Nodal
Analysis  KCL and Matlab
A nodal analysis can be performed by examining each node in a circuit.
The goal is to find out what the voltages are in each node with respect
to our reference node. We need to know the currents flowing in the
circuit and the resistances between each nodes. This is just an
application of the Ohm's Law.
Kirchhoff’s current law (KCL) states that for any electrical
circuit, the algebraic sum of all the currents at any node in the
circuit is
zero.
In this type of analysis, if there are n
nodes in a circuit, and we select a reference node (node 0), the other
nodes can
be numbered from V_{1}
through V_{n1}.
With one node selected as
node 0 (reference), there will be n1
independent equations. If we assume
that the admittance between nodes i
and j is given as Y_{ij}, we can write the
following
equations including all of the nodes in the circuit:
where:
m
= n  1
V_{1},
V_{2},...,
V_{m}
are voltages from all the nodes with respect to node
0
is
the algebraic sum
of current sources at node m.
The above system of equations can be expressed in matrix
form as:
Y
V = I
And naturally, its
solution is:
V
= Y^{ 1}I, which in Matlab can be easily solved
using the backslash
operator (\), like this:
V
= Y\I
The following example
illustrates the use of Matlab for solving
nodal voltages of electrical circuits.
Nodal Analysis  find the voltages in a circuit
For the circuit shown below, resistors are in ohms and
current sources are in amps. We’re asked to find the nodal voltages V_{1}, V_{2},
and V_{3}.
Using KCL, and forming
our matrices Y and I,
let’s see...
For node V_{1
}we
have,
and the first row of our Y
matrix is going to be the coefficients for the voltages. This means
that we can form our matrix in Matlab like this:
Y(1,:) =
[(1/1 + 1/2) 1/1 1/2];
and naturally I(1)
= 5;
At node V_{2},
the second row of our Y
matrix is going to be the new set of
coefficients for the voltages. Letting Matlab work out the operations,
we
express
Y(2,:) =
[1/1 (1/1  1/4 
1/5) 1/4];
and I(2) = 0;
Finally,
at node V_{3
}we have,
thus, our
third row of the Y matrix is
Y(3,:) =
[1/2 1/4 (1/2 + 1/4)];
and I(3)
= 2;
Summarizing our findings,
we fillin our matrices Y and I, like this:
Y = 1.5
1.00
0.50
1.0 1.45
0.25
0.5 0.25
0.75
I = 5
0
2
The solution is easy
using Matlab: just type V
= Y\I. The
answer is matrix V,
with three rows (our vector of unknowns contains three
variables in a column: V_{1}, V_{2}
and V_{3})
V = 40.4286
35.0000
41.2857
An analysis with more components
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