chap11

chap11 - Chapter 11 Solution 1 v t = 160 cos(50t i t =-20...

• Homework Help
• davidvictor
• 60

This preview shows pages 1–7. Sign up to view the full content.

Chapter 11, Solution 1. ) t 50 cos( 160 ) t ( v = ) 90 180 30 t 50 cos( 2 ) 30 t 50 sin( 20 - ) t ( i ° ° + ° = ° = ) 60 t 50 cos( 20 ) t ( i ° + = ) 60 t 50 cos( ) t 50 cos( ) 20 )( 160 ( ) t ( i ) t ( v ) t ( p ° + = = [ ] W ) 60 cos( ) 60 t 100 cos( 1600 ) t ( p ° + ° + = = ) t ( p W ) 60 t 100 cos( 1600 800 ° + + ) 60 cos( ) 20 )( 160 ( 2 1 ) cos( I V 2 1 P i v m m ° = θ θ = = P W 800 Chapter 11, Solution 2. First, transform the circuit to the frequency domain. ° → 0 30 ) t 500 cos( 30 , 500 = ω 150 j L j H 3 . 0 = ω → 100 j - ) 10 )( 20 )( 500 ( j - C j 1 F 20 6 - = = ω → µ I I 1 I 2 + 30 0 ° V j150 -j100 200 2 . 0 j - 90 2 . 0 150 j 0 30 1 = ° = ° = I ) t 500 sin( 2 . 0 ) 90 t 500 cos( 2 . 0 ) t ( i 1 = ° = 06 . 0 j 12 . 0 56 . 26 1342 . 0 j 2 3 . 0 100 j 200 0 30 2 + = ° = = ° = I

This preview has intentionally blurred sections. Sign up to view the full version.

) 56 . 25 t 500 cos( 1342 . 0 ) t ( i 2 ° + = ° = = + = 49.4 - 1844 . 0 14 . 0 j 12 . 0 2 1 I I I ) 35 t 500 cos( 1844 . 0 ) t ( i ° = For the voltage source, ] ) 35 t 500 cos( 1844 . 0 [ ] ) t 500 cos( 30 [ ) t ( i ) t ( v ) t ( p ° × = = At , s 2 t = ) 35 1000 cos( ) 1000 cos( 532 . 5 p ° = ) 935 . 0 )( 5624 . 0 )( 532 . 5 ( p = = p W 91 . 2 For the inductor, ] ) t 500 sin( 2 . 0 [ ] ) t 500 cos( 30 [ ) t ( i ) t ( v ) t ( p × = = At , s 2 t = ) 1000 sin( ) 1000 cos( 6 p = ) 8269 . 0 )( 5624 . 0 )( 6 ( p = = p W 79 . 2 For the capacitor, ° = = 63.44 - 42 . 13 ) 100 j - ( 2 c I V ) 56 . 25 t 500 cos( 1342 . 0 [ ] ) 44 . 63 500 cos( 42 . 13 [ ) t ( i ) t ( v ) t ( p ° + × ° = = At , s 2 t = ) 56 . 26 1000 cos( ) 44 . 63 1000 cos( 18 p ° + ° = ) 1329 . 0 )( 991 . 0 )( 18 ( p = = p W 37 . 2 For the resistor, ° = = 56 . 25 84 . 26 200 2 R I V ] ) 56 . 26 t 500 cos( 1342 . 0 [ ] ) 56 . 26 t 500 cos( 84 . 26 [ ) t ( i ) t ( v ) t ( p ° + × ° + = = At , s 2 t = ) 56 . 25 1000 ( cos 602 . 3 p 2 ° + = 2 1329 . 0 )( 602 . 3 ( p = = p W 0636 . 0
Chapter 11, Solution 3. 10 , ° → ° + 30 10 ) 30 t 2 cos( 2 = ω 2 j L j H 1 = ω → -j2 C j 1 F 25 . 0 = ω → 4 2 I I 1 I 2 + 10 30 ° V j2 -j2 2 j 2 2 ) 2 j 2 )( 2 j ( ) 2 j 2 ( || 2 j + = = ° = + + ° = 565 . 11 581 . 1 2 j 2 4 30 10 I ° = = = 565 . 101 581 . 1 j 2 2 j 1 I I I ° = = 565 . 56 236 . 2 2 2 j 2 2 I I For the source, ) 565 . 11 - 581 . 1 )( 30 10 ( 2 1 * ° ° = = I V S 5 . 2 j 5 . 7 18.43 905 . 7 + = ° = S The average power supplied by the source = W 5 . 7 For the 4- resistor, the average power absorbed is = = = ) 4 ( ) 581 . 1 ( 2 1 R 2 1 P 2 2 I W 5 For the inductor, 5 j ) 2 j ( ) 236 . 2 ( 2 1 2 1 2 L 2 2 = = = Z I S The average power absorbed by the inductor = W 0

This preview has intentionally blurred sections. Sign up to view the full version.

For the 2- resistor, the average power absorbed is = = = ) 2 ( ) 581 . 1 ( 2 1 R 2 1 P 2 2 1 I W 5 . 2 For the capacitor, 5 . 2 j - ) 2 j - ( ) 581 . 1 ( 2 1 2 1 2 c 2 1 = = = Z I S The average power absorbed by the capacitor = W 0 Chapter 11, Solution 4. 20 10 I 2 I 1 + -j10 50 V j5 For mesh 1, 2 1 10 j ) 10 j 20 ( 50 I I + = 2 1 j ) j 2 ( 5 I I + = (1) For mesh 2, 1 2 10 j ) 10 j 5 j 10 ( 0 I I + + = 1 2 2 j ) j 2 ( 0 I I + = (2) In matrix form, = 2 1 j 2 2 j j j 2 0 5 I I 4 j 5 = , ) j 2 ( 5 1 = , -j10 2 = ° = = = 1 . 12 746 . 1 4 j 5 ) j 2 ( 5 1 1 I ° = = = 66 . 128 562 . 1 j4 - 5 j10 - 2 2 I For the source, ° = = 12.1 - 65 . 43 2 1 * 1 I V S
The average power supplied = ° = ) 1 . 12 cos( 65 . 43 W 68 . 42 For the 20- resistor, = = R 2 1 P 2 1 I W 48 . 30 For the inductor and capacitor, = P W 0 For the 10- resistor, = = R 2 1 P 2 2 I W 2 . 12 Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 2 + j6 –j2 8 –40 ˚ W 4159 . 1 1 2 6828 . 1 P 38 . 25 6828 . 1 2 j 2 6 j ) 2 j 2 ( 6 j 1 40 8 I 2 1 1 = = ° = + + ° = P 3H = P 0.25F = 0 W 097 . 5 2 2 258 . 2 P 258 . 2 38 . 25 6828 . 1 2 j 2 6 j 6 j I 2 2 2 = = = ° + =

This preview has intentionally blurred sections. Sign up to view the full version.

Chapter 11, Solution 6.
This is the end of the preview. Sign up to access the rest of the document.
• '
• NoProfessor

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern