DeCarlo Ch2 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
S OLUTIONS C HAPTER 2 S OLUTION 2.1. Using KCL at the center node of each circuit: (a) I 3 = I 2 - I 1 = - 1 - 2 = - 3 A (b) I 3 = I 1 + I 2 - I 4 = 2 - 1 - 0.5 = 0.5 A S OLUTION 2.2. KCL at the bottom node gives I 1 = - 7 - 8 = - 15 A , and at the right node I 4 =- 6 - 8 = - 14 . From these, KCL at the top node gives I 3 = I 4 - 5 = - 19 A , and finally at the central node gives I 2 = 6 + I 3 - 7 = - 20 A , S OLUTION 2.3. Use a gaussian surface on the top triangle. Performing KCL around this surface yields 1 A - 2 A + 3 A + 4 A - 5 A = I = 1 A . S OLUTION 2.4. Use a gaussian surface around the bottom rectangle. KCL yields I 1 = 2 A + 10 A + 3 A = 15 A . S OLUTION 2.5. Using KVL, V 1 = 55 V - 15 V + 105 V - 100 V - 30 V = 15 V . S OLUTION 2.6. Using KVL, V x = 5 V - 1 V - 1 V - 1 V + 1 V - 1 V = 2 V . S OLUTION 2.7. Using KVL once again. v 1 = 7 + 6 + 5 = 18 V v 2 = 6 + 7 - 8 = 5 V v 3 = - 5 - 6 = - 11 V v 4 = 8 - 7 = 1 V S OLUTION 2.8. KVL is used to find the voltage across each current source, and KCL to find the current through each voltage source. I 3 V = 6 A - 7 A = - 1 A I 4 V = I 3 V + 8 A = 7 A I 5 V =- 8 A - 6 A = - 14 A V 7 A = 4 V + 3 V = 7 V V 8 A = - 4 V + 5 V = 1 V V 6 A = V 8 A - 3 V = - 2 V
Image of page 1

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

View Full Document Right Arrow Icon
Chap 2 Probs P2 - 2 © R. A. DeCarlo, P. M. Lin S OLUTION 2.9. Using the same method as before, the current and voltages are found through and across each sources. I 5 V = 9 + 8 - 7 = 10 P = 50 W I 4 V = - 6 - I 5 V = - 16 P =- 64 W I 2 V = 6 - 7 = - 1 A P = - 2 W I 3 V = - I 2 V - 9 = - 8 A P =- 24 W V 8 A = 4 - 5 = - 1 V P =- 8 W V 9 A = 3 + V 8 A = 2 V P = 18 W V 7 A = 2 - V 9 A = 0 P = 0 W V 6 A = 5 - V 7 A = 5 P = 30 W Summing all the power give 0W, hence conservation of power. S OLUTION 2.10. Doing KVL around the right loop does not balance out. Changing 8V to 5V would fix this. S OLUTION 2.11. Using KVL to determine the voltages, and KCL to determine the currents: V y = 8 V V x = V y - 4 = 4 V I a = 4 A I y = 4 - 14 + 2 I a = - 2 A I x = I a - I y = 6 A S OLUTION 2.12. First V in = I 2 8 Ω = 24 V . Then I 1 = V in / 3 Ω = 8 A and I 3 = 12 A - I 1 - I 2 = 1 A . Therefore R L = V in / I 3 = 24 Ω ⇒ P = I 3 V in = 24 W S OLUTION 2.13. (a) First, from current division, get I 1 = 1/ 3 1/ 3 + 1/ 6 + 1/ R L 12 - aI 1 ( 29 I 1 = 12 / 3 1 + a ( 29 / 3 + 1/ 6 + 1/ R L . (b) Using the previous equation and solving for 1/ R L = 12/ 3 I 1 ( 29 - 1/ 6 - 1 + a ( 29 / 3 = 0.5 S or R L = 2 . The power P = I 3 2 R L = 1/ R L 1/ 3 + 1/ 6 + 1/ R L 12 - aI 1 ( 29 R L = 18 W S OLUTION 2.14. For the power delivered by the source to be 60W, the voltage across it should be V = P / 2 A = 30 V . Therefore the current through the 20 must be I 20 = 30 / 20 = 1.5 A , and by KCL the current through I R L = 2 - I 20 = 0.5 A . From this, R L = V / I R L = 60 .
Image of page 2
Chap 2 Probs P2 - 3 © R. A. DeCarlo, P. M. Lin S OLUTION 2.15. Writing KVL around the loop 25 V - 4 I - 15 V - 5 I - I = 0 I = 1 A , and P 5 = I 2 R 5 = 5 W S OLUTION 2.16. The total power supplied by the source is P = 50 V 0.5 A = 25 W . The power absorbed by the resistor is P 60 = 0.5 A ( 29 2 60 Ω = 15 W . Therefore by conservation of power, the power absorbed by X is 10W.
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern