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HW4solutionsWIN10

# HW4solutionsWIN10 - ECE35 Win10 Homework 4 Solutions...

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ECE35 Win’10 Homework 4 Solutions Problems 5.2-3, 5.2-5, 5.3-11, 5.4-1, 5.4-5, 5.4-6, 5.4-16, 5.5-2, 5.5-6, 5.5-8, 5.6-1, 5.6-5 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: Finally, apply KVL to loop

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P5.2-5 P5.3-11 (a)
(b) From (a), we require n =4, i.e. R 2 = 4 R 1 and . For example P5.4-1

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P5.4-5 Find v oc : Notice that v oc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: v a = v oc Apply KCL at node a: Find R t : We’ll find i sc and use it to calculate R t . Notice that the short circuit forces v a = 0 Apply KCL at node a: P5.4-6 Find v oc : Apply KCL at the top, middle node: The voltage across the right-hand 3 Ω resistor is zero so: v a = v oc = 18 V Find i sc :
Apply KCL at the top, middle node:

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HW4solutionsWIN10 - ECE35 Win10 Homework 4 Solutions...

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