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EE 302 - HW Solution - 2

# EE 302 - HW Solution - 2 - 3 and the 10-volt source we get...

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EE302 Homework #2 Chapter 2, Problem 12. In the circuit in Fig. 2.76, obtain v 1, v 2 , and v 3 . Chapter 2, Solution 12. For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35v For loop 2, -10 +15 -v 2 = 0 v 2 = 5v For loop 3, -v 1 +v 2 +v 3 = 0 v 3 = 30v + v 1 - + v 2 - + v 3 - 25v + + 10v - + 15v - + 20v - loop 1 loop 2 loop 3

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Chapter 2, Problem 16. Determine V o in the circuit in Fig. 2.80. Figure 2.80 For Prob. 2.16. Chapter 2, Solution 16. Apply KVL, -9 + (6+2)I + 3 = 0, 8I = 9-3=6 , I = 6/8 Also, -9 + 6I + V o = 0 V o = 9- 6I = 4.5 V 6 2 + _ + _ + _ V o 9 V 3 V circle5 circle5
Chapter 2, Problem 17. Obtain v 1 through v 3 in the circuit Chapter 2, Solution 17. Applying KVL around the entire outside loop we get, –24 + v 1 + 10 + 12 = 0 or v 1 = 2V Applying KVL around the loop containing v 2 , the 10-volt source, and the 12-volt source we get, v 2 + 10 + 12 = 0 or v 2 = –22V

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Unformatted text preview: 3 and the 10-volt source we get, –v 3 + 10 = 0 or v 3 = 10V Chapter 2, Problem 18. Find I and V ab in the circuit. Chapter 2, Solution 18. Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A-V ab + 5I + 8 = 0 V ab = 28V Chapter 2, Problem 20. Determine i o in the circuit Chapter 2, Solution 20. Applying KVL around the loop, -36 + 4i + 5i = 0 i = 4A Chapter 2, Problem 22. Find V o in the circuit in Fig. 2.86 and the power dissipated by the controlled source. Chapter 2, Solution 22. KCL requires that v 2 10 4 v + + = 0 v = –4.444V The current through the controlled source is i = 2V = -8.888A v = (6 + 4) i (where i = v /4) = 10 111 . 11 4 v-= Hence, Power = (-8.888)(-11.111) = 98.75 W...
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