Nodal Analysis Collection 2

8.6 Nodal Analysis

In the previous section we applied Kirchhoff’s voltage law to arrive at loop

currents in a network. In this section we will apply Kirchhoff’s current law

to determine the potential difference (voltage) at any node with respect to

some arbitrary reference point in a network. Once the potentials of all nodes

are known, it is a simple matter to determine other quantities such as current

and power within the network.

The steps used in solving a circuit using nodal analysis are as follows:

1. Arbitrarily assign a reference node within the circuit and indicate this

node as ground. The reference node is usually located at the bottom of

the circuit, although it may be located anywhere.

2. Convert each voltage source in the network to its equivalent current

source. This step, although not absolutely necessary, makes further calcu-

lations easier to understand.

3. Arbitrarily assign voltages (V1, V2, . . . , Vn) to the remaining nodes in

the circuit. (Remember that you have already assigned a reference node,

so these voltages will all be with respect to the chosen reference.)

4. Arbitrarily assign a current direction to each branch in which there is no

current source. Using the assigned current directions, indicate the corre-

sponding polarities of the voltage drops on all resistors.

5. With the exception of the reference node (ground), apply Kirchhoff’s cur-

rent law at each of the nodes. If a circuit has a total of n� 1 nodes (includ-

ing the reference node), there will be n simultaneous linear equations.

6. Rewrite each of the arbitrarily assigned currents in terms of the potential

difference across a known resistance.

7. Solve the resulting simultaneous linear equations for the voltages (V1, V2,

. . . , Vn).

288 Chapter 8 ■ Methods of Analysis

EXAMPLE 8–14 Given the circuit of Figure 8–30, use nodal analysis to

solve for the voltage Vab.

I2 = 50 mA

20 �

R1

30 �R3I3

200 mA

I1

R2

40 �

R3

E

30 �

a

b

6 V

200 mA

FIGURE 8–30

Section 8.6 ■ Nodal Analysis 289

Steps 3 and 4: Arbitrarily assign node voltages and branch currents. Indicate

the voltage polarities across all resistors according to the assumed current

directions.

Step 5: We now apply Kirchhoff’s current law at the nodes labelled as V1 and

V2:

Node V1: Ientering Ileaving

200 mA � 50 mA I1 � I2

Node V2: Ientering Ileaving

200 mA � I2 50 mA � I3

Step 6: The currents are rewritten in terms of the voltages across the resistors

as follows:

I1 !20

V1

�!

I2 !V

4

1

0

"

�

V2!

I3 !30

V2

�!

The nodal equations become

200 mA � 50 mA !20

V1

�!�!

V

4

1

0

"

�

V2!

200 mA �!V

4

1

0

"

�

V2! 50 mA �!

30

V2

�!

FIGURE 8–31

R1

V1 V2

200 mA 200 mA

50 mA

(Reference)

20 � R3 30 �

40 �

I1

I2

! "

!

"

I3

!

"

R2

Solution

Step 1: Select a convenient reference node.

Step 2: Convert the voltage sources into equivalent current sources. The

equivalent circuit is shown in Figure 8–31.


Substituting the voltage expressions into the original nodal equations, we

have the following simultaneous linear equations:

�!20

1

�!�!

40

1

�! V1 " �!40

1

�! V2 250 mA

"�!40

1

�! V1 � �!30

1

�!�!

40

1

�! V2 150 mA

These may be further simplified as

(0.075 S)V1 " (0.025 S)V2 250 mA

"(0.025 S)V1 � (0.0583!)V2 150 mA

Step 7: Use determinants to solve for the nodal voltages as

" "0.250 "0.025

0.150 0.0583!

!0

0

.

.

0

0

0

1

3

8

7

3!5! 4.89 V

(0.250)(0.0583!) " (0.150)(" 0.025)!!!!(0.075)(0.0583!) " ("0.025)("0.025)

V1

" "0.075 "0.025

0.025 0.0583!

If we go back to the original circuit of Figure 8–30, we see that the voltage V2

is the same as the voltage Va, namely

Va 4.67 V 6.0 V � Vab

Therefore, the voltage Vab is simply found as

Vab 4.67 V " 6.0 V "1.33 V

" "0.075 0.250

"0.025 0.150V2

" "

!0

0

.

