Thevenins norton theorem, Lecture notes of Circuit Theory

Download Thevenins norton theorem and more Lecture notes Circuit Theory in PDF only on Docsity! ECE 211 Workshop: Thevenin’s and Norton’s Theorems Academic Resource Center Agenda • Background: Thevenin’s Theorems Review • Thevenin's Analysis o How to find Equivalent Thevenin's Voltage Source and relating problems o How to find Equivalent Thevenin's Resistor and relating problems • Transformation between two Theorems • Practice Problems and Solutions Thevenin's Voltage Example • Find equivalent voltage source in new circuit • Solution: Between terminals A and B, we need to find out V. Since it's open circuit and there is no current going through R1.Treat R1 as wire. ciucuit become simple three series resistor and a voltage source. Secondly, find the current. Thirdly, find the sum voltage across R3 and R2. That's the answer we're looking for. Using Ohm’s law, we find: I = V/R, where V = 15V R = R4+R3+R2 = 4Ohm I = 15/4 = 3.75A Ohm's law again, Vab = 2*3.75 = 7.5V Practice Problems 1A Find the Thevenin equivalent at terminals a-b of the circuit m Fig. 4.107. + — Vo Vy vA 10 kQ 20 kQ Ja ov 2) l+ 4V, ob THE ARC. Practice Problems 1B There is only one path in the circuit, so the ratio of the two resistors are the ratio of their voltages(same current). V0 + 2V0(20K Ohms) +4V0 = 70V V0 = 10V Two ways finding Vab: Vab = 70V- V0 =2V0 + 4V0 = 60V Thevenin's Resistor Example A • Find equivalent resistor in new circuit • Solution: Original method: short terminals A and B as shown in the picture. Find the current I going through A to B. Rth can be found by V/I,where V is the voltage we get from last problem. R2,3 = R2+ R3 = 2 Ohm R2,3,1 = R2,3 || R1 = 2/3 ohm Rsigma = R1,2,3,4= R2,3,1 + R4 = 8/3 ohm Isigma = V1/Rsigma = 45/8 A IAB = Isigma * (R2+R3) / (R1+R2+R3) = 15/4 A Rth = VAB/IAB = 2 Ohm Thevenin's Resistor Example B • Find equivalent resistor in new circuit • Solution: Alternatively method: leave terminals A and B open,instead of short Volage source V1, shown in the picture. Since no dependent source apprearing in the graph, we just need to find Rth by series and parallel theory. R2,3 = R2+ R3 = 2 Ohm R2,3,4 = R2,3 || R4 = 1 ohm Rsigma = R1,2,3,4= R2,3,4 + R1 = 2 ohm Which matches the previous results. Thevenin's Example summary 1 The graph shown in the right hand side gives the final result for Thevenin's theory. compared with original circuit, it looks a lot easier to further analyze. 2 Generally speaking, out of the two ways in findng equivalent resistor. A is more suitable for graph containing denpendent source. B is more useful in the situation of simply parallel and series resistors. Practice Problems 4A 0.25y, 4. NY 22 6Q ANN NAA Oa + < isv (= 305% — oO 6 THE ARC. Practice Problems 4B Short indenpendet voltage source, then you have lower left circuit. In order to find out the equivalent resistor, we need an additional indentpent source apply to terminnals a and b. Figure show lower. Practice Problems 4B Assumption: current direction in 2 Ohm is a to N. 6 Ohm and 3 Ohm share the same voltage(V0). For node N, current going out are (V0/6 + V0/3). Assumption: current direction in 2 Ohm is a to N. Total current going in: (0.25v0 + i). From KCL we equal (V0/6 + V0/3) with (0.25v0 + i), we get i = v0/4. From KVL v0/4 * 2 + v0 = 12 V v0 = 8 V Iab = 0.25v0 + v0/4(2 Ohm) = 0.5 v0 = 4A Rth = Vab/Iab = 12/4 = 3 Ohm Since we assume the voltage source has a value of 12V, we need one more parameter, current I(through voltage source) to figure out resistance using equation R= U/I. Practice Problems 1kQ -—_H A, Wh -—————_- sh 8 Ty QO 404, 3 3V (2 2V, v,= 502 _7 ob THE ARC. Practice Problems 30 20 WW NWN Oa + ‘e sov ©) 6Q2% 05», = 102 ob THE ARC. A Why NAN Solutions to Practice Problems Vth: (original figure) 1k * I0 + 2 Vx = 3V Vx = 40I0 * 50 solve the above: Vx = Vth = 1.2V I0 = 0.6 mA We short the votage source and add an additional votage source between a and b.(lower figure) Rth: So, we have: 1000 * I0 = -2*Vx Vx = 3V solve the above: I0 = -6 mA -40 I0 - Vx/50 = Iab = 0.18A direction: a to b Rth = 3V/ Iab = -16.67 Ohm