--- title: "Parallel RLC with exponential current source, Laplace" subject: "Electrical Circuit Theory" chapter: 4 key: "semester_3/electrical_circuit_theory:parallel-rlc-exponential-current-laplace" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/parallel-rlc-exponential-current-laplace site: BCT Engineering --- # Parallel RLC with exponential current source, Laplace Electrical Circuit Theory · Chapter 4: Transient analysis in RLC circuit by Laplace Transform - Asked in: 2075 Asoj, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/parallel-rlc-exponential-current-laplace - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4 ## Question 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. ## 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. ## Answer 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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which node equation is used for the parallel RLC in the s-domain? A. V(s)[1/R + sC + 1/(sL)] = I(s) with I(s) = 20/(s+4) B. V(s)[R + sL + 1/(sC)] = I(s) C. V(s)[1/R + sC + 1/(sL)] = 20/s D. V(s)[R + sC + 1/(sL)] = 20/(s+4) Answer: A. Parallel elements add admittances 1/R, sC and 1/(sL), and the exponential source transforms to 20/(s+4). (Key point: Node equation) 2. Why is the circuit overdamped, and what are the poles of the characteristic part? A. alpha is less than w0; complex poles B. alpha = 1/(2RC) = 2x10^6 is much greater than w0 = 6.3x10^3; poles about -10 and -4x10^6 C. alpha equals w0; repeated poles D. alpha is much greater than w0; poles about +10 and +4x10^6 Answer: B. alpha >> w0 gives two real negative poles. (Key point: Overdamped poles) 3. Which term of v(t) is the particular (forced) part? A. 3.333 e^(-10t) B. -2 e^(-4x10^6 t) C. -1.333 e^(-4t), which has the same exponent as the source D. All three terms Answer: C. The pole at s = -4 comes from the source term 1/(s+4); the other two are from the circuit poles (natural response). 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