grossman/melkonian chapter 3 resistive network analysis

Grossman/Melkonian

Chapter 3

Resistive Network Analysis

Grossman/Melkonian

COMBINING INDEPENDENT SOURCES: Voltage sources in series add algebraically:

Equivalent

Grossman/Melkonian

COMBINING INDEPENDENT SOURCES:

Current sources in parallel add algebraically:

3mA 5mA -4mA1mA R1 R2

5mA R1R2

Equivalent

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.

+ R1 - + R3 -

KCL at Node B: i1 - i2 - i3 = 0

Applying Ohm’s Law:VA - VB

VB - VC

VB - VRef

i1 i2 i3

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 V4, V6, i1, and i2.

2A 3A4 6

i1 i2+

Grossman/Melkonian

2A 3A4 6i1

Example 1 cont.:

1. Identify and label a reference node. All other nodes will be referenced to this node.

Grossman/Melkonian

Example 1 cont.:

3. Apply KCL to each N-1 node. Writing equation in terms of node voltage.

2A 3A4 6i1

-2A +VA - 0

VA - 0

6+ + 3A = 0

Grossman/Melkonian

Example 1 cont.:

2A 3A4 6i1

4. Solve equation for VA.

VA = -2.4V

Grossman/Melkonian

2A 3A4 6i1

5. Solve for i1 and i2.

Example 1 cont.:

Using Ohm’s Law and the calculated value for VA:

i1 = VA/4 = -2.4V/4

i1 = -0.6A

i2 = VA/6 = -2.4V/6

i2 = -0.4A

Grossman/Melkonian

2A 3A4 6i1

Referring to the circuit in example 1, calculate VA by transforming the circuit to an equivalent circuit with one current source and one resistor:

Original Circuit

Combining Independent Sources:

Grossman/Melkonian

Since the current sources are in parallel, they can be combined into a single equivalent current source:

1A 2.4iT

VA = iT • 2.4 = -1A • 2.4

VA = -2.4V

Note: iT is equal to the sum of i1 and i2 of original circuit.

i1 = -0.6A

i2 = -0.4Aand iT = -1A

Grossman/Melkonian

Example 2:

4mA 6mA

200 400

NODE VOLTAGE – CURRENT SOURCES:

Using node voltage analysis calculate i1 and i2:

Grossman/Melkonian

4mA 6mA

200 400

Example 2 cont.:

Grossman/Melkonian

4mA 6mA

200 400

Example 2 cont.:

Node 1: -4mA +V1 - V2

200+ = 0

Grossman/Melkonian

4mA 6mA

200 400

Example 2 cont.:

Node 2:V2 – V1

400+ 6mA = 0+

Grossman/Melkonian

4mA 6mA

200 400

Example 2 cont.:

Solving equations for V1 and V2:

V1 = -88.89mV V2 = -622.22mV

Grossman/Melkonian

Example 2 cont.:

i1 = V1/200 = -88.89mV/200 = -0.445mA

i2 = 0 – V2/400 = 622.22mV/400 = 1.556mA

4mA 6mA

200 400

Grossman/Melkonian

NODE VOLTAGE – VOLTAGE SOURCES:

3mA 6V

2k 1.5k

Example 3:

Use node voltage analysis to calculate VA and VB:

Grossman/Melkonian

3mA 6V

2k 1.5k

Example 3 cont.:

Node A: -3mA +VA - VB

2k+ = 0

Grossman/Melkonian

3mA 6V

2k 1.5k

Example 3 cont.:

Node B: VB = 6V

Grossman/Melkonian

3mA 6V

2k 1.5k

Example 3 cont.:

Solving equations for VA and VB:

VA = 6V VB = 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; 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 Supernode is needed when neither the positive nor the negative terminal of a voltage source is connected to the reference node.

