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Electric Circuit - Lecture 04
1. Electric Circuit Analysis
Lec4: Nodal analysis
Ahsan Khawaja
Ahsan_khawaja@comsats.edu.pk
Lecturer
Room 102
Department of Electrical Engineering
2. Steps of Nodal Analysis
1. 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.
3. Common symbols for indicating a reference node,
(a) common ground, (b) ground, (c) chassis.
4. 1. Reference Node
The reference node is called the ground node
where V = 0
+
–
V 500
500
1k
500
500
I1 I2
5. Steps of Nodal Analysis
1. 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.
6. 2. Node Voltages
V1, V2, and V3 are unknowns for which we solve
using KCL
500
500
1k
500
500
I1 I2
1 2 3
V1 V2 V3
7. Steps of Nodal Analysis
1. 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.
11. 3. KCL at Node 3
2
323
500500
I
VVV500
500
I2
V2 V3
12. Steps of Nodal Analysis
1. 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.
25. • 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.
26. 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
27. • 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
28. • 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