Questions Question 1 Animation: Applying Thévenin`s theorem
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Questions Question 1 Animation: Applying Thévenin’s theorem This question consists of a series of images (one per page) that form an animation. Flip the pages with your fingers to view this animation (or click on the ”next” button on your viewer) frame-by-frame. The following animation shows the steps involved in ”Thévenizing” a circuit. 1 18 V 14 kΩ R1 R3 R2 12 kΩ 2 10 kΩ Rload This is our original circuit: 18 V 14 kΩ R1 R3 R2 12 kΩ 3 10 kΩ Rload We may use Thevenin’s theorem to simplify this portion of the circuit . . . 18 V 14 kΩ R1 R3 R2 12 kΩ 4 10 kΩ Rload We may use Thevenin’s theorem to simplify this portion of the circuit . . . 18 V 14 kΩ R1 R3 R2 12 kΩ 5 10 kΩ Rload We may use Thevenin’s theorem to simplify this portion of the circuit . . . 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ 6 Rload We may use Thevenin’s theorem to simplify this portion of the circuit . . . 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ 7 Rload 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ 8 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ To this Thevenin equivalent circuit . . . 9 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ To this Thevenin equivalent circuit . . . RTH VTH 10 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ . . . to which we may attach the same load and analyze. RTH VTH 11 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ . . . toRwhich we may attach load the same load and analyze. RTH VTH 12 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ . . . to which we may attach the same load and analyze. Rload RTH VTH 13 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ . . . to which we may attach the same load and analyze. RTH Rload VTH 14 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ . . . to which we may attach the same load and analyze. RTH Rload VTH 15 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ RTH Rload VTH 16 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ RTH Rload VTH 17 First we disconnect the load resistor. 18 V 14 kΩ R1 R3 First we disconnect Rload the load resistor. 10 kΩ R2 12 kΩ RTH VTH 18 18 V 14 kΩ R1 R3 First we disconnect the load resistor. Rload 10 kΩ R2 12 kΩ RTH VTH 19 18 V 14 kΩ R1 R3 First we disconnect the load resistor. 10 kΩ R2 12 kΩ Rload RTH VTH 20 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload RTH VTH 21 Then we calculate how much voltage appears across the open load terminals. 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload RTH VTH 22 Then we calculate how much voltage appears + across the open load terminals. V 18 V 14 kΩ R1 R3 10 kΩ - R2 12 kΩ Rload RTH VTH 23 Then we calculate how much voltage appears + across the open load terminals. V 18 V 14 kΩ R1 R3 10 kΩ - R2 12 kΩ Rload RTH VTH 24 Then we calculate how much voltage appears + across the open load terminals. V 18 V 14 kΩ R1 R3 10 kΩ 10 kΩ - (18 volts) (14 kΩ + 12 kΩ + 10 kΩ) R2 12 kΩ Rload RTH VTH 25 Then we calculate how much voltage appears + across the open load terminals. V 18 V 14 kΩ R1 R3 10 kΩ 10 kΩ - (18 volts) (14 kΩ + 12 kΩ + 10 kΩ) R2 = 5 volts 12 kΩ Rload RTH VTH 26 18 V + 14 kΩ R1 R3 V 5V - 10 kΩ R2 12 kΩ Rload RTH VTH 27 18 V 14 kΩ R1 R3 This voltage becomes our Thevenin source + V 5 V voltage . . . 10 kΩ - R2 12 kΩ Rload RTH VTH 28 18 V 14 kΩ R1 R3 This voltage becomes our Thevenin source + V 5 V voltage . . . 10 kΩ - R2 12 kΩ Rload RTH 5V 29 18 V 14 kΩ R1 R3 This voltage becomes our Thevenin source + V 5 V voltage . . . 10 kΩ - R2 12 kΩ Rload RTH 5V . . . in the Thevenin equivalent circuit. 30 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload RTH 5V 31 18 V 14 kΩ R1 Now we replace each source in the original circuit with its own internal resistance. R3 10 kΩ R2 12 kΩ Rload RTH 5V 32 18 V 14 kΩ R1 Now we replace each source in the original circuit with its own internal resistance. R3 10 kΩ For voltage sources, this means a short-circuit. R2 12 kΩ Rload RTH 5V 33 18 V Now we replace each source in the original circuit with its own internal resistance. 