LOOP ANALYSIS
The second systematic technique
to determine all currents and
voltages in a circuit
IT IS DUAL TO NODE ANALYSIS - IT FIRST DETERMINES ALL CURRENTS IN A CIRCUIT
AND THEN IT USES OHM’S LAW TO COMPUTE NECESSARY VOLTAGES
THERE ARE SITUATION WHERE NODE ANALYSIS IS NOT AN EFFICIENT TECHNIQUE
AND WHERE THE NUMBER OF EQUATIONS REQUIRED BY THIS NEW METHOD IS
SIGNIFICANTLY SMALLER
In the following we shall solve using loop
analysis two circuits that had previously been
solved using node analysis
This is one circuit.
we recap first the node analysis
approach and then we solve using
loop analysis
LEARNING EXAMPLE
FIND THE VOLTAGE Vo
RECAP OF NODE ANALYSIS
@V4 : V4  4V
AT SUPER NODE
V1  V2  2VX
V2 V2  V3 V1  V3 V1  4V

2
mA




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  V3  6V
V1  VX  2V3  2V
VARIABLE OF INTEREST
VO  V1  V3
USING LOOP ANALYSIS
DETERMINE Vo
Write loop equations
Loop 1 : I 1  2mA
Loop 3 : I 3  2mA
Loop 2 :  2VX  1kI 2  1k ( I 2  I 3 )  0
Loop 4 : 4V  1k ( I 4  I 3  I 1 )  2VX  1kI 4  0
Controlling variable:VX  1k ( I1  I 3  I 4 )
2kI 2  2kI 4  6 
  I 2  1mA, I 4  2mA
4kI 4  8 
I1
I2
I4
I3
START SELECTION USING MESHES
SELECT A GENERAL LOOP TO AVOID
SHARING A CURRENT SOURCE
Variable of Interest VO  1kI 2
LEARNING EXAMPLE
RECAP OF NODE ANALYSIS
Find the current Io
@V2 : V2  12V
@V3 : V3  2VX
@ super node:
V4  V1  6V (constraint eq.)
V4  V3 V4  V5 V4
V1  V2 V1  V3

 2I X 


1k
1k
1k
1k
1k
0
@V5 :  2 I X 
V5  V4 V5

0
1k
1k
CONTROLLING VARIABLES
VX  V1  V2
7 eqs in 7 variables
IX 
V4
1k
VARIABLE OF INTEREST
IO 
V5
1k
Find the current Io using mesh analysis
Write loop/mesh equations
Loop1: 1kI1  1k ( I1  I 2 )  1k ( I 1  I 4 )  0
Loop 2: 1k ( I 2  I1 )  6V  1k ( I 2  I 5 )  0
Loop 3: I 3  2 I X
Loop 4:  12V  1k ( I 4  I 1 )  2VX  0
Loop 5: 1k ( I 5  I 2 )  1k ( I 5  I 6 )  2VX  0
Loop 6: 1k ( I 6  I 3 )  1kI 6  1k ( I 6  I 5 )  0
Select mesh currents
Controlling variables
VX  1kI 1
I X  I5  I6
8 eqs in 8 unknowns
Variable of interest:
IO  I6