--- title: "Source current expression and time to reach 25 mA" subject: "Electrical Circuit Theory" chapter: 3 key: "semester_3/electrical_circuit_theory:source-current-25ma-time-classical" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/source-current-25ma-time-classical site: BCT Engineering --- # Source current expression and time to reach 25 mA Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution - Asked in: 2070 Asar (old course), 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/source-current-25ma-time-classical - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3 ## Question Using classical method, find the expression for the current supplied by the source in the network shown in figure. Also find the time taken by the source current to reach 25 mA. [Figure: 10 V source with a switch (t = 0) in the top wire; a 500 Ω resistor as a shunt; then a 700 Ω resistor in series with a 100 μF capacitor in the right branch] ## Short answer Source current i_s(t) = 20 + 14.286e^(−14.286t) mA. Setting 25 = 20 + 14.286e^(−14.286t) gives e^(−14.286t)=0.35, so t = ln(1/0.35)/14.286 = 0.0735 s. ## Answer **Assumption:** ideal 10 V source; switch closes at t = 0; capacitor initially uncharged. The 500 Ω resistor and the 700 Ω + 100 μF branch are both connected directly across the source. #### Branch currents Because the source is ideal, the two branches are independent. - Resistor branch: i₁ = 10/500 = 20 mA (constant, from t = 0) - RC branch: loop equation 700 i₂ + (1/C)∫i₂ dt = 10, i.e. di₂/dt + i₂/(RC) = 0 ```text τ = RC = 700 × 100×10⁻⁶ = 0.07 s i2(0⁺) = 10/700 = 14.286 mA (capacitor initially a short) i2(t) = 14.286 e^(−t/0.07) mA = 14.286 e^(−14.286 t) mA ``` #### Source current **i_s(t) = i₁ + i₂ = 20 + 14.286 e^(−14.286t) mA, t ≥ 0** At t = 0, i_s = 34.29 mA; it falls to 20 mA as t → ∞. #### Time to reach 25 mA ```text 25 = 20 + 14.286 e^(−14.286 t) e^(−14.286 t) = 5/14.286 = 0.35 t = ln(1/0.35)/14.286 = 1.0498/14.286 ``` **Answer: t = 0.0735 s ≈ 73.5 ms** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the 500 ohm branch and the RC branch be analysed separately? A. The ideal source holds 10 V across both branches B. The two branches are in series C. The capacitor is initially charged D. The 500 ohm resistor is removed by the switch Answer: A. With an ideal source each branch sees the full 10 V, so branch currents are independent. (Key point: Independent branches) 2. What is the RC branch current expression? A. 20 e^(-14.286t) mA B. 14.286 e^(-14.286t) mA C. 14.286 (1 - e^(-14.286t)) mA D. 14.286 e^(-t/700) mA Answer: B. i2(0+) = 10/700 = 14.286 mA and tau = RC = 700 x 100 uF = 0.07 s, so 1/tau = 14.286. (Key point: RC branch current) 3. Setting 25 = 20 + 14.286 e^(-14.286t), what is t? A. about 70 ms B. about 14.3 ms C. about 73.5 ms D. about 1.05 s Answer: C. e^(-14.286t) = 0.35, so t = ln(1/0.35)/14.286 = 1.0498/14.286 = 0.0735 s. 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