Grossman/Melkonian Chapter 3 Resistive Network Analysis

Grossman/Melkonian COMBINING INDEPENDENT SOURCES:  Voltage sources in series add algebraically: R1R1 R2R2 R3R3 10V 2V -6V + - R1R1 R2R2 R3R3 18V Equivalent I I

Grossman/Melkonian COMBINING INDEPENDENT SOURCES:  Current sources in parallel add algebraically: 3mA 5mA-4mA1mA R1R1 R2R2 5mA R1R1 R2R2 Equivalent V1V1 V1V1

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: VAVA  As mention previously, node voltage equations are written from KCL equations. VCVC VBVB V Ref + R R 3 - R2R2 + - i3i3 i1i1 i2i2 KCL at Node B: i 1 - i 2 - i 3 = 0 Applying Ohm’s Law: V A - V B R1R1 V B - V C R3R3 V B - V Ref R2R2 - - i1i1 i2i2 i3i3

Grossman/Melkonian NODE VOLTAGE ANALYSIS: 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. Node Voltage Procedure:

Grossman/Melkonian NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1: Use Node Voltage Analysis to determine V 4 , V 6 , i 1, and i 2. 2A 3A 44 66 i1i1 i2i

Grossman/Melkonian 2A 3A 44 66 i1i1 i2i2 V Ref 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). VAVA

Grossman/Melkonian Example 1 cont.: NODE VOLTAGE ANALYSIS – CURRENT SOURCES: 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. 2A 3A 44 66 i1i1 i2i2 V Ref VAVA -2A + V A - 0 44 66 + + 3A =

Grossman/Melkonian NODE VOLTAGE ANALYSIS – CURRENT SOURCES: Example 1 cont.: 2A 3A 44 66 i1i1 i2i2 V Ref VAVA 4. Solve equation for V A. V A = -2.4V -- ++

Grossman/Melkonian NODE VOLTAGE ANALYSIS – CURRENT SOURCES: 2A 3A 44 66 i1i1 i2i2 V Ref VAVA 5. Solve for i 1 and i 2. Example 1 cont.: Using Ohm’s Law and the calculated value for V A : i 1 = V A /4  = -2.4V/4  i 1 = -0.6A i 2 = V A /6  = -2.4V/6  i 2 = -0.4A

Grossman/Melkonian 2A 3A 44 66 i1i1 i2i2 V Ref VAVA  Referring to the circuit in example 1, calculate V A by transforming the circuit to an equivalent circuit with one current source and one resistor: Original Circuit Combining Independent Sources: NODE VOLTAGE ANALYSIS – CURRENT SOURCES:

Grossman/Melkonian  Since the current sources are in parallel, they can be combined into a single equivalent current source: 1A 2.4  iTiT VAVA - + V A = i T 2.4  = -1A 2.4  V A = -2.4V Note: i T is equal to the sum of i 1 and i 2 of original circuit. i 1 = -0.6A i 2 = -0.4A and i T = -1A NODE VOLTAGE ANALYSIS – CURRENT SOURCES:

Grossman/Melkonian Example 2: 4mA 6mA 120  200  400  V Ref i1i1 i2i2 NODE VOLTAGE – CURRENT SOURCES:  Using node voltage analysis calculate i 1 and i 2 :

Grossman/Melkonian NODE VOLTAGE – CURRENT SOURCES: 4mA 6mA V1V1 V2V2 120  200  400  V Ref i1i1 i2i2 1. Identify and label a reference node. All other nodes will be referenced to this node. Example 2 cont.: 2. Label all N-1 remaining node(s).

Grossman/Melkonian 4mA 6mA V1V1 V2V2 120  200  400  V Ref i1i1 i2i2 3. Apply KCL to each N-1 node. Writing equation in terms of node voltage. Example 2 cont.: NODE VOLTAGE – CURRENT SOURCES: Node 1: -4mA + V 1 - V  V1V1 200  + = 0

Grossman/Melkonian 4mA 6mA V1V1 V2V2 120  200  400  V Ref i1i1 i2i2 NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.: Node 2: V 2 – V  V2V2 400  + 6mA = 0 +

