NODAL ANALYSIS

In this method, one of the nodes is taken as the reference node and the other as independent nodes. The voltages at the different independent nodes are assumed and the equations are written for each node as per KCL. After solving these equations, the node voltages are determined. Then, the branch currents are determined.

Consider a circuit shown in Figure 2.49, where D and B are the two independents nodes. Let D be the reference node and the voltage of node B be V_B.

Fig. 2.49  Network with Node B and D

According to KCL,

 

I₁+ I₂ = I₃        (2.40)

In mesh ABDA, the potential difference across R₁ is E₁− V_B

In mesh BCDB, the potential difference across R₂ is E₂− V_B

Further, current, 

Substituting these values in Equation (2.40), we get

Rearranging the terms,

Since all other value are known, except V_B, calculate the value of V_B. Then, determine the value of I₁, I₂, and I₃. This method is faster as the result are obtained by solving lesser number of equations.

Example 2.17

Find the current I₁ and I₂ in the passive elements of the network shown in Figure 2.50.

(U.P.T.U. Tut.)

Fig. 2.50  Given network

Solution:

The independent nodes are A, B, and C. Let C be the reference node and V_A and V_B be the voltages at node A and B, respectively. Let us assume the direction of flow of current as in Figure 2.51.

Fig. 2.51  Assumed direction of flow of current in various branches

For node A,  and , assuming V_A > V_B

Similarly for node B,    and   

Now, by applying KCL to node A, we get

 

I₁ = I₄+ I₃

or

By applying KCL at node B, we get

 

I₂ + I₄ = I₅

or

 

Solving Equations (2.41) and (2.42), we get

 

V_A = 9.25 V   and   V_B = 11 V

Current 

Example 2.18

Two batteries A and B are connected in parallel to a load of 10 Ω. Battery A has an emf of 12 V and an internal resistance of 2 Ω and battery B has an emf of 10 V and internal resistance of 1 Ω. Using nodal analysis, determine the currents supplied by each battery and load current.

(U.P.T.U. Dec. 2003)

Fig. 2.52  Given network

Solution:

Considering node Z as reference node and the potentials of nodes X and Y be V_X and V_Y, respectively. The assumed current distribution is shown in Figure 2.52.

For node X,

For node Y,

and

By applying KCL to node B, we get

I₁+ I₂ = I₃    or           (2.43)

Moreover,

V_X = V_Y        (2.44)

Solving Equations (2.43) and (2.44), we get

 

V_X = V_Y = 10 V

Thus, current supplied by battery 

Current supplied by battery 

Load current, 

Example 2.19

Using nodal analysis, find current I through 10-Ω resistor in Figure 2.53.

Fig. 2.53  Given network

Solution:

The independent nodes are A, B, and C. Let C be the reference node and V_A and V_B be the voltages at node A and B, respectively. Let us assume the direction of flow of current is as marked in Figure 2.54. By applying KCL at node A, we get

Fig.2.54  Assumed direction of flow of current in various branches

I₁+ I₂ = I

or

 

−5V_A + 60 − 4V_A = 2V_A − 2V_B   or   11V_A −2V_B = 60         (2.45)

By applying KCL at node B, we get

 

I = I₄+ I₃

or

 

12V_A − 12V_B = 30V_B − 900 + 20V_B

or

 

12V_A − 62V_B = −900        (2.46)

Solving Equation (2.45) and (2.46), we get

 

V_A = 8.39 V   and   V_B = 16.14 V

Current,

I = 0.775 A (from B to A)

Example 2.20

Calculate currents in all the resistors of the circuit shown in Figure 2.55 using node analysis method.

(U.P.T.U. 2006–07)

Fig. 2.55  Given network

Solution:

The independent nodes are A, B, and C. Let C be the reference node and V_A and V_B be the voltages at node A and B, respectively. Let us assume the direction of flow of current is marked as in Figure 2.56.

Fig. 2.56  Assumed direction of flow of current in various branches

Here,

V_A = 6 V

By applying KCL at node B, we get

Current in 12-Ω resistor,

Current in 2-Ω resistor,

Current in 3-Ω resistor,

Example 2.21

Use nodal analysis to find the current in various resistors of the circuit shown in Figure 2.57.

(U.P.T.U. 2005–06)

Fig. 2.57  Given network

Solution:

The independent nodes are A, B, C, and D. Let D be the reference node and V_A, V_B, and V_C be the voltages at nodes A, B, and C, respectively, The current flowing through various branches are marked in Figure 2.58.

