--- title: "Laplace: i(t) with 12 V, 4 Ω, 1 F, 2 Ω" subject: "Electrical Circuit Theory" chapter: 4 key: "semester_3/electrical_circuit_theory:laplace-current-12v-8ohm-4ohm-1f" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-current-12v-8ohm-4ohm-1f site: BCT Engineering --- # Laplace: i(t) with 12 V, 4 Ω, 1 F, 2 Ω Electrical Circuit Theory · Chapter 4: Transient analysis in RLC circuit by Laplace Transform - Asked in: 2070 Asar (old course), 8 marks - Video: [Using Laplace Transform, determine current i(t).](https://www.youtube.com/watch?v=fwkZGlAQDZA) — Engg-Course-Made-Easy - 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-12v-8ohm-4ohm-1f - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4 ## Question Using Laplace transform method, find the current i(t) for t > 0 in the circuit shown in the figure below. [Figure: 12 V source with an 8 Ω resistor in the top wire; a 4 Ω resistor as a shunt branch carrying i(t) (downward); a 1 F capacitor as a shunt branch; and a switch (t = 0) in series with a 2 Ω resistor as the last shunt branch] ## Short answer Initial capacitor voltage is 4 V. After switch closes, s-domain node equation gives V(s). Current through 4 Ω is V/4: i(t) = 0.4286 + 0.5714 e^(−0.875t) A, or 3/7 + (4/7)e^(−7t/8) A, t ≥ 0. ## Answer **Assumption:** the switch is open for a long time before t = 0, so the capacitor is charged by the 12 V source through 8 Ω and the 4 Ω resistor. The branch with 4 Ω carries i(t) downward. #### Initial condition (t = 0⁻) No current flows in the capacitor, so 8 Ω and 4 Ω form a divider: v_C(0⁻) = 12 × 4/(8 + 4) = 4 V = v_C(0⁺) #### For t > 0 (switch closed, extra 2 Ω in parallel) Let V be the common node voltage. KCL (C = 1 F): ```text (12 − V)/8 = V/4 + V/2 + C dV/dt dV/dt + 0.875 V = 1.5 ``` s-domain (including the initial condition V(0) = 4): ```text (s + 0.875) V(s) = 1.5/s + 4 V(s) = (4 s + 1.5) / [ s (s + 0.875) ] = 1.7143/s + 2.2857/(s + 0.875) ``` V(t) = 1.7143 + 2.2857 e^(−0.875t) V #### Current i(t) in the 4 Ω resistor i = V/4: **i(t) = 0.4286 + 0.5714 e^(−0.875t) A = 3/7 + (4/7) e^(−7t/8) A, t ≥ 0** Check: i(0) = 1 A = 4 V/4 Ω ✓; i(∞): with C open the final node voltage is 12 × (4∥2)/(8 + 4∥2) = 12 × 1.333/9.333 = 1.714 V, so i = 1.714/4 = 0.4286 A ✓. τ = R_eq C = (8 ∥ 4 ∥ 2) × 1 = 1.143 s ✓ (1/0.875). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Before t = 0 the capacitor is fully charged and carries no current. What is v_C(0-) in this solution? A. 4 V B. 12 V C. 6 V D. 8 V Answer: A. With no capacitor current, 8 ohm and 4 ohm form a divider: 12 x 4/(8+4) = 4 V, and v_C(0+) = v_C(0-). (Key point: Initial capacitor voltage) 2. For t > 0 the extra 2 ohm branch is connected. Which equation correctly gives the node voltage V? A. (12 - V)/8 = V/4 + C dV/dt B. (12 - V)/8 = V/4 + V/2 + C dV/dt C. (12 - V)/4 = V/8 + V/2 + C dV/dt D. 12 = 8V + 4V + 2V + C dV/dt Answer: B. KCL: current through 8 ohm equals the sum of the currents in 4 ohm, 2 ohm and the capacitor. (Key point: Node equation for t > 0) 3. What are the time constant and final current i(infinity) for t > 0? A. tau = 1.143 s, i(inf) = 1 A B. tau = 0.875 s, i(inf) = 3 A C. tau = 1.143 s, i(inf) = 0.4286 A D. tau = 8 s, i(inf) = 0.4286 A Answer: C. tau = (8||4||2) x 1 = 1.143 s (the pole at 0.875), and with C open V = 1.714 V so i = V/4 = 0.4286 A. (Key point: Time constant and final value) 4. Once V(t) = 1.7143 + 2.2857 e^(-0.875t) is found, how is i(t) in the 4 ohm resistor obtained and what is i(0)? A. i = V/8, so i(0) = 0.5 A B. i = V/4, so i(0) = 0.4286 A C. i = (12 - V)/4, so i(0) = 2 A D. i = V/4, so i(0) = 1 A Answer: D. The 4 ohm branch current is V/4; at t = 0, V = 4 V so i(0) = 1 A, matching the initial condition. 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