+A BVs

Grossman/Melkonian

Example 4:

Using node voltage analysis, calculate i1, i2, and V4A:

Grossman/Melkonian

Example 4 cont.:

Grossman/Melkonian

VB+-2V

Supernode

Supernode:

Example 4 cont.:

2A + VA/1 + VB /2 + 4A = 0

VA + 0.5 VB = - 6A (one equation, two unknowns)

Grossman/Melkonian

But, we know VA - VB = 2V (gives us second equation)

VB+-2V

Supernode

Example 4 cont.:

Grossman/Melkonian

VB+-2V

Supernode

Example 4 cont.:

VA + 0.5 VB = - 6A

VA - VB = 2V

VA = -3.334V

VB = -5.334V

Grossman/Melkonian

NODE VOLTAGE ANALYSIS - SUPERNODE:Example 4 cont.:

VB+-2V

Supernode

Calculate i1, i2, and V4A:

i1 = VA/1 = -3.334V/1

i2 = VB/2 = -5.334V/2

i1 = -3.334A

i2 = -2.667A

V4A = VB VB = -5.334V

Grossman/Melkonian

Example 5:

Calculate i1 and V1mA using node voltage analysis:

+- 4VV1mA

Grossman/Melkonian

Example 5 cont.:

1mAi1150

Identify and label all nodes.

Grossman/Melkonian

1mAi1150

Supernode

Supernode: -1mA + VB/150 + (VA - VC)/ 200 = 0

Node C: VC = 4V

VA - VB = 2V

Grossman/Melkonian

VA = 2.943V VB = 0.943V VC = 4V

1mAi1150

Supernode

i1 = VB/150 = 0.943V/150 i1 = 6.286mA

Grossman/Melkonian

VBV1mA

KVL: -V1mA + 2V + V150 = 0

-V1mA + 2V + 0.943V = 0

V1mA = 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:

Use Mesh Analysis to solve for ix, is, V1, and V2:

75 100

Grossman/Melkonian

75 100

1. Identify and label mesh currents for each mesh.

Example 6 cont.:

Grossman/Melkonian

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.:

75 100

Mesh 1: -10V + 75(i1) + 50(i1 - i2) + 4V = 0

mesh 1 i1 – i2

Grossman/Melkonian

Mesh 2: 50(i2 - i1) + 100(i2) - 2V = 0

Example 6 cont.:

75 100

mesh 2 i2 – i1

Grossman/Melkonian

75 100

Example 6 cont.:

Solving for mesh currents i1 and i2:

i1 = 61.54mA i2 = 33.85mANote: These are mesh currents.

Grossman/Melkonian

Example 6 cont.:

75 100

Calculate ix and is:

is = i2 = 33.85mA

ix = (i1 - i2) = 61.54mA - 33.85mA ix = 27.69mA

Grossman/Melkonian

75 100

-i1 i2

Example 6 cont.:

Solving for V1, and V2:

V1 = 75•i1 = 75•61.54mA V1 = 4.62V V2 = 50(i2 – i1) = 50(33.85mA - 61.54mA) V2 = -1.38V

Grossman/Melkonian

75 100

Example 6 cont.:

Check results using KVL:

Mesh 1: -10V + V1 - V2 + 4V = 0

-10V + 4.62V - (-1.38V) + 4V = 0

Grossman/Melkonian

Example 6 cont.:

75 100

KVL KVL

Mesh 2: V2 + V3 - 2V = 0

-1.38V + 100•33.85mA - 2V = 0

Grossman/Melkonian

MESH CURRENT ANALYSIS – CURRENT SOURCES:

1.5k 1k

2k2mA20V

Use Mesh Analysis to solve for ix, V1, and V2:

Example 7:

1. Identify and label mesh currents for each mesh.

Grossman/Melkonian

Example 7 cont.:

1.5k 1k

2k 2mA20V

KVL KVL

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

Grossman/Melkonian

Example 7 cont.:

1.5k 1k

2k 2mA20V

KVL KVL

Mesh 2: i2 = -2mA

Solving for mesh currents i1 and i2:

i1 = 4.57mA i2 = -2mA

Grossman/Melkonian

Example 7 cont.:

1.5k 1k

2k 2mA20V

KVL KVL

Calculate:

ix = i2 – i1 = (-2mA) – (4.57mA) ix = -6.57mA

V1 = 1.5k(i1) = 1.5k(4.57mA) V1 = 6.86V

V2 = 2k(-ix) = 2k(6.57mA) V2 = 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 ia and ib. We then write mesh equations for the

remaining meshes in the usual way.

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: ia – ib = is

Supermesh

Grossman/Melkonian

4V 800

++ 2mA 300

Example 8:

Calculate mesh currents and solve for V1 and V2:

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

i1 i2i3V2

Example 8 cont.:

Grossman/Melkonian

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.

