hw 25

# hw 25 - ECE 201 Spring 2010 Homework 25 Solutions Problem...

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Unformatted text preview: ECE 201 Spring 2010 Homework 25 Solutions Problem 43 (a) To find the initial conditions at t=0-, we can write the following KVL equa- tions, 20 + 40 i L (0- )- 60(0 . 1- i L (0- )) = 0 ⇒ i L (0- ) = i L (0+) = i C (0+) =- . 14 A ⇒ v C (0+) = v C (0- ) = 20- . 14 × 40 = 14 . 4 V After t=0, the circuit is a series RLC circuit with the following characteristic equation, s 2 + R L s + 1 LC = 0 s 2 + 15 s + 50 = 0 ⇒ s 1 =- 5 , s 2 =- 10 ⇒ v C ( t ) = c 1 e- 5 t + c 2 e- 10 t ⇒ c 1 + c 2 = 14 . 4 (5 × 10- 3 )(5 c 1 + 10 c 2 ) = 0 . 14 ⇒ c 1 = 23 . 2 , c 2 =- 8 . 8 ⇒ v C ( t ) = (23 . 2 e- 5 t- 8 . 8 e- 10 t ) V The plot of v C ( t ) for 0 < t ≤ 1 is shown on the next page. 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 5 10 15 t(seconds) v C (t)(Volts) Plot of v C (t) (part (a)) for 0<t<=1s (b) In this case, the situation after t=0 can be visualized as a series RLC circuit with a constant input voltage source (after source transformation) which will have the value 0 . 5 × 60 = 30...
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## This note was uploaded on 11/06/2010 for the course ECE 201 taught by Professor All during the Spring '08 term at Purdue.

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hw 25 - ECE 201 Spring 2010 Homework 25 Solutions Problem...

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