Electrical
Circuits  Analytical Methods
Introduction
Goals
Setting Up Circuit
Equations (Node Equations)
Writing
Node Equations
Setting
Up Simultaneous Equations
Circuits
Containing Current Sources
Introduction
You have learned how to use conductance matrix analysis in simple circuits.
However, the discussion in the first lesson did not give a formal algorithm
for solution of these circuits. Here are our goals for this lesson.

Given
an electrical circuit composed of resistors, voltage sources and current
sources,

Be able to write the conductance
matrix representation for the circuit,

Be able to solve the conductance
matrixsource voltage equation.
A
General Method Of Setting Up Circuit Equations.  Writing The Node Equations
In this section we will consider how circuit equations are generated and
how different elements affect those circuit equations. We will examine
how node equations are written.
We start by looking at a typical node  extracted from some larger circuit.
That node is shown below.
Note the following about this circuit:

The node shown  Node
"n"  is part of a larger circuit. The resistors connected to node "n"
are connected to other nodes through other resistors  shown by dotted
lines in the figure.

We define currents flowing
out of node "n" as positive when flowing away from the node. Obviously
they can't all be positive, but using this convention helps in developing
a formal methodology for setting up the node equations.
The
methodology that we use is as follows.

Define a voltage at every
node in the circuit. A typical voltage can be denoted
V_{n}.

That voltage symbol, V_{n},
denotes the voltage at node "n" measured with respect to ground.

Note, that at this point
we have to realize that every node voltage might not be an unknown in the
system of equations that we are developing. For example, in the circuit
below, the voltage at the green node is known to be V_{s},
whereas the voltage, V_{x},
is not known a priori.

When you are setting up
a group of equations to solve, it is important to know how many unknowns
you have (because the level of difficulty of solution is related to the
number of unknowns) and that information is important.

Having defined a voltage
at every node, it is now possible to write KCL at nodes where the voltage
is not known. The KCL equations have a form that we can discover
from the general node above.

Then we write KCL at the
node where V_{x}
appears.

Finally, we solve whatever
equation results from writing KCL.
We
will look at these two steps separately, and we will work first on the
problem of writing KCL at a node.
Writing
Node Equations
Here is that node extracted from a larger circuit.
Examining
the sample node  node "n"  it is relatively easy to write the KCL equations
for the node. Using KCL for the sample node, we would have:
I_{x}
+ I_{y} + I_{z} = 0
And, each of the currents can be expressed
in terms of the voltage at the node and the voltage at one of the neighboring
nodes. For example:
I_{x}
= (V_{n} V_{x})/R_{x}
Express every current in that manner, and
put those expressions into the KCL equation:
(V_{n}
V_{x})/R_{x} + (V_{n} V_{y})/R_{y}
+ (V_{n} V_{z})/R_{z} = 0
Notice how this equation plays out. By defining currents as positive
when they are flowing away from the node (i.e., the current arrows defining
current polarity point away from the node.), every term that involves V_{n}
turns out to have a plus sign in the resultant equation. That gives
you a way to know when you have written the equation correctly. There
is a payoff to being systematic and consistent when you write the KCL equations.
You will find almost every textbook that discusses this topic writes the
equations this way! You should adopt that practice as well, and you
should never go wrong.
The case of nodes with only resistors connected is simple enough, but there
are special situations that you will need to take care of. Here are
those special cases.

One of the adjacent nodes
might be connected to a voltage source. Actually, there are two special
cases here because the other end of the voltage source might be connected
go ground  the first case  or it might be connected to still another
node  the second case.

There might be a current
source connected to the node at which you write KCL.
Let's
consider the first case. Here is a picture of the situation with
an adjacent node connected to a grounded voltage
source.
In this situation, KCL is written exactly
as it was before. However, the voltage source serves to fix a node
voltage and that node voltage is no longer a variable with an unknown value.
Instead, the voltage at node 'x' in the figure is known to be V_{s}.
After you have written KCL at the node, when you go to solve the KCL equations
simultaneously, that known value will appear in a term on the right hand
side (the knowns) of the equation. Thus:

If there is a grounded
source connected to node "n", then:

No KCL equation is written
at node "n".

We have V_{n}
= V_{s}.
There
is another case in which there is a voltage source
connected between two nodes  neither of which is ground.
In this case the voltage source establishes a voltage difference between
two nodes, and the way this situation is handled is slightly different.

If there is an ungrounded
source connected between node "n" and node "z", then

If V_{z}
can be found, then V_{z}
is known from this equation.

One less KCL equation
will be written for this circuit.

Then, the approach would
be to write a KCL equation at either node "n" or node "z", and use the
equation V_{n} = V_{z}
+ V_{s} to find the voltage
at the other node.

