Electric Circuits I - Philadelphia University Circuits I Course/4 Nodal... · Dr. Firas Obeidat...

Electric Circuits INodal Analysis

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Dr. Firas Obeidat

Dr. Firas Obeidat – Philadelphia University

Nodal analysis, which is based on a systematic application ofKirchhoff’s current law (KCL).

A node is defined as a junction of two or more branches.

define one node of any network as a reference (that is, a pointof zero potential or ground), the remaining nodes of thenetwork will all have a fixed potential relative to thisreference.

For a network of N nodes, therefore, there will exist (N-1)nodes with a fixed potential relative to the assigned referencenode.

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Nodal Analysis Without Voltage Source


Steps to Determine Node Voltages

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1- Determine the number of nodes within the network.2- Select a node as the reference node. Assign voltages v1,v2,v3,…,vn-1 to theremaining nodes. The voltages are referenced with respect to the referencenode.3- Apply KCL to each of the non-reference nodes. Use Ohm’s law to expressthe branch currents in terms of node voltages. Assume that all unknown currents leave the node for each

application of Kirchhoff’s current law. In other words, for eachnode, don’t be influenced by the direction that an unknowncurrent for another node may have had. Each node is to be treatedas a separate entity, independent of the application of Kirchhoff’scurrent law to the other nodes.

Current flows from a higher potential to a lower potential in a resistor.


Steps to Determine Node Voltages

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This principle can be expressed as

� =�� − ��

�4- Solve the resulting simultaneous equations to obtain theunknown node voltages.

Example: Calculate the node voltages in the circuit?

−5 +��

�+

��

�=0 (1)

Node 1

5 +��

�+

��

�-10=0 (3)

Node 2

Multiplying each term by 4, we obtain

3��−��= 20(2)

Dr. Firas Obeidat – Philadelphia University 5


Multiplying each term by 12, we obtain

-3��+5��= 60(4)

Using the elimination technique to solve equations (2) and (4) to get v1 and v2

3��−��= 20(2)

-3��+5��= 60(4)

4�� = 80

�� = 20�

Substituting v2 in equation (2)

3�� − 20 = 20(2)

�� =40

3= 13.33�




−3 +��

�+

��

�=0 (1)

Node 1

Multiplying by 4 and rearranging terms

3�� −2�� −��= 12(2)

��

�+

��

�+

��

�=0 (3)

Node 2


−4�� +7�� −��= 0(4)

2�� +��

�+

��

�=0 (5)

Node 3




6�� −9�� +3��= 0

�� =��

�(6)

Substitute eq.(6) in eq.(5)

2(��

�) +

��

�+

��

�=0 (7)

Solve equations (2), (4), and (8) usingelimination technique to get v1, v2 and v3

Add eq.(2) to eq.(8)

3�� −2�� −��= 12(2)

2�� −3�� +��= 0(8)

5��−5��= 12

2�� −3�� +��= 0(8)

��−��=12

5= 2.4(9)

Add eq.(4) to eq.(8)

2�� −3�� +��= 0(8)

−4�� +7�� −��= 0(4)

−2��+4��= 0

�� = 2�� (10)

Substitute eq.(10) in eq.(9)

2��−��= 2.4

�� = 2.4� ( 11)

�� = 2��= 2 × 2.4 = 4.8�(12)

Substitute eq.(11) in eq.(12) in eq.(8)

2 × 4.8 − 3 × 2.4+��= 0

�� = −2. 4�(8)




−5 +��

�.�+

��

�.�+

��

�.�=0 (1)

Node 1

9.5�� −2.5�� −5��= 5(2)

5 +��

�.�+

��

�.�+

��

�.�=0 (3)

Node 2

−2.5�� +14.5�� −10��= −5(4)

−12 +��

�.��+

��

�.�+

��

�.�=0 (5)

Node 2

−5�� −10�� +19��= 12(6)

Solve eq.(2), eq.(4) and eq.(6) to get v1, v2, and v3.


Nodal Analysis With Voltage Source

CASE 1: If a voltage source is connectedbetween the reference node and a non-reference node, we simply set the voltage atthe non-reference node equal to the voltageof the voltage source. In the figure forexample, v1=10V.

CASE 2: If the voltage source (dependentor independent) is connected between twonon-reference nodes, the two non-referencenodes form a generalized node orsupernode; we apply both KCL and KVL todetermine the node voltages.

A supernode is formed by enclosing a (dependent or independent) voltagesource connected between two non-reference nodes and any elementsconnected in parallel with it.



In the figure shown, nodes 2 and 3 form asupernode.

��−��= 5

��

�+

��

�+

��

�+

��

�=0

Applying KCL on supernode, then we get

Applying KVL on supernode, then we get

Note the following properties of asupernode:1. The voltage source inside the supernodeprovides a constraint equation needed tosolve for the node voltages.2. A supernode has no voltage of its own.3. A supernode requires the application ofboth KCL and KVL.



Example: For the circuit shown, find the node voltages.

The supernode contains the 2V source, nodes1 and 2, and the 10Ω resistor.

-2+��

�+

��

�+ 7 = 0

-8+2�� + �� + 28 = 0

�� = −20 − 2�� (1)

��−��=2 or �� = �� +2 (2)

Applying KCL to the supernode gives

Applying KVL to the supernode gives

From eq.(1) and eq.(2)

�� = �� +2 = −20 − 2��

3��=-22 ��=-7.33V

��== �� +2=-7.33+2=-5.33V

Note that the 10- resistor does notmake any difference because it isconnected across the supernode.



Example: For the circuit shown, find the node voltages.

8 + 3 +��

�+

��

�=0

Node 1

0.5833�� −0.3333�� −0.25��= −11(1)

−3 +��

�+

��

�− 25 +

��

�+

��

�=0

Supernode

−0.5833�� +1.3333�� +0.45��= 28(2)

��−��= 22(3)

Solve eq.(1), eq.(2) and eq.(3) to get v1, v2, and v3.



Example: For the circuit shown, find thenode voltages.

Node 1

Supernode

�� = 18�(2)

Node 2

��−��= 12(3)

9.5�� −2.5�� −5��= 5(1)

�� − 18 = 12

�� = 30�

Substitute v2 and v3 in eq.(1)

9.5×�� − 2.5 × 18 − 5 × 30 = 5

�� = 21.5�



Example: For the circuit shown, find thenode voltages.

−14 +��

�.�+

��

�=0 (2)

Node 1

−0.5�� +��

�+

��

�+

��

�.�=0 (3)

Supernode

��−��= 0.2��(4)

�� = −12�(1)

Node 2

��−��= 0.2(��−��)(4)

−0.5(��−��) +��

�+

��

�+

��

�.�=0 (3)

Rearrange these four equations thenwe have the following equations

−2�� + 2.5��−0.5��= 14(6)

0.1�� − �� +0.5�� +1.4��= 0(7)

�� = −12�(5)

0.2�� +�� −1.2��= 0(8)

�� = −12� �� = −4�

�� = 0� �� = −42�

Electric Circuits I - Philadelphia University Circuits I Course/4 Nodal... · Dr. Firas Obeidat...

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