4V 800

i1 i2i3V2

Supermesh

Grossman/Melkonian

Example 8 cont.:

4V 800

i1 i2i3V2

Supermesh

Supermesh:

-4V + 200•i1 + 600•i2 + 800•(i2 - i3) = 0

Grossman/Melkonian

4V 800

i1 i2i3V2

MESH CURRENT ANALYSIS – SUPERMESH:Example 8 cont.:

Mesh 3: 800•(i3 - i2) + 300•i3 = 0

Common Current Source: i1 – i2 = 2mA

Grossman/Melkonian

4V 800

i1 i2i3V2

Example 8 cont.:

Solving for mesh currents i1, i2, and i3:

i1 = 5.536mA i2 = 3.536mA i3 = 2.571mA

Grossman/Melkonian

4V 800

i1 i2i3V2

Example 8 cont.:

V1: KVL Mesh 1 -4V + 200i1 + V1 = 0

V1 = 2.89V

Grossman/Melkonian

4V 800

i1 i2i3V2

Example 8 cont.:

V2: V2 = 800(i2 – i3)

V2 = 0.772V

Grossman/Melkonian

Example 9:

8mA 300

Calculate ix.

Use KCL: -8mA + 2mA - ix = 0

ix = -6mA

Grossman/Melkonian

SUPERPOSITION:Section 3.5

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:

iT = i1 + i2

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.

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.

Current source set equal to zero VS +

Grossman/Melkonian

SUPERPOSITION:Example 10:

5V 250

Calculate VR using superposition:

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:

Voltage source set equal to zero

Grossman/Melkonian

i1 = 5mA

VR due to current source only (VR1):

400 + 200 + 250Current divider

Vi = i1250 = 2.353mA250 = 0.588V

VR1 = -Vi VR1 = -0.588V

Grossman/Melkonian

VR due to voltage source only (VR2):

VR2 = 5V 250

400 + 200 + 250= 1.471V

VR2 = 1.471V

Voltage divider

Grossman/Melkonian

VR = VR1 + VR2 = -0.588V + 1.471V

VR = 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 VT in series with an equivalent resistance RT.

-Thevenin Circuit

Norton CircuitRN

Grossman/Melkonian

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 (VT):

The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed.

VOC RLVT = 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).

R2VS RL

R2REQ = RTVS set equal

to zero

Grossman/Melkonian

Find the Thevenin equivalent circuit at terminals ‘A’ and ‘B’:

VT = VOC = VAB = V12 = io12

Thevenin Voltage:

Using mesh analysis: iO = 211.27mA

Example 11:

Grossman/Melkonian

VT = VOC = 211.3mA12 VT = 2.54V

RT = 5.92 RT = [(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

Find the Thevenin equivalent circuit seen by the load RL:

Example 12:

VT: VT = VOC = V700 = 4mA700

VT = 2.8V

Grossman/Melkonian

Example 12 cont.:

RT: RT = 500 + 700 RT = 1.2k

VOCV700

Thevenin Resistance: Set all sources equal to zero:

Grossman/Melkonian

10mA 200

5004mA

Example 12 cont.:

Thevenin Resistance: RT =iSC

iSC = 4mA700

700 + 500= 2.33mA Current Divider

RT = 2.8V/2.33mA RT = 1.2k

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.

IN = iSC

Grossman/Melkonian

Norton Resistance (RN):

R2REQ = RNVS set equal

to zero

The Norton resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero).

Grossman/Melkonian

Summary:

VT = VOC = VAB (with load removed)

RT = VT/iSC = REQ as seen by RL

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 VT/iSC.

Grossman/Melkonian

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 VT/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 RL. 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+-

Source Transformation

Rest of

Circuit

Grossman/Melkonian

Rest of

Circuit+-VS = ISR1

Source Transformation

IS = VS/R1

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:

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.

IS = VS/R1

R1 R2 R4

Voltages across and currents through R2, R3, and R4 are the same for both circuits!

Grossman/Melkonian

SOURCE TRANSFORMATION:Example 13:

Use source transformation and current divider rule to calculate io:

Grossman/Melkonian

Example 13 cont.:

SOURCE TRANSFORMATION:

8V/1.5k = 5.33mA

1.5k 300

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

1/1.3k1/1.5k + 1/2k + 1/1.3k

5.33mA

iO = 2.12mA

grossman/melkonian chapter 3 resistive network analysis

node voltage equations

node n

node b

node voltage method

reference node

node voltage procedure

voltage sources

use node voltage analysis

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