hmwk 4 sol - EECS 314 Fall 2006 HW 04 Problem 1 Student's...

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EECS 314 Fall 2006 HW 04 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) nd ams, 2 ight how your work. The voltage source V S in series with a resistor R S , as shown on the left top a on the center top diagr is equivalent to the current source I S in parallel with the same resistor R S , shown on the right top diagram, provided that V S = I S R S . Replacing a source with its equivalent is called “source transformation”; it is a good way to solve many circuit problems, especially the “ladder” circuits such as shown on the bottom diagram on this page. When you apply the source transformation, it is a good idea to redraw the circuit each time you replace a source with its equivalent. Note that, according to KCL, a parallel combination of current sources I 1 and I 2 with parallel resistors R 1 and R 2 is equivalent to one current source I 1 +I 2 and one resistor R 1 || R 2 . Also, according to KVL, a series combination of voltage sources V 1 and V with series resistors R 1 and R 2 is equivalent to one voltage source V 1 +V 2 and one resistor R 1 +R 2 . Use the source transformation to reduce the circuit shown on this diagram to its Thevenin equivalent (top left circuit on this page), and to its Norton equivalent (top r diagram on this page) . S Alexander Ganago September 22, 2005
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Kevin Klein Homework 4 Problem 1 Solution This problem asks you to find the Thevenin and Norton equivalent circuits of the circuit above using source transformations and parallel/series reductions. Begin by transforming the current source and resistor circled above. Doing this gives the following circuit: Here, the current source in parallel with the 2 resistor has been transformed into a voltage source in series with the same resistor. The value of the voltage source is given by V = IR, where I is the value of the current source. Note that the convention is to have the current Solution Page 1/6
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source’s arrow pointing from the negative terminal of the voltage source to the positive terminal. Next, combine the resistors circled in the above diagram in series: Next, transform the voltage source and resistor circled on the diagram above into a current source in parallel with a resistor: Here, the current in the transformed current source is given by I = V/R, where V is the voltage of the transformed voltage source, and R is the resistor it was in series with. Note that the arrow on the transformed current source points from the negative to the positive terminal of the original voltage source. For the next step, combine the two current sources and two resistors in parallel, which have been circled in the diagram above. Solution Page 2/6
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hmwk 4 sol - EECS 314 Fall 2006 HW 04 Problem 1 Student's...

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