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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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- Fall '06
- MCCANN
- loop
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