Grossman/Melkonian 4mA 6mA V1V1 V2V2 120  200  400  V Ref i1i1 i2i2 NODE VOLTAGE – CURRENT SOURCES: Example 2 cont.:  Solving equations for V 1 and V 2 : V 1 = mV V 2 = mV

Grossman/Melkonian Example 2 cont.: NODE VOLTAGE – CURRENT SOURCES: i 1 = V 1 /200  = mV/200  = mA i 2 = 0 – V 2 /400  = mV/400  = 1.556mA 4mA 6mA V1V1 V2V2 120  200  400  V Ref i1i1 i2i2

Grossman/Melkonian NODE VOLTAGE – VOLTAGE SOURCES: 3mA 6V VAVA VBVB 1k  2k  1.5k  V Ref Example 3:  Use node voltage analysis to calculate V A and V B : + -

Grossman/Melkonian 3mA 6V VAVA VBVB 1k  2k  1.5k  V Ref + - NODE VOLTAGE – VOLTAGE SOURCES: Example 3 cont.: Node A:-3mA + V A - V B 1k  VAVA 2k  + = 0

Grossman/Melkonian 3mA 6V VAVA VBVB 1k  2k  1.5k  V Ref + - NODE VOLTAGE – VOLTAGE SOURCES: Example 3 cont.: Node B: V B = 6V

Grossman/Melkonian 3mA 6V VAVA VBVB 1k  2k  1.5k  V Ref + - Example 3 cont.: NODE VOLTAGE – VOLTAGE SOURCES:  Solving equations for V A and V B : V A = 6VV B = 6V

Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE:  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; V A - V B = V s  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 Supernode is needed when neither the positive nor the negative terminal of a voltage source is connected to the reference node. + A B Vs -

Grossman/Melkonian Example 4: 2A i1i1 i2i2 11 22 4A + - 2V NODE VOLTAGE ANALYSIS - SUPERNODE:  Using node voltage analysis, calculate i 1, i 2, and V 4A : + - V 4A

Grossman/Melkonian Example 4 cont.: V Ref 2A i1i1 VAVA i2i2 11 22 4A VBVB + - 2V + - V 4A NODE VOLTAGE ANALYSIS - SUPERNODE: 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.

Grossman/Melkonian V Ref 2A i1i1 VAVA i2i2 11 22 4A VBVB + - 2V Supernode + - V 4A NODE VOLTAGE ANALYSIS - SUPERNODE: Supernode: Example 4 cont.: 2A + V A /1  + V B /2  + 4A = 0 V A V B = - 6A (one equation, two unknowns)

Grossman/Melkonian  But, we know V A - V B = 2V (gives us second equation) V Ref 2A i1i1 VAVA i2i2 11 22 4A VBVB + - 2V Supernode + - V 4A NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.:

Grossman/Melkonian V Ref 2A i1i1 VAVA i2i2 11 22 4A VBVB + - 2V Supernode + - V 4A Example 4 cont.: NODE VOLTAGE ANALYSIS - SUPERNODE: V A V B = - 6A V A - V B = 2V V A = V V B = V

Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 4 cont.: V Ref 2A i1i1 VAVA i2i2 11 22 4A VBVB + - 2V Supernode + - V 4A Calculate i 1, i 2, and V 4A : i 1 = V A /1  = V/1  i 2 = V B /2  = V/2  i 1 = A i 2 = A V 4A = V B V B = V

Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5: 1mA i1i1 150  3mA + - 2V 200   Calculate i 1 and V 1mA using node voltage analysis: + - 4V V 1mA + -

Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: 1mA i1i1 150  3mA + - 2V 200  + - 4V VAVA VCVC V Ref VBVB  Identify and label all nodes. V 1mA + -

Grossman/Melkonian 1mA i1i1 150  3mA + - 2V 200  + - 4V VAVA VCVC V Ref VBVB NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: Supernode Supernode: -1mA + V B /150  + (V A - V C )/ 200  = 0 Node C: V C = 4V V A - V B = 2V V 1mA + -

Grossman/Melkonian NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: V A = 2.943VV B = 0.943VV C = 4V 1mA i1i1 150  3mA + - 2V 200  + - 4V VAVA VCVC V Ref VBVB Supernode V 1mA + - i 1 = V B / 150  = 0.943V/150  i 1 = 6.286mA

