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 matrix-source 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.

• 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 Vn.
• That voltage symbol, Vn, 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 Vs, whereas the voltage, Vx, 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 Vx 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:

Ix + Iy + Iz = 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:

Ix = (Vn -Vx)/Rx

Express every current in that manner, and put those expressions into the KCL equation:

(Vn -Vx)/Rx + (Vn -Vy)/Ry + (Vn -Vz)/Rz = 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 Vn 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 Vs.  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 Vn = Vs.
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
• Vn = Vz + Vs.
• If Vz can be found, then Vz 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 Vn = Vz + Vs 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 Is is positive.  Now, consider writing the current equations for this circuit.

Ix +Iy + Iz = 0

becomes:

(Vn -Vx)/Rx + (Vn -Vy)/Ry - Is = 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 nVg grounded voltage sources, that reduces the number of unknown node voltages by nVg to n - nVg.
• 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 - nV, where nV 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. Vx, Vy and Vz).

Considere a typical node voltage equation (taken from earlier in this lesson):

(Vn -Vx)/Rx + (Vn -Vy)/Ry + (Vn -Vz)/Rz = 0

Separate all of the terms that involve Vx, etc. and combine, and the result is:

(1/Rx + 1/Ry + 1/Rz)Vn - Vx/Rx - Vy/Ry  - Vz/Rz = 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:

Vn = Vs

If there is an ungrounded voltage source, there is one less KCL equation.

Here there are two possibilities.

We could write:

Vn = Vs + Vz

Then we would write a KCL equation at node "z" that included Ix and Iy 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:
• Vx = V1
• 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 Vy as the unknown voltage, so:
• Vy = Vz + V2
• 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:

Ix + Iy + Iz = 0

Then, note that:
Ix = (Vy -Vx)/Rx

Iy = Vy / Ry

Iz = Vz / Rz= (Vy - V2)/ Rz

And the KCL equation for the supernode becomes:

(Vy -Vx)/Rx + Vy / Ry + (Vy - V2)/ Rz = 0

And, for this circuit that is the one independent KCL equation that can be written (unless you choose to keep Vz as your unknown, but there is still only one equation.)  That means writing this in the standard form, we would have:

(1/Rx + 1 / Ry + 1/ Rz)Vy  - Vx/Rx = V2/ Rz

The coefficient of Vy 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.