Question

Initially, a switch S open in the circuit shown in the figure, a capacitor of capacitance 2C carries the electric charge q0, a capacitor of capacitance C is uncharged, and there are no electric currents in both coils of inductance L and 2L, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch S is instantly shorted. If the maximum current Imax through the switch S thereafter is

• A. q0/(3√(LC))
• B. q0/√(LC)
• C. q0/√(2LC)
• D. q0/(3√(2LC))

Answer 1

Answer: (D)

q0/(3√(2LC))

Answer 2

Let's analyze the circuit in two phases:

Phase 1: Switch S is open.

Initially, only the left branch (inductor L and capacitor 2C) is active. This is an LC circuit with initial charge $q_0$ on the capacitor 2C and zero current in the inductor L. The angular frequency of oscillation is $\omega_1 = \frac{1}{\sqrt{L(2C)}} = \frac{1}{\sqrt{2LC}}$. The charge on the capacitor 2C oscillates as $q_1(t) = q_0 \cos(\omega_1 t)$. The current in the inductor L is $i_1(t) = -\frac{dq_1}{dt} = q_0 \omega_1 \sin(\omega_1 t)$. The maximum current in the inductor L is $I_{1,max} = q_0 \omega_1 = \frac{q_0}{\sqrt{2LC}}$. This maximum current occurs when $\sin(\omega_1 t) = 1$. At this time, the charge on the capacitor 2C is 0. The energy initially stored in the capacitor 2C ($U_{C1} = \frac{q_0^2}{2(2C)} = \frac{q_0^2}{4C}$) is completely transferred to the inductor L ($U_L = \frac{1}{2} L I_{1,max}^2$). $\frac{1}{2} L I_{1,max}^2 = \frac{q_0^2}{4C} \implies I_{1,max}^2 = \frac{q_0^2}{2LC} \implies I_{1,max} = \frac{q_0}{\sqrt{2LC}}$.

Phase 2: At $t_1$, the switch S is instantly shorted.

At this moment, the state of the circuit is:

• Capacitor 2C: charge 0
• Inductor L: current $I_{1,max}$
• Capacitor C: charge 0
• Inductor 2L: current 0

The circuit consists of two parallel branches.

• Branch 1: L in series with 2C.
• Branch 2: 2L in series with C.

The switch S connects the junction between L and 2C in Branch 1 to the junction between 2L and C in Branch 2.

Let $i_L$ and $i_{2L}$ be the currents flowing through L and 2L respectively. Let $i_{2C}$ and $i_C$ be the currents flowing through 2C and C respectively. Let $i_S$ be the current through the switch S.

Applying Kirchhoff's Current Law (KCL) at the junctions:

• $i_L = i_{2C} + i_S$
• $i_{2L} + i_S = i_C$

Also, $q_{2C} = 2q_C$, which implies $i_{2C} = 2i_C$. Using the initial conditions, we find that $i_S = \frac{I_{1,max}}{3} = \frac{q_0}{3\sqrt{2LC}}$. This current remains constant.

Therefore, the maximum current through the switch S thereafter is $\frac{q_0}{3\sqrt{2LC}}$.