.

0

0

0

1

3

7

7

5

5! 4.67 V

(0.075)(0.150) " ("0.025)(0.250)!!!!

0.00375

0.075 0.0255

"0.025 0.0583!

and


EXAMPLE 8–15 Determine the nodal voltages for the circuit shown in

Figure 8–32.

Solution By following the steps outlined, the circuit may be redrawn as

shown in Figure 8–33.

2 A5 �R1

R2 = 3 �

3 A6 �

4 �R3

R46 �

18 V

FIGURE 8–32

FIGURE 8–33

V1 V2

2 A 3 A

(reference)

6 �R3 4 �

R2 = 3 �

I2I4

I3

R1 5 �

I1

Applying Kirchhoff’s current law to the nodes corresponding to V1 and V2,

the following nodal equations are obtained:

Ileaving Ientering

Node V1: I1 � I2 2 A

Node V2: I3 � I4 I2 � 3 A

The currents may once again be written in terms of the voltages across the

resistors:

I1 !5

V

�

1!

I2 !V1

3

"

�

V2!

I3 !4

V

�

2!

I4 !6

V

�

2!


The nodal equations become

Node V1: !5

V

�

1!�!

(V1

3

"

�

V2)! 2 A

Node V2: !4

V

�

2!�!

6

V

�

2! !

(V1

3

"

�

V2)!� 3 A

These equations may now be simplified as

Node V1: �!51

�!�!

3

1

�! V1 " �!3

1

�! V2 2 A

Node V2: "�!31

�! V1 � �!4

1

�!�!

6

1

�!�!

3

1

�! V2 3 A

The solutions for V1 and V2 are found using determinants:

" "2 "0.333

3 0.750V1 !2

0

.

.

5

2

0

8

0

9! 8.65 V

" "0.533 "0.333

"0.333 0.750

" "0.533 2

"0.333 3V2 !

2

0

.

.

2

2

6

8

7

9! 7.85 V

" "0.533 "0.333

"0.333 0.750

In the previous two examples, you may have noticed that the simultaneous

linear equations have a format similar to that developed for mesh analysis.

When we wrote the nodal equation for node V1 the coefficient for the vari-

able V1 was positive, and it had a magnitude given by the summation of the

conductance attached to this node. The coefficient for the variable V2 was

negative and had a magnitude given by the mutual conductance between

nodes V1 and V2.

Format Approach

A simple format approach may be used to write the nodal equations for any

network having n � 1 nodes. Where one of these nodes is denoted as the ref-

erence node, there will be n simultaneous linear equations which will appear

as follows:

G11V1 " G12V2 " G13V3 "… " R1nVn I1

"G21V1 � G22V2 " G23V3 "… " R2nVn I2

"Gn1V1 " Gn2V2 " Gn3V3 "… � RnnVn In

The coefficients (constants) G11, G22, G33, . . . , Gnn represent the sum-

mation of the conductances attached to the particular node. The remaining

coefficients are called the mutual conductance terms. For example, the

mutual conductance G23 is the conductance attached to node V2, which is

common to node V3. If there is no conductance that is common to two nodes,

then this term would be zero. Notice that the terms G11, G22, G33, . . . , Gnn

are positive and that the mutual conductance terms are negative. Further, if

the equations are written correctly, then the terms will be symmetrical about

the principal diagonal, e.g., G23 G32.

The terms V1, V2, . . . , Vn are the unknown node voltages. Each voltage

represents the potential difference between the node in question and the ref-

erence node.

The terms I1, I2, . . . , In are the summation of current sources entering

the node. If a current source has a current such that it is leaving the node,

then the current is simply assigned as negative. If a particular current source

is shared between two nodes, then this current must be included in both

nodal equations.

The method used in applying the format approach of nodal analysis is as

follows:

1. Convert voltage sources into equivalent current sources.

2. Label the reference node as . Label the remaining nodes as V1, V2, . . . , Vn.

3. Write the linear equation for each node using the format outlined.

4. Solve the resulting simultaneous linear equations for V1, V2, . . . , Vn.

The next examples illustrate how the format approach is used to solve

circuit problems.


EXAMPLE 8–16 Determine the nodal voltages for the circuit shown in

Figure 8–34.

