--- title: "Initial values with 20 Ω shorted by switch, 4 H, 2 F" subject: "Electrical Circuit Theory" chapter: 2 key: "semester_3/electrical_circuit_theory:initial-values-switch-open-20-30" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-values-switch-open-20-30 site: BCT Engineering --- # Initial values with 20 Ω shorted by switch, 4 H, 2 F Electrical Circuit Theory · Chapter 2: Initial Conditions - Asked in: 2068 Baisakh, 8 marks - Video: [For the network shown below switch is closed at t=0. Determine 𝒊, 𝒅𝒊/𝒅𝒕](https://www.youtube.com/watch?v=py-HwjhEwq8) — 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/initial-values-switch-open-20-30 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2 ## Question The switch has been opened for a long time as shown in the figure below. At time t = 0, it is suddenly closed. At t = 0⁺, find current through inductor, voltage across capacitor, charge across capacitor, current and voltage across each resistor. [Figure: 100 V source in series with a 20 Ω resistor, which has a switch (t = 0) connected in parallel across it. After that, two parallel branches: a 30 Ω resistor in series with a 4 H inductor, and a 2 F capacitor in series with a 10 Ω resistor.] ## Short answer iL = 2 A, vC = 60 V, q = 120 C. 30 Ω: 2 A, 60 V; 10 Ω: 4 A, 40 V; 20 Ω: 0, 0; vL = 40 V. Switch shorts 20 Ω, using continuity of iL and vC. ## Answer **Circuit:** 100 V source in series with 20 Ω, which has a switch across it. The remaining circuit is two parallel branches: 30 Ω + 4 H, and 2 F + 10 Ω. The switch has been open for a long time, then closes at t = 0. #### At t = 0⁻ (dc steady state, switch open) - Inductor = short, capacitor = open circuit. - Current: i_L = 100/(20 + 30) = 2 A. - No current flows in the 2 F + 10 Ω branch, so the capacitor voltage equals the voltage across the 30 Ω resistor: v_C = 30 × 2 = 60 V. #### At t = 0⁺ (switch closed) The inductor current and capacitor voltage cannot change suddenly. The closed switch short-circuits the 20 Ω resistor, so the full 100 V now appears across the two parallel branches. | Quantity | Value | |---|---| | Inductor current i_L(0⁺) | 2 A | | Capacitor voltage v_C(0⁺) | 60 V | | Capacitor charge q = C v_C | 2 × 60 = 120 C | | 30 Ω resistor | i = 2 A, v = 60 V | | Inductor voltage | 100 − 60 = 40 V | | 10 Ω resistor | i = (100 − 60)/10 = 4 A, v = 40 V | | Capacitor current | 4 A | | 20 Ω resistor | v = 0 V, i = 0 A (shorted by the switch) | | Switch / source current | 2 + 4 = 6 A | **Answer: i_L = 2 A, v_C = 60 V, q = 120 C; 30 Ω: 2 A, 60 V; 10 Ω: 4 A, 40 V; 20 Ω: 0 A, 0 V (shorted); v_L = 40 V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At t = 0−, what is i_L? A. 100/30 A B. 100/(20 + 30) = 2 A C. 0 D. 5 A Answer: B. Inductor is a short and the capacitor is open. (Key point: Steady state) 2. What is v_C(0+)? A. 100 V B. 40 V C. 60 V D. 0 V Answer: C. Equal to the 30 Ω drop, 30 × 2. (Key point: Capacitor voltage) 3. Effect of closing the switch? A. Opens inductor B. Capacitor discharges instantly C. 20 Ω carries 5 A D. Shorts the 20 Ω so 20 Ω has 0 V and 100 V appears across the branches Answer: D. i_L and v_C stay the same. (Key point: Switch effect) 4. What is the capacitor branch current at 0+? A. (100 − 60)/10 = 4 A B. 6 A C. 10 A D. 0 Answer: A. Voltage across the 10 Ω is 40 V. 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