LC circuits: A charged capacitor is connected to an ideal inductor. At time t = 0, the charge on the capacitor is equal to 6.00 μC. At time t = 2.00 ms the charge on the capacitor is zero for the first time. What is the amplitude of the current at that same instant?
A) 2.38 mA
B) 4.71 mA
C) 0.780 mA
D) 1.53 mA
E) 1.19 mA

Question

Quizplus · Accepted Answer

We can use the equation for the charge on a capacitor in an LC circuit:
q = q0 cos(ωt)
where q is the charge on the capacitor, q0 is the initial charge, ω is the angular frequency (2πf) of the circuit, and t is the time.

At t = 0, q = q0 = 6.00 μC. At t = 2.00 ms, q = 0. Thus, we can solve for ω:
0 = q0 cos(ωt)
cos(ωt) = 0
ωt = π/2
ω = π/(2t) = π/(2×2.00×10^-3) = 785.4 rad/s

The current in the circuit is given by:
i = dq/dt = -ωq0 sin(ωt)

Thus, at t = 2.00 ms, the current is:
i = -ωq0 sin(ωt) = -785.4 × 6.00 × 10^-6 sin(π/2) = -4.71 mA

Since the current is alternating, we take the absolute value to get the amplitude:
| i | = 4.71 mA

Therefore, the best choice is B.

LC Circuits: a Charged Capacitor Is Connected to an Ideal