Solution The circuit has a total of three nodes: the reference node (at a

potential of zero volts) and two other nodes, V1 and V2.

By applying the format approach for writing the nodal equations, we get

two equations:

Node V1: �!31

�!�!

5

1

�! V1 " �!5

1

�! V2 "6 A � 1 A

Node V2: "�!51

�! V1 � �!5

1

�!�!

4

1

�! V2 "1 A " 2 A

R1

V1 V2I2

I1 6 A I3 2 A

1 A

3 � R3 4 �

R2 = 5 �

FIGURE 8–34


EXAMPLE 8–17 Use nodal analysis to find the nodal voltages for the cir-

cuit of Figure 8–35. Use the answers to solve for the current through R1.

Solution In order to apply nodal analysis, we must first convert the voltage

source into its equivalent current source. The resulting circuit is shown in

Figure 8–36.

2 mA 5 k�

3 mA

3 k�5 k�

10 V

2 k�R4

2 mA

I4R2

R3 = 4 k�

R1

I2

FIGURE 8–35

On the right-hand sides of the above, those currents that are leaving the nodes

are given a negative sign.

These equations may be rewritten as

Node V1: (0.533 S)V1 " (0.200 S)V2 "5 A

Node V2: "(0.200 S)V1 � (0.450 S)V2 "3 A

Using determinants to solve these equations, we have

" ""5 "0.200

"3 0.450V1 !

"

0.

2

2

.

0

8

0

5! "14.3 V

" "0.533 "0.200

"0.200 0.450

" "0.533 "5

"0.200 "3V2 !

"

0.

2

2

.

0

6

0

0! "13.0 V

" "0.533 "0.200

"0.200 0.450


R4

V2V1

2 k�R2

R3 = 4 k�

3 k�5 k�

I4 2 mAI2 3 mA

2 mA

FIGURE 8–36

Labelling the nodes and writing the nodal equations, we obtain the following:

Node V1: �!5 k

1

�!�!

3 k

1

�!�!

4 k

1

�! V1 " �!4 k

1

�! V2 2 mA " 3 mA

Node V2: "�!4 k

1

�! V1 � �!4 k

1

�!�!

2 k

1

�! V2 2 mA

Because it is inconvenient to use kilohms and milliamps throughout our cal-

culations, we may eliminate these units in our calculations. You have already

seen that any voltage obtained by using these quantities will result in the units

being “volts.” Therefore the nodal equations may be simplified as

Node V1: (0.7833)V1 " (0.2500)V2 "1

Node V2: "(0.2500)V1 � (0.750)V2 2

The solutions are as follows:

" ""1 "0.250

2 0.750V1 !

"

0

0

.5

.2

2

5

5

0! "0.476 V

" "0.7833 "0.250

"0.2503 0.750

" "0.7833 "1

"0.2503 2V2 !

1

0

.3

.5

1

2

6

5

7! 2.51 V

" "0.7833 "0.250

"0.2503 0.750

Using the values derived for the nodal voltages, it is now possible to solve for

any other quantities in the circuit. To determine the current through resistor

R1 5 k�, we first reassemble the circuit as it appeared originally. Since the

node voltage V1 is the same in both circuits, we use it in determining the

desired current. The resistor may be isolated as shown in Figure 8–37.

V1 = "0.476 V

10 V

5 k�R1I

FIGURE 8–37

A common mistake is that the

current is determined by using

the equivalent circuit rather than

the original circuit. You must

remember that the circuits are

only equivalent external to the

conversion.

NOTES...

Answers: V1 3.00 V, V2 6.00 V, V3 "2.00 V


The current is easily found as

I 10 V "!("0

5

.4

k

7

�

6 V)! 2.10 mA (upward)

PRACTICE

PROBLEMS 4

Use nodal analysis to determine the node voltages for the circuit of Figure 8–38.

1 �

9 V

1 �2 �

8 �

4 A

3 A

V2

V3V1

FIGURE 8–38

60 �

10 �

90 �

15 �

30 �30 V

30 �

I

FIGURE 8–39

8.7 Delta-Wye (Pi-Tee) Conversion

Delta-Wye Conversion

You have previously examined resistor networks involving series, parallel,

and series-parallel combinations. We will next examine networks which

cannot be placed into any of the above categories. While these circuits

may be analyzed using techniques developed earlier in this chapter, there

is an easier approach. For example, consider the circuit shown in Figure

8–39.

