BG_4991_Chapter3

BG_4991_Chapter3 - Chapter 3 Resistive Network Analysis...

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Unformatted text preview: Chapter 3 Resistive Network Analysis Grossman/Melkonian COMBINING INDEPENDENT SOURCES: Voltage sources in series add algebraically: 10V + - R1 2V I R2 R3 + - -6V Equivalent + 18V + - R1 R2 I R3 Grossman/Melkonian COMBINING INDEPENDENT SOURCES: Current sources in parallel add algebraically: V1 3mA 1mA R1 5mA -4mA R2 Equivalent V1 5mA R1 R2 Grossman/Melkonian NODE VOLTAGE ANALYSIS: Section 3.2 The Node Voltage Method is based on defining the voltage at each node as an independent variable. A reference node is selected and all other node voltages are referenced to this node. The Node Voltage Method defines each branch current in terms of one or more node voltages. This is done by using Ohm's Law and KCL. Since branch currents are defined in terms of node voltages, currents do not explicitly enter into the equations. Grossman/Melkonian NODE VOLTAGE ANALYSIS: As mention previously, node voltage equations are written from KCL equations. KCL at Node B: i1 - i 2 - i 3 = 0 VA - VB R1 i1 VA + R1 + R2 VRef i2 VB Applying Ohm's Law: - VB - VRef R2 i2 + R3 i3 - VB - V C R3 i3 VC i1 Grossman/Melkonian NODE VOLTAGE ANALYSIS: Node Voltage Procedure: 1. Select and label a reference node. All other nodes are referenced to this node. 2. Label all N-1 remaining node(s). 3 Apply KCL to each node (N-1 node(s)). Writing equations in terms of node voltages. 4. Solve N-1 equations for V's. 5. Calculate I's using V, I, and R relationships. Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Example 1: Use Node Voltage Analysis to determine V4, V6, i1, and i2. i1 2A + 4 i2 + 6 3A - - Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Example 1 cont.: 1. Identify and label a reference node. All other nodes will be referenced to this node. 2. Label all N-1 remaining node(s). VA 2A i1 + i2 4 + 6 3A VRef - Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Example 1 cont.: 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. -2A + VA - 0 4 VA + VA - 0 6 + 3A = 0 2A i1 + 4 i2 + 6 3A VRef - Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Example 1 cont.: 4. Solve equation for VA. VA = -2.4V VA 2A i1 + 4 i2 + 6 3A VRef - Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Example 1 cont.: 5. Solve for i1 and i2. Using Ohm's Law and the calculated value for VA: i1 = VA/4 = -2.4V/4 i1 = -0.6A VA i2 = VA/6 = -2.4V/6 i2 = -0.4A 2A i1 + 4 i2 + 6 3A VRef - Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Combining Independent Sources: Referring to the circuit in example 1, calculate VA by transforming the circuit to an equivalent circuit with one current source and one resistor: VA Original Circuit 2A i1 + i2 4 + 6 3A VRef - Grossman/Melkonian NODE VOLTAGE ANALYSIS CURRENT SOURCES: Since the current sources are in parallel, they can be combined into a single equivalent current source: VA = iT 2.4 = -1A 2.4 VA iT + 2.4 VA = -2.4V 1A - Note: iT is equal to the sum of i1 and i2 of original circuit. i1 = -0.6A and iT = -1A i2 = -0.4A Grossman/Melkonian NODE VOLTAGE CURRENT SOURCES: Example 2: Using node voltage analysis calculate i1 and i2: 120 i1 4mA 200 400 6mA i2 VRef Grossman/Melkonian NODE VOLTAGE CURRENT SOURCES: Example 2 cont.: 1. Identify and label a reference node. All other nodes will be referenced to this node. 2. Label all N-1 remaining node(s). V1 120 6mA V2 i1 4mA 200 i2 400 VRef Grossman/Melkonian NODE VOLTAGE CURRENT SOURCES: Example 2 cont.: 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. V1 V1 - V 2 = 0 Node 1: + -4mA + 200 120 V1 i1 4mA 120 6mA V2 200 i2 400 VRef Grossman/Melkonian NODE VOLTAGE CURRENT SOURCES: Example 2 cont.: Node 2: V2 V 1 120 + V2 400 + 6mA = 0 V1 i1 4mA 120 V2 200 i2 400 6mA VRef Grossman/Melkonian NODE VOLTAGE CURRENT SOURCES: Example 2 cont.: Solving equations for V1 and V2: V1 = -88.89mV V1 i1 4mA 120 6mA V2 = -622.22mV V2 200 i2 400 VRef Grossman/Melkonian NODE VOLTAGE CURRENT SOURCES: Example 2 cont.: i1 = V1/200 = -88.89mV/200 = -0.445mA i2 = 0 V2/400 = 622.22mV/400 = 1.556mA