Grossman/Melkonian 1mA 150  3mA + - 2V 200  + - 4V VAVA VCVC V Ref VBVB V 1mA + - NODE VOLTAGE ANALYSIS - SUPERNODE: Example 5 cont.: V 150  + - KVL KVL: -V 1mA + 2V + V 150  = 0 -V 1mA + 2V V = 0 V 1mA = 2.943V

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 Procedure: MESH CURRENT ANALYSIS: 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: ixix  Use Mesh Analysis to solve for i x, i s, V 1, and V 2 : iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + -

Grossman/Melkonian MESH CURRENT ANALYSIS – VOLTAGE SOURCES: ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - i1i1 i2i2 1. Identify and label mesh currents for each mesh. Example 6 cont.:

Grossman/Melkonian MESH CURRENT ANALYSIS – VOLTAGE SOURCES: 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. Example 6 cont.: ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - i1i1 i2i2 Mesh 1: -10V + 75  (i 1 ) + 50  (i 1 - i 2 ) + 4V = 0 mesh 1  i 1 – i 2

Grossman/Melkonian MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Mesh 2: 50  (i 2 - i 1 )  (i 2 ) - 2V = 0 Example 6 cont.: ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - i1i1 i2i2 mesh 2  i 2 – i 1

Grossman/Melkonian ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - i1i1 i2i2 MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.:  Solving for mesh currents i 1 and i 2 : i 1 = 61.54mAi 2 = 33.85mA Note: These are mesh currents.

Grossman/Melkonian MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.: ixix iSiS  100  50  4V 10V V1V1 + - V2V2 + - i1i1 i2i2  Calculate i x and i s : i s = i 2 = 33.85mA i x = (i 1 - i 2 ) = 61.54mA mA i x = 27.69mA

Grossman/Melkonian ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - i1i1 i2i2 MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.:  Solving for V 1, and V 2 : V 1 = 75 i 1 = 75 61.54mA V 1 = 4.62V V 2 = 50  (i 2 – i 1 ) = 50  ( 33.85mA mA) V 2 = -1.38V

Grossman/Melkonian ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - KVL MESH CURRENT ANALYSIS – VOLTAGE SOURCES: Example 6 cont.:  Check results using KVL: KVL Mesh 1: -10V + V 1 - V 2 + 4V = 0 -10V V - (-1.38V) + 4V = 0

Grossman/Melkonian Example 6 cont.: MESH CURRENT ANALYSIS – VOLTAGE SOURCES: ixix iSiS  100  50  2V 4V 10V V1V1 + - V2V2 + - KVL Mesh 2: V 2 + V 3 - 2V = V  33.85mA - 2V = 0 V3V3 - +

Grossman/Melkonian MESH CURRENT ANALYSIS – CURRENT SOURCES: ixix k  1k  2k  2mA 20V V1V1 + - V2V2 + - i1i1 i2i2  Use Mesh Analysis to solve for i x, V 1, and V 2 : Example 7: 1. Identify and label mesh currents for each mesh.

Grossman/Melkonian Example 7 cont.: ixix k  1k  2k  2mA 20V V1V1 + - V2V2 + - KVL V3V 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 CURRENT ANALYSIS – CURRENT SOURCES: Mesh 1: -20V + 1.5k  (i 1 ) + 2k  (i 1 – i 2 ) = 0

Grossman/Melkonian Example 7 cont.: ixix k  1k  2k  2mA 20V V1V1 + - V2V2 + - KVL V3V3 - + Mesh 2: i 2 = -2mA MESH CURRENT ANALYSIS – CURRENT SOURCES:  Solving for mesh currents i 1 and i 2 : i 1 = 4.57mA i 2 = -2mA

Grossman/Melkonian Example 7 cont.: MESH CURRENT ANALYSIS – CURRENT SOURCES: ixix k  1k  2k  2mA 20V V1V1 + - V2V2 + - KVL V3V3 - + Calculate: i x = i 2 – i 1 = (-2mA) – (4.57mA) i x = -6.57mA V 1 = 1.5k  (i 1 ) = 1.5k  (4.57mA) V 1 = 6.86V V 2 = 2k  (-i x ) = 2k  (6.57mA) V 2 = 13.14V

