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An inductor of 2.0 mH is connected across a charged capacitor of 5.0 microF, forming an LC circuit oscillating at its natural frequency. The maximum charge on the capacitor is Qmax = 200 microC. Find: (a) |dI/dt| when Q = 100 microC, (b) I when Q = 200 microC, (c) the maximum current Imax, (d) |Q| when I equals half its maximum value.

1. (a) 1.0*10⁴ A/s, (b) 0, (c) 2.0 A, (d) 173 microC
2. (a) 2.0*10⁴ A/s, (b) 2.0 A, (c) 1.0 A, (d) 100 microC
3. (a) 1.0*10⁴ A/s, (b) 2.0 A, (c) 2.0 A, (d) 200 microC
4. (a) 0.5*10⁴ A/s, (b) 0, (c) 4.0 A, (d) 50 microC

Correct answer: (a) 1.0*10⁴ A/s, (b) 0, (c) 2.0 A, (d) 173 microC

Solution

omega = 1/sqrt(LC) = 1/sqrt(2e-3 * 5e-6) = 10⁴ rad/s. (a) From L*dI/dt = Q/C, |dI/dt| = Q/(LC) = 100e-6/(1e-8) = 1.0*10⁴ A/s. (b) At Q = Qmax all energy is electrostatic, so I = 0. (c) Imax = Qmax*omega = 200e-6 * 10⁴ = 2.0 A. (d) Q = Qmax*sqrt(1 - (I/Imax)²) = 200*sqrt(1 - 1/4) = 200*(sqrt3/2) = 173 microC.

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