Nodal analysis using KCL and KVL for a simple circuit

1. NODAL AND LOOP ANALYSIS TECHNIQUES
LEARNING GOALS
NODAL ANALYSIS
LOOP ANALYSIS
Develop systematic techniques to determine all the voltages
and currents in a circuit

2. NODE ANALYSIS
• One of the systematic ways to
determine every voltage and
current in a circuit
The variables used to describe the circuit will be “Node Voltages”
-- The voltages of each node with respect to a pre-selected
reference node

3. LEARNING EXAMPLE
V1 V1 − V2
@ V1 : − 4mA + +
=0
6k 12k
@ V2 : 2mA +
V2 V2 − V1
+
=0
6k
12k
BY “INSPECTION”
1 
1
 1
+
V1 −
V2 = 4mA

÷
12k
 6k 12k 
1
1 
 1
−
V1 +  +
÷V2 = − 2mA
12k
 6k 12k 
USING KCL

4. LEARNING EXAMPLE
SUPERNODE
V1 = 6V
V4 = −4V
SOURCES CONNECTED TO THE
REFERENCE
CONSTRAINT EQUATION
KCL @ SUPERNODE
V3 − V2 = 12V
V2 − 6 V2 V3 V3 − ( −4)
+ +
+
= 0 * / 2k
2k
1k 2k
2k
V2 IS NOT NEEDED
FOR I O 3V2 + 2V3 = 2V
− V2 + V3 = 12V * / 3 and add
5V3 = 38V
V
OHM'S LAW I O = 3 = 3.8mA
2k

5. CIRCUITS WITH DEPENDENT SOURCES
PRESENT NO SIGNIFICANT ADDITIONAL
COMPLEXITY. THE DEPENDENT SOURCES
ARE TREATED AS REGULAR SOURCES
WE MUST ADD ONE EQUATION FOR EACH
CONTROLLING VARIABLE

6. LEARNING EXAMPLE
FIND I O
VOLTAGE SOURCE CONNECTED TO REFERENCE
V1 = 3V
V − V1 V2
KCL@V2 : 2
+
− 2I x = 0
3k
6k
CONTROLLING VARIABLE IN
TERMS OF NODE VOLTAGES
Ix =
V2 − V1 V2
V
+
− 2 2 = 0 * / 6k
3k
6k
6k
V2 − 2V1 = 0 ⇒ V2 = 6V
IO =
V1 − V2
= −1mA
3k
REPLACE
V2
6k

7. SUPER NODE WITH DEPENDENT SOURCE
VOLTAGE SOURCE CONNECTED TO REFERENCE
V3 = 6V
SUPERNODE CONSTRAINT
V1 − V2 = 2Vx
CONTROLLING VARIABLE IN TERMS OF NODES
KCL AT SUPERNODE
Vx = V2 ⇒ V1 = 3V2
* / 12k
2(V1 − 6) + V1 + 2V2 + V2 − 6 = 0
3V1 + 3V2 = 18 ⇒ 4V1 = 18

8. LEARNING EXAMPLE
FIND THE VOLTAGE Vo
@V4 : V4 = 4V
AT SUPER NODE
V1 − V2 = 2VX
V V − V3 V1 − V3 V1 − 4V
−2mA + 2 + 2
+
+
=0
1k ×
1k
1k
1k
1k
1k × @V : − 2mA + V − V + V − V = 0
3
3
2
1k
CONTROLLING VARIABLE
3
1
1k
VX = V2
SOLVE EQUATIONS NOW
V1 = 3VX
2V1 + 2VX − 2V3 = 6V
−V1 − VX + 2V3 = 2V
VARIABLE OF INTEREST
VO = V1 − V3

9. LEARNING EXAMPLE
Find the current Io
@ V2 : V2 = 12V
@ V3 : V3 = 2VX
@ super node:
V4 − V1 = 6V (constraint eq.)
FIND NODES – AND SUPER NODES
V − V3 V4 − V5 V4
V1 − V2 V1 − V3
+
+ 2I X + 4
+
+
1k
1k
1k
1k
1k
=0
V5 − V4 V5
@ V5 : − 2 I X +
+
=0
1k
1k
CONTROLLING VARIABLES
VX = V1 − V2
IX =
V4
1k
7 eqs in 7 variables
VARIABLE OF INTEREST
IO =
V5
1k

10. LEARNING EXAMPLE
Find the current Io
@ V2 : V2 = 12V
@ V3 : V3 = 2VX
@ super node:
V4 − V1 = 6V (constraint eq.)
FIND NODES – AND SUPER NODES
V − V3 V4 − V5 V4
V1 − V2 V1 − V3
+
+ 2I X + 4
+
+
1k
1k
1k
1k
1k
=0
V5 − V4 V5
@ V5 : − 2 I X +
+
=0
1k
1k
CONTROLLING VARIABLES
VX = V1 − V2
IX =
V4
1k
7 eqs in 7 variables
VARIABLE OF INTEREST
IO =
V5
1k