--- title: "Derivatives at t=0+: 2 H, 100 V, 1 μF" subject: "Electrical Circuit Theory" chapter: 2 key: "semester_3/electrical_circuit_theory:initial-values-2h-100v-1uf" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-values-2h-100v-1uf site: BCT Engineering --- # Derivatives at t=0+: 2 H, 100 V, 1 μF Electrical Circuit Theory · Chapter 2: Initial Conditions - Asked in: 2074 Asoj, 8 marks - Video: [The switch is closed at t =0.With the capacitor uncharged, find value for I, di/dt ,d2i/dt2at t = 0+](https://www.youtube.com/watch?v=k1rvMCTIymI) — 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/initial-values-2h-100v-1uf - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2 ## Question In the given network of the figure below, both the energy storing elements are initially relaxed i.e. no current is flowing through the inductor and no charge is accumulated across the capacitor before application of voltage. The switch K is closed at t = 0. Find the values of i1, i2, di1/dt, di2/dt, d²i1/dt², d²i2/dt² at t = 0⁺. [Figure: 100 V source; switch K and a 2 H inductor in the top wire of the left mesh (i1); a 10 Ω resistor in the middle vertical branch; a 10 Ω resistor in the top wire of the right mesh (i2); a 1 μF capacitor on the right] ## Short answer At t = 0+, i1 = i2 = 0; di1/dt = 50 A/s, di2/dt = 25 A/s; d²i1/dt² = −125 A/s², d²i2/dt² = −1.25×10⁶ A/s². All results from mesh equations with zero initial inductor current and capacitor voltage. ## Answer **Assumptions:** both elements relaxed. Left mesh (current i₁): 100 V, switch K, L = 2 H, common resistor 10 Ω. Right mesh (current i₂): common 10 Ω, another 10 Ω, C = 1 μF. #### Mesh equations ```text Mesh 1: 100 = 2 di₁/dt + 10(i₁ − i₂) Mesh 2: 10(i₂ − i₁) + 10 i₂ + (1/C)∫i₂ dt = 0 → 20 i₂ − 10 i₁ + v_C = 0 ``` #### At t = 0⁺ (i_L = 0, v_C = 0) ```text i₁(0⁺) = 0 20 i₂ − 0 + 0 = 0 → i₂(0⁺) = 0 ``` #### First derivatives ```text Mesh 1: 100 = 2 i₁′ + 0 → i₁′(0⁺) = 50 A/s Mesh 2 differentiated: 20 i₂′ − 10 i₁′ + i₂/C = 0 20 i₂′ = 10 × 50 → i₂′(0⁺) = 25 A/s ``` #### Second derivatives ```text Mesh 1 differentiated: 0 = 2 i₁″ + 10(i₁′ − i₂′) i₁″ = −10(50 − 25)/2 = −125 A/s² Mesh 2 differentiated twice: 20 i₂″ − 10 i₁″ + i₂′/C = 0 i₂″ = [10(−125) − 25/10⁻⁶]/20 = (−1250 − 2.5×10⁷)/20 = −1.25006×10⁶ A/s² ``` | Quantity | Value at 0⁺ | |---|---| | i₁ | 0 | | i₂ | 0 | | di₁/dt | 50 A/s | | di₂/dt | 25 A/s | | d²i₁/dt² | −125 A/s² | | d²i₂/dt² | −1.25×10⁶ A/s² | **Answer: i₁ = i₂ = 0; di₁/dt = 50, di₂/dt = 25; d²i₁/dt² = −125, d²i₂/dt² = −1.25×10⁶ (SI units).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are i1(0+) and i2(0+) zero? A. Inductor current and capacitor voltage are continuous and both start at zero B. The switch is open C. The source is zero D. The resistors are shorted Answer: A. Relaxed elements keep i_L = 0 and v_C = 0. (Key point: Initial currents) 2. What is di1/dt at 0+? A. 100 A/s B. 50 A/s = 100/2 C. 200 A/s D. 0 Answer: B. 100 = 2 i1′ with no resistor drop. (Key point: First derivative) 3. Why is d²i2/dt² so large (about −1.25×10^6)? A. Because L is large B. Because i1 is large C. The term i2′/C with C = 1 μF is huge D. Because R is large Answer: C. i2′/C = 25/10^−6 = 2.5×10^7 dominates. 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