Lecture 5
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INC 112 Basic Circuit Analysis Week 5 Thevenin’s Theorem Special Techniques • Superposition Theorem • Thevenin’s Theorem • Norton’s Theorem • Source Transformation Linearity Characteristic 6Ω 42V IL + RL V RL - 4Ω 10V If RL change its value , how will it effect the current and voltage across it? I ISC Voc = Voltage open-circuit Isc = Current short-circuit V VOC For any circuit constructed from only linear components Not just RL, all resistors have this property. Thevenin’s Theorem When we are interested in current and voltage across RL, we can simplify other parts in the circuit. 6Ω 4Ω RL 42V 10V Equivalent circuit RTH VTH RL 6Ω 4Ω RL 42V 10V I ISC Slope = 1/R RTH VTH RL V VOC Voc = Voltage open-circuit Isc = Current short-circuit R = R equivalent Thevenin’s Equivalent Circuit Thevenin’s equivalent circuit RTH VTH RL VTH = Voc (by removing RL and find the voltage difference between 2 pins) RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.) Why do we need equivalent circuit? • To analyze a circuit with several values of RL • For circuit simplification (source transformation) • To find RL that gives maximum power (maximum power transfer theorem) Procedure 1. Remove RL from the circuit 2. Find voltage difference of the 2 opened connections. Let it equal VTH. 3. From step 2 find RTH by 3.1 short-circuit voltage sources 3.2 open-circuit current sources 3.3 Look into the 2 opened connections. Find equivalent resistance. Example Find Thevenin’s equivalent circuit and find the current that passes through RL when RL = 1Ω 2Ω 10V 10Ω 3Ω RL 2Ω Find VTH 10V 6V 6V 2Ω 10Ω 3Ω 10V 2Ω 0V 0V VTH 3 10 6V 23 0V Find RTH 2Ω 10Ω 3Ω 10V 2Ω RTH 10 2 || 3 2 Short voltage source 2Ω 10Ω 3Ω 23 10 2 23 13.2 RTH 2Ω Thevenin’s equivalent circuit 13.2Ω 6V If RL = 1Ω, the current is RL 6 0.423 A 13 .2 1 Example Find Thevenin’s equivalent circuit 2Ω 1A 10Ω 3Ω RL 2Ω Find VTH 5V 2Ω 1A 3V 3V 10Ω 3Ω 2Ω 0V 0V VTH 1 3 3V 0V Find RTH 2Ω 1A 10Ω 3Ω 2Ω RTH 10 3 2 Open circuit current source 2Ω 15 10Ω 3Ω RTH 2Ω Thevenin’s equivalent circuit 15Ω 3V RL Example: Bridge circuit Find Thevenin’s equivalent circuit R1=2K R3=4K RL=1K 10V + R2=8K R4=2K 10V Find VTH R1=2K 10V R3=4K 8V 2V R4=1K R2=8K 0V VTH = 8-2 = 6V Find RTH R1=2K R3=4K RTH R2=8K R4=1K R1=2K R3=4K R1=2K R3=4K R2=8K R4=1K R2=8K R4=1K R1=2K R3=4K R2=8K R4=1K RTH 2 K || 8K 4 K || 1K 1.6 K 0.8K 2.4 K Thevenin’s equivalent circuit 2.4K 6V RL Special Techniques • Superposition Theorem • Thevenin’s Theorem • Norton’s Theorem • Source Transformation I ISC V VOC For any point in linear circuit Thevenin’s Equivalent Circuit RTH VTH RL Norton’s Equivalent Circuit IN RN RL In= Isc from replacing RL with an electric wire (resistance = 0) and find the current Rn = RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.) Example Find Norton’s equivalent circuit and find the current that passes through RL when RL = 1Ω 2Ω 10V 10Ω 3Ω RL 2Ω Find In 2Ω 10Ω 3Ω 10V Isc 2Ω Find R total 3 12 2 3 || (10 2) 2 4.4 3 12 Find I total Current divider V 10 I 2.27 A R 4.4 I SC 3 2.27 0.45 A 3 12 Find Rn 2Ω 10Ω 3Ω 10V 2Ω RTH 10 2 || 3 2 Short voltage source 2Ω 10Ω 3Ω 23 10 2 23 13.2 RTH 2Ω Norton’s equivalent circuit 0.45 If RL = 1Ω, the current is 13.2 RL 13.2 0.45 0.418 A 13.2 1 Relationship Between Thevenin’s and Norton’s Circuit RTH RN VTH I N RTH I ISC Slope = - 1/Rth V VOC Norton’s equivalent circuit Thevenin’s equivalent circuit 13.2 6V RL 0.45 13.2 Same R value RTH RN VTH I N RTH 6 0.45 13 .2 RL Example Find Norton’s equivalent circuit 2Ω 1A 10Ω 3Ω RL 2Ω Find In 2Ω 1A 10Ω 3Ω Isc 2Ω Current divider I SC 3 1 0.2 A 3 12 Find RTH 2Ω 1A 10Ω 3Ω 2Ω RTH 10 3 2 Open circuit current source 2Ω 15 10Ω 3Ω RTH 2Ω Norton’s equivalent circuit 0.2 15 RL Norton’s equivalent circuit Thevenin’s equivalent circuit 15 0.2 15 RL 3V 0.2 x 15 = 3 RL Equivalent Circuits with Dependent Sources We cannot find Rth in circuits with dependent sources using the total resistance method. But we can use RTH VOC I SC Example 250 1V 2K 4K + Vx - 80 + 100Vx + RL - Find Thevenin and Norton’s equivalent circuit 250 Find Voc 1V KVL loop1 KVL loop2 I1 4K 2K + Vx - + 80 I2 + 100Vx - 1 250I1 4000( I1 I 2) 0 4250I1 4000I 2 1 4000( I 2 I1) 2000I 2 80I 2 100Vx 0 Vx 4000( I1 I 2) 404000I1 406080I 2 0 250 1V I1 4K 2K + Vx - + 80 I2 + 100Vx - Solve equations I1 = 3.697mA I2 = 3.678mA VOC 80I 2 100Vx 80I 2 400000( I1 I 2) 80(3.678m A) 400000(3.697 3.678) 7.3V 250 Find Isc 1V KVL loop1 KVL loop2 KVL loop3 I1 4K 2K + Vx - 80 I2 I3 Isc + 100Vx 1 250I1 4000( I1 I 2) 0 4250I1 4000I 2 1 4000( I 2 I1) 2000I 2 80( I 2 I 3) 100Vx 0 Vx 4000( I1 I 2) 404000I1 406080I 2 80I 3 0 80( I 3 I 2) 100Vx 0 400000I1 400080I 2 80I 3 0 250 Find Isc 1V I1 4K I1 = 0.632mA I2 = 0.421mA I3 = -1.052 A Isc = I3 = -1.052 A 2K + Vx - 80 I2 I3 + 100Vx Isc RTH VOC 7.28 6.94 I SC 1.052 Norton’s equivalent circuit Thevenin’s equivalent circuit 6.94 -7.28V RL -1.052 6.94 RL
