lesson06

# lesson06 - Lesson 6 Circuit Reduction (Section 2-6)(CLO...

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Lesson 6 – Circuit Reduction (Section 2-6)(CLO 2-6) Two analysis techniques are applicable to ladder-type circuits – source transformation and circuit reduction. We will use the following circuit for demonstrating these techniques. This problem asks the students to calculate three things: the power supplied by the source P S , the output voltage V O , and the current I X . Let us start by finding the source power. To find the power supplied by the source we need to find the current supplied and then use P S = I S x 24. The easiest way may be to collapse or crunch the resistors into one equivalent resistor R EQ and use Ohm’s law to find the current. By observation R EQ = 100 . Ω Hence I S = – 24/100 = – 240 mA. P S = – 0.24 x 24 = – 5.76 W . Finding V O is a not as simple. The approach here is do a series of source transformations (Flip) and combine appropriate parallel and series resistors (Crunch) until we can reduce the circuit to a voltage divider problem. To find

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## This note was uploaded on 10/12/2010 for the course MAE 140 taught by Professor Mauriciodeoliveria during the Spring '08 term at UCSD.

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lesson06 - Lesson 6 Circuit Reduction (Section 2-6)(CLO...

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