--- title: "Classical: i1, i2 with two R-L branches" subject: "Electrical Circuit Theory" chapter: 3 key: "semester_3/electrical_circuit_theory:classical-i1-i2-100v-10ohm-1h" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-i1-i2-100v-10ohm-1h site: BCT Engineering --- # Classical: i1, i2 with two R-L branches Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution - Asked in: 2075 Asoj, 8 marks - Video: [Transient Analysis: Solved Examples on First order RC and RL Circuits](https://www.youtube.com/watch?v=zDcXt9Vx34o) — ALL ABOUT ELECTRONICS - 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-i1-i2-100v-10ohm-1h - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3 ## Question In the network shown, the switch is closed at t = 0, with the network previously unenergised. For the element values shown on the diagram, find i1(t) and i2(t) by classical method for t > 0. [Figure: 100 V source with a switch (closes at t = 0) and a 10 Ω resistor in series, feeding two parallel branches: each is a 10 Ω resistor in series with a 1 H inductor; the branch currents are i1 and i2] ## Short answer The two identical branches share current equally. KVL gives di/dt + 30i = 100 with i(0)=0. Solution: i1(t) = i2(t) = 3.33(1 − e^(−30t)) A. ## Answer **Assumptions:** 100 V source, switch, and a 10 Ω resistor in series feed two identical parallel branches, each 10 Ω + 1 H. The network is initially unenergised (i = 0). Because the two branches are identical, i₁ = i₂ = i, and the current through the series 10 Ω is 2i. #### Differential equation (KVL) ```text 100 = 10(2i) + 10 i + 1·di/dt di/dt + 30 i = 100 ``` #### Complete solution - Particular (steady state): i_p = 100/30 = 3.333 A - Complementary: characteristic equation s + 30 = 0, so i_c = K e^(−30t) ```text i = 3.333 + K e^(−30t) i(0⁺) = 0 → K = −3.333 ``` #### Result ```text i₁(t) = i₂(t) = 3.333 (1 − e^(−30t)) A, t > 0 ``` Time constant τ = L/R_eq = 1/30 s = 33.3 ms. Check: substituting back, di/dt + 30i = 100 e^(−30t) + 100(1 − e^(−30t)) = 100 ✓. **Answer: i₁(t) = i₂(t) = 3.333(1 − e^(−30t)) A.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can both branch currents be written as the same current i? A. The two parallel branches are identical B. The source is dc C. The inductors are initially charged D. The 10 ohm series resistor carries i1 only Answer: A. Identical branches share current equally, so i1 = i2 = i and the series resistor carries 2i. (Key point: Symmetry) 2. Which differential equation results from KVL? A. di/dt + 20 i = 100 B. di/dt + 30 i = 100 C. di/dt + 10 i = 100 D. di/dt + 30 i = 50 Answer: B. 100 = 10(2i) + 10i + 1 di/dt = 30i + di/dt. (Key point: KVL equation) 3. What are the steady state value and time constant? A. 10 A and 1/30 s B. 3.333 A and 30 s C. 3.333 A and 1/30 s D. 5 A and 1/20 s Answer: C. i_p = 100/30 = 3.333 A and the root s = -30 gives tau = 1/30 s. (Key point: Complete solution) 4. Why is K = -3.333? A. Because the final current is zero B. Because the inductor acts as a short at 0+ C. Because K equals the time constant D. i(0+) = 0 for the unenergised network, so 3.333 + K = 0 Answer: D. The inductor current cannot jump, so i(0+) = 0 fixes K. 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