V1 i1 4mA 120 6mA V2 200 i2 400 VRef Grossman/Melkonian NODE VOLTAGE VOLTAGE SOURCES: Example 3: Use node voltage analysis to calculate VA and VB: VA 1k 3mA + 2k 1.5k VB - 6V VRef Grossman/Melkonian NODE VOLTAGE VOLTAGE SOURCES: Example 3 cont.: Node A: -3mA + VA 2k + VA - V B 1k = 0 VA 1k 3mA + 2k 1.5k VB - 6V VRef Grossman/Melkonian NODE VOLTAGE VOLTAGE SOURCES: Example 3 cont.: Node B: VB = 6V VA 1k 3mA + 2k 1.5k VB - 6V VRef Grossman/Melkonian NODE VOLTAGE VOLTAGE SOURCES: Example 3 cont.: Solving equations for VA and VB: VA = 6V VB = 6V VB 1k 3mA + 2k 1.5k 6V VA - VRef Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: A Supernode is needed when neither the positive nor the negative terminal of a voltage source is connected to the reference node. We use the fact that KCL applies to the currents penetrating this boundary to write a node equation at the supernode. We then write node equations at the remaining non-reference nodes in the usual way. We now have regular node equations plus one supernode equation, leaving us one equation short of the N - 1 required. Using the fundamental property of node voltages, we can write; VA - VB = Vs The voltage source inside the supernode constrains the difference between the node voltages at nodes A and B. The voltage source constraint provides the additional relationship needed to write N - 1 independent equations. A Vs + - B Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4: Using node voltage analysis, calculate i1, i2, and V4A: 2V +i1 2A 1 2 i2 4A + V4A - Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: 1. Identify and label a reference node. All other nodes will be referenced to this node. 2. Label all N-1 remaining node(s). 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. VA i1 2A 1 2 +2V i2 4A + V4A VB - Grossman/Melkonian VRef NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: Supernode: 2A + VA/1 + VB /2 + 4A = 0 VA + 0.5 VB = - 6A Supernode VA i1 2A 1 2 (one equation, two unknowns) +- VB 2V i2 4A + V4A - VRef Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: But, we know VA - VB = 2V (gives us second equation) Supernode VA i1 +- VB 2V i2 4A 2 + V4A 2A 1 - VRef Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: VA + 0.5 VB = - 6A VA - VB = 2V VA = -3.334V VB = -5.334V VB 2V i2 4A 1 2 + V4A Supernode VA i1 +- 2A - VRef Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: Calculate i1, i2, and V4A: i1 = -3.334A i2 = -2.667A VB = -5.334V VA i1 2A 1 2 +VB 2V i2 4A + V4A i1 = VA/1 = -3.334V/1 i2 = VB/2 = -5.334V/2 V4A = VB Supernode - Grossman/Melkonian VRef NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5: Calculate i1 and V1mA using node voltage analysis: 200 2V + + 1mA V1mA - 3mA 150 + - 4V - i1 Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: Identify and label all nodes. VA VC 200 2V + - + 1mA V1mA VB 150 i1 3mA + - 4V - VRef Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: Supernode: Node C: -1mA + VB/150 + (VA - VC)/ 200 = 0 VC = 4V VA - VB = 2V VA 200 2V + VC Supernode + V1mA 1mA VB i1 3mA + - 4V - 150 Grossman/Melkonian VRef NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: VA = 2.943V VB = 0.943V VC = 4V i1 = 6.286mA 200 2V + VC i1 = VB/150 = 0.943V/150 VA Supernode + V1mA 1mA VB i1 3mA + - 4V - 150 Grossman/Melkonian VRef NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: KVL: -V1mA + 2V + V150 = 0 -V1mA + 2V + 0.943V = 0 V1mA = 2.943V VA 2V + 200 VC + 1mA V1mA KVL - + V150 VB 150 3mA + - 4V - - VRef Grossman/Melkonian MESH CURRENT ANALYSIS: Section 3.3 The Mesh Current Method is based on writing independent mesh current equations with mesh currents as the independent variable. Each mesh is identified and a direction for the mesh current is selected (e.g. clockwise). KVL is applied to each mesh containing an unknown mesh current. Using Ohm's law, the voltage across each resistor is written in terms of one or more mesh currents and a resistance. Since element voltages are defined in terms of mesh currents, voltages do not explicitly enter into the equations as variables. The equations are solved to find the mesh currents. Individual branch currents are then calculated using the mesh currents. Grossman/Melkonian MESH CURRENT ANALYSIS: Mesh Current Procedure: 1. Identify and label mesh currents for each mesh. Define each mesh current consistently. Typically, mesh currents are defined in the clockwise direction. 