Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15EE201 Lecture 13 P. 1Thevenin’s and Norton’s Theorems (cont.)More on circuits with dependent sources_+ 0.2 iy50 100iy+_ 20 ixixiSCIn Lecture 12, we saw that combinations of resistors and dependent source have Thevenin and Norton equivalent networks that consist of a resistor (RTH) only. To determine RTH, we added a variable current or voltage source to the circuit.Consider the circuit below. Let’s short the external terminals (equivalent to adding a deactivated voltage source).EE201 Lecture 13 P. 2Find the Thevenin and Norton equivalent circuits.To find iSC, short circuit external terminals. Then, iy = 0From this, ix can be calculated. 100 ix + 20 ix = 0 ix = 0 Aand, iSC = 0 AThis is a quantitative verification of the statement made on P.1.Since iSC = 0 A, vOC = 0 VTherefore, the Thevenin and Norton equivalent circuits will be,RTHEE201 Lecture 13 P. 3Conclusion: combination of dependent sources and resistors behave similar to a resistor.To find RTH apply a variable source to the circuit. We will solve for RTH using a voltage source and a current source._+ 0.2 iy50 100iy+_ 20 ixix1 V+_iVoltage source addedBy adding a voltage source, we only need to solve for the current “i” to calculate RTH.EE201 Lecture 13 P. 4From KVL,1 -20 ix - 100 ix = 0ix = (1/120) ANow solve for iy using nodal voltage VB = 1. iy = (1/50) ATo find i, use KCL at supernode below._+ 0.2 iy50 100iy+_ 20 ixix1 V+_iABi + (1/250) = (1/50) + (1/120)i = 0.024 AEE201 Lecture 13 P. 5Calculate RTH,RTH = (1 / 0.024) = 41. 1 _+ 0.2 iy50 100iy+_ 20 ixix1 AiABKCL@A: (1-iy) + 0.2iy = 1 - 0.8iy = ixKVL: 20ix -50iy + 100ix = 0Solving for ix and iy, ix = 0.342 A; iy = 0.822 AvCurrent source addedEE201 Lecture 13 P. 6v equals drop over 50 resistor, v = 50 (0.822) = 41.1 Vand the Thevenin resistance,RTH = (41.1 / 1 ) = 41. 1 EE201 Lecture 13 P. 7Circuits with independent and dependent sourcesExample:Find the Thevenin and Norton equivalent circuits.Step 1: Find vOC. Apply KCL at node A.4 - VA vOC 2 4vOC = 8 VNote: VA = voc because no current flows through 3 resistor!_A vOC/42vOC+_ 4 V3+= 0+EE201 Lecture 13 P. 8_A vOC/42vOC = 0 V; current from dependent current source is zero.+_ 4 V3+Step 2: Find iSC.iSC 4 V 2 + 3iSC = 0.8 AiSC =EE201 Lecture 13 P. 9Step 3: Calculate RTH. vOC iSCRTH = 10 RTH = 8 V 0.8 A=_+vOC vOC = 8 V+__+Thevenin EquivalentNorton Equivalent iSC = 0.8 A= 10 EE201 Lecture 13 P. 10Example: For a complex circuit, two sets of measurement data are shown below. Find the Thevenin and Norton equivalent circuits.v = 10 V, iL = 0 Av = -10 V, iL = 2 AEquation of line representing iL-v relation:iL = -(1 /10) v + 1Step 1: Find vOC.If iL = 0 (open circuit), then vOC = v.vOC = 10 VStep 2: Find RTH by comparing with theorem. Thevenin’s theorem: v = RTH i + vOC Norton’s theorem: i = (1/RTH) v - iSCEE201 Lecture 13 P. 11iL = -iiL = -(1 /RTH) v + iSCCompare this with iL-v equation on P.9 RTH = 10 ; iSC = 1 A_+ 10 V+__+Thevenin EquivalentNorton Equivalent 1 A10 10 EE201 Lecture 13 P. 12Example: Find the Thevenin and Norton equivalent circuits for the circuit below._+1/3 0.5 +_ 4AABvStep 1: Define supernode and connect to variable current source.iA 0.25 iA _+1/3 0.5 +_ 4AiiA 0.25 iAEE201 Lecture 13 P. 13Step 2: Apply KCL at supernode.4 + 2 (v + 0.25 i) + 3 v = ii = 10 v + 8Step 3: Compare with theory.RTHvOC_+_+vi v = vOC-RTH(-i) i = (1 /RTH) v - iSCRTH = 0.1 ; iSC = -8 AvOC = - 0.8 V_+ 0.8 V+__+ 8 A0.10.1EE201 Lecture 13 P. 14Strategies for solving for Thevenin and Norton Equivalent Circuits1. Are two of three parameters (vOC, iSC, RTH) known? If so, find unknown parameter using, vOC iSCRTH =2. Are two datapoints of i-v relationship known? If so, then compare line defined by data to theory. v = RTH(i) + vOC i = (1 /RTH) v - iSCiL = -(1 /RTH) v + iSCEE201 Lecture 13 P. 153. Are dependent sources absent from circuit? If so, is deactivated circuit series-parallel arrangement of resistors? If both true, RTH is found by evaluating equivalent resistance,RTH = Reqand vOC is found by KCL, KVL, source transformation, or any other general method discussed in the course.4. If dependent sources are present, or if deactivated circuit is not series-parallel, then solve for Thevenin/Norton equivalents by applying external sources or by general
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