14 kΩ R1 R3 10 kΩ For voltage sources, this means a short-circuit. R2 12 kΩ Rload RTH 5V 34 18 V Now we replace each source in the original circuit with its own internal resistance. 14 kΩ R1 R3 10 kΩ For voltage sources, this means a short-circuit. R2 12 kΩ Rload RTH 5V 35 Now we replace each source in the original circuit with its own internal resistance. 14 kΩ R1 R3 10 kΩ For voltage sources, this means a short-circuit. R2 12 kΩ Rload RTH 5V 36 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload RTH 5V 37 14 kΩ R1 R3 . . . and we calculate resistance across the open load terminals. 10 kΩ R2 12 kΩ Rload RTH 5V 38 14 kΩ R1 R3 . . . and we calculate resistance across the open Ω load terminals. 10 kΩ R2 12 kΩ Rload RTH 5V 39 14 kΩ R1 R3 . . . and we calculate resistance across the open Ω load terminals. 10 kΩ R2 12 kΩ Rload RTH 5V 40 (14 kΩ + 12 kΩ) // 10 kΩ 14 kΩ R1 R3 . . . and we calculate resistance across the open Ω load terminals. 10 kΩ R2 12 kΩ Rload RTH 5V 41 14 kΩ R1 R3 (14 kΩ + 12 kΩ) // 10 kΩ = 7.22 kΩ . . . and we calculate resistance across the open Ω load terminals. 10 kΩ R2 12 kΩ Rload RTH 5V 42 14 kΩ R1 R3 Ω 7.22 kΩ 10 kΩ R2 12 kΩ Rload RTH 5V 43 This resistance becomes our Thevenin source resistance . . . 14 kΩ R1 R3 Ω 7.22 kΩ 10 kΩ R2 12 kΩ Rload RTH 5V 44 This resistance becomes our Thevenin source resistance . . . 14 kΩ R1 R3 Ω 7.22 kΩ 10 kΩ R2 12 kΩ Rload 7.22 kΩ 5V 45 This resistance becomes our Thevenin source resistance . . . 14 kΩ R1 R3 Ω 7.22 kΩ 10 kΩ R2 12 kΩ Rload 7.22 kΩ 5V . . . in the Thevenin equivalent circuit. 46 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload 7.22 kΩ 5V 47 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload 7.22 kΩ 5V 48 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Rload 7.22 kΩ 5V 49 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ 7.22 kΩ Rload 5V 50 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ 7.22 kΩ Rload 5V 51 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ 7.22 kΩ Rload 5V 52 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Calculate: 7.22 kΩ Rload 5V 53 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V 54 Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V 55 Iload Now that we have an equivalent circuit to work with, we may insert the load there to see what happens! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 56 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 57 These load calculations will reflect what happens in the original circuit! 18 V 14 kΩ R1 R3 10 kΩ R2 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 58 These load calculations will reflect what happens in the original circuit! 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 59 These load calculations will reflect what happens in the original circuit! 18 V 14 kΩ R1 R3 Rload 10 kΩ Vload R2 (same) 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 60 These load calculations will reflect what happens in the original circuit! 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 Vload Iload 12 kΩ (same) Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 61 These load calculations will reflect what happens in the original circuit! 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 Vload Iload Pload 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 62 (same) These load calculations will reflect what happens in the original circuit! 18 V 14 kΩ R1 R3 Rload 10 kΩ R2 Vload Iload Pload 12 kΩ Calculate: Vload 7.22 kΩ Rload 5V Iload Pload 63 18 V Vload 14 kΩ R1 R3 Rload 10 kΩ R2 Iload Pload 12 kΩ Vload 7.22 kΩ Rload 5V Iload Pload 64 18 V Vload 14 kΩ R1 R3 Rload 10 kΩ R2 Iload Pload 12 kΩ Vload 7.22 kΩ Rload 5V Iload Pload 65 18 V Vload 14 kΩ R1 R3 Rload 10 kΩ R2 Iload Pload The load cannot ‘‘tell’’ any difference between the original circuit and the Thevenin equivalent circuit. Vload Rload Iload 12 kΩ 7.22 kΩ 5V Pload file 03261 66 Answers Answer 1 Nothing to note here. 67 Notes Notes 1 The purpose of this animation is to let students see how Thévenin’s theorem may be applied to the simplification of a resistor network. 68
