Question

A capacitor of capacitance 2C is charged with charge $q_0$ is connected in series with an uncharged capacitor and inductor of inductance L. Find maximum value of current through the inductor.

• A. $\frac{q_0}{\sqrt{3LC}}$
• B. $\frac{q_0}{\sqrt{6LC}}$
• C. $\frac{q_0}{\sqrt{2LC}}$
• D. $\frac{q_0}{\sqrt{LC}}$

Answer 1

Answer: (B)

$\frac{q_0}{\sqrt{6LC}}$

Answer 2

The initial energy stored in the charged capacitor ($C_1 = 2C$) is $E_{initial} = \frac{q_0^2}{2(2C)} = \frac{q_0^2}{4C}$.

When the capacitors ($C_1 = 2C$ and $C_2 = C$) are in series, the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(2C)(C)}{2C+C} = \frac{2C^2}{3C} = \frac{2C}{3}$.

The maximum current in an LC circuit occurs when all the initial energy is converted into magnetic energy in the inductor, i.e., when the charge on the capacitors is zero. However, in this case, the capacitors are in series, and the charge distribution changes.

The total energy is conserved: $E_{initial} = \frac{q_0^2}{4C}$. At maximum current, the energy is purely magnetic: $E_{max\_current} = \frac{1}{2}LI_{max}^2$. This occurs when the charge distribution across the capacitors is such that the total energy stored in the capacitors is minimized, and the remaining energy is in the inductor.

Alternatively, consider the total energy in the system. When the current is maximum, the energy stored in the capacitors is minimal. The charge on $C_1$ is $q_1$ and on $C_2$ is $q_2$. Since they are in series, $q_1 = q_2 = q_{series}$. The total energy is $E = \frac{q_1^2}{2C_1} + \frac{q_2^2}{2C_2} + \frac{1}{2}LI^2 = \frac{q_{series}^2}{2(2C)} + \frac{q_{series}^2}{2C} + \frac{1}{2}LI^2 = \frac{3q_{series}^2}{4C} + \frac{1}{2}LI^2$. The initial energy is $\frac{q_0^2}{4C}$. This initial charge $q_0$ is on $C_1$. When connected, charge redistributes. The total charge on the series combination is $q_{series}$. The initial energy is $\frac{q_0^2}{4C}$. The maximum current occurs when the total energy is $\frac{1}{2}LI_{max}^2$. By conservation of energy, $\frac{q_0^2}{4C} = \frac{1}{2}LI_{max}^2$. This assumes the final charge on both capacitors is zero, which is not always true.

A more rigorous approach: The total initial energy is $U_i = \frac{q_0^2}{4C}$. At maximum current, the energy is $U_{max\_I} = \frac{1}{2}LI_{max}^2$. This energy comes from the initial potential energy. The total charge on the series combination is $Q$. The voltage across $C_1$ is $V_1 = Q/(2C)$ and across $C_2$ is $V_2 = Q/C$. The initial charge $q_0$ is on $2C$. So the initial voltage is $V_0 = q_0/(2C)$. When connected, $V_1 = V_2$ if no inductor is present. With an inductor, the total energy is conserved. $E = \frac{q_1^2}{4C} + \frac{q_2^2}{2C} + \frac{1}{2}LI^2$. Initially, $q_1 = q_0$, $q_2 = 0$, $I = 0$. $E = \frac{q_0^2}{4C}$. At maximum current, $I = I_{max}$, and the charge on capacitors is such that the total potential energy is minimal. The charge on $C_1$ is $q_0 - q$ and on $C_2$ is $q$. $U_{cap} = \frac{(q_0-q)^2}{4C} + \frac{q^2}{2C}$. $dU_{cap}/dq = -\frac{q_0-q}{2C} + \frac{q}{C} = \frac{-q_0+q+2q}{2C} = \frac{3q-q_0}{2C}$. Setting $dU_{cap}/dq = 0$, we get $q = q_0/3$. Then $q_1 = q_0 - q_0/3 = 2q_0/3$ and $q_2 = q_0/3$. The minimum capacitor energy is $U_{cap,min} = \frac{(2q_0/3)^2}{4C} + \frac{(q_0/3)^2}{2C} = \frac{4q_0^2/9}{4C} + \frac{q_0^2/9}{2C} = \frac{q_0^2}{9C} + \frac{q_0^2}{18C} = \frac{3q_0^2}{18C} = \frac{q_0^2}{6C}$. By conservation of energy: $E_{initial} = U_{cap,min} + \frac{1}{2}LI_{max}^2$. $\frac{q_0^2}{4C} = \frac{q_0^2}{6C} + \frac{1}{2}LI_{max}^2$. $\frac{1}{2}LI_{max}^2 = \frac{q_0^2}{4C} - \frac{q_0^2}{6C} = \frac{3q_0^2 - 2q_0^2}{12C} = \frac{q_0^2}{12C}$. $I_{max}^2 = \frac{q_0^2}{6LC}$. $I_{max} = \frac{q_0}{\sqrt{6LC}}$.