitor of capacitance C break through, if its breakdown occurs at voltage |
| V ? | | V ? |
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| ### Solution | | ### Solution |
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| 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 |
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| $V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$ | | $V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$ |
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| The circuit equation is obtained from the mesh: | | The circuit equation is obtained from the mesh: |
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| $V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$ | | $V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$ |
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| Substituting the above expressions, | | Substituting the above expressions, |
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| $\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$ | | $\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$ |
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| Rearranging, we obtain the differential equation of the forced harmonic oscillator: | | Rearranging, we obtain the differential equation of the forced harmonic oscillator: |
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| $L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$ | | $L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$ |
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| The natural angular frequency of the circuit is | | The natural angular frequency of the circuit is |
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| $\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$ | | $\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$ |
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| 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. |
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| 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 |
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| $q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$ | | $q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$ |
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| Therefore, the voltage on capacitor C evolves as | | Therefore, the voltage on capacitor C evolves as |
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| $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)$ |
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| The maximum value that V_C can reach is | | The maximum value that V_C can reach is |
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| $V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$ | | $V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$ |
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| 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 $V_C(\tau) = V$ |
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| Solving for the cosine and then for time: | | Solving for the cosine and then for time: |
