EENG 2009 FS 2006 Part 5: Nodal Analysis 146 EENG 2009 Part 5. Nodal Analysis 5.1 Nodes and Node...

EENG 2009 FS 2006 Part 5: Nodal Analysis

1

EENG 2009

Part 5. Nodal Analysis

5.1 Nodes and Node Voltages

5.2 Nodal Analysis By Example

5.3 Supernodes

5.4 Dependent Sources


2

Recall that nodes are the connected segments of conductor that remain when we remove the circuit elements (which at present for us are either resistors or sources).

9 A

16

8 12 3 A

Identify the nodes for the given circuit.

Redraw the circuit with the circuit elements removed and note the connected segments of conductors. We see that there are 3 nodes for this circuit:

5.1 Nodes and Node VoltagesNodes

Example 1

Solution:

A single node!


3

A node voltage associated with a given node is defined to be the voltage difference between the given node and a reference node, which has been chosen from among the nodes. For a circuit with N essential nodes, there are N–1 node voltages.

Once the set of node voltages is determined, all the other voltages and currents can be obtained in a straightforward manner.

Identify a reference node and corresponding node voltages for the given circuit (whose nodes we found in the previous example).

9 A

16

8 12 3 A

Node Voltages

Example 2


4

node 1

v1 v2

Reference node, ground, earth, “sea level”

Using the results of Example 1, we draw the circuit with the nodes emphasized:

9 A

16

8 12 3 A

Next choose the bottom node as the reference node, and designate the node voltages for the other two

nodes as v1 and v2:

9 A

16

8 12 3 A

node 2

Example 2 Solution:


5

9 A

16

8 12 3 A

1 2

v1 v2

64 V 48 V

16 V

The voltage across the 16- branch is not a node voltage. It is a branch voltage, and is actually the difference between the

two node voltages v1 and v2.

voltmeter voltmeter

voltmeter

+ – + –

+ –The voltage being measured is a branch voltage.

Node Voltages!

Measurement of Branch and Node Voltages


6

1 2v1 v2

3

+ v12 –

+

v23

–

+

v13

–

Find the relationships among the branch voltages and the node voltages.

There are three branch voltages, v13 , v12 , and v23, and two

node voltages, v1 and v2 (not counting v3) .

The branch voltages v13 and v23 are clearly equal to the

node voltages v1 and v2:

v13 = v1

v23 = v2

The branch voltage v12 is a combination of the node

voltages v1 and v2.

Branch Voltages In Terms of Node Voltages

Example 3

Solution:


7

v12 = v13 – v23 = v1 – v2

Note the correlation with the order of the subscripts:

The subscripts of v12 are 1 & 2 and the subscripts

of the two node voltages being subtracted are 1 & 2.

In general,

We can write the branch voltage v12 in terms of the

node voltages v13 and v23 by applying KVL, as

follows:

1 2v1 v2

3

+ v12 –

+

v23

–

+

v13

–

node j node kvj vk

Memorizeme!

Solution (cont.):

vjk = vj – vk


8

Rv1 v2

The branch current flowing from node 1 to node 2 is:

i12 = ( v1 – v2 ) / R

The branch current flowing from node 2 to node 1 is:i21 = ( v2 – v1 ) / R

In general:

node 1 node 2

reference node

Find the relationships among the branch current and the node voltages. Note that neither voltage-polarity markings nor a current reference arrow is shown on the diagram!

Branch Currents In Terms of Node Voltages

Example 4

Solution:

node j node kvj vk

ijk = ( vj – vk ) / R

ikj = ( vk – vj ) / R

RNote that to avoid confusion, ijk, ikj, and their

corresponding reference directions are not shown on the diagram.


9

The rationale for nodal analysis is that once the node voltages are determined, all the other voltages and currents can be obtained in a simple manner.

The reference node is chosen by the circuit analyst. In electronic circuits, we frequently choose the node to which lots of branches are connected. In power systems, we usually choose “ground” or “earth.” Look for the associated symbol on the circuit:

Basic Procedure For Nodal Analysis:

1. Identify the nodes and the node voltages.

2. Write KCL at the proper nodes. (Here’s where we use the branch-current / node-voltage relationship we just developed.)

3. Solve for the node voltages.

4. Use the node voltages to solve for any other required quantities.

5.2 Nodal Analysis By Example


10

is1R1 R3 is2

v1v2

R2

Find the values of the node voltages.