Fig. 2.58  Assumed direction of flow of current in various branches

By applying KCL at different nodes, different node voltage equations are obtained as follows:

Node A

 

I₁ + I₂ + I₃ = I

15V_A+ 10(V_A−V_B) + 6(V_A−V_C) = 300   or   31V_A − 10V_B − 6V_C = 300        (2.47)

Node B

 

I₂ − I₄ − I₅ = 0

5(V_A − V_B) − 15(V_B − V_C) − 3V_B= 0   or   5V_A − 23V_B + 15V_C = 0        (2.48)

Node C

 

I₃ + I₄ − I₆− I₇ = 0

4(V_A − V_c) + 20(V_B − V_C) − 5V_c − 40 = 0   or   4V_A + 20V_B − 29V_c = 40        (2.49)

The three equations in matrices form are:

Current in various resistors:

Example 2.22

Using nodal analysis, determine current in each branch of the network as shown in Figure 2.59. Further, find total power loss in the network.

(U.P.T.U. Feb. 2002)

Fig. 2.59  Given network

Solution:

Redraw the circuit and mark the arbitrary assumed values of currents in various branches as shown in Figure 2.60. Let G (or D) be the reference node.

Fig. 2.60  Assumed direction of flow of current in various branches

By applying KCL at different nodes, we get

At node A

 

I₁ = I₂ + I₃

or

At node B

I₃ = I₄+ I₅   or   

or

2V_A − 4V_B + V_C = 10         (2.51)

 

At node C

 

I₅ + I₇ = I_C

or

Solving Equations (2.50), (2.51), and (2.52), we get

 

V_A = 6 V, V_B = 2 V, and V_C = 6 V

Current through different branches:

Current through current source, I₁ = 1 A

Current through branch AG (10-Ω resistor), 

Current through branch AB (10-Ω resistor), 

Current through branch BG (20-Ω resistor), 

Current through branch BC (20-Ω resistor), 

Current through branch CG (20-Ω resistor), 

Current through current source, I₇ = 0.5 A

Total power loss = (0.6)² × 10 + (0.4)² × 10 + (0.6)² × 20 + (0.2)² × 20 + (0.3)² × 20

= 3.6 + 1.6 + 7.2 + 0.8 + 1.8 = 15 W

Example 2.23

Use the node voltage method to solve the mesh currents in the network shown in Figure 2.61.

(U.P.T.U. June 2001)

Fig. 2.61  Given network

Solution:

Redraw the circuit and mark the arbitrary assumed values of currents in various branches, as shown in Figure 2.62. Let C be the reference node.

Fig. 2.62  Assumed direction of flow of current in various branches

By applying KCL at different nodes, we get

At node A:  considering V_A > V_B

or

 

8V_A − V_B = 50        (2.53)

At node B: 

or

 

2V_A − 17 V_B = −500        (2.54)

Multiplying Equation (2.54) by 4 and subtracting it from Equation (2.53), we get

 

V_A = 10.0746 V

Substituting the value of V_A in Equation (2.53), we get

 

V_B = 30.597 V

Various currents of the network

and

Example 2.24

By applying KCL, determine current I_s in the electric circuit at Figure 2.63. Take V₀ = 16 V

(U.P.T.U. Feb. 2001)

Fig. 2.63  Given network

Solution:

Let the current flowing through the various branches of the circuit be as shown in Figure 2.64.

Fig. 2.64  Assumed direction of flow of current in various branches

By applying KCL to node B, we get

 

I₂ + I_S = I₁        (2.55)

By applying KCL to node C, we get

Voltage at node C = V₀ = 16 V

 

4I₂ + V₁ = 16        (2.57)

 

In branch BG,    or   V₁ = 6I₁        (2.58)

In branch DE, 

Substituting the value of I₃ and V₁ in Equation (2.56), we get

Substituting the value of V₁ = 6I₁, we get

 

4I₂ + 6I₁ = 16   or   3I₁ + 2I₂ = 8         (2.60)

Solving Equations (2.59) and (2.60), we get

 

6I₁ = 12   or   I₁ = 2 A   and   I₂ = 1 A

From Equation (2.55), we get I_S = I₁ – I₂ = 2 – 1 = 1 A

whereas V₁ = 6I₁ = 6 × 2 = 12 V