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Nodal Circuit Analysis Using KCL
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Most useful for when we have mostly current sources
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Node analysis uses KCL to establish the currents
Procedure
(1) Choose one node as the common (or datum) node
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Number (label) the nodes
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Designate a voltage for each node number
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Each node voltage is with respect to the common or datum node
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Number of nodes used = number of nodes – 1 = n-1
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Note: number of nodes = branches – 1 = b-1
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Thus less equations with node analysis than mesh analysis (2) For each node write the KCL for current flows in each node
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Use differences in the node voltages to calculate currents
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Assume the current directions and write the KCL
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Generally assume the node is a positive V relative to all others
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Current directions different for same branch in each node
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Often better to use conductance equations
⎥⎦⎤⎢⎣⎡+=+=+=
211211121
11
R RV RV RV I I I
R R
(3) Solve the equations for the node voltages
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Get currents in each branch from the voltage differences
Example Nodal Circuit Analysis
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Consider the 2 node, 3 loop circuit below (1) Setting the base node, and node voltages
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Set the common node to ground
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Label voltages on the others (2) For each node write KCL for current flows
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For node 1,
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Defining the current directions for I
1
as into the node
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Use differences in the node voltages for currents
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Collect each voltage into one term
( )
13231132111
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I RV R RV RV V RV
=−⎥⎦⎤⎢⎣⎡+=−+
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Or in the conductance form
( ) ( )
13231132111
I GV GGV GV V GV
=−+=−+
Example Nodal Circuit Analysis continued
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For node 2,
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Defining the current directions for I
2
as into the node
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This means current in R
3
different from node 1
( )
23223131222
11
I R RV RV RV V RV
=⎥⎦⎤⎢⎣⎡++−=
−+
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Or if changing to conductance form
( ) ( )
23223131222
I GGV GV GV V GV
=++−=−+
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Now have two equations and unknowns (V’s)
132311
11
I RV R RV
=−⎥⎦⎤⎢⎣⎡+
232231
I R1 R1V RV
=⎥⎦⎤⎢⎣⎡++−
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Could solve this algebraically
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Instead use the numerical methods
Example Nodal Circuit Analysis
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Putting the equations in numerical form
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for node 1:
( ) ( )
013.000025.0V 00125.0V
013.04000V 4000110001V RV R1 R1V
212132311
=−=−⎥⎦⎤⎢⎣⎡+=−⎥⎦⎤⎢⎣⎡+
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For node 2:
( ) ( )
003.000075.000025.0
003.04000120001400011
212132231
=+−=⎥⎦⎤⎢⎣⎡++−=
⎥⎦⎤⎢⎣⎡++−
V V V V R RV RV
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Using substitution method for node 1
( )
00125.000025.0V 013.0
V
21
+=
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Thus using node 2 for the solution
( )( )
003.000075.0V 00025.0
00125.000025.0V 013.0
22
=+⎥⎦⎤⎢⎣⎡ +
( )
0026 .0003.000075.000005.0V
2
+=+−
V 8007 .00056 .0V
2
==

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