Lecture 23

1 mf vc 13ut2 025 h 1 0 2 2 since circuit

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Unformatted text preview: • Thus, = 1/2, and = −2 3/5/2012 ECE 201-4, Prof. Bermel Solution 20 Ω 0.1 mF + VC - [-1+3u(t)]/2 0.25 H 1 ≥ 0 = −2 + 2 • Since circuit elements are in parallel: +1 ≥0 = = 0.25 −2 −100 = 50 3/5/2012 − ECE 201-4, Prof. Bermel 1 + −400 2 Solution Voltage (VC) 25 20 15 10 5 0 -5 0 5 10 15 20 Time t (ms) 2/27/2012 ECE 201-4, Prof. Bermel 25 30 35 40 Example 2 • Given the RLC voltage data below, a 1 Ω resistor, and an input of 10 V turned on at t=0, what are the approximate values of L and C? Voltage (VC) 10 R 5 L 0 -0.1 0 0.1 0.2 0.3 0.4 -5 C Vs +- -10 3/5/2012 0.5 ECE 201-4, Prof. Bermel Solution Voltage (VC) 10 8 6 4 2 0 -0.1 -2 0 -4 -6 -8 -10 R 0.1 0.2 0.3 0.4 0.5 C L Vs +- • Recall that underdamped RLC circuits obey: = From inspection, completes 8 periods in about 0.5 seconds, implying = 3/5/2012 ECE 201-4, Prof. Bermel Solution Voltage (VC) 10 8 6 4 2 0 -0.1 -2 0 -4 -6 -8 -10 R 0.1 0.2 0.3 0.4 0.5 C L Vs +- • Also decays by a factor of 10 in about 0.5 seconds. , Since we estimate...
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This note was uploaded on 11/22/2013 for the course ECE 201 taught by Professor All during the Spring '08 term at Purdue.

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