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Mesh-current and nodal analysis 551

V1  Vx V1 V2  V1 C D ZA ZD ZB which if rearranged is seen to be exactly the same as equation (1).) Rearranging equations (1) and (2) gives: 





1 1 1 C C ZA ZB ZD

1 ZB



V1 C





V1 



1 ZB

1 1 1 C C ZB ZC ZE





V2 

V2 C

1 Vx D 0 ZA

⊲3⊳

1 ZC

⊲4⊳









VY D 0

Equations (3) and (4) may be rewritten in terms of admittances (where admittance Y D l/Z ): ⊲Y A C YB C YD ⊳V1  YB V2  YA Vx D 0

⊲5⊳

YB V1 C ⊲Y B C YC C YE ⊳V2 C YC VY D 0

⊲6⊳

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

1    Y B   ⊲Y B C YC C YE ⊳

V

2 D   ⊲Y A C YB C YD ⊳   YB

YA  YC 



YA  YC 

1   ⊲Y C Y C Y ⊳ YB B D   A   YB ⊲Y B C YC C YE ⊳

D 

Current equations, and hence voltage equations, may be written at each principal node of a network with the exception of a reference node. The number of equations necessary to produce a solution for a circuit is, in fact, 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 circuit contains, Basically, the method that requires the least number of equations is used. The method of nodal analysis is demonstrated in the following problems. Problem 4. For the network shown in Figure 31.8, determine the voltage VAB , by using nodal analysis.

Figure 31.8

552 Electrical Circuit Theory and Technology

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 equation for node 1 is obtained as follows. Applying Kirchhoff’s current law to node 1 gives: IX C IY D I V1 V1 C ⊲4 C j3⊳ D 206 0 16

i.e.,

Thus

1 1 C 16 4 C j3



D 20

4  j3 0.0625 C 2 4 C 32



D 20

V1

V1





V1 ⊲0.0625 C 0.16  j0.12⊳ D 20 V1 ⊲0.2225  j0.12⊳ D 20 20 20 D 0.25286 28.34 ⊲0.2225  j0.12⊳

from which, V1 D

i.e., voltage V1 D 79.16 28.34 V The current through the ⊲4 C j3⊳ branch, Iy D V1 /⊲4 C j3⊳ Hence the voltage drop between points A and B, i.e., across the 4  resistance, is given by: VAB D ⊲I y ⊳⊲4⊳ D

79.16 28.34 V1 ⊲4⊳ D ⊲4⊳ D 63.36 −8.53° V ⊲4 C j3⊳ 56 36.87

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

The 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. Vx  86 90 VX  86 0 C D0 ⊲5 C 4⊳ ⊲3 C j6⊳

Thus

from which,

VX

VX

1 1 C 9 3 C j6



D

8 j8 C 9 3 C j6

3  j6 1 C 2 3 C 62 9



D

8 j8⊲3  j6⊳ C 32 C 62 9





Mesh-current and nodal analysis 553

48 C j24 45 VX ⊲0.22226 36.86 ⊳ D 1.9556 C j0.5333

VX ⊲0.1778  j0.1333⊳ D 0.8889 C

D 2.0276 15.25 Since point Y is the reference node, voltage VX D VXY D

2.0276 15.25 0.22226 36.86

D 9.126 52.11° V

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

This is the same problem as problem 1 of Chapter 30, page 536, which was solved using Kirchhoff’s laws. A comparison of methods can be 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 V1  1006 0

i.e.,

i.e.,

25 

C

V1  506 90 V1 C D0 20 10

1006 0 506 90 1 1 1 D0 C 10 V1  25  C 10 25 20 0.19 V1 D 4 C j5 

4 C j5 D 33.706 51.34 V Thus the voltage at node 1, V1 D 0.19 or ⊲21.05 C j26.32⊳V Hence the current in the 25  resistance, I1 D

