Nodal analysis involves looking at a circuit and determining all the node voltages in the circuit.

**NODAL ANALYSIS:**

Nodal analysis involves looking at a circuit and determining all the node voltages in the circuit. The voltage at any given node of a circuit is the voltage drop between that node and a reference node (usually ground). Once the node voltages are known any of the currents flowing in the circuit can be determined. The node method offers an organized way of achieving this.

**Approach:**

Firstly all the nodes in the circuited are counted and identified. Secondly nodes at which the voltage is already known are listed. A set of equations based on the node voltages are formed and these equations are solved for unknown quantities. The set of equations are formed using KCL at each node. The set of simultaneous equations that is produced is then solved. Branch currents can then be found once the node voltages are known. This can be reduced to a series of steps:

Step 1: Identify the nodes

Step 2: Choose a reference node

Step 3: Identify which node voltages are known if any Step 4: Identify the BRANCH currents

Step 5: Use KCL to write an equation for each unknown node voltage Step 6: Solve the equations

This is best illustrated with an example. Find all currents and voltages in the following circuit using the node method. (In this particular case it can be solved in other ways as well)

**Step 1:**

There are four nodes in the circuit. A, B, C and D

**Step 2:**

Ground, node D is the reference node.

**Step 3:**

Node voltage B and C are unknown. Voltage at A is V and at D is 0

**Step 4:**

The currents are as shown. There are 3 different currents

**Step 5:**

I need to create two equations so I apply KCL at node B and node C

The statement of KCL for node B is as follows:

The statement of KCL for node C is as follows:

**Step 6:**

We now have two equations to solve for the two unknowns VB and VC. Solving the above two equations we get:

**Further Calculations**

The node voltages are know all known. From these we can get the branch currents by a simple application of Ohm's Law:

**Problem 1: **Find the current through each resistor of the circuit shown in fig, using nodal** **analysis

**Solution:**

**At node1,**

-I1-I2-I3 = 0 -[V1-15/1]-[V1/1][V1-V2/0.5] = 0

-V1+15-V1-2V1+2V2 = 0

4V1-2V2 = 15 --------------------- (1)

**At node2,**

I3-I4-I5 = 0

V1-V2/0.5 â€“V2/2 â€“V2-20/1 = 0

2V1-2V2-0.5V2-V2 + 20 = 0

2V1-3.5V2 = -20 ------------------ (2)

Multiplying (2) by 2 & subtracting from (1)

5V2 = 55 V2 = 11V V = 9.25V

I1 = V1-5/1 = 9.25-15 = -5.75A = 5.75 I2 = V1/1 = 9.25A

I3 = V1-V2/0.5 = -3.5A = 3.5A ÃŸ I4 = V2/2 = 5.5A

I5 = V2-20/I = 11-20/1 = -9A=9A.

**Problem 2: **For the bridge network shown in Figure determine the currents in each of the resistors.

Let the current in the 2 resistor be I1, and thens currentby law,Kirchhoffâ€™thecurrentinthe14 resistor is (I -I1). Let the current in the 32 resistor be I2 as shown in Figure Then the current in the 11 resistor is (I1 - I2) and that in the 3 resistor is (I - I1 + I2). Applying Kirchhoffâ€™s and moving in a clockwise direction as shown in Figure gives:

54 = 2I1 +11(I1 _-I2)

i.e. 13I1 - 11I2 = 54

Applying Kirchhoffâ€™s voltage law to loop 2 and direction as shown in Figure gives:

0 = 2I1 + 32I2 -14(I - I1)

However I = 8 A

Hence 0 = 2I1 + 32I2 â€“14(8 -_I1) i.e. 16I1 + 32I2 = 112

Equations (1) and (2) are simultaneous equations with two unknowns, I1 and I2.

16 * (1) gives: 208I1 _-176I2 = 864

16 * (1) gives: 208I1 _-176I2 = 864

13 * (2) gives: 208I1 + 416I2 = 1456

(4) - (3) gives: 592I2 = 592, I2 = 1 A

Substituting for I2 in (1) gives:

13I1 - 11 = 54

I1 = 65/13 = 5 A

Hence,

the current flowing in the 2 resistor = I1 = 5 A

the current flowing in the 14 resistor = I -I1 = 8 -5 = 3 A

the current flowing in the 32 resistor = I2 = 1 A

the current flowing in the 11 resistor = I1 - I2 = 5 -1 = 4 A and

the current flowing in the 3 resistor = I - I1 + I2 = 8 - 5 + 1 = 4 A

Problem 3: Determine the values of currents I, I1 and I2 shown in the network of Figure

Total circuit impedance,

ZT = 5 + (8)(j6)/8 + j6

= 5 + (j48)(8 - j6)/82 + 62

= 5 + (j384 + 288)/100

= (7.88 + j3.84) or 8.776 25.98Â° A Current I = V/ZT

= 50âˆŸ 0Â°/8.77âˆŸ 25.98Â°

= 5.7066 *âˆ’*25.98Â° A

Current I1 = I (j6/8 + j6)

= (5.702âˆŸ5.98Â°) (6âˆŸ 90Â°)/10âˆŸ 36.87Â°

= 3.426âˆŸ27.15Â° A

Current I2 = I (8/ (8 + j6)

= (5-.2570âˆŸ.98Â°)*8âˆŸ 0Â°/10âˆŸ 36.87Â°

= 4.5666 *âˆ’*62.85Â° A

[Note: I = I1 + I2-62.=85Â° 3.42 âˆŸ27.15Â° + 4.56âˆŸ

= 3.043 + j1.561 + 2.081 - j4.058

= 5.124 - j2.497 A = 5.706 -25.98Â° A

**Problem 4: **For the a.c. network shown in Figure, determine the current flowing in each branch** **using Kirchhoffâ€™s laws.

**Problem 5: **For the a.c. network shown in Figure determine, using mesh-current analysis, (a) the** **mesh currents I1 and I2 (b) the current flowing in the capacitor, and (c) the active power delivered by the 1006 0Â° V voltage source.

**Problem 6: **In the network of Figure use nodal analysis to determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 Î© inductance, (c) of the active power dissipated-10) in the 2.5 Î©

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Electrical and electronics : Circuit Theory : Nodal Analysis |

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