2. Apply KVL to each mesh containing an unknown current. Using Ohm's Law, express the voltages in terms of one or more mesh currents. 3. Solve the linear equations for the mesh currents. Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6: Use Mesh Analysis to solve for ix, is, V1, and V2: 75 + V1 + 100 - ix 50 iS 10V - V2 + + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: 1. Identify and label mesh currents for each mesh. 75 + V1 + i1 100 V2 + ix 50 i2 iS 10V - + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: 2. Apply KVL to each mesh containing an unknown current. Use Ohm's Law to express the voltages in terms of one or more mesh currents. Mesh 1: -10V + 75(i1) + 50(i1 - i2) + 4V = 0 mesh 1 i1 i2 75 + V1 + i1 100 V2 + ix 50 i2 iS 10V - + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: Mesh 2: 50(i2 - i1) + 100(i2) - 2V = 0 mesh 2 i2 i1 75 + V1 + i1 100 V2 + ix 50 i2 iS 10V - + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: Solving for mesh currents i1 and i2: i1 = 61.54mA 75 + V1 + i1 i2 = 33.85mA 100 Note: These are mesh currents. V2 + ix 50 i2 iS 10V - + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: Calculate ix and is: ix = (i1 - i2) = 61.54mA - 33.85mA is = i2 = 33.85mA 75 + V1 + i1 100 ix = 27.69mA V2 + ix 50 i2 iS 10V - + 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: Solving for V1, and V2: V1 = 75i1 = 7561.54mA V1 = 4.62V V2 = -1.38V V2 = 50(i2 i1) = 50(33.85mA - 61.54mA) 75 + V1 + i1 100 V2 + ix 50 i2 iS 10V - + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: Check results using KVL: Mesh 1: -10V + V1 - V2 + 4V = 0 -10V + 4.62V - (-1.38V) + 4V = 0 75 + V1 + KVL 100 V2 + ix 50 KVL iS 10V - + 2V 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS VOLTAGE SOURCES: Example 6 cont.: Mesh 2: V2 + V3 - 2V = 0 -1.38V + 10033.85mA - 2V = 0 75 + V1 + KVL 100 V2 + ix + V3 + iS 2V 10V - 50 KVL 4V - + Grossman/Melkonian MESH CURRENT ANALYSIS CURRENT SOURCES: Example 7: Use Mesh Analysis to solve for ix, V1, and V2: 1. Identify and label mesh currents for each mesh. 1.5k + V1 + i1 1k + V2 ix 2k i2 20V - - 2mA Grossman/Melkonian MESH CURRENT ANALYSIS CURRENT SOURCES: Example 7 cont.: 2. Apply KVL to each mesh containing an unknown currents. Use Ohm's Law to express the voltages in terms of one or more mesh currents. Mesh 1: -20V + 1.5k(i1) + 2k(i1 i2) = 0 1.5k + V1 + KVL 1k + V2 ix + V3 - 20V - 2k KVL 2mA - Grossman/Melkonian MESH CURRENT ANALYSIS CURRENT SOURCES: Example 7 cont.: Mesh 2: i2 = -2mA Solving for mesh currents i1 and i2: i1 = 4.57mA 1.5k i2 = -2mA 1k + V1 KVL + V2 ix + V3 - 20V + - 2k KVL 2mA - Grossman/Melkonian MESH CURRENT ANALYSIS CURRENT SOURCES: Example 7 cont.: Calculate: ix = i2 i1 = (-2mA) (4.57mA) V1 = 1.5k(i1) = 1.5k(4.57mA) V2 = 2k(-ix) = 2k(6.57mA) 1.5k 1k + V1 KVL ix = -6.57mA V1 = 6.86V V2 = 13.14V - + V2 ix + V3 20V + - 2k KVL 2mA - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: When a current source is contained in two meshes, we create a Supermesh by excluding the current source and any elements connected in series with the current source. We write one mesh equation around the supermesh in terms of the currents ia and ib. We then write mesh equations for the remaining meshes in the usual way. Supermesh ia is ib ic Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: This leaves us one equation short since meshes "a" and "b" are included in the single supermesh equation. However, one equation is gained by the fundamental property of currents: ia i b = i s Supermesh ia is ib ic Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8: Calculate mesh currents and solve for V1 and V2: 200 600 + 4V + + V2 800 300 - 2mA V1 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: 1. Identify and label mesh currents for each mesh. Define each mesh current consistently. Typically, mesh currents are defined in the clockwise direction. 