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In an LC circuit, the capacitor carries a maximum charge q0. What is the maximum rate of change of current (dI/dt)_max?

1. q0 / (LC)
2. q0 / sqrt(LC)
3. q0 / (2LC)
4. 2*q0 / (LC)

Correct answer: q0 / (LC)

Solution

The current in an LC circuit: I = (q0/sqrt(LC)) * sin(omega*t) where omega = 1/sqrt(LC). Then dI/dt = (q0/sqrt(LC)) * omega * cos(omega*t) = q0 * omega² * cos(omega*t) = (q0/LC)*cos(omega*t). Maximum value: (dI/dt)_max = q0/(LC).

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