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:

+-

+ -

+-

R1 R2

R3

10V2V

-6V

+-

R1R2

R3

18V

Equivalent

I

I

Grossman/Melkonian

COMBINING INDEPENDENT SOURCES:

Current sources in parallel add algebraically:

3mA 5mA -4mA1mA R1 R2

5mA R1R2

Equivalent

V1

V1

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:

VA

As mention previously, node voltage equations are written from KCL equations.

VCVB

VRef

+ R1 - + R3 -

R2

+

-

i3i1

i2

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

Applying Ohm’s Law:VA - VB

R1

VB - VC

R3

VB - VRef

R2

- -

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

i2

VRef


Example 1 cont.:

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


VA

--

++

Grossman/Melkonian

Example 1 cont.:


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

2A 3A4 6i1

i2

VRef

VA

-2A +VA - 0

4

VA - 0

6+ + 3A = 0

--

++

Grossman/Melkonian


Example 1 cont.:

2A 3A4 6i1

i2

VRef

VA

4. Solve equation for VA.

VA = -2.4V

--

++

Grossman/Melkonian


2A 3A4 6i1

i2

VRef

VA

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

i2

VRef

VA

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

-

+

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

120

200 400

VRef

i1

i2

NODE VOLTAGE – CURRENT SOURCES:

Using node voltage analysis calculate i1 and i2:

Grossman/Melkonian


4mA 6mA

V1V2

120

200 400

VRef

i1

i2


Example 2 cont.:


Grossman/Melkonian

4mA 6mA

V1 V2

120

200 400

VRef

i1

i2


Example 2 cont.:


Node 1: -4mA +V1 - V2

120

V1

200+ = 0

Grossman/Melkonian

4mA 6mA

V1 V2

120

200 400

VRef

i1

i2


Example 2 cont.:

Node 2:V2 – V1

120

V2

400+ 6mA = 0+

Grossman/Melkonian

4mA 6mA

V1 V2

120

200 400

VRef

i1

i2


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

V1 V2

120

200 400

VRef

i1

i2

Grossman/Melkonian

NODE VOLTAGE – VOLTAGE SOURCES:

3mA 6V

VA VB

1k

2k 1.5k

VRef

Example 3:

Use node voltage analysis to calculate VA and VB:

+-

Grossman/Melkonian

3mA 6V

VA VB

1k

2k 1.5k

VRef

+-


Example 3 cont.:

Node A: -3mA +VA - VB

1k

VA

2k+ = 0

Grossman/Melkonian

3mA 6V

VA VB

1k

2k 1.5k

VRef

+-


Example 3 cont.:

Node B: VB = 6V

Grossman/Melkonian

3mA 6V

VA VB

1k

2k 1.5k

VRef

+-

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:

2A

i1i2

1 2

4A

+-2V


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

+

-

V4A

Grossman/Melkonian

Example 4 cont.:

VRef

2A

i1

VA

i2

1 2

4A

VB

+-2V

+

-

V4A





Grossman/Melkonian

VRef

2A

i1

VA

i2

1 2

4A

VB+-2V

Supernode

+

-

V4A


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)

VRef

2A

i1

VA

i2

1 2

4A

VB+-2V

Supernode

+

-

V4A


Example 4 cont.:

Grossman/Melkonian

VRef

2A

i1

VA

i2

1 2

4A

VB+-2V

Supernode

+

-

V4A

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.:

VRef

2A

i1

VA

i2

1 2

4A

VB+-2V

Supernode

+

-

V4A

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:

1mA

i1

150

3mA

+-2V

200

Calculate i1 and V1mA using node voltage analysis:

+- 4VV1mA

+

-

Grossman/Melkonian


Example 5 cont.:

1mAi1150

3mA

+-2V

200

+- 4V

VA VC

VRef

VB

Identify and label all nodes.

V1mA

+

-

Grossman/Melkonian

1mAi1150

3mA

+-2V

200

+- 4V

VAVC

VRef

VB


Supernode

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

Node C: VC = 4V

VA - VB = 2V

V1mA

+

-

Grossman/Melkonian


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

1mAi1150

3mA

+-2V

200

+- 4V

VAVC

VRef

VB

Supernode

V1mA

+

-

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

Grossman/Melkonian

1mA

150

3mA

+-2V

200

+- 4V

VAVC

VRef

VBV1mA

+

-


V150

+

-

KVL

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:

ix

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

iS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

Grossman/Melkonian


ixiS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

i1 i2

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.:

ix iS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

i1 i2

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.:

ixiS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

i1 i2

mesh 2 i2 – i1

Grossman/Melkonian

ixiS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

i1 i2


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.:

ixiS

-+

+

+

-

-

75 100

50

4V

10V

V1+ -

V2

+

-

i1 i2

Calculate ix and is:

is = i2 = 33.85mA

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

Grossman/Melkonian

ixiS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-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

ixiS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

KVL


Example 6 cont.:

Check results using KVL:

KVL

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

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

Grossman/Melkonian

Example 6 cont.:


ixiS

-+

+

+

-

-

75 100

50 2V

4V

10V

V1+ -

V2

+

-

KVL KVL

Mesh 2: V2 + V3 - 2V = 0

-1.38V + 100•33.85mA - 2V = 0

V3-+

Grossman/Melkonian

MESH CURRENT ANALYSIS – CURRENT SOURCES:

ix

-+

1.5k 1k

2k2mA20V

V1+ -

V2

+

-

i1 i2

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.:

ix

-+

1.5k 1k

2k 2mA20V

V1+ -

V2

+

-

KVL KVL

V3-+

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.:

ix

-+

1.5k 1k

2k 2mA20V

V1+ -

V2

+

-

KVL KVL

V3-+

Mesh 2: i2 = -2mA


Solving for mesh currents i1 and i2:

i1 = 4.57mA i2 = -2mA

Grossman/Melkonian

Example 7 cont.:


ix

-+

1.5k 1k

2k 2mA20V

V1+ -

V2

+

-

KVL KVL

V3-+

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.

ia ib

icis

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

ia ib

icis

Supermesh


Grossman/Melkonian

4V 800

600

V2

++ 2mA 300

200

--


Example 8:

Calculate mesh currents and solve for V1 and V2:

V1

+

-

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

i1 i2i3V2

++

2mA

300

200

--


Example 8 cont.:

V1

+

-

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

600

i1 i2i3V2

++

2mA

300

200

--

Supermesh

V1

+

-

Grossman/Melkonian

Example 8 cont.:


4V 800

600

i1 i2i3V2

++

2mA

300

200

--

Supermesh

Supermesh:

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

V1

+

-

Grossman/Melkonian

4V 800

600

i1 i2i3V2

++

2mA

300

200

--

MESH CURRENT ANALYSIS – SUPERMESH:Example 8 cont.:

Mesh 3: 800•(i3 - i2) + 300•i3 = 0

Common Current Source: i1 – i2 = 2mA

V1

+

-

Grossman/Melkonian

4V 800

600

i1 i2i3V2

++

2mA

300

200

--

Example 8 cont.:


Solving for mesh currents i1, i2, and i3:

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

V1

+

-

Grossman/Melkonian

4V 800

600

i1 i2i3V2

++

2mA

300

200

--

V1

+

-

Example 8 cont.:


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

V1 = 2.89V

Grossman/Melkonian

4V 800

600

i1 i2i3V2

++

2mA

300

200

--

V1

+

-


Example 8 cont.:

V2: V2 = 800(i2 – i3)

V2 = 0.772V

Grossman/Melkonian


Example 9:

8mA 300

500

2mA

1k

Calculate ix.

ix

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

ix = -6mA

Grossman/Melkonian

SUPERPOSITION:Section 3.5

V2

iT

V1+

+

-

-

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:

i1

V1+-

RV2

i2

+-

R

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.

VS R3

R2

+ iS

R1

-

R3

R2

iS

R1

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.

VS R3

R2

+ iS

R1

-

R3

R2R1

Current source set equal to zero VS +

-

Grossman/Melkonian

SUPERPOSITION:Example 10:

5V 250

200

VR

+

+ 5mA

400

-

-

Calculate VR using superposition:

Grossman/Melkonian

250

200

VR

+

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


250

200

VR

+

5mA

400

-

i1 = 5mA

VR due to current source only (VR1):

i1

400

400 + 200 + 250Current divider

Vi

+

-

Vi = i1250 = 2.353mA250 = 0.588V

VR1 = -Vi VR1 = -0.588V

Grossman/Melkonian


250

200

VR

+

400

-VR2

+

-

VR due to voltage source only (VR2):

+-5V

VR2 = 5V 250

400 + 200 + 250= 1.471V

VR2 = 1.471V

Voltage divider

Grossman/Melkonian

5V250

200

VR

+

+ 5mA

400

-

-


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.

VT+

RT

-Thevenin Circuit

A

B

IN

A

B

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.

+-

R3R1

R2VS

+

-

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).

+-

R3R1

R2VS RL

R3R1

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:

+-

520

1015V

+

-VOC

A

B

12io

Grossman/Melkonian


+-

520

1015V

+

-VOC

A

B

12io

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

400

700

500


RL

4mA

Find the Thevenin equivalent circuit seen by the load RL:

Example 12:

+

-

VOC

VT: VT = VOC = V700 = 4mA700

V700

+

-

Check

VT = 2.8V

Grossman/Melkonian

Example 12 cont.:

RT: RT = 500 + 700 RT = 1.2k

200

400

700

500

RL

+

-

VOCV700

+

-


Thevenin Resistance: Set all sources equal to zero:

Grossman/Melkonian

10mA 200

400

700

5004mA

V700

+

-

Example 12 cont.:


Thevenin Resistance: RT =iSC

VT

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.

+-

R3R1

R2VS

IN = iSC

iSC

Grossman/Melkonian

Norton Resistance (RN):


R3R1

R2REQ = RNVS set equal

to zero

+-

R3R1

R2

VSiSC

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+-

VS

R1

A

B

Source Transformation

R1

IS A

B

Rest of

Circuit

Grossman/Melkonian

Rest of

Circuit+-VS = ISR1

R1 A

B

Source Transformation

R1

IS = VS/R1

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:

-+VS

R1

R2R4

R3

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

R3

Voltages across and currents through R2, R3, and R4 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 io:

1k

2k

iO

Grossman/Melkonian

Example 13 cont.:

SOURCE TRANSFORMATION:

8V/1.5k = 5.33mA

1.5k 300

1k

2k

iO

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

iO

iO =

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

5.33mA

iO = 2.12mA

grossman/melkonian chapter 3 resistive network analysis

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