Step 1. There are 3 nodes. Choose the bottom node as the reference node, and designate the other two nodes as v1 and v2.

is1 R1 R3 is2

With 3 nodes and one of them as the reference node, there are two nodal equations needed.

R2

node 1 node 2

node 3

Example 1

Solution:


11

v1 v2

is1R1

R3 is2

Step 2. Write KCL at nodes 1 and 2.

KCL at node 1 (Summing the currents flowing out of node 1):

v1 / R1 + (v1 – v2) / R2 – is1 = 0

At node 2:

(v2 – v1) / R2 + v2 / R3 + is2 = 0

These are the equations to solve for v1 and v2.

We can put them in orderly form (and discover a shortcut to writing them), as we show next.

R2

node 1 node 2

Solution (cont.):


12

Consider the previous circuit and its nodal equations:

v1 / R1 + (v1 – v2) / R2 – is1 = 0

(v2 – v1) / R2 + v2 / R3 + is2 = 0

Collect terms in v1 and v2 and put the

independent-source terms on the RHS:

(1 / R1 + 1 / R2) v1 – 1/ R2 v2 = is1

– 1/ R2 v1 + (1 / R2 + 1 / R3) v2 = – is2

v1 v2

is1 R1 R3 is2

R2

Development of the Algorithmic Method for Writing Nodal- Analysis Equations


13

Now, using conductances:

(G1 + G2) v1 – G2 v2 = is1 (1)

– G2 v1 + (G2 + G3) v2 = – is2 (2)

Observe that the following algorithm is valid:

In (1), the coefficient of v1 = (conductances connected to node 1)

In (1), the coefficient of v2 =

– (conductances between nodes 1 & 2)


(conductances connected to node 2)


– (conductances between nodes 1 & 2)In (1) and (2) the RHS = (current sources entering the node)

v1 v2

is1 R1 R3 is2

R2


14

9 A

+

V13 A

Find v1 and v2.

8

16

12

Solution:

9 A 3 A8 12

v1 v2

KCL @ node 1:

v1 / 8 + (v1 – v2) / 16 – 9 = 0KCL @ node 2:

(v2 – v1) / 16 + v2 /12 – 3 = 0

Collecting terms:

(1/8 + 1/16) v1 – 1/16 v2 = 9

–1/16 v1 + (1/16 + 1/12) v2 = 3

The solution is v1 = 64 V, v2 = 48 V.

16 Electrically the same as the node 2

node 1 node 2

Example 2

+

V2


15

Note that these equations could have been written directly by using the algorithm!

9 A 3 A8 12

v1 v2

(1/8 + 1/16)v1 - 1/16 v2 = 9

–1/16 v1 + (1/16 + 1/12) v2 = 3

The TI-85 & TI-86 keystrokes for solving this equation are:2nd SIMULTNumber = 2 ENTERa1,1 = 8 –1 + 16 –1 ENTERa1,2 = – 16 –1 ENTERb1 = 9 ENTER

a2,1 = – 16 –1 ENTER a2,2 = 16 –1 + 12 –1 ENTERb2 = 3 ENTER

SOLVEx1 = 64.000x2 = 48.000

16

Solution (cont.)


16

Find the voltage differences across the sources.

3 A

4 5 A

2

4

Solution:

3 A

4 5 A

2

v2

v1

KCL @ nodes 1 and 2:

(v1 – v2) / 4 + 5 + v1 / 2 – 3 = 0

(v2 – v1) / 4 – 5 – 6 + v2 / 4 = 0

Solution: v1 = – 4 V, v2 = 20 V

4

6 A

6 Anode 1

node 2

Example 3.


17

We still have to find the voltage differences across the three sources, as follows.*

3 A

4 5 A

2

v2 = 20 V

v1 = – 4 V 4 6 A+v3A

–

+v5A

–+v6A

–

v3A = v1 = – 4 V

v6A = v2 = 20 V

v5A = v2 – v1 = 24 V

Example 3. (cont.)

* At this point it is up to us to choose the polarity markings for the sources, as none were specified.


18

Find the current i and the power absorbed by the 6 resistor.

