Key F06 Homework 10

# Key F06 Homework 10 - ECET157 Homework Set 10 2 mH = 2 µs...

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Unformatted text preview: ECET157 Homework Set 10 2 mH = 2 µs, b. VL limit = 20 V, iL limit = 0 mA with L modeled as open, 1 kO c. VL ss = 0 V, iL ss = 20 mA with L modeled as short, 16-11 a. τ = d. VL (t ) = 0 V + (20 − 0 ) V × e − t/2 µs = 20 V·e− t / 2 µs, iL (t ) = 20 mA + (0 mA − 20 mA ) × e − t/2 µs = 20 mA − 20 mA·e− t / 2 µs, e. VL (2τ ) = 20V × e −2 = 2.71 V, iL (2τ) = 20 mA − 20 mA × e −2 = 17.3 mA, f. spike 0 V to 20 V then exponential fall 20 V to 0 V in 10 µs, exponential rise from 0 mA to 20 mA in 10 µs 2 mH = 2 µs, b. VR = 0 V, iR = 0 mA with L modeled as open, 1 kO 20 V c. VR = esupply = 20 V, iR = = 20 mA L modeled as short, 1 Ok 16-12 a. τ = d. VR (t ) = 20 V − 20 V × e − t/2 µs = 20 V − 20 V·e− t / 2 µs, iR (t ) = 20 mA − 20 mA × e − t/2 µs = 20 mA − 20 mA·e− t / 2 µs, e. VR (2τ ) = 20 V − 20V × e −2 = 17.3 V, iR (2τ ) = 20 mA − 20 mA × e −2 = 17.3 mA, f. exponential rise 0 V to 20 V in 10 µs, exponential rise 0 mA to 20 mA in 10 µs 2 mH = 2 µs b. VL limit = 20 V − (2 mA × 1 kO ) = 18 V, I L limit = 2 mA with L modeled as 1 kO 20 V supply I of 2 mA, c. 0 V, I L limit = = 20 mA with L modeled as a short, 1 kO 16-13 a. τ = d. VL (t ) = 0 V + (18 - 0 )V × e − t/2 µs = 18 V e− t / 2µs, iL (t ) = 20 mA + (2 mA − 20 mA ) × e − t/2 µs = 20 mA – 18 mA e− t / 2µs, e. VL (3 µs ) = 18 V × e −3 µs/2 µs = 4.02 V, iL (3 µs ) = 20 mA − 18 mA × e −3 µs/2 µs = 16.0 mA, f. spike 0 V to 18 V then exponential fall 18 V to 0 V in 10 µs, exponential rise 2 mA to 20 mA in 10 µs 16-16 a. τ = 1 mH = 1 µs 1 kO b. T ≥ 10τ ? 10 µs ≥ 10 µs ? yes. c. 0-5 µs exponential rise 0 V to 4 V 5-10 µs exponential fall 4 V to 0 V d. Use Ohm’s Law: iR(t) = vR(t) / R 0-5 µs exponential rise 0 mA to 4 mA 5-10 µs exponential fall 4 mA to 0 mA e. 0-5 µs spike to 4 V then exponential fall 4 V to 0 V in 5 µs 5-10 µs spike to –4 V then exponential rise −4 V to 0 V in 5 µs 1 mA − 4 mA × −1 µs = 693 ns − 4 mA f. t = 1n 2 ...
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