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Maximum current in LC circuit

Short answer

Using energy conservation ½CV₀² = ½LI_max² gives I_max = V₀√(C/L). With C=2μF, L=50mH, V₀=50V, I_max = 50√(2×10⁻⁶/50×10⁻³) ≈ 0.316 A.

A 2μF capacitor is charged up to 50V. The battery is disconnected and a 50mH coil is connected across the capacitor so that LC oscillation occurs. Calculate the maximum value of the current in the circuit.

Given: C = 2 μF = 2 × 10⁻⁶ F, V0 = 50 V, L = 50 mH = 50 × 10⁻³ H.

Method (energy conservation). When the coil is connected, all the energy is in the capacitor: U = ½CV0². At the instant the current is maximum, the capacitor is empty and all energy is in the inductor: ½LI_max². There is no resistance, so the two are equal.

½ C V0² = ½ L I_max²
I_max = V0 √(C/L)

Steps

  1. Energy initially stored:
U = ½ × 2×10⁻⁶ × (50)² = 2.5 × 10⁻³ J
  1. C/L = 2×10⁻⁶ / 50×10⁻³ = 4 × 10⁻⁵ F/H.
  2. √(C/L) = 6.325 × 10⁻³ A/V.
  3. I_max = 50 × 6.325 × 10⁻³ = 0.316 A.

Check: ½ × 0.05 × (0.316)² = 2.5 × 10⁻³ J.

Answer: I_max ≈ 0.316 A (about 0.32 A).

✓ Answer checked by Claude Opus 5.5 (high effort)

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