Solution:

6

6

12

5 A

18

4 i

6

12

5 A

18

4 i

v2

v3 v1

KCL @ 1: (v1 – v3)/4 + (v1 – v2)/12 + v1/6 = 0

KCL @ 2: (v2 – v1)/12 + (v2 – v3)/6 – 5 = 0

KCL @ 3: v3/18 + (v3 – v1)/4 + (v3 – v2)/6 = 0

Solving for the node voltages gives:

v1 = 21 V, v2 = 45 V, v3 = 27 V

6

node 1

node 2

node 3

Example 4.


19

To calculate i we apply Ohm’s Law (which requires calculating the branch voltage across the 4 resistor):

i = ( v3 – v1) / 4

= (27 – 21) / 4

= 3/2 A*

* Now we know that the true current thru the 4 resistor really is flowing from left-to-right. At node 1 we wrote the expression for the current flowing from right to left, and at node 3 from left-to-right.

To find the power absorbed by the 6 resistor:

v3 v1

4 i

6

12

5 A

18

4

v2 = 45 V

v3 = 27 Vv1

p6 = (v2 – v3)2 / 6

= (45 – 27)2 / 6

= 54 W6

Solution (cont.):


20

Find va

6 V

3 k1 mA

2 k

+–

1 k

+

va

–

(Note that in this circuit the 6 V voltage source has one of its nodes connected to the reference node. This makes solution easier!)

va is the only

unknown node voltage, so write KCL at node a:

6 V

3 k1 mA

2 k

+–

1 k

+

va

–

6 V

va

va/1000 + va/3000 + (va – 6)/2000 – 10–3 = 0

(6 + 2 + 3) va = 18 + 6

va = 2.18 V

node A

Example 5

Solution:

Node voltage already known!


21

6 V

3 k1 mA

2 k

+–

1 k

+

va

–

6 V

va

In this circuit, the independent current source produces mA, the independent voltage source produces volts, and the resistors have values in k. Currents in the circuit will be in the mA range. For example,

i1k = 2.18 V / 1 k

= 2.18 mA.

In these situations it is convenient when writing KCL to express the currents in mA and write the equation as follows:

va / 1 + va/ 3 + (va – 6) / 2 – 1 = 0

Solution (cont.):

i1k


22

3 A 5 14 A4

13

+ –

10 Vsupernode

Note that the supernode includes the component(s) in parallel with the voltage source. (It includes all of the circuit elements connected between the two

nodes with the node voltages v1 and v2.)

v1 v2

5.3 Supernodes

Example 1

Solution:

A supernode is a set of nodes connected to each other by voltage sources, but not to the reference node by a path of voltage sources.

Identify the supernode.


23

Step 1. Write KCL equation for a surface enclosing the supernode.

Step 2. Write a KVL equation relating the nodal voltages in the supernode (a constraint equation).

Step 3. Solve the equations.

Find v1 and v2.

3 A 5 14 A4

13

+ –

10 Vsupernode

1. KCL @ supernode: 3 + v1/5 + v2/4 – 14 = 0

2. Constraint equation: v1 – v2 = 10

3. Re-writing these two equations:

4 v1 + 5 v2 = 220

– v1 + v2 = – 10

Solving gives: v1 = 30 V, v2 = 20 V

v1 v2

Solution Procedure for Circuits with Supernodes

Example 2.

Solution:


24

Find v1

14 mA6 k 2 mA3 k

12 k+ –

6 V v1

In this circuit there are two regular nodes and one supernode:

14 mA6 k 2 mA3 k

12 k+ –

6 V v1

The solution procedure in this case is:

Step 1. Write the KCL equation for the supernode and the KCL equation for the regular node that was not chosen as the supernode.

Step 2. Write the constraint equation.

Step 3. Solve.

v3v2

supernoderegular node

regular node

Example 3

Solution:


25

Solution (cont.)

Step 1: KCL @ the supernode:

(v2 – v3) / (12x103 ) + v1 / (3x103 ) + 2x10-3 = 0

Step 1 (cont.): KCL @ the non-reference node:

v3 / (6x103 ) + (v3 – v2) / (12x103 ) – 14x10-3 = 0

Step 2: Constraint equation:

v2 – v1 = 6

Step 3: Solving the 3 equations gives:

v1 = 6 V

v2 = 12 V

v3 = 60 V

14 mA 2 mA3 k

12 k 6 V v1v3 v2

6 k

+ –


26

Example 4

2 k4 k

30 V +–

5 k

20 V +– +

vo

–

Find vo

Solution:

2 k

4 k30 V +

–

5 k

20 V +–

+

vo

–

vo

v1 v2

Step 1: KCL @ the supernode:

v1 / 2x103 + v2 / 5x103 + vo / 4x103 = 0

Step 2: Constraint equations:

vo – v1 = 30

vo – v2 = 20

Step 3: Solving the 3 equations gives:

vo = 20 V

v1 = – 10 V

v2 = 0 V


27

5.4 Dependent SourcesDependent sources require constraint equations.

Find v1, v2,

and v3.