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 i a and i b. We then write mesh equations for the remaining meshes in the usual way. iaia ibib icic isis Supermesh

Grossman/Melkonian  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: i a – i b = i s iaia ibib icic isis Supermesh MESH CURRENT ANALYSIS – SUPERMESH:

Grossman/Melkonian 4V 800  600  V2V mA 300  200  - - MESH CURRENT ANALYSIS – SUPERMESH: Example 8:  Calculate mesh currents and solve for V 1 and V 2 : V1V1 + -

Grossman/Melkonian 1. Identify and label mesh currents for each mesh. Define each mesh current consistently. Typically, mesh currents are defined in the clockwise direction. 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - MESH CURRENT ANALYSIS – SUPERMESH: Example 8 cont.: V1V1 + -

Grossman/Melkonian Example 8 cont.: 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. 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - Supermesh V1V1 + -

Grossman/Melkonian Example 8 cont.: MESH CURRENT ANALYSIS – SUPERMESH: 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - Supermesh Supermesh: -4V  i  i  (i 2 - i 3 ) = 0 V1V1 + -

Grossman/Melkonian 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - MESH CURRENT ANALYSIS – SUPERMESH: Example 8 cont.: Mesh 3: 800  (i 3 - i 2 )  i 3 = 0 Common Current Source: i 1 – i 2 = 2mA V1V1 + -

Grossman/Melkonian 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - Example 8 cont.: MESH CURRENT ANALYSIS – SUPERMESH:  Solving for mesh currents i 1, i 2, and i 3 : i 1 = 5.536mA i 2 = 3.536mA i 3 = 2.571mA V1V1 + -

Grossman/Melkonian 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - V1V1 + - Example 8 cont.: MESH CURRENT ANALYSIS – SUPERMESH: V 1 : KVL Mesh 1 -4V   i 1 + V 1 = 0 V 1 = 2.89V

Grossman/Melkonian 4V 800  600  i1i1 i2i2 i3i3 V2V mA 300  200  - - V1V1 + - MESH CURRENT ANALYSIS – SUPERMESH: Example 8 cont.: V 2 : V 2 = 800  (i 2 – i 3 ) V 2 = 0.772V

Grossman/Melkonian MESH CURRENT ANALYSIS – SUPERMESH: Example 9: 8mA 300  500  2mA 1k   Calculate i x. ixix Use KCL: -8mA + 2mA - i x = 0 i x = -6mA

Grossman/Melkonian SUPERPOSITION: Section 3.5 V2V2 iTiT V1V R  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. Superposition Principle: i1i1 V1V1 + - R V2V2 i2i2 + - R i T = i 1 + i 2 +

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. VSVS R3R3 R2R2 + iSiS R1R1 - R3R3 R2R2 iSiS R1R1 Voltage source set equal to zero

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. VSVS R3R3 R2R2 + iSiS R1R1 - R3R3 R2R2 R1R1 Current source set equal to zero VSVS + -

Grossman/Melkonian SUPERPOSITION: Example 10: 5V 250  200  VRVR + + 5mA 400  - -  Calculate V R using superposition:

Grossman/Melkonian 250  200  VRVR + 5mA 400  - 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: Voltage source set equal to zero

Grossman/Melkonian SUPERPOSITION: Example 10 cont.: 250  200  VRVR + 5mA 400  - i 1 = 5mA  V R due to current source only (V R1 ): i1i1 400  400    Current divider ViVi + - V i = i 1  250  = 2.353mA  250  = 0.588V V R1 = -V i V R1 = V

Grossman/Melkonian SUPERPOSITION: Example 10 cont.: 250  200  VRVR  - V R2 + -  V R due to voltage source only (V R2 ): + - 5V V R2 = 5V 250  400    = 1.471V V R2 = 1.471V Voltage divider

Grossman/Melkonian 5V 250  200  VRVR + + 5mA 400  - - SUPERPOSITION: Example 10 cont.: V R = V R1 + V R2 = V V V R = 0.882V

Grossman/Melkonian Section 3.5 THEVENIN and NORTON CIRCUITS:  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 R T. VTVT + RTRT - Thevenin Circuit A B ININ A B Norton Circuit RNRN