This circuit could be analyzed using mesh analysis. However, you

see that the analysis would involve solving four simultaneous linear

equations, since there are four separate loops in the circuit. If we were

to use nodal analysis, the solution would require determining three node

voltages, since there are three nodes in addition to a reference node. Unless

a computer is used, both techniques are very time-consuming and prone to

error.

As you have already seen, it is occasionally easier to examine a circuit

after it has been converted to some equivalent form. We will now develop a

technique for converting a circuit from a delta (or pi) into an equivalent wye

(or tee) circuit. Consider the circuits shown in Figure 8–40. We start by

making the assumption that the networks shown in Figure 8–40(a) are equiv-

alent to those shown in Figure 8–40(b). Then, using this assumption, we will

determine the mathematical relationships between the various resistors in the

equivalent circuits.

The circuit of Figure 8–40(a) can be equivalent to the circuit of Figure

8–40(b) only if the resistance “seen” between any two terminals is exactly

550 Electrical Circuit Theory and Technology

31.2 Nodal analysis A node of a network is defined as a point where two or more branchesare joined. If three or more branches join at a node, then that node iscalled a principal node or junction. In Figure 31.5, points 1, 2, 3, 4 and5 are nodes, and points 1, 2 and 3 are principal nodes.A node voltage is the voltage of a particular node with respect to a

node called the reference node. If in Figure 31.5, for example, node 3 ischosen as the reference node then V13 is assumed to mean the voltageat node 1 with respect to node 3 (as distinct from V31). Similarly, V23

would be assumed to mean the voltage at node 2 with respect to node 3,and so on. However, since the node voltage is always determined withrespect to a particular chosen reference node, the notation V1 for V13 andV2 for V23 would always be used in this instance.The object of nodal analysis is to determine the values of voltages at

all the principal nodes with respect to the reference node, e.g., to findvoltages V1 and V2 in Figure 31.5. When such voltages are determined,the currents flowing in each branch can be found.Kirchhoff’s current law is applied to nodes 1 and 2 in turn in

Figure 31.5 and two equations in unknowns V1 and V2 are obtained whichmay be simultaneously solved using determinants.

Figure 31.5 Figure 31.6

The branches leading to node 1 are shown separately in Figure 31.6.Let us assume that all branch currents are leaving the node as shown.Since the sum of currents at a junction is zero,

V1 Vx

ZACV1

ZDCV1 V2

ZBD 0 ⊲1⊳

Similarly, for node 2, assuming all branch currents are leaving the nodeas shown in Figure 31.7,

V2 V1

ZBCV2

ZECV2 C VY

ZCD 0 ⊲2⊳

In equations (1) and (2), the currents are all assumed to be leaving thenode. In fact, any selection in the direction of the branch currents maybe made— the resulting equations will be identical. (For example, if fornode 1 the current flowing in ZB is considered as flowing towards node 1instead of away, then the equation for node 1 becomesFigure 31.7

Mesh-current and nodal analysis 551

V1 Vx

ZACV1

ZDDV2 V1

ZB

which if rearranged is seen to be exactly the same as equation (1).)Rearranging equations (1) and (2) gives:

(

1

ZAC

1

ZBC

1

ZD

)

V1

(

1

ZB

)

V2

(

1

ZA

)

Vx D 0 ⊲3⊳

(

1

ZB

)

V1 C

(

1

ZBC

1

ZCC

1

ZE

)

V2 C

(

1

ZC

)

VY D 0 ⊲4⊳

Equations (3) and (4) may be rewritten in terms of admittances (whereadmittance Y D l/Z ):