4 vx3/2 A4

1

7 A

v3v1v2

2

1/2

– vx +1 2 3+

v1–

KCL @ node 1: v1 / 2 + (v1–v2) / 1 – 4vx + 7 = 0

KCL @ node 2: (v2–v1) / 1 + v2 / 4 + (v2–v3) /(1/2) = 0

KCL @ node 3: (v3–v2) /(1/2) – 3/2 – 7 = 0

constraint equation: vx = v3 – v2

Four equations in four unknowns. The solution is:

v1 = 24 V, v2 = 30.25 V,

v2 = 26 V, vx = 4.25 V

Example 1.

Solution:


28

15 A3

2 i1

v1v2

2 4 i2i1

Find i1 and i2.

KCL @ node 1: v1 / 2 + 2 i1 + (v1–v2) / 3 = 15

KCL @ node 2: (v2–v1) / 3 + v2 / 4 – 2 i1 = 0

constraint: v1 / 2 = i1

Solving the 3 simultaneous equations in v1, v2, i1,

and then using the relationship i2 = v2 / 4 gives:

i1 = 7 A

i2 = 8 A

Example 2.

Solution:


29

6 A

1

1/2 – vx +

1/2 1/3

3 vx

+ –

30 V

1/4 Find vx.

First, redraw to emphasize the nodes. Then choose a reference node and label the node voltages. Then write the KCL and constraint equations.

+ –

30 V

1/2

1/2 1/3

3 vx

v1– 30 V

v2

– vx

+

1 6 A

1/4 Note that for the chosen reference node there is no supernode present, but one of the node voltages is now known to be –30 V.

Example 3.

Solution:


30

+ –

30 V

1/2

1/2 1/3

3 vx

v1–30 V

v2–

vx +

16 A

1/4

Solution (cont.):

node 1: – vx / 1 + v1 / (1/2) + 3 vx = 0

node 2: vx / 1 + v2 / (1/2) + (v2+30)/(1/4) – 6 = 0

constraint: v2 – v1 = vx

Solving: v1 = –12 V

v2 = 18 v

vx = 6 v

node 2

node 1


31

(Contains dependent sources and a supernode, too!)

Find v and v1.

6 A

6 V

+ –

4 i1

1

1 +

v

–

– +

4 1.5 v1

i1

+

v1–

2

6 A

6 V

+ –

4 i1

1

1 +

v

–

– +

4 1.5 v1

i1

+

v1

–

2

First, identify the nodes and supernode. There is one supernode and two regular nodes (including the reference node).

v va vb

vc

supernode

node c

reference node

Example 5.

Solution:


32

Solution (cont.):

KCL @ supernode:

v / 1 + va / 4 – 1.5v1 + (vb –vc) / 1 – 6 = 0

KCL @ node c:

(vc –vb) / 1 + vc / 2 + 6 = 0

Constraint equations: v – va = 6vb – va = 4 i1

i1 = – v / 1vc – vb = v1

Solving:

v = – 2 V, v1 = – 4 V

6 unknowns:

v

v1

va

vb

vc

i1

Could also

write –i1 here

6 A

6 V

+ –

4 i1

1

1 +

v

–

– +

4 1.5 v1

i1

+

v1–

2

v va vb

vc

supernode

node c

reference node


33

Find v

Example 6.

8 V

+ –

6 i– +

2

i

4 V

4 A

2

+–

2

+ v –

Try working this out on your own. If you select the reference node carefully you can save yourself some effort.

Solution:

Date post:	30-Dec-2015
Category:	Documents
Upload:	abel-lambert-york
View:	227 times
Download:	1 times

EENG 2009 FS 2006 Part 5: Nodal Analysis 146 EENG 2009 Part 5. Nodal Analysis 5.1 Nodes and Node...

Documents