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**Unformatted text preview: **SOLUTIONS CHAPTER 2 SOLUTION 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 SOLUTION 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 , SOLUTION 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 . SOLUTION 2.4. Use a gaussian surface around the bottom rectangle. KCL yields I 1 = 2 A + 10 A + 3 A = 15 A . SOLUTION 2.5. Using KVL, V 1 = 55 V- 15 V + 105 V- 100 V- 30 V = 15 V . SOLUTION 2.6. Using KVL, V x = 5 V- 1 V- 1 V- 1 V + 1 V- 1 V = 2 V . SOLUTION 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 SOLUTION 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 Chap 2 Probs P2 - 2 © R. A. DeCarlo, P. M. Lin SOLUTION 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 = ⇒ P = W V 6 A = 5- V 7 A = 5 ⇒ P = 30 W Summing all the power give 0W, hence conservation of power. SOLUTION 2.10. Doing KVL around the right loop does not balance out. Changing 8V to 5V would fix this. SOLUTION 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 SOLUTION 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 SOLUTION 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 Ω . 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 SOLUTION 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 Ω must be I 20 Ω = 30 / 20 = 1.5 A , and by KCL the current through I R L = 2- I 20 Ω = 0.5 A . From this, R L = V / I R L = 60 Ω . Chap 2 Probs P2 - 3 © R. A. DeCarlo, P. M. Lin SOLUTION 2.15. Writing KVL around the loop 25 V...

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