--- title: "Classical: capacitor voltage, switch opened, 25 V and 60 V" subject: "Electrical Circuit Theory" chapter: 3 key: "semester_3/electrical_circuit_theory:classical-capacitor-switch-opened-25v-60v" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-capacitor-switch-opened-25v-60v site: BCT Engineering --- # Classical: capacitor voltage, switch opened, 25 V and 60 V Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution - Asked in: 2071 Chaitra, 8 marks - Video: [The switch is closed and steady-state. At t = 0, the switch is opened. Determine the current thro...](https://www.youtube.com/watch?v=wjlECJtAKMA) — MANTHAN FOUNDATION - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-capacitor-switch-opened-25v-60v - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3 ## Question Keeping the switch closed for a long time, if the switch is opened at t = 0 in the circuit shown in figure below, find expression for voltage across capacitor in the circuit shown below using classical method of solution. [Figure: 25 V source in series with a 4 H inductor, a 3 Ω resistor and a 15 Ω resistor along the top wire; a 3 F capacitor as a shunt from the junction of 3 Ω and 15 Ω to the bottom wire; at the right end, a 60 V source in series with a switch (t = 0) at the end of the 15 Ω resistor] ## Short answer At t=0⁻, vC=30.83 V, iL=−1.944 A. After opening, the series RLC with 25 V source has overdamped response vC(t)=25+6.132e^{-0.1356t}−0.299e^{-0.6144t} V for t>0. ## Answer **Assumptions:** the 25 V source (+ up), 4 H inductor, 3 Ω and 15 Ω are in series along the top wire; the 3 F capacitor is a shunt from the 3 Ω/15 Ω junction to ground; at the right end a 60 V source (+ up) is connected through the switch to the 15 Ω resistor. The switch was closed for a long time and opens at t = 0. #### Before t = 0 (dc steady state) Inductor short, capacitor open; the 25 V and 60 V sources oppose each other through 3 + 15 = 18 Ω: ```text i_L(0⁻) = (25 − 60)/18 = −1.944 A v_C(0⁻) = 25 − 3(−1.944) = 30.83 V ``` #### After t = 0 (switch open) The 15 Ω and 60 V source are disconnected. The circuit becomes series RLC: 25 V, 4 H, 3 Ω, 3 F. ```text LC d²v/dt² + RC dv/dt + v = 25 12 v″ + 9 v′ + v = 25 s² + 0.75 s + 0.0833 = 0 s = [−0.75 ± √(0.5625 − 0.3333)]/2 = −0.1356, −0.6144 ``` Real, distinct roots (overdamped). #### Particular and complementary solution v_p = 25 V, and ```text v_C(t) = 25 + K₁ e^(−0.1356t) + K₂ e^(−0.6144t) ``` #### Constants ```text v_C(0⁺) = 30.83: K₁ + K₂ = 5.833 dv_C/dt(0⁺) = i_L(0⁺)/C = −1.944/3 = −0.648 V/s: −0.1356 K₁ − 0.6144 K₂ = −0.648 ``` Solving: K₁ = 6.132, K₂ = −0.2989. #### Result ```text v_C(t) = 25 + 6.132 e^(−0.1356t) − 0.2989 e^(−0.6144t) V, t > 0 ``` Numerical integration of the circuit confirms this (for example v_C(5 s) = 28.098 V). **Answer: v_C(t) = 25 + 6.132e^(−0.1356t) − 0.299e^(−0.6144t) V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is i_L(0-)? A. (25 - 60)/18 = -1.944 A B. 25/18 = 1.389 A C. (25 + 60)/18 = 4.72 A D. 0 Answer: A. Opposing sources through 3 + 15 ohm. (Key point: Before switching) 2. What is v_C(0-)? A. 25 V B. 25 + 3(1.944) = 30.83 V C. 60 V D. 19.17 V Answer: B. v_C = 25 - 3(-1.944) = 30.83 V. (Key point: Initial capacitor voltage) 3. What are the characteristic roots and type? A. Complex roots; underdamped B. Equal roots; critically damped C. -0.1356 and -0.6144; overdamped D. Positive roots; unstable Answer: C. s^2 + 0.75 s + 0.0833 = 0 has discriminant 0.2292 > 0. (Key point: Damping) 4. What is the particular solution? A. 60 V B. 0 V C. 30.83 V D. 25 V, as the capacitor ends charged to the remaining source Answer: D. After the switch opens the 15 ohm and 60 V are disconnected, leaving 25 V. 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