The implication is that
one less KCL equation will be written.
Considering
Current Sources
Now
consider what happens when a current source
is connected to a node. You might suspect that this situation is
the most complex of all, but, in fact, it is one of the simpler ones.
In the example below the current source is deliberately set so that the
current is actually flowing into the node if I_{s}
is positive. Now, consider writing the current equations for this
circuit.
I_{x}
+I_{y} + I_{z} = 0
becomes:
(V_{n}
V_{x})/R_{x} + (V_{n} V_{y})/R_{y}
 I_{s} = 0
Given all of the above, there is one last question. That is, "How
many independent equations result when KCL is written at all of the nodes?",
Consider the following:

Assume that the circuit
is composed of resistors, voltage sources and current sources. Assume
that the circuit has n + 1 nodes.

One of the nodes in the
circuit must be chosen as a reference.
Usually, if there is a node connected to ground,
then that will be the reference.

Otherwise, choose any
other node as the reference.

All
other node voltages will be measured from the voltage at the reference
node. That reduces the number of voltages
that must be found from n + 1
to n.

Every other node voltage
is potentially an unknown voltage that must be found  n node voltages
in all.

If there is a grounded
voltage source, then one node voltage is determined
by that voltage source  at the ungrounded end of the voltage source.

Every grounded voltage
source reduces the number of unknown node voltages by one. If there
are n_{Vg}
grounded voltage sources, that reduces the number of unknown node voltages
by n_{Vg}
to n  n_{Vg}.

If there are ungrounded
voltage sources, then those ungrounded sources
fix the difference in voltage between two nodes. Essentially, if
you know one of those node voltages, you can compute the other. Thus,
each ungrounded voltage source also reduces the number of unknown node
voltages by one.

If there are current
sources in the network, they have
no effect on the number of unknown node voltages,
although it will most definitely have an effect on the value of node voltages
in the network.

Thus, the number of node
equations is n  n_{V},
where n_{V}
is the number of voltage sources in the network.

You get those equations
by writing KCL at every node not connected to a voltage source (grounded
or ungrounded) and one of the nodes connected to every ungrounded voltage
source.
After writing the node equations, the next problem is to put them in a
form that is easier to solve, and that is the next thing you have to consider.
Getting
A Set Of Simultaneous Equations
Once the KCL equations are written for a network, the next step is to cast
the equations into a form amenable to solution. The form you should
be looking for is a set of simultaneous linear equations. In other
words, you are looking for something like the following:
a*x + b*y + c*z = j
d*x + e*y + f*z = k
g*x + h* y + i*z =
l
This is a standard form for simultaneous
linear equations, and for circuit equations the variables (x, y and z)
would probably be node voltages (e.g. V_{x}, V_{y}
and V_{z}).
Considere a typical node voltage equation (taken from earlier in this lesson):
(V_{n}
V_{x})/R_{x} + (V_{n} V_{y})/R_{y}
+ (V_{n} V_{z})/R_{z} = 0
Separate all of the terms that involve V_{x},
etc. and combine, and the result is:
(1/R_{x}
+ 1/R_{y} + 1/R_{z})V_{n} 
V_{x}/R_{x}  V_{y}/R_{y}
 V_{z}/R_{z} = 0
And this is the form we want. This
particular equation has four unknown node voltages, but that would be the
case for the node shown just above.
Similar equations occur for the other possibilities.
If there is a grounded voltage source, no
KCL equation is written at that node because the node voltage is known:
V_{n} =
V_{s}
If there is an ungrounded
voltage source, there is one less KCL equation.
Here there are two possibilities.
We could write:
V_{n}
= V_{s} + V_{z}
_{}
Then we would write
a KCL equation at node "z" that included I_{x}
and I_{y} flowing away from
node "z'. Here is an example to illustrate the point.
Example
E1
In this circuit, note the following:

The voltage at node "x"
is known:

We can select either node
"y" or node "z" as the node with the unknown voltage. Select one,
and the other is determined. We will take V_{y}
as the unknown voltage, so:

We can write KCL at node
"y". Actually, it may be better to think of a supernode
that includes both node "y" and node "z". The green shading shows
the supernode. The sum of the currents leaving the supernode must
sum to zero  for the same reason that the sum of the currents leaving
a node sum to zero  conservation of charge!
Write the KCL equations for the entire super node:
I_{x}
+ I_{y} + I_{z} = 0
Then, note that:
I_{x}
= (V_{y} V_{x})/R_{x}
I_{y}
= V_{y }/ R_{y}
I_{z}
= V_{z }/ R_{z}= (V_{y} 
V_{2})/ R_{z}
And the KCL equation
for the supernode becomes:
(V_{y}
V_{x})/R_{x} + V_{y }/ R_{y}
+ (V_{y}  V_{2})/ R_{z} =
0
And, for this circuit
that is the one independent KCL equation that can be written (unless you
choose to keep V_{z}
as your unknown, but there is still only one equation.) That means
writing this in the standard form, we would have:
(1/R_{x
}+
1_{ }/ R_{y} + 1/ R_{z})V_{y}
 V_{x}/R_{x} = V_{2}/ R_{z}
The coefficient of
V_{y}
is the sum of the conductances connected to the supernode, the source appears
on the right hand side, and the negative reciprocal of the shared resistor
shows up just where we might expect it.