200 600 4V + - i1 2mA + V1 + i2 V2 800 i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: When a current source is contained in two meshes, we create a supermesh by excluding the current source and any elements connected in series with the current source. 200 600 Supermesh 4V + - i1 2mA + V1 + i2 V2 800 i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: Supermesh: -4V + 200i1 + 600i2 + 800(i2 - i3) = 0 200 600 Supermesh 4V + - i1 2mA + V1 + i2 V2 800 i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: Mesh 3: 800(i3 - i2) + 300i3 = 0 i1 i2 = 2mA Common Current Source: 200 600 4V + + V1 i1 2mA i2 + V2 800 - i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: Solving for mesh currents i1, i2, and i3: i1 = 5.536mA 200 i2 = 3.536mA 600 i3 = 2.571mA 4V + + V1 i1 2mA i2 + V2 800 - i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: V1 : KVL Mesh 1 -4V + 200 i1 + V1 = 0 V1 = 2.89V 200 600 + 4V + + V1 i2 V2 800 - i1 2mA i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 8 cont.: V2: V2 = 800(i2 i3) V2 = 0.772V 200 600 + 4V + + V1 i2 V2 800 - i1 2mA i3 300 - - Grossman/Melkonian MESH CURRENT ANALYSIS SUPERMESH: Example 9: Calculate ix. Use KCL: -8mA + 2mA - ix = 0 ix = -6mA 1k 500 8mA 2mA ix 300 Grossman/Melkonian SUPERPOSITION: Section 3.5 Superposition Principle: The output of a circuit can be found by finding the contribution from each source acting alone and then adding the individual responses to obtain the total response. iT = i1 + i2 V1 V2 + + iT R V1 + i1 R + V2 + - i2 R Grossman/Melkonian SETTING SOURCES EQUAL TO ZERO: Voltage Source: In order to set a voltage source to zero, it is replaced by a short circuit. R1 VS + R2 iS R3 - R1 Voltage source set equal to zero R2 iS R3 Grossman/Melkonian SETTING SOURCES EQUAL TO ZERO: Current Source: In order to set a current source to zero, it is replaced by an open circuit. R1 VS + R2 iS R3 - Current source set equal to zero VS R1 + R2 R3 - Grossman/Melkonian SUPERPOSITION: Example 10: Calculate VR using superposition: 400 200 + 5V + - 5mA VR 250 - Grossman/Melkonian SUPERPOSITION: Example 10 cont.: 1. Turn off all independent sources except one and find response due to that source acting alone. Turning off voltage source: 400 200 + Voltage source set equal to zero 5mA VR 250 - Grossman/Melkonian SUPERPOSITION: Example 10 cont.: VR due to current source only (VR1): i1 = 5mA 400 400 + 200 + 250 Current divider Vi = i1250 = 2.353mA250 = 0.588V VR1 = -Vi 400 VR1 = -0.588V 200 + 5mA i1 VR 250 + Vi - Grossman/Melkonian SUPERPOSITION: Example 10 cont.: VR due to voltage source only (VR2): VR2 = 5V 250 400 + 200 + 250 VR2 = 1.471V 400 200 + 5V + + 250 VR2 = 1.471V Voltage divider - VR - - Grossman/Melkonian SUPERPOSITION: Example 10 cont.: VR = VR1 + VR2 = -0.588V + 1.471V VR = 0.882V 400 200 + 5V + - 5mA VR 250 - Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Section 3.5 Thevenin and Norton circuits deal with the concept of equivalent circuits. Even the most complicated circuits can be transformed into an equivalent circuit containing a single source and resistor. When viewed from the load, any network composed of ideal voltage and current sources, and of linear resistors, may be represented by an equivalent circuit consisting of an ideal voltage source V T in series with an equivalent resistance RT. RT VT + A Thevenin Circuit A IN RN Norton Circuit - B B Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Thevenin Equivalent Circuits: At this point you should be asking yourself two questions; how do we calculate the Thevenin voltage and the Thevenin resistance? Thevenin Voltage (VT): The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed. R1 R3 + VS + - R2 VOC RL VT = VOC - Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Thevenin Resistance (RT): The Thevenin resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero). R1 VS + R3 R2 RL - R1 VS set equal to zero R3 R2 REQ = RT Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Example 11: Find the Thevenin equivalent circuit at terminals `A' and `B': Thevenin Voltage: VT = VOC = VAB = V12 = io12 Using mesh analysis: 20 iO = 211.27mA 5 A io + VOC 15V + - 10 12 B Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Example 11 cont.: VT = VOC = 211.3mA12 VT = 2.54V Thevenin Resistance: Setting all sources equal to zero and looking back into the circuit from terminals "A" and "B": RT = [(20 10) + 5] 20 12 5 RT = 5.92 A 15V + 10 12 io + VOC - B Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Example 12: Find the Thevenin equivalent circuit seen by the load R L: VT: VT = VOC = V700 = 4mA700 4mA Check VT = 2.8V 400 500 + + 10mA 200 700 V700 VOC RL - - Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Example 12 cont.: Thevenin Resistance: Set all sources equal to zero: RT: RT = 500 + 700 400 RT = 1.2k 500 + + 200 700 V700 VOC RL - - Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Example 12 cont.: Thevenin Resistance: iSC = 4mA 700 RT = VT iSC = 2.33mA Current Divider 700 + 500 RT = 2.8V/2.33mA RT = 1.2k 400 10mA 200 4mA 700 500 + V700 iSC - Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Norton Equivalent Circuits: Norton Current (IN): The Norton current is equal to the short-circuit current at the load terminals with the load removed. R1 R3 VS + - R2 iSC IN = iSC Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Norton Resistance (RN): The Norton resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero). R1 R3 VS + - R2 iSC R1 VS set equal to zero R3 R2 REQ = RN Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Summary: Thevenin Equivalent Circuit: The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed. The Thevenin resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero) or V T/iSC. VT = VOC = VAB (with load removed) RT = VT/iSC = REQ as seen by RL Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Summary: Norton Equivalent Circuit: The Norton current is equal to the short-circuit current at the load terminals with the load removed. The Norton resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero) or V T/IN. IN = ISC (with load removed) RN = RT VOC/IN = REQ as seen by RL Grossman/Melkonian SOURCE TRANSFORMATION: Source transformation allows for the conversion of an ideal voltage source in series with a resistor to an ideal current source in parallel with a resistor and vice versa. As previously seen, any circuit can be transformed to its Thevenin or Norton equivalent circuit at the load resistance R L. Therefore, a voltage source in series with a resistor (Thevenin) can be transformed to a current source in parallel with a resistor (Norton) and the V-I characteristics at the terminals "A" "B" will be the same. R1 VS + A Rest of Circuit IS Source Transformation A R1 B Rest of Circuit - B Grossman/Melkonian SOURCE TRANSFORMATION: Source transformation allows for the conversion of an ideal voltage source in series with a resistor to an ideal current source in parallel with a resistor and vice versa. R1 A + VS = ISR1 - Rest of Circuit B Source Transformation A IS = VS/R1 R1 B Rest of Circuit Grossman/Melkonian SOURCE TRANSFORMATION: Source transformation allows for the conversion of an ideal voltage source in series with a resistor to an ideal current source in parallel with a resistor and vice versa. R1 R2 R3 R4 Voltages across and currents through R2, R3, and R4 are the same for both circuits! VS + - IS = VS/R1 R1 R2 R3 R4 Grossman/Melkonian SOURCE TRANSFORMATION: Example 13: Use source transformation and current divider rule to calculate io: 1.5k 1k iO 8V + - 2k 300 Grossman/Melkonian SOURCE TRANSFORMATION: Example 13 cont.: Converting the voltage source in series with the 1.5k resistor to a current source in parallel with a resistor we have the following circuit: Same V-I characteristics 1k 8V/1.5k = 5.33mA 1.5k 2k 300 iO Grossman/Melkonian SOURCE TRANSFORMATION: Example 13 cont.: iO = 5.33mA 1/1.3k 1/1.5k + 1/2k + 1/1.3k iO = 2.12mA 1k 5.33mA 1.5k 2k 300 iO Grossman/Melkonian ...
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This note was uploaded on 02/11/2012 for the course ECE 4991 taught by Professor Crandel during the Fall '07 term at FIT.

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