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Electric Circuit Analysis Lec4: Nodal analysis Ahsan Khawaja [email protected]. pk Lecturer Room 102 Department of Electrical Engineering

Electric Circuit - Lecture 04

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Page 1: Electric Circuit - Lecture 04

Electric Circuit Analysis

Lec4: Nodal analysis

Ahsan [email protected] 102Department of Electrical Engineering

Page 2: Electric Circuit - Lecture 04

Steps of Nodal Analysis1. Choose a reference (ground) node.2. Assign node voltages to the other nodes.3. Apply KCL to each node other than the reference

node; express currents in terms of node voltages.4. Solve the resulting system of linear equations for

the nodal voltages.

Page 3: Electric Circuit - Lecture 04

Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis.

Page 4: Electric Circuit - Lecture 04

1. Reference Node

The reference node is called the ground node where V = 0

+

V 500W

500W

1kW

500W

500WI1 I2

Page 5: Electric Circuit - Lecture 04

Steps of Nodal Analysis1. Choose a reference (ground) node.2. Assign node voltages to the other nodes.3. Apply KCL to each node other than the reference

node; express currents in terms of node voltages.4. Solve the resulting system of linear equations for

the nodal voltages.

Page 6: Electric Circuit - Lecture 04

2. Node Voltages

V1, V2, and V3 are unknowns for which we solve using KCL

500W

500W

1kW

500W

500WI1 I2

1 2 3

V1 V2 V3

Page 7: Electric Circuit - Lecture 04

Steps of Nodal Analysis1. Choose a reference (ground) node.2. Assign node voltages to the other nodes.3. Apply KCL to each node other than the reference

node; express currents in terms of node voltages.4. Solve the resulting system of linear equations for

the nodal voltages.

Page 8: Electric Circuit - Lecture 04

Currents and Node Voltages

500W

V1

500WV1 V2

50021 VV

5001V

Page 9: Electric Circuit - Lecture 04

3. KCL at Node 1

500W

500WI1

V1 V2

500500

1211

VVVI

Page 10: Electric Circuit - Lecture 04

3. KCL at Node 2

500W

1kW

500W V2 V3V1

0500k1500

32212

VVVVV

Page 11: Electric Circuit - Lecture 04

3. KCL at Node 3

2323

500500I

VVV

500W

500W

I2

V2 V3

Page 12: Electric Circuit - Lecture 04

Steps of Nodal Analysis1. Choose a reference (ground) node.2. Assign node voltages to the other nodes.3. Apply KCL to each node other than the reference

node; express currents in terms of node voltages.4. Solve the resulting system of linear equations for

the nodal voltages.

Page 13: Electric Circuit - Lecture 04

Typical circuit for nodal analysis

Page 14: Electric Circuit - Lecture 04

• Find the node voltages in the circuit shown below.

Page 15: Electric Circuit - Lecture 04

• At node 1

2

0

45 121

321

vvv

iii

Page 16: Electric Circuit - Lecture 04

• At node 2

6

0

45 212

5142

vvv

iiii

Page 17: Electric Circuit - Lecture 04

Practice

Page 18: Electric Circuit - Lecture 04

• Determine the voltage at the nodes in Fig. below

Page 19: Electric Circuit - Lecture 04

• At node 1,

243

3

2131

1

vvvv

ii x

Page 20: Electric Circuit - Lecture 04

• At node 2

4

0

8223221

32

vvvvv

iiix

Page 21: Electric Circuit - Lecture 04

• At node 3

2

)(2

84

2

213231

21

vvvvvv

iii x

Page 22: Electric Circuit - Lecture 04

• In matrix form:

0

0

3

8

3

8

9

4

38

1

8

7

2

14

1

2

1

4

3

3

2

1

v

v

v

Page 23: Electric Circuit - Lecture 04

Spot quiz…

Page 24: Electric Circuit - Lecture 04

Example:Use nodal analysis to find the voltage at each node of this circuit.

Page 25: Electric Circuit - Lecture 04

• Step 1:Identify and label, each node in the circuit. Ground has been chosen for you.

• Step 2:Write the Nodal equation for each node identified in Step 1. • Node 1:Because the voltage at node 1 is known with respect to our

reference point, the equation is:V1 = 71 volt

• Node 2:At node 2, assume all currents are leaving the node as shown in Figure 5.

Page 26: Electric Circuit - Lecture 04

All currents are assumed to be leaving node V2,

so all terms are positive.

Node 2 I (1) + I (2) + I (3) = 0

Ohm's Law I (1) = (V2 - V1)/2

I (2) = (V2 – 0)/11

I (3) = (V2 - V3)/10

Substituting: (V2 - V1)/2 + V2/11 + (V2 - V3)/10 = 0

Page 27: Electric Circuit - Lecture 04

• I (4) and I (5) are assumed to be leaving node V3, so these terms are positive. But Is (the 2A source) is entering, so it is assigned a negative sign.

Node 3 -Is + I (4) + I (5) = 0

Ohm's Law I (4) = (V3 - V2)/10

I (5) = V3/5

Substituting:-2 + (V3 - V2)/10 + V3/5 = 0

Page 28: Electric Circuit - Lecture 04

• The three equations are shown below:

Node 1: V1 = 71v

Node 2 : (V2 - V1)/2 + V2/11 + (V2 - V3)/10 = 0

Node 3: -2 + (V3 - V2)/10 + V3/5 = 0

• After solving we find that the node voltages are:

V1 = 71v

V2 = 55v

V3 = 25v

Page 29: Electric Circuit - Lecture 04

Example:Use nodal analysis to find the voltage at each node of this circuit.