V1  1006 0 25

D

21.05 C j26.32  100 25

D

78.95 C j26.32 25

D 3.336 161.56° A flowing away from node 1 ⊲or 3.336 ⊲161.56  180 ⊳A D 3.336 −18.44° A flowing toward node 1) The current in the 20  resistance, 33.706 51.34 V1 D 1.696 51.34° A I2 D 20 D 20 flowing from node 1 to node 2

554 Electrical Circuit Theory and Technology

The current in the 10  resistor, I3 D

V1  506 90 10

D

21.05  j23.68 21.05 C j26.32  j50 D 10 10

D 3.176 −48.36° A away from node 1 ⊲or 3.176 ⊲48.36  180 ⊳ D 3.176 228.36 A D 3.176 131.64° A toward node 1) Problem 7. In the network of Figure 31.11 use nodal analysis to determine (a) the voltage at nodes 1 and 2, (b) the current in the j4  inductance, (c) the current in the 5  resistance, and (d) the magnitude of the active power dissipated in the 2.5  resistance.

Figure 31.11

(a) At node 1,

V1  256 0 2

C

V1 V1  V2 C D0 j4 5

Rearranging gives: 

1 1 1 V1  C C 2 j4 5 

i.e.,

1 V  256 0 2 D0 5 2

 

⊲0.7 C j0.25⊳V1  0.2V2  12.5 D 0

At node 2,

V2  256 90 2.5

C

⊲1⊳

V2 V2  V1 C D0 j4 5

Rearranging gives: 1 V1 C  5  

i.e.,



256 90 1 1 1 D0 C C 5 V2  2.5 2.5 j4 

0.2V1 C ⊲0.6  j0.25⊳V2  j10 D 0

⊲2⊳

Thus two simultaneous equations have been formed with two unknowns, V1 and V2 . Using determinants, if ⊲0.7 C j0.25⊳V1  0.2V2  12.5 D 0

⊲1⊳

Mesh-current and nodal analysis 555

and then

0.2V1 C ⊲0.6  j0.25⊳V2  j10 D 0 V

⊲2⊳ V

1 2   D    0.2 12.5    ⊲0.7 C j0.25⊳     ⊲0.6  j0.25⊳ j10   0.2

D 



1

12.5   j10 

    ⊲0.7 C j0.25⊳ 0.2    0.2 ⊲0.6  j0.25⊳ 

i.e.,

V2 V1 D ⊲j2 C 7.5  j3.125⊳ ⊲j7 C 2.5  2.5⊳ D

and

1 ⊲0.42  j0.175 C j0.15 C 0.0625  0.04⊳

V2 V1 1 D D 0.4436 3.23 7.5846 8.53 76 90

Thus voltage, V1 D

7.5846 8.53 D 17.126 5.30 V 0.4436 3.23

D 17.16 −5.3° V, correct to one decimal place, and voltage, V2 D

76 90 D 15.806 93.23 V 0.4436 3.23

D 15.86 93.2° V, correct to one decimal place. (b) The current in the j4  inductance is given by: 15.806 93.23 V2 D D 3.956 3.23° A flowing away from node 2 j4 46 90 (c) The current in the 5  resistance is given by: 17.126 5.30  15.806 93.23 V1  V2 D 5 5 ⊲17.05  j1.58⊳  ⊲0.89 C j15.77⊳ I5 D 5 24.966 44.04 17.94  j17.35 D D 5 5 D 4.996 −44.04° A flowing from node 1 to node 2

I5 D i.e.,

(d) The active power dissipated in the 2.5  resistor is given by P2.5 D ⊲I 2.5 ⊳2 ⊲2.5⊳ D D



V2  256 90 2 2.5

⊲2.5⊳

⊲9.2736 95.51 ⊳2 ⊲0.89 C j15.77  j25⊳2 D 2.5 2.5

556 Electrical Circuit Theory and Technology

D

85.996 191.02

2.5 D 34.46 169 W

by de Moivre’s theorem

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 the voltage VXY using nodal analysis.

