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HW4solutions - ECE35 Win09 Homework 4 Solutions P5.2-3 Source transformation at left equivalent resistor for parallel 6 and 3 resistors Equivalents

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ECE35 Win’09 Homework 4 Solutions 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)
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(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 :
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Apply KCL at the top, middle node: Apply Ohm’s law to the right-hand 3
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This note was uploaded on 04/15/2010 for the course ECE ece35 taught by Professor Xiejiang during the Winter '10 term at UCSD.

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HW4solutions - ECE35 Win09 Homework 4 Solutions P5.2-3 Source transformation at left equivalent resistor for parallel 6 and 3 resistors Equivalents

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