--- title: "Laplace: current through 2 Ω, exponential source" subject: "Electrical Circuit Theory" chapter: 4 key: "semester_3/electrical_circuit_theory:laplace-current-through-2ohm-exponential-source" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-current-through-2ohm-exponential-source site: BCT Engineering --- # Laplace: current through 2 Ω, exponential source Electrical Circuit Theory · Chapter 4: Transient analysis in RLC circuit by Laplace Transform - Asked in: 2070 Asar (old course), 8 marks - Video: [L85 | Response of First Order System Excited by Exponential Source | Laplace Transform | ECM |TU-IOE](https://www.youtube.com/watch?v=ixTtohW6-v4) — Pidits - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-current-through-2ohm-exponential-source - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4 ## Question Using Laplace transform method, find the expression for current through 2 Ω resistor for t > 0 in the circuit shown in figure. [Figure: source V = 1/10 e^(−3t) with a switch (t = 0), a 1 Ω resistor and a 0.1 H inductor in series; then a capacitor (labelled 0.5 Ω in the scan, presumably 0.5 F) and the 2 Ω resistor as parallel shunt branches] ## Short answer Transform exponential source, find parallel impedance. Current in 2 Ω resistor is I2(s)=1/[(s+3)(s+5)(s+6)]. Inverse transform: i2(t) = (1/6)e^(−3t) − (1/2)e^(−5t) + (1/3)e^(−6t) A, t>0. ## Answer **Assumption:** the capacitor labelled "0.5 Ω" in the scan is 0.5 F, in parallel with the 2 Ω resistor. All initial conditions are zero. Source V(s) = 0.1/(s + 3). #### Parallel branch impedance Y = sC + 1/R = 0.5 s + 0.5, so Z_p = 1/(0.5 s + 0.5) = 2/(s + 1). #### Source current ```text I(s) = V(s) / (R + sL + Z_p) = [0.1/(s + 3)] / [ 1 + 0.1 s + 2/(s + 1) ] ``` #### Voltage across the parallel part and current in 2 Ω Simplify the total impedance: 1 + 0.1s + 2/(s + 1) = (0.1 s² + 1.1 s + 3)/(s + 1) = 0.1 (s + 5)(s + 6)/(s + 1), so ```text V_p(s) = 0.1/(s + 3) × 2/[0.1 (s + 5)(s + 6)] = 2 / [ (s + 3)(s + 5)(s + 6) ] I₂(s) = V_p/2 = 1 / [ (s + 3)(s + 5)(s + 6) ] ``` #### Partial fractions ```text I₂(s) = (1/6)/(s + 3) − (1/2)/(s + 5) + (1/3)/(s + 6) ``` (K₁ = 1/((2)(3)) = 1/6, K₂ = 1/((−2)(1)) = −1/2, K₃ = 1/((−3)(−1)) = 1/3) **i₂(t) = (1/6) e^(−3t) − (1/2) e^(−5t) + (1/3) e^(−6t) A, t > 0** Check: i₂(0) = 1/6 − 1/2 + 1/3 = 0 ✓ (the capacitor is initially a short across the 2 Ω). At t = 0.5 s, i₂ = 0.01274 A, equal to the numerical inverse Laplace value. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With zero initial conditions, what is the Laplace transform of the source (1/10) e^(-3t)? A. 0.1/(s + 3) B. 0.1/s C. 0.1/(s - 3) D. 0.3/(s + 3) Answer: A. L{e^(-at)} = 1/(s+a), so (1/10)e^(-3t) becomes 0.1/(s+3). (Key point: Source in s-domain) 2. The 0.5 F capacitor is in parallel with the 2 ohm resistor. What is their combined impedance Z_p? A. 1/(0.5s + 2) B. 2/(s + 1) C. 2s/(s + 1) D. 0.5/(s + 2) Answer: B. Y = sC + 1/R = 0.5s + 0.5, so Z_p = 1/(0.5s+0.5) = 2/(s+1). (Key point: Parallel branch impedance) 3. How is the 2 ohm resistor current I2(s) obtained, and what is the result? A. I2 = V_p x 2 = 4/[(s+3)(s+5)(s+6)] B. I2 = source current I(s) itself C. I2 = V_p/2 = 1/[(s+3)(s+5)(s+6)] D. I2 = V_p/2 = 1/[(s+3)(s+1)(s+6)] Answer: C. V_p = 2/[(s+3)(s+5)(s+6)] is the voltage across the parallel part; dividing by 2 ohm gives I2. (Key point: Finding current in 2 ohm) 4. What is the final time-domain current and what check does i2(0) satisfy? A. (1/6)e^(-3t) + (1/2)e^(-5t) + (1/3)e^(-6t); i2(0) = 1 B. (1/3)e^(-3t) - (1/2)e^(-5t) + (1/6)e^(-6t); i2(0) = 0 C. (1/6)e^(-3t) - (1/2)e^(-5t) + (1/3)e^(-6t); i2(0) = 1/6 D. (1/6)e^(-3t) - (1/2)e^(-5t) + (1/3)e^(-6t); i2(0) = 0 Answer: D. The residues are 1/6, -1/2, 1/3 and they sum to 0, since the capacitor initially shorts the 2 ohm resistor. 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