Новое решение

Alexphysics правка от 26 июн. 2026 г., 16:03

правка #19281 позже →

en/11.4.21.md +56 −0

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		+	### Statement
		+
		+	$11.4.21.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	a) Conservation of$ L_1 I_1 + L_2 I_2$
		+
		+	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:
		+
		+	$L_1 \frac{dI_1}{dt} = L_2 \frac{dI_2}{dt}$
		+
		+	Integrating over time from an initial instant to any later time, and assuming that initially both currents are zero, we obtain:
		+
		+	$L_1 I_1(t) + L_2 I_2(t) = \text{constant}$.
		+
		+	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:
		+
		+	$\boxed{L_1 I_1 + L_2 I_2 = L_1 I_0}$
		+
		+
		+	b) Maximum current in $L_2$
		+
		+	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:
		+
		+	$I_0 = \frac{V_0}{\omega L_1} = V_0 \sqrt{\frac{C}{L_1}}$
		+
		+	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:
		+
		+	$L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2}$
		+
		+	$\omega' = \frac{1}{\sqrt{L_{\text{eq}} C}} = \frac{1}{\sqrt{\frac{L_1 L_2}{L_1 + L_2} C}}$
		+
		+	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:
		+
		+	$I_1 - I_2 = I_0 \cos \omega' t$
		+
		+	Combining this equation with the conservation law from part (a):
		+
		+	$L_1 I_1 + L_2 I_2 = L_1 I_0$
		+
		+	we solve the system for I_2:
		+
		+	$I_2(t) = \frac{L_1}{L_1 + L_2} I_0 \bigl(1 + \cos \omega' t\bigr)$.
		+
		+	The maximum value of this current occurs when $\cos \omega' t = 1$:
		+
		+	$I_{2,\text{max}} = \frac{2 L_1}{L_1 + L_2} I_0
		+	= \frac{2 L_1}{L_1 + L_2} \cdot V_0 \sqrt{\frac{C}{L_1}}
		+	= 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}$
		+
		+	$\boxed{I_{2,\text{max}} = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,56 @@
			### Statement

			$11.4.21.$ [Insert the problem statement]

			### Solution

			a) Conservation of$ L_1 I_1 + L_2 I_2$

			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:

			$L_1 \frac{dI_1}{dt} = L_2 \frac{dI_2}{dt}$

			Integrating over time from an initial instant to any later time, and assuming that initially both currents are zero, we obtain:

			$L_1 I_1(t) + L_2 I_2(t) = \text{constant}$.

			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:

			$\boxed{L_1 I_1 + L_2 I_2 = L_1 I_0}$


			b) Maximum current in $L_2$

			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:

			$I_0 = \frac{V_0}{\omega L_1} = V_0 \sqrt{\frac{C}{L_1}}$

			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:

			$L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2}$

			$\omega' = \frac{1}{\sqrt{L_{\text{eq}} C}} = \frac{1}{\sqrt{\frac{L_1 L_2}{L_1 + L_2} C}}$

			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:

			$I_1 - I_2 = I_0 \cos \omega' t$

			Combining this equation with the conservation law from part (a):

			$L_1 I_1 + L_2 I_2 = L_1 I_0$

			we solve the system for I_2:

			$I_2(t) = \frac{L_1}{L_1 + L_2} I_0 \bigl(1 + \cos \omega' t\bigr)$.

			The maximum value of this current occurs when $\cos \omega' t = 1$:

			$I_{2,\text{max}} = \frac{2 L_1}{L_1 + L_2} I_0
			= \frac{2 L_1}{L_1 + L_2} \cdot V_0 \sqrt{\frac{C}{L_1}}
			= 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}$

			$\boxed{I_{2,\text{max}} = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}}$

			#### Answer

			[Insert a concise answer or boxed result]