Figure 31.12 Node 3 is taken as the reference node. At node 1,

256 0 D 

i.e., or

1 4  j3 1 V1  V2  25 D 0 C 5 5 25

or i.e.,



⊲0.3796 18.43 ⊳V1  0.2V2  25 D 0

⊲1⊳

V2 V2 V2  V1 C C D0 j10 j20 5

At node 2,

i.e.,

V1  V2 V1 C 4 C j3 5

0.2V1 C



1 1 1 V2 D 0 C C 5 j10 j20



0.2V1 C ⊲j0.1  j0.05 C 0.2⊳V2 D 0 0.2V1 C ⊲0.256 36.87 ⊳V2 C 0 D 0

⊲2⊳

Simultaneous equations (1) and (2) may be solved for V1 and V2 by using determinants. Thus,

Mesh-current and nodal analysis 557

V

1    0.2    0.256 36.87

V

2  D  6 18.43 0.379      0.2

0

D 

0.3796

  

i.e.,

6.256



25 

25 

18.43

0.2



0 

1     6 0.25 36.87 

0.2

V2 V1 1 D D 0.094756 55.30  0.04 36.87 5 D

Hence voltage, V1 D and voltage, V2 D

1 0.0796 79.85 6.256 36.87

0.0796 79.85

D 79.116 42.98° V

5 D 63.296 79.85° V 0.0796 79.85

The current flowing in the ⊲4 C j3⊳ branch is V1 /⊲4 C j3⊳. Hence the voltage between point X and node 3 is: ⊲79.116 42.98 ⊳⊲36 90 ⊳ V1 ⊲j3⊳ D ⊲4 C j3⊳ 56 36.87 D 47.476 96.11 V Thus the voltage VXY D VX  VY D VX  V2 D 47.476 96.11  63.296 79.85 D 16.21  j15.10 D 22.156 −137° V Problem 9. Use nodal analysis to determine the voltages at nodes 2 and 3 in Figure 31.13 and hence determine the current flowing in the 2  resistor and the power dissipated in the 3  resistor. This is the same problem as Problem 2 of Chapter 30, page 537, which was solved using Kirchhoff’s laws. In Figure 31.13, the reference node is shown at point A. At node 1, i.e., Figure 31.13

At node 2,

V1  V2 V1 V1  8  V3 D0 C C 5 1 6 1.367V1  V2  0.2V3  1.6 D 0 V2 V2  V1 V2  V3 C C D0 2 1 3

⊲1⊳

558 Electrical Circuit Theory and Technology

i.e.,

V1 C 1.833V2  0.333V3 C 0 D 0

At node 3,

V3 V3  V2 V3 C 8  V1 C D0 C 4 5 3

i.e.,

⊲2⊳

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 using determinants. Hence   1    1.833   0.333

D      

V1 0.2 0.333 0.783

D   1.367 1.6      1 0     0.2  1.6

V3 1.367 1 1 1.833 0.2 0.333

Solving for V2 gives:

V2 0.31104

D

1

D   1.367 1 0.2    1 1. 833 0 .333   0.2 0.333 0.783

1.6   0  1.6 

V2 1.6⊲0.8496⊳ C 1.6⊲0.6552⊳ D

hence

V2  0.2 1.6   0.333 0   0.783 1.6 

      

1 1.367⊲1.3244⊳ C 1⊲0.8496⊳  0.2⊲0.6996⊳

0.31104 1 from which, voltage, V 2 D 0.82093 0.82093 D 0.3789 V

V2 0.3789 D 0.19 A, Thus the current in the 2 Z resistor D D 2 2 flowing from node 2 to node A. Solving for V3 gives: hence

V3 1.2898

D

1 V3 D 1.6⊲0.6996⊳ C 1.6⊲1.5057⊳ 0.82093

1 1.2898 from which, voltage,V3 D 0.82093 0.82093 D −1.571 V

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



V2  V3 3

2

⊲3⊳

⊲0.3789  ⊲1.571⊳⊳2 D 1.27 W 3

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