⊲YA C YB C YD⊳V1 YBV2 YAVx D 0 ⊲5⊳

YBV1 C ⊲YB C YC C YE⊳V2 C YCVY D 0 ⊲6⊳

Equations (5) and (6) may be solved for V1 and V2 by using determinants.Thus

V1∣

∣

∣

∣

YB YA⊲YB C YC C YE⊳ YC

∣

∣

∣

∣

D V2

∣

∣

∣

∣

⊲YA C YB C YD⊳ YA YB YC

∣

∣

∣

∣

D1

∣

∣

∣

∣

⊲YA C YB C YD⊳ YB YB ⊲YB C YC C YE⊳

∣

∣

∣

∣

Current equations, and hence voltage equations, may be written at eachprincipal node of a network with the exception of a reference node. Thenumber of equations necessary to produce a solution for a circuit is, infact, always one less than the number of principal nodes.Whether mesh-current analysis or nodal analysis is used to determine

currents in circuits depends on the number of loops and nodes the circuitcontains, Basically, the method that requires the least number of equationsis used. The method of nodal analysis is demonstrated in the followingproblems.

Problem 4. For the network shown in Figure 31.8, determine thevoltage VAB, by using nodal analysis.

Figure 31.8


Figure 31.8 contains two principal nodes (at 1 and B) and thus only one

nodal equation is required. B is taken as the reference node and the equa-

tion for node 1 is obtained as follows. Applying Kirchhoff’s current law

to node 1 gives:

IX C IY D I

i.e.,V1

16C

V1

⊲4C j3⊳D 20 6 0°

Thus V1

(

1

16C

1

4C j3

)

D 20

V1

(

0.0625C4 j3

42 C 32

)

D 20

V1⊲0.0625C 0.16 j0.12⊳ D 20

V1⊲0.2225 j0.12⊳ D 20

from which, V1 D20

⊲0.2225 j0.12⊳D

20

0.2528 6 28.34°

i.e., voltage V1 D 79.1 6 28.34° V

The current through the ⊲4C j3⊳� branch, Iy D V1/⊲4C j3⊳

Hence the voltage drop between points A and B, i.e., across the 4 �

resistance, is given by:

VAB D ⊲Iy⊳⊲4⊳ DV1⊲4⊳

⊲4C j3⊳D

79.1 6 28.34°

5 6 36.87°⊲4⊳ D 63.36 6 −8.53° V

Problem 5. Determine the value of voltage VXY shown in thecircuit of Figure 31.9.

Figure 31.9The circuit contains no principal nodes. However, if point Y is chosen as

the reference node then an equation may be written for node X assuming

that current leaves point X by both branches.

ThusVX 8 6 0°

⊲5C 4⊳CVx 8 6 90°

⊲3C j6⊳D 0

from which, VX

(

1

9C

1

3C j6

)

D8

9C

j8

3C j6

VX

(

1

9C

3 j6

32 C 62

)

D8

9Cj8⊲3 j6⊳

32 C 62


VX⊲0.1778 j0.1333⊳ D 0.8889C48C j24

45

VX⊲0.2222 6 36.86°⊳ D 1.9556C j0.5333

D 2.027 6 15.25°

Since point Y is the reference node,

voltage VX D VXY D2.027 6 15.25°

0.2222 6 36.86°D 9.12 6 6 52.11° V

Problem 6. Use nodal analysis to determine the current flowingin each branch of the network shown in Figure 31.10.

Figure 31.10This is the same problem as problem 1 of Chapter 30, page 536, whichwas solved using Kirchhoff’s laws. A comparison of methods canbe made.There are only two principal nodes in Figure 31.10 so only one nodal

equation is required. Node 2 is taken as the reference node.

The equation at node 1 is I1 C I2 C I3 D 0

i.e.,V1 100 6 0°

25CV1

20CV1 50 6 90°

10D 0

i.e.,

(

1

25C

1

20C

1

10

)

V1 100 6 0°

25

50 6 90°

10D 0

0.19 V1 D 4C j5

Thus the voltage at node 1, V1 D4C j5

0.19D 33.70 6 51.34° V

or ⊲21.05C j26.32⊳V

Hence the current in the 25 � resistance,

I1 DV1 100 6 0°

25D

21.05C j26.32 100

25

D 78.95C j26.32

25

D 3.33 6 6 161.56° A flowing away from node 1

⊲or 3.33 6 ⊲161.56° 180°⊳A D 3.33 6 6 −18.44° A flowing towardnode 1)

The current in the 20 � resistance,

I2 DV1

20D

33.70 6 51.34°

20D 1.69 6 6 51.34° A

flowing from node 1 to node 2


The current in the 10 � resistor,

I3 DV1 50 6 90°

10D

21.05C j26.32 j50

10D

21.05 j23.68

10

D 3.17 6 6 −48.36° A away from node 1

⊲or 3.17 6 ⊲ 48.36° 180°⊳ D 3.17 6 228.36° A D 3.17 6 6 131.64° A

toward node 1)

Problem 7. In the network of Figure 31.11 use nodal analysis todetermine (a) the voltage at nodes 1 and 2, (b) the current in thej4 � inductance, (c) the current in the 5 � resistance, and (d) themagnitude of the active power dissipated in the 2.5 � resistance.

