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| The two coils are connected in parallel, so they share the same terminal voltage. The voltage across an ideal inductor is $V = L\frac{dI}{dt}$ Being in parallel: | | The two coils are connected in parallel, so they share the same terminal voltage. The voltage across an ideal inductor is $V = L\frac{dI}{dt}$ Being in parallel: |
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| $L_1 \frac{dI_1}{dt} = L_2 \frac{dI_2}{dt}$ | | $L_1 \frac{dI_1}{dt} = L_2 \frac{dI_2}{dt}$ |
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| Integrating over time from an initial instant to any later time, and assuming that initially both currents are zero, we obtain: | | Integrating over time from an initial instant to any later time, and assuming that initially both currents are zero, we obtain: |
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| $L_1 I_1(t) + L_2 I_2(t) = \text{constant}$. | | $L_1 I_1(t) + L_2 I_2(t) = \text{constant}$. |
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| At the moment switch K is closed, the current in $L_1 $is maximum$ (I_0)$ and that in$ L_2$ is zero Therefore, the constant is $L_1 I_0$and the invariance is demonstrated: | | At the moment switch K is closed, the current in $L_1 $is maximum$ (I_0)$ and that in$ L_2$ is zero Therefore, the constant is $L_1 I_0$and the invariance is demonstrated: |
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| $\boxed{L_1 I_1 + L_2 I_2 = L_1 I_0}$ | | $\boxed{L_1 I_1 + L_2 I_2 = L_1 I_0}$ |
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| b) Maximum current in $L_2$ | | b) Maximum current in $L_2$ |
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| The capacitor C charged to $V_0$ initially discharges only through$ L_1$ The frequency of that $L_1 $C circuit is $\omega = 1/\sqrt{L_1 C}$and the maximum current reached in L_1 (when the capacitor is fully discharged) is: | | The capacitor C charged to $V_0$ initially discharges only through$ L_1$ The frequency of that $L_1 $C circuit is $\omega = 1/\sqrt{L_1 C}$and the maximum current reached in L_1 (when the capacitor is fully discharged) is: |
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| $I_0 = \frac{V_0}{\omega L_1} = V_0 \sqrt{\frac{C}{L_1}}$ | | $I_0 = \frac{V_0}{\omega L_1} = V_0 \sqrt{\frac{C}{L_1}}$ |
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| Exactly at that instant, switch K is closed, connecting $L_2 $in parallel with $L_1 fFrom that moment on, the combination oscillates with a new frequency determined by the equivalent inductance of both coils in parallel: | | Exactly at that instant, switch K is closed, connecting $L_2 $in parallel with $L_1 fFrom that moment on, the combination oscillates with a new frequency determined by the equivalent inductance of both coils in parallel: |
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| $L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2}$ | | $L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2}$ |
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| $\omega' = \frac{1}{\sqrt{L_{\text{eq}} C}} = \frac{1}{\sqrt{\frac{L_1 L_2}{L_1 + L_2} C}}$ | | $\omega' = \frac{1}{\sqrt{L_{\text{eq}} C}} = \frac{1}{\sqrt{\frac{L_1 L_2}{L_1 + L_2} C}}$ |
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| The total current$ I_1 - I_2 $$(the difference of currents at the common node) oscillates cosinusoidally with this frequency, starting from its maximum value I_0: | | The total current$ I_1 - I_2 $$(the difference of currents at the common node) oscillates cosinusoidally with this frequency, starting from its maximum value I_0: |
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| $I_1 - I_2 = I_0 \cos \omega' t$ | | $I_1 - I_2 = I_0 \cos \omega' t$ |
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| Combining this equation with the conservation law from part (a): | | Combining this equation with the conservation law from part (a): |
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| $L_1 I_1 + L_2 I_2 = L_1 I_0$ | | $L_1 I_1 + L_2 I_2 = L_1 I_0$ |
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| we solve the system for I_2: | | we solve the system for I_2: |
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| $I_2(t) = \frac{L_1}{L_1 + L_2} I_0 \bigl(1 + \cos \omega' t\bigr)$. | | $I_2(t) = \frac{L_1}{L_1 + L_2} I_0 \bigl(1 + \cos \omega' t\bigr)$. |
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| The maximum value of this current occurs when $\cos \omega' t = 1$: | | The maximum value of this current occurs when $\cos \omega' t = 1$: |
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| $I_{2,\text{max}} = \frac{2 L_1}{L_1 + L_2} I_0 | | $I_{2,\text{max}} = \frac{2 L_1}{L_1 + L_2} I_0 |
| = \frac{2 L_1}{L_1 + L_2} \cdot V_0 \sqrt{\frac{C}{L_1}} | | = \frac{2 L_1}{L_1 + L_2} \cdot V_0 \sqrt{\frac{C}{L_1}} |
| = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}$ | | = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}$ |
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| $\boxed{I_{2,\text{max}} = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}}$ | | $\boxed{I_{2,\text{max}} = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}}$ |
