Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13EE201 Lecture 12 P. 1Thevenin’s and Norton’s TheoremsThevenin’s Theorems for Passive Networks_+Linear resistive networkv_+vFor any linear resistive network,There exists a Thevenin equivalent network+_RTHvOCiiEE201 Lecture 12 P. 2Where: vOC is the open-circuit voltage appearing across the terminals of the network, and RTH is the Thevenin equivalent resistance when all independent sources are deactivated._+vThere also exists a Norton equivalent networkRTHiSCWhich is equivalent to the Thevenin network if,iSC = vOC / RTHiSC is the short-circuit current through a short (wire) between the terminals of the networki_+Linear resistive networkvConcept of modifying configuration at external terminals to obtain parameter of interest. For any linear resistive network with an attached load resistor:To find voc, remove the load resistor. Now, there is an open circuit condition at the external terminals.EE201 Lecture 12 P. 3RL_+Linear resistive networkvoci_+Linear resistive networkv=0VTo find isc, replace the load resistor with a wire. Now, there is a short circuit condition at the external terminals. Note that v = 0 V because the wire brings the terminals to the same potential!EE201 Lecture 12 P. 4iscFor circuits with dependent current or voltage sources, the portion of the circuit to be replaced by a Thevenin or Norton equivalent must include the controlling circuit element for the dependent source.EE201 Lecture 12 P. 5Example: Find the Thevenin and Norton equivalent seen by the 1 k load resistor._+ 2mA2kvL 1k+_ 4 V3kSolution: Use source transformations to find Thevenin equivalent. (Note: this is not the only way to solve the problem)._+ 2mA2kvL 1k 2 mA3k4V source/ 2k resistor transformed to 2mA source in parallel with 2k resistor.EE201 Lecture 12 P. 6_+2kvL 1k 4 mA3kCombine current sources defined between same nodes.Use source transformations to replace current source with voltage source._+2 kvL 1k 8 V3k+_EE201 Lecture 12 P. 7_+5 kvL 1k 8 V+_Add series resistances._+RTH = 5 kvL vOC = 8 V+__+vLThevenin EquivalentNorton Equivalent iSC = 1.6 mARTH = 5 kNow, either short or open the external terminals (see pages 3 and 4) to obtain isc or voc, respectively.EE201 Lecture 12 P. 8Example: Find the Thevenin and Norton equiv. cir.+__+vOC1 A31 V7 5 2 6 i1i2RTHSolution: Step 1: Find RTH by deactivating all independent voltage sources (short) and all independent current sources (open).12 2 6 RTH = 4 + 2 = 6 EE201 Lecture 12 P. 9 Step 2: Find vOC using mesh analysis.M1: 31 - 7i1 - 6i1 - 5(i1-i2) = 0M2: i2 = 1 A from M1, i1 = 2 AvOC = 6i1 + 2i2 = 14 VStep 3: Combine vOC and RTH to define Thevenin equivalent._+RTH = 6 vL vOC = 14V+_EE201 Lecture 12 P. 10Step 4: Obtain Norton equivalent using source transformations._+vL iSC = 2.33ARTH = 6 iSC = vOC / RTHiSC = 14/6 = 2.33 AEE201 Lecture 12 P. 11_+1.5vOC+ 1.5 v116Example: Find the Thevenin and Norton equivalent for circuits with dependent sources._6+_v1General principle: Since there are no independent sources, vOC = 0, iSC = 0. How to evaluate RTH? Attach a variable current or voltage source._+1.5+ 1.5 v116_6+_RTHv1 1 AEE201 Lecture 12 P. 12Use mesh analysis to find voltage across terminals._+1.5+ 1.5 v116_6+_v1i1i2M1: 1.5 v1 - 1.5i1 - 6(i1+i2) = 0M2: i2 = 1 A v1 = -16 Vand from M1, i1 = -4 AComputing the voltage drop across terminals: v - 16 (1) - 6(i1+1) = 0 v = -2 Vand the Thevenin equivalent resistance is RTH = v / i2 = -2 1 AvEE201 Lecture 12 P. 13RTH = -2 The Thevenin and Norton equivalent circuits are shown
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