Figure 31.11

(a) At node 1,V1 25 6 0°

2C

V1

j4CV1 V2

5D 0

Rearranging gives:

(

1

2C

1

j4C

1

5

)

V1

(

1

5

)

V2 25 6 0°

2D 0

i.e., ⊲0.7C j0.25⊳V1 0.2V2 12.5 D 0 ⊲1⊳

At node 2,V2 25 6 90°

2.5CV2

j4CV2 V1

5D 0

Rearranging gives:

(

1

5

)

V1 C

(

1

2.5C

1

j4C

1

5

)

V2 25 6 90°

2.5D 0

i.e., 0.2V1 C ⊲0.6 j0.25⊳V2 j10 D 0 ⊲2⊳

Thus two simultaneous equations have been formed with twounknowns, V1 and V2. Using determinants, if

⊲0.7C j0.25⊳V1 0.2V2 12.5 D 0 ⊲1⊳


and 0.2V1 C ⊲0.6 j0.25⊳V2 j10 D 0 ⊲2⊳

thenV1

∣

∣

∣

∣

∣

0.2 12.5

⊲0.6 j0.25⊳ j10

∣

∣

∣

∣

∣

D V2

∣

∣

∣

∣

∣

⊲0.7C j0.25⊳ 12.5

0.2 j10

∣

∣

∣

∣

∣

D1

∣

∣

∣

∣

∣

⊲0.7C j0.25⊳ 0.2

0.2 ⊲0.6 j0.25⊳

∣

∣

∣

∣

∣

i.e.,V1

⊲j2C 7.5 j3.125⊳D

V2

⊲ j7C 2.5 2.5⊳

D1

⊲0.42 j0.175C j0.15C 0.0625 0.04⊳

andV1

7.584 6 8.53°D

V2

7 6 90°D

1

0.443 6 3.23°

Thus voltage, V1 D7.584 6 8.53°

0.443 6 3.23°D 17.12 6 5.30° V

D 17.1 6 6 −5.3° V, correct to one decimal place,

and voltage, V2 D7 6 90°

0.443 6 3.23°D 15.80 6 93.23° V

D 15.8 6 6 93.2° V, correct to one decimal place.

(b) The current in the j4 � inductance is given by:

V2

j4D

15.80 6 93.23°

4 6 90°D 3.95 6 6 3.23° A flowing away from node 2

(c) The current in the 5 � resistance is given by:

I5 DV1 V2

5D

17.12 6 5.30° 15.80 6 93.23°

5

i.e., I5 D⊲17.05 j1.58⊳ ⊲ 0.89C j15.77⊳

5

D17.94 j17.35

5D

24.96 6 44.04°

5

D 4.99 6 6 −44.04° A flowing from node 1 to node 2

(d) The active power dissipated in the 2.5 � resistor is given by

P2.5 D ⊲I2.5⊳2⊲2.5⊳ D

(

V2 25 6 90°

2.5

)2

⊲2.5⊳

D⊲0.89C j15.77 j25⊳2

2.5D⊲9.273 6 95.51°⊳2

2.5


D85.99 6 191.02°

2.5by de Moivre’s theorem

D 34.4 6 169° W

Thus the magnitude of the active power dissipated in the 2.5 Z

resistance is 34.4 W

Problem 8. In the network shown in Figure 31.12 determine thevoltage VXY using nodal analysis.

Figure 31.12

Node 3 is taken as the reference node.

