HW2solns - HW #2 Solutions E245B Problem 2.10 Find I1 and I...

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Problem 2.10 Find 1 I and 2 I in the circuit shown. Suggested Solution By KCL, 14 6 mA II == Also, 62 4 I += or 2 0.003 0.006 I 2 3 mA I Þ = Problem 2.14 Find x I , y I and z I in the network shown. Suggested Solution I 3 = 5 mA I S I 4 = 6 mA I 1 I 6 = 3 mA I 2 12 mA 3mA 2mA 4mA I x I y I z HW #2 Solutions E245B
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0.003 0.012 x I += 9 mA x I Þ = 0.012 0.002 0.004 y I ++= 10 mA y I Þ =− 0.004 0.002 z I 2 mA z I Þ Problem 2.25 Find x V in the circuit shown. Suggested Solution 12 mA 3mA 2mA 4mA I x I y I z 24 V 6 V 2k 5k + V x _
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Using voltage division, () 5000 24 6 10 V 2000 5000 2000 x V æö =− = ç÷ ++ èø Problem 2.34 Find o I in the network shown. Suggested Solution Combining the 3-k and 6-k resistors in parallel yields the following equivalent circuit. 1 4000 0.012 6 mA 4000 2000 2000
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This note was uploaded on 10/12/2009 for the course ECE E 245B taught by Professor Uftureli during the Fall '01 term at Stevens.

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HW2solns - HW #2 Solutions E245B Problem 2.10 Find I1 and I...

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