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Parallel RLC with exponential current source, Laplace

Short answer

Complete voltage solution: v(t)=−1.333e^(−4t)+3.333e^(−10t)−2e^(−4×10^6 t) V. The forced term −1.333e^(−4t) matches the source, while the natural terms have fast and slow decays.

An exponential current i(t) = 20e^(−4t) Amp is suddenly applied at time t = 0 to a parallel RLC circuit comprising resistor R = 1/10 Ω, inductor L = 10 mH and capacitor C = 2.5 μF. Obtain the complete particular solution for voltage v(t) across the network, by Laplace transform method. Assume zero initial current through inductor and zero initial charge across the capacitor before application of the current.

Zero initial conditions, so the s-domain admittances are 1/R, sC and 1/(sL). The source is I(s) = 20/(s + 4).

Node equation

V(s) [ 1/R + sC + 1/(sL) ] = I(s)
R = 0.1 Ω → 1/R = 10;  C = 2.5×10⁻⁶ F;  L = 0.01 H → 1/L = 100

V(s) = 20 /(s + 4) × s / (C s² + 10 s + 100)
     = 8×10⁶ s / [ (s + 4)(s² + 4×10⁶ s + 4×10⁷) ]

Roots

s² + 4×10⁶ s + 4×10⁷ = 0 → s = −10.00 and s = −3.99999×10⁶ ≈ −4×10⁶ (overdamped: α = 1/2RC = 2×10⁶ ≫ ω₀ = 1/√(LC) = 6.3×10³).

V(s) = 8×10⁶ s / [(s + 4)(s + 10)(s + 4×10⁶)] (approximately)

Partial fractions (residues)

  • At s = −4: 8×10⁶(−4) / [(6)(4×10⁶)] = −1.3333
  • At s = −10: 8×10⁶(−10) / [(−6)(4×10⁶)] = 3.3333
  • At s = −4×10⁶: 8×10⁶(−4×10⁶) / [(−4×10⁶)(−4×10⁶)] = −2.000

Time domain

v(t) = −1.333 e^(−4t) + 3.333 e^(−10t) − 2 e^(−4×10⁶ t) V, t ≥ 0

  • The term −1.333e^(−4t) is the forced (particular) part, which has the shape of the source.
  • The other two terms are the natural response (very fast decay at 4×10⁶ s⁻¹ and slow decay at 10 s⁻¹).

Check: v(0) = −1.333 + 3.333 − 2 = 0 ✓ (no initial energy; the capacitor voltage cannot jump). Values at t = 0.01 s: v = 1.735 V and at t = 0.1 s: v = 0.3325 V, identical to a numerical inverse Laplace transform and to a direct numerical solution of the RLC differential equations.

✓ Answer checked by Claude Opus 5.5 (high effort)

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