At node 1, 25 6 0° DV1

4C j3CV1 V2

5

i.e.,

(

4 j3

25C

1

5

)

V1 1

5V2 25 D 0

or ⊲0.379 6 18.43°⊳V1 0.2V2 25 D 0 ⊲1⊳

At node 2,V2

j10CV2

j20CV2 V1

5D 0

i.e., 0.2V1 C

(

1

j10C

1

j20C

1

5

)

V2 D 0

or 0.2V1 C ⊲ j0.1 j0.05C 0.2⊳V2 D 0

i.e., 0.2V1 C ⊲0.25 6 36.87°⊳V2 C 0 D 0 ⊲2⊳

Simultaneous equations (1) and (2) may be solved for V1 and V2 by using

determinants. Thus,


V1∣

∣

∣

∣

∣

0.2 25

0.25 6 36.87° 0

∣

∣

∣

∣

∣

D V2

∣

∣

∣

∣

∣

0.379 6 18.43° 25

0.2 0

∣

∣

∣

∣

∣

D1

∣

∣

∣

∣

∣

0.379 6 18.43° 0.2

0.2 0.25 6 36.87°

∣

∣

∣

∣

∣

i.e.,V1

6.25 6 36.87°D V2

5D

1

0.09475 6 55.30° 0.04

D1

0.079 6 79.85°

Hence voltage, V1 D6.25 6 36.87°

0.079 6 79.85°D 79.11 6 6 42.98° V

and voltage, V2 D5

0.079 6 79.85°D 63.29 6 6 79.85° V

The current flowing in the ⊲4C j3⊳� branch is V1/⊲4C j3⊳. Hence thevoltage between point X and node 3 is:

V1

⊲4C j3⊳⊲j3⊳ D

⊲79.116 42.98°⊳⊲3 6 90°⊳

5 6 36.87°

D 47.47 6 96.11° V

Thus the voltage

VXY D VX VY D VX V2 D 47.47 6 96.11° 63.29 6 79.85°

D 16.21 j15.10 D 22.156 6 −137° V

Problem 9. Use nodal analysis to determine the voltages at nodes2 and 3 in Figure 31.13 and hence determine the current flowingin the 2 � resistor and the power dissipated in the 3 � resistor.

This is the same problem as Problem 2 of Chapter 30, page 537, whichwas solved using Kirchhoff’s laws.In Figure 31.13, the reference node is shown at point A.

At node 1,V1 V2

1CV1

6CV1 8 V3

5D 0

i.e., 1.367V1 V2 0.2V3 1.6 D 0 ⊲1⊳

At node 2,V2

2CV2 V1

1CV2 V3

3D 0

Figure 31.13


i.e., V1 C 1.833V2 0.333V3 C 0 D 0 ⊲2⊳

At node 3,V3

4CV3 V2

3CV3 C 8 V1

5D 0

i.e., 0.2V1 0.333V2 C 0.783V3 C 1.6 D 0 ⊲3⊳

Equations (1) to (3) can be solved for V1, V2 and V3 by usingdeterminants. Hence

V1∣

∣

∣

∣

∣

∣

∣

1 0.2 1.6

1.833 0.333 0

0.333 0.783 1.6

∣

∣

∣

∣

∣

∣

∣

D V2

∣

∣

∣

∣

∣

∣

∣

1.367 0.2 1.6

1 0.333 0

0.2 0.783 1.6

∣

∣

∣

∣

∣

∣

∣

DV3

∣

∣

∣

∣

∣

∣

∣

1.367 1 1.6

1 1.833 0

0.2 0.333 1.6

∣

∣

∣

∣

∣

∣

∣

D 1

∣

∣

∣

∣

∣

∣

∣

1.367 1 0.2

1 1.833 0.333

0.2 0.333 0.783

∣

∣

∣

∣

∣

∣

∣

Solving for V2 gives: V2

1.6⊲ 0.8496⊳C 1.6⊲ 0.6552⊳

D 1

1.367⊲1.3244⊳C 1⊲ 0.8496⊳ 0.2⊲0.6996⊳

hence V2

0.31104D

1

0.82093from which, voltage,V 2 D

0.31104

0.82093

D 0.3789 V

Thus the current in the 2 Z resistor DV2

2D

0.3789

2D 0.19 A,

flowing from node 2 to node A.

