CH5sec2 - Problems Section 5-2: Source Transformations...

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Problems Section 5-2: Source Transformations P5.2-1 (a)
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= 2 = 0.5 V t t R v Ω (b) 9 4 2 ( 0.5) 0 9 ( 0.5) 1.58 A 42 ii i −−−+ − = −+− == + 9 4 9 4( 1.58) 2.67 V vi =+ =+ − = (c) 1.58 A a (checked using LNAP 8/15/02) P5.2-2 Finally, apply KVL: 16 10 3 4 0 2.19 A 3 aa a i −+ + − = ∴= (checked using LNAP 8/15/02)
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P5.2-3 Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left:
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Finally, apply KVL to loop o 6 ( 91 9 )3 6 0 iv −+ + − − = o 5/ 2 42 28 (5/ 2) 28 V =⇒ = + = (checked using LNAP 8/15/02) P5.2-4 4 2000 4000 10 2000 3 0 375 A aa a a ii i i μ −− +− = ∴= (checked using LNAP 8/15/02)
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P5.2-5 12 6 24 3 3 0 1 A aa a ii i −− + − − = ⇒ = (checked using LNAP 8/15/02) P5.2-6 A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor: Source transformations on both the right side and the left side of the circuit:
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This note was uploaded on 04/16/2009 for the course ECE 2040 taught by Professor Yili during the Spring '08 term at Georgia Tech.

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CH5sec2 - Problems Section 5-2: Source Transformations...

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