Basic Electric Circuits Session 8B

Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 2
Session 8B: Focus
 Example Problems
◦ Writing Nodal Equations
◦ Nodal Analysis
 Summary of Steps: Nodal Analysis
 Super Node
◦ Example with Super Node
 Home Work Problems

Nodal Equations: Examples

Example 1: Nodal Equations
 KCL for node V1 :
 Write the nodal equations:
 KCL for node V2 :
Note: This is problem E3.1 on page 110, Figure E3.1 in the Ref 2 book (by Irwin)

Example 2: Nodal Analysis
KCL for node v1 :
 Find the power supplied by the dependent source :
KCL for node v2 :
3i1 = v2 /3 + (v2 - v1)/ 115 = (v1 – v2 )/1 + v1 /2
i1 = v1 /2
3(v1 /2) = v2 /3 + (v2 - v1)/ 1
9v1 = 2v2 + 6v2 - 6v1
15v1 = 8 v2
30 = 3v1 – 2v2
v1 = -40 V
v2 = –75V
i1 = -40/2 = -20A
P3i1 = (-60) *(-75)
P3i1 = 4.5 kW
P3i1 = 4.5 kW
P3i1 = (3i1) *(v2)
Note: This is solved problem 4.4 on page 88, Figure 4.7 in the Ref 1 book (by Hayt)

Summary of Basic Nodal Analysis Procedure
1. Count the number of nodes (N).
2. Designate a reference node. The number of terms in your nodal
equations can be minimized by selecting the node with the greatest
number of branches connected to it.
3. Label the nodal voltages (there are N − 1 of them).
4. Write a KCL equation for each of the non-reference nodes. Sum the
currents flowing into a node from sources on one side of the equation.
On the other side, sum the currents flowing out of the node through
resistors. Pay close attention to “−” signs.
5. Express any additional unknowns such as currents or voltages other
than nodal voltages in terms of appropriate nodal voltages. This
situation can occur if voltage sources or dependent sources appear in
our circuit.
6. Organize the equations. Group terms according to nodal voltages.
7. Solve the system of equations for the nodal voltages (N − 1 of them).

Nodal Analysis: Super Node

Example 1: Super Node
 Nodal analysis of a circuit with independent voltage source(s)
 The nodes of independent voltage source are combined for KCL
 It is called a super node.
 There will be one super node per independent voltage source in
the circuit
Super Node
Note: This is Practice problem 4.4 on page 90, Figure 4.10 in the Ref 1 book (by Hayt)

Example 1: Super Node contd.
 Find voltages across current sources:
KCL for the super node [v1, v2]:
4 + 9 = v1 / (1/2) + (v1 - v2 )/ (1/3) + (v2 - v1 )/ (1/3) + v2 / (1/6)
v1 v2
4 + 9 = v1 / (1/2) + v2 / (1/6)
13 = 2v1 + 6v2
5 = v1 - v2
v1 = 5 + v2
Voltage across nodes [v1, v2]:
13 = 2(5 + v2 )+ 6v2 3 = 8v2
v2 = 3/8
v2 = 0.375 V
v1 = 5.375 V
v1 = 5.375 V, v2 = 375 mV
Note: There is only one KCL equation here.
Note: This is Practice problem 4.4 on page 90, Figure 4.10 in the Ref 1 book (by Hayt)

Super Node Explained
 In circuit theory, a supernode is a theoretical construct
that can be used to solve a circuit.
 This is done by viewing a voltage source on a wire as a
point source voltage in relation to other point voltages
located at various nodes in the circuit, relative to
a ground node assigned a zero or negative charge.
 Each supernode contains two nodes, one a non-reference
node and
◦ Another node that may be a second non-reference node or the
reference node.
◦ Supernodes containing the reference node have one node voltage
variable.
 For nodal analysis, the supernode construct is only
required between two non-reference nodes.

Nodal Analysis: Home Work Problems

S8B_HW_Problem1:
 Find the current i1 in the circuit:
Ref
v1
v2
i1 = 742 mA
Note: This is Exercise problem 17 on page 112, Figure 4.43 in the Ref 1 book (by Hayt)

S8B_HW_Problem2:
 Find the voltage V5 and power dissipated P7: v5 = 8.833 V P7 = 10.65 W
v1
v3v2
Note: This is Exercise problem 13 on page 111, Figure 4.39 in the Ref 1 book (by Hayt)

S8B_HW_Problem3:
No KCL for node v1 :
 Find nodal voltages in the circuit:
KCL for node v2 :
14 = (v2 - v1)/ 0.5 + (v2 - v3)/ 2
KCL for super node [v3 v4 ]:
Additional equations :
v1 = -12 V v3 – v4 = 0.2vy
vy = v4 - v1
vx = v2 - v1
0.5vx= (v3- v2)/2 +( v4/ 1)+ (v4- v1)/2.5
14 = 2 (v2 - v1) + 0.5 (v2 - v3)
-2v1 + 2.5 v2 – 0.5 v3 = 14
0.2vy = 0.2(v4 - v1 )
v3 – v4 = 0.2(v4 - v1 )
0.2v1 + v3 - 1.2 v4 = 0
0.5 (v2 - v1 ) = 0.5 (v3- v2) + v4+ 0.4(v4- v1)
0.1 v1 - v2 + 0.5 v3 + 1.4v4 = 0
Note: This is solved problem on page 91, Figure 4.11 in the Ref 1 book (by Hayt)

Session 8B: Summary
 Example Problems
◦ Writing Nodal Equations
◦ Nodal Analysis
 Summary of Steps: Nodal Analysis
 Super Node
◦ Example with Super Node
 Home Work Problems

References
Ref 1 Ref 2

Basic Electric Circuits Session 8B

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