Solving for V3 gives:V3

1.6⊲0.6996⊳C 1.6⊲1.5057⊳D

1

0.82093

henceV3

1.2898D

1

0.82093from which, voltage,V3 D

1.2898

0.82093

D −1.571 V

Power in the 3 Z resistor D ⊲I3⊳2⊲3⊳ D

(

V2 V3

3

)2

⊲3⊳

D⊲0.3789 ⊲ 1.571⊳⊳2

3D 1.27 W

Further problems on nodal analysis may be found in Section 31.3following, problems 10 to 15, page 560.


31.3 Further problemson mesh-current and

nodal analysis

Mesh-current analysis

1 Repeat problems 1 to 10, page 542, of Chapter 30 using mesh-

current analysis.

2 For the network shown in Figure 31.14, use mesh-current analysis

to determine the value of current I and the active power output of

the voltage source. [6.966 49.94° A; 644 W]

3 Use mesh-current analysis to determine currents I1, I2 and I3 for the

network shown in Figure 31.15.

[I1 D 8.73 6 1.37° A, I2 D 7.02 6 17.25° A,

I3 D 3.05 6 48.67° A]Figure 31.14

Figure 31.15

4 For the network shown in Figure 31.16, determine the current

flowing in the ⊲4C j3⊳� impedance. [0]

Figure 31.16

5 For the network shown in Figure 31.17, use mesh-current analysis

to determine (a) the current in the capacitor, IC, (b) the current in

the inductance, IL, (c) the p.d. across the 4 � resistance, and (d) the

total active circuit power.

[(a) 14.5 A (b) 11.5 A (c) 71.8 V (d) 2499 W]

Figure 31.17

6 Determine the value of the currents IR, IY and IB in the network

shown in Figure 31.18 by using mesh-current analysis.

[IR D 7.84 6 71.19° AI IY D 9.04 6 37.50° A;

IB D 9.89 6 168.81° A]

7 In the network of Figure 31.19, use mesh-current analysis to

determine (a) the current in the capacitor, (b) the current in the 5 �

resistance, (c) the active power output of the 15 6 0° V source, and

(d) the magnitude of the p.d. across the j2 � inductance.

[(a) 1.03 A (b) 1.48 A(c) 16.28 W (d) 3.47 V]

8 A balanced 3-phase delta-connected load is shown in Figure 31.20.

Use mesh-current analysis to determine the values of mesh currents


Figure 31.18

I1, I2 and I3 shown and hence find the line currents IR, IYand IB.

[I1 D 83 6 173.13° A, I2 D 83 6 53.13° A,I3 D 83 6 66.87° A IR D 143.86 143.13° A,IY D 143.8 6 23.13° A, IB D 143.8 6 96.87° A]

9 Use mesh-circuit analysis to determine the value of currents IA to IEin the circuit shown in Figure 31.21.

[IA D 2.40 6 52.52° A; IB D 1.02 6 46.19° A;IC D 1.39 6 57.17° A; ID D 0.67 6 15.57° A;

IE D 0.996 6 83.74° A]



Nodal analysis

10 Repeat problems 1, 2, 5, 8 and 10 on page 542 of Chapter 30, andproblems 2, 3, 5, and 9 above, using nodal analysis.


Figure 31.23

11 Determine for the network shown in Figure 31.22 the voltage atnode 1 and the voltage VAB

[V1 D 59.0 6 28.92° V; VAB D 45.3 6 10.89° V]

12 Determine the voltage VPQ in the network shown in Figure 31.23.[VPQ D 55.87 6 50.60° V]

13 Use nodal analysis to determine the currents IA, IB and IC shown inthe network of Figure 31.24.

[IA D 1.21 6 150.96° AI IB D 1.06 6 56.32° A;IC D 0.55 6 32.01° A]

Figure 31.24

14 For the network shown in Figure 31.25 determine (a) the voltages atnodes 1 and 2, (b) the current in the 40 � resistance, (c) the currentin the 20 � resistance, and (d) the magnitude of the active powerdissipated in the 10 � resistance

[(a) V1 D 88.12 6 33.86° V, V2 D 58.72 6 72.28° V(b) 2.20 6 33.86° A, away from node 1,(c) 2.80 6 118.65° A, away from node 1, (d) 223 W]


15 Determine the voltage VAB in the network of Figure 31.26, usingnodal analysis. [VAB D 54.23 6 102.52° V]