LC circuits: A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be 3.0 μC and the current in the circuit is equal to 75 μA. What is the maximum charge of the capacitor?
A) 6.0 µC
B) 10 µC
C) 8.0 µC
D) 2.0 µC
E) 4.0 µC

Question

Quizplus · Accepted Answer

The LC circuit has a resonant frequency given by f = 1/(2π√(LC))
We are given f = 1.6 Hz, so we can solve for the product of LC:
LC = (1/(2πf))^2 = (1/(2π*1.6))^2 = 6.25*10^-5 H F

We are also given the charge on the capacitor, Q = 3.0 μC, and the current in the circuit, I = 75 μA.

At any given instant in time, the charge on the capacitor is given by Q = CV, where V is the voltage across the capacitor and C is the capacitance.

The current in the circuit is given by I = dQ/dt = C(dV/dt).

Since the circuit is purely resistless, the voltage across the capacitor is equal to the voltage across the inductor, and is given by V = L(di/dt), where di/dt is the rate of change of current in the circuit.

Combining these equations, we can write:
I = C(dV/dt) = C(d/dt)(L(di/dt)) = CL(d^2i/dt^2)

Substituting the values we know, we get:
75*10^-6 = C*(6.25*10^-5)*(d^2i/dt^2)

Solving for d^2i/dt^2, we get:
d^2i/dt^2 = (75*10^-6)/(6.25*10^-5*C) = 1.2*(10^3)/C

At maximum charge, the current in the circuit is zero, so d^2i/dt^2 reaches its maximum value. Using this value, we can solve for C:
1.2*(10^3)/Cmax = 75*10^-6/(6.25*10^-5)
Cmax = 8*10^-6 F

Finally, we can solve for the maximum charge on the capacitor:
Qmax = CVmax = (8*10^-6)*(di/dt)_max = (8*10^-6)*(2πf*Imax) = 8.5 μC

Therefore, the correct answer is (C) 8.0 µC.

LC Circuits: a Charged Capacitor Is Connected to an Ideal