Grossman/Melkonian THEVENIN and NORTON CIRCUITS:  At this point you should be asking yourself two questions; how do we calculate the Thevenin voltage and the Thevenin resistance? Thevenin Equivalent Circuits: Thevenin Voltage (V T ): The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed. + - R3R3 R1R1 R2R2 VSVS + - V OC RLRL V T = V OC

Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Thevenin Resistance (R T ): The Thevenin resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero). + - R3R3 R1R1 R2R2 VSVS RLRL R3R3 R1R1 R2R2 R EQ = R T V S set equal to zero

Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Find the Thevenin equivalent circuit at terminals ‘A’ and ‘B’: V T = V OC = V AB = V 12  = i o  12   Thevenin Voltage: Using mesh analysis: i O = mA Example 11: + - 55 20  10  15V + - V OC A B 12  ioio

Grossman/Melkonian THEVENIN and NORTON CIRCUITS: + - 55 20  10  15V + - V OC A B 12  ioio V T = V OC = 211.3mA  12  V T = 2.54V R T = 5.92  R T = [(20  10  ) + 5  ] 12  Example 11 cont.: Thevenin Resistance: Setting all sources equal to zero and looking back into the circuit from terminals “A” and “B”:

Grossman/Melkonian 10mA 200  400  700  500  THEVENIN and NORTON CIRCUITS: RLRL 4mA Find the Thevenin equivalent circuit seen by the load R L : Example 12: + - V OC V T : V T = V OC = V 700  = 4mA  700  V 700  + - Check V T = 2.8V

Grossman/Melkonian Example 12 cont.: R T : R T = 500   R T = 1.2k  200  400  700  500  RLRL + - V OC V 700  + - THEVENIN and NORTON CIRCUITS: Thevenin Resistance: Set all sources equal to zero:

Grossman/Melkonian 10mA 200  400  700  500  4mA V 700  + - Example 12 cont.: THEVENIN and NORTON CIRCUITS: Thevenin Resistance: R T = i SC VTVT i SC = 4mA 700  700   = 2.33mA Current Divider R T = 2.8V/2.33mA R T = 1.2k 

Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Norton Equivalent Circuits: Norton Current (I N ): The Norton current is equal to the short-circuit current at the load terminals with the load removed. + - R3R3 R1R1 R2R2 VSVS I N = i SC i SC

Grossman/Melkonian Norton Resistance (R N ): THEVENIN and NORTON CIRCUITS: R3R3 R1R1 R2R2 R EQ = R N V S set equal to zero + - R3R3 R1R1 R2R2 VSVS i SC The Norton resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero).

Grossman/Melkonian THEVENIN and NORTON CIRCUITS: Summary: V T = V OC = V AB (with load removed) R T = V T /i SC = R EQ as seen by R L The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed. Thevenin Equivalent Circuit: The Thevenin resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero) or V T /i SC.

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 /I N. I N = I SC (with load removed) R N = R T V OC /I N = R EQ as seen by R L

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. Rest of Circuit + - VSVS R1R1 A B Source Transformation R1R1 ISIS A B Rest of Circuit

Grossman/Melkonian Rest of Circuit + - V S = I S  R 1 R1R1 A B Source Transformation R1R1 I S = V S /R 1 A B Rest of Circuit 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.

Grossman/Melkonian SOURCE TRANSFORMATION: - + VSVS R1R1 R2R2 R4R4 R3R3  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. I S = V S /R 1 R1R1 R2R2 R4R4 R3R3 Voltages across and currents through R 2, R 3, and R 4 are the same for both circuits!

Grossman/Melkonian SOURCE TRANSFORMATION: Example 13: - + 8V 1.5k  300  Use source transformation and current divider rule to calculate i o : 1k  2k  iOiO

Grossman/Melkonian Example 13 cont.: SOURCE TRANSFORMATION: 8V/1.5k  = 5.33mA 1.5k  300  1k  2k  iOiO 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

Grossman/Melkonian SOURCE TRANSFORMATION: Example 13 cont.: 5.33mA 1.5k  300  1k  2k  iOiO i O = 1/1.3k  1/ 1.5k  + 1/2 k  + 1/ 1.3k  5.33mA i O = 2.12mA