NODAL ANALYSIS
Nodal analysis is a technique used to analyze circuits with more than one voltage
source. Based on KCL, it is used to determine the voltages of the nodes in a circuit.
Once you determine the node voltages, it is possible to find all branch voltages,
currents and the power supplied or absorbed by each circuit element.
C
A
An example…
D
8.2
R1
12V 
5.6
R3
10
R2
 6V
B
Find
VA
Step 1: Choose a reference node (usually ground).
In this case, we choose node B as reference and assign node B a value of 0V.
Step 2: Identify known node voltages with respect to the reference node.
By inspection:
VA = ?
VB = 0V
VC = 12V
VD = 6V
Step 3: Assume a direction for the current passing through each resistor that is adjacent
to a node with unknown voltage. The directions are arbitrary. Label the currents with an
arrow and a common sense numbering scheme (I1, IR1, etc.) on your circuit schematic.
Here we assumed that each current is going OUT of Node A.
C
R1
- VR +
1
I R3
I R1
I R2
R3
D
+ VR -
A
3
+
12V 
R2
V R2
B
 6V
Step 4: Write KCL at the node with the unknown Voltage
In this case VA is unknown, so write KCL for node A.
- VR +
1
I R3
I R1
R1
C
D
+ VR -
A
I R2
R3
3
+
12V 
 6V
V R2
R2
B
KCLAT NODE A  I R1  I R 2  I R 3  0
Step 5: Express each branch current in terms of Ohm’s Law.
I R1 
VR1
I R2 
R1
VR2
R2
IR3 
VR3
R3
Step 6: Express each element’s voltage as a difference between node voltages.
To get the current, simply subtract the voltage at the tip of each current arrow from
the voltage at the tail of the arrow, and then divide by the value of the associated
resistor.
Use : I X 
VTail  VTip
RX
I R1 
VA  VC
R1
IR2 
VA  VB
R2
I R3 
VA  VD
R3
Step 7: Substitute the known node voltages from Step 2.
I R1 
VA  12V
R1
IR 2 
VA  0V
R2
I R3 
VA  6V
R3
Step 8: Substitute these equations back into the KCL equation from Step 3.
KCLAT NODE A  I R1
VA  12V
R1

 IR 2
VA
R2
 IR3

0
VA  6V
R3
 0
Step 9: Plug in the resistor values.
VA  12
8.2

VA
10

VA  6
5.6
 0
Drop the units here
Step 10: Solve the equation for VA
VA  6.33V
Now that you know the node voltage VA . . . .
. . . . You know the voltage across each element
. . . . You know the current through each element
. . . . Power calculations are straightforward.
If you have two or more unknown nodes, you will have more equations to
solve.
** You must always have one equation for each unknown node. **
(2 unknown nodes
2 KCL equations)
Another example:
Given:
R1
A
I R1
1K
3V
I R2
R2
+
-
I R3 R3
B
Step 1 Reference node is “D”
Step 2
VA  3V
VB  ??
V C  ??
VD  0V
VE  4V
I R4
4V
D
Find VB, VC
1.5K
2.2K
VS1
I R5
C
R4
3.3K
+
-
E
VS 2
R5
820
D
Step 3 See drawing above for currents
2 unknown nodes (B, C)
KCLB :
2 KLC equations
KCLC :
I R1  I R 2  I R 3  0
I R 4  I R5  I R3
Step 4,5
I R1 
VB  VA
R1
I R4 
VC  VE
R4
I R2 
VB  VD
R2
I R5 
VC  VD
R5
I R3 
VB  VC
R3
I R1 
VB  3V
R1
I R4 
VC  4V
R4
I R2 
VB  0V
R2
I R5 
VC  0V
R5
I R3 
VB  VC
R3
Step 6
Step 7
VB  3V
R1
VB
R2

I R1
VB  VC
 0
R3

I R2
VB  VC
R3

I R3
KCLB
I R3
VC  4V
R4

I R4
VC
R5
KCLC
I R5
Step 8
VB  3

1000
VC  VB
1500

VB
2200
VB  VC
 0
1500

VC  4
3300

VC
 0
820
KCLB
KCLC
Step 9
You should get
VB  1.76V
VC  1.09V
These values are with
respect to your Reference
Node (D).