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Unformatted text preview: EE 40: Introduction to Microelectronic Circuits Spring 2008: HW 3 Solution Venkat Anantharam February 9, 2008 Referenced problems from Hambley, 4th edition. 1. P2.78 To find the Thevenin equivalent, we first find Thevenin equivalent voltage V TH as the open circuit voltage across the terminals a and b . By the current divider principle, the current flowing through the series connection of the 5 resistors is 1 A and it flows in a direction such that the voltage across a and b is 5 V . Hence V TH = 5 V . To find R TH , we zero the independent source. Hence R TH = 5  15 = 15 4 = 3 . 75. The Thevenin equivalent circuit is depicted in Figure 1. b + 5 V 3 . 75 a Figure 1: Thevenin Equivalent Circuit For the Norton equivalent, we have R N = R TH = 3 . 75 I N = V TH R TH = 5 V 3 . 75 = 4 3 A The Norton equivalent circuit is depicted in Figure 2. We could also have found I N by computing the short circuit current through a short circuit across terminals a and b . By the current divider principle this current would be 2 A 1 5 1 5 + 1 10 = 4 3 A 1 b 3 . 75 4 3 A a Figure 2: Norton Equivalent Circuit where the negative sign comes because the current would flow from b to a . This matches the earlier expression. 2. Find the Thevenin and Norton equivalent circuits across terminals a and b for the circuit in Figure 3. With an open circuit across terminals a and b the current through the + 30 V 5 i x . 5 i x a 20 b Figure 3: Circuit 1 20 resistor must be 0 . 5 i x and the voltage across it is then 10 i x . By KVL we also get 30 V = 5 i x + 10 i x So i x = 2 A and we calculate V TH = 10 i x = 20 V To find R TH we zero the independent source to get the circuit in Figure 4. We wish to determine what resistance this circuit is equivalent to across 5 i x . 5 i x a 20 b Figure 4: Circuit 1 with zeroed voltage source the terminals a and b . One way to do this is to imagine connecting an independent voltage source of voltage V S across the terminals and finding the current drawn from this voltage source. We therefore consider the 2 5 i x . 5 i x a 20 i 1 b + V S i Figure 5: Circuit 1 with test source circuit in Figure 5....
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This note was uploaded on 04/07/2008 for the course EE 40 taught by Professor Changhasnain during the Spring '07 term at University of California, Berkeley.
 Spring '07
 ChangHasnain
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