EE302Lecture18

# EE302Lecture18 - Source Transformation Example Problems...

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Chapter 4, Problem 20. Use source transformations to reduce the circuit in Fig. 4.88 to a single voltage source in series with a single resistor. Figure 4.88 For Prob. 4.20. This problem has only one node at the top of the diagram. We assume the bottom node is the ground reference at 0 volts. We are to transform the above circuit into the equivalent circuit below: Any two of the three quantities, Voc (open circuit voltage), Isc (short circuit current), and Req (equivalent resistance), could be used to develop the above equivalent. The Voc (open circuit voltage) is found using KCL: The sum of currents is zero for the top node: –3 + V / 10 + ( V – 12) / 20 + ( V – 16) / 40 = 0 Solving for Voc: V ( 1/10 + 1/20 + 1/40) = 3 + 12/20 + 16/40 V = 22.857 V therefore Voc = 22.857 V 1 + R eq V oc Voc and Req looking into the Thevenin Equivalent circuit. 10

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EE302Lecture18 - Source Transformation Example Problems...

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