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Unformatted text preview: Problem 2.17 Determine currents I 1 to I 4 in the circuit of Fig. P2.17. I 1 I 2 I 3 I 4 2 Ω 6 A 4 Ω 2 Ω 4 Ω Figure P2.17: Circuit for Problem 2.17. Solution: The same voltage exists across all four resistors. Hence, 2 I 1 = 4 I 2 = 2 I 3 = 4 I 4 . Also, KCL mandates that I 1 + I 2 + I 3 + I 4 = 6 It follows that I 1 = 2 A, I 2 = 1 A, I 3 = 2 A, and I 4 = 1 A. Problem 2.18 Determine the amount of power dissipated in the 3k Ω resistor in the circuit of Fig. P2.18. 10 3 V 2 k Ω 3 k Ω 10 mA V + _ Figure P2.42: Circuit for Problem 2.18. Solution: In the left loop, V = 10 × 10 − 3 × 2 × 10 3 = 20 V . The dependent current source is I = 10 − 3 V = 20 mA. The power dissipated in the 3k Ω resistor is p = I 2 R = ( 20 × 10 − 3 ) 2 × 3 × 10 3 = 1 . 2 W . Problem 2.20 Find V ab in the circuit in Fig. P2.20. I a b 2 Ω 2 Ω 2 Ω 6 V 12 V + _ + _ V ab + _ Figure P2.20: Circuit for Problem 2.20....
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This note was uploaded on 10/31/2009 for the course ECE 0031 at Pittsburgh.
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