Practice_Homework_Prob_Sol_EE210_Ch-10

# Practice_Homework_Prob_Sol_EE210_Ch-10 - 1 PROBLEM...

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Unformatted text preview: 1 PROBLEM SOLUTIONS CHAPTER 10 SOLUTION 10.1 Using KCL, we can write C dv C dt + v C R = i in t ( 29 Dividing by C: dv C dt + 1 RC v C = i in t ( 29 C We know that i in t ( 29 = 20sin 400 t ( 29 mA, which can be represented by a complex exponential, i in t ( 29 = Re 20 e j 400 t e- j π /2 [ ] mA. For convenience we will simply let i in t ( 29 = 20 e j 400 t e- j π /2 mA, knowing that we must take the real part to complete our solution. The output voltage will also be reparesented as a complex exponential: v C t ( 29 = V m e j 400 t + ( 29 = V m e j 400 t e j Substituting this expression into the differential equation and canceling e j 400 t : j 400 V m e j + V m RC e j = 20 × 10- 3 e- j π /2 C Thus V m e j 1 RC + j 400 = 20 × 10- 3 C ⇒ V m e j =- j 4000 1000 + j 400 = 3.714 ∠- 111.8 o where the values for R = 100 Ω and C = 5 mF were substituted in. Thus, V m = 3.714 =- tan- 1 400 1000 = - 111.8 ° Taking into account a 90 o phase shift we obtain v C t ( 29 = 3.714cos 400 t- 111.8 ° ( 29 = 3.714sin 400 t- 21.8 ° ( 29 V and i out t ( 29 = 18.57sin 400 t- 21.8 ° ( 29 mA SOLUTION 10.2 From KCL and component definitions: 2 i in t ( 29 - v L 25- i L = 0 ⇒ 0.1 25 di L dt + i L = i in t ( 29 ⇒ di L dt + 250 i L = 250 i in t ( 29 We represent the input signal by the complex exponential: i in t ( 29 = 0.2 e j 250 t A and the unknown current can be represented as i L ( t ) = I L e j 250 t + ( 29 . Substituting this into the differential equation and canceling e j 250 t : j 250 I L e j + 250 I L e j = 50 Thus I L e j j 250 + 250 ( 29 = 50 ⇒ I L e j = 50 250 + j 250 = 0.14142 ∠ - 45 o and I L = 0.141, = - 45 ° ⇒ i L t ( 29 = 0.141cos 250 t- 45 ° ( 29 A SOLUTION 10.3. Construct differential equation by KVL and device definitions: v in t ( 29 - 0.5 di L dt- 200 i L = 0 ⇒ di L dt + 400 i L = 2 v in t ( 29 We represent v in t ( 29 as the complex exponential function, v in t ( 29 = 20 e j 400 t V. The current in the inductor has the form: i L = I L e j 400 t + ( 29 . Substituting into the differential equation and canceling e j 400 t : j 400 I L e j + 400 I L e j = 40 Thus I L e j j + 400 ( 29 = 40 ⇒ I L e j = 40 400 + j 400 = 0.070711 ∠- 45 o and I L = 0.0707, = - 45 ° , ⇒ i L t ( 29 = 70.7cos 400 t- 45 ° ( 29 mA Hence, v out t ( 29 = 14.14cos 400 t- 45 ° ( 29 V SOLUTION 10.4. Construct differential equation using KVL and device definitions: 3 v in t ( 29 - v C- C dv C dt R = 0 ⇒ RC dv C dt + v C = v in t ( 29 The output voltage is defined as: v out t ( 29 = v in t ( 29 - v C t ( 29 This means that finding v C is enough to be able to obtain the output voltage. The input voltage is represented by the complex exponential: v in t ( 29 = 20 e j 250 t e- j π /2 V and v C t ( 29 = V m e j 250 t + ( 29 . Substituting into the differential equation, dividing by e j 250 t , and rearranging: j 250 RCV C e j + V C e j = - j 20 ⇒ V C e j j 250 RC + 1 ( 29 = - j 20 ⇒ V C e j =- j 20 1 + j = 14.142 ∠-...
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## This note was uploaded on 04/30/2008 for the course ECE 2100 taught by Professor N/a during the Fall '06 term at Missouri (Mizzou).

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Practice_Homework_Prob_Sol_EE210_Ch-10 - 1 PROBLEM...

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