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F%2007%20HW%2004%20all-s - EECS 314 Fall 2007 HW 04...

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Problem 1 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 16V voltage source and 8 resistor circled above into Norton equivalent circuit: Rn = Rt = 8 and Is = Vs/R = 2A flowing downward Next, combine the two Norton circuits in parallel in the blue circle: Rn = (2 ||8 Is = Is1 + Is2 = (2A) + (3A) = 5A flowing downward Note it’s Thevenin voltage = Is*Rn = 8V, where the downward node is positive. Third steps, let’s transform the current source and resistor in the green circle into Thevenin circuit: Rt = Rn = 2 and Vs = IsR = 2V, where its left node is positive. After these transformations, the reduced circuit is draw below: We may immediately combine the resistors in blue circle: Rt = 5 Ω , Vt = 10 V, where its downward node is positive. EECS 314 Fall 2007 HW 04 solutions Problems 1 and 2 Page 1 of 3
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Since the blue and red circles are connected in parallel, we would like to transform them into Norton circuit. Blue: Rn = Rt = 5 and Is = Vs/R = 2 A, flowing downward Red: Rn = Rt = 5 and Is = Vs/R = 5 A, flowing upward Combine them to get one Norton Circuit: Rn = 5||5 and Is = - Is1 + Is2 = -2 + 5 A = 3 A, flowing upward There is a 2.5 resistor left. Finally, we transform the Norton circuit above into Thevenin circuit so that we may combine the 2.5 resistor easily. Rt = Rn = 2.5 and Vs = Is*R = 7.5 V Combine the 2.5 resistor, we get the final Thevenin equivalent circuit : Rt = 2.5+2.5 = 5 and Vs = 7.5V Note that node a is positive and node b is negative. The Norton equivalent circuit : Rn = Rt = and Is = Vs/R = 1.5 A Note that the source current flows from node b to node a. Thevenin Equivalent Circuit Norton Equivalent Circuit 5 Ω EECS 314 Fall 2007 HW 04 solutions Problems 1 and 2 Page 2 of 3
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Problem 2 Part. 1 Part. 2 Again, we have two unknowns Vs and Rs and would like to set up two equations to solve them. Note that the current passing through the resistors can be easily solved since we know how much power them absorb.
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