The node voltage method simplifies the writing of equations for a circuit because you need only write Kirchhoff's Current Law. Kirchhoff's Voltage Law is implicit in the writing of the other equations. Consider the resistor below.
The current is given by i = v/R by Ohm's Law. But v is vab which
is va - vb, so
i = (va - vb)/R.
Stated in words, this is: "The current flowing from a to b is the voltage at a minus the voltage atb divided by the resistance". Similarly, we can say the current that flows from b to a is
(vb - va)/R.
We don't have to know in advance which voltage has a bigger magnitude in order to choose which current we want (a to b or b to a) because the sign of the answer will take care of that, and
iab = -iba, just as
vab = -vba.
Using Ohm's Law we can write Kirchhoff's Current Law at any node in terms of the voltage at the nodes. Note that it is not necessary to write an equation at every node. You never write an equation at the reference node (this is, by definition, zero volts), and it is not necessary to write an equation at a node where the voltage is known.
An example is given below.
First, we must identify the nodes. Note that the line between the 7 Ω resistor and the 3 Ω resistor is one node. There are 4 nodes. One of those must be chosen as the reference node. A good choice is the one at the bottom. This is convenient because it gives us an immediate answer for the node on the right (5 V). Now we can label all the nodes:
We can immediately assign 5 V to the node on the right, because we observe that it is 5 V higher than the reference node.
The "node" at the upper right labeled "VB + 9" is a non-essential node. It is labeled for convenience. We see that the voltage at this point is 9 V higher than VB.
We now write KCL at the remaining two nodes:
-2 + (VA - VB)/3 + (VA - VB)/6 +
(VA - 5)/7 + VA/4 = 0
(VB - VA)/3 + (VB - VA)/6 + (VB + 9 - 5)/8 = 0
When writing these equations, we assume that the current out of each node is
positive. In the first equation, this gives us terms like:
(VA - VB)/3,
and in the second, we have:
(VB - VA)/3.
This is not inconsistent, because
(VA - VB)/3 = -(VB - VA)/3.
(VB + 9 - 5)/8
will take a little explanation. We want a term for the current between the nodes labeled VB and 5 V. The same current flows through the 9 V battery as flows through the 8 Ω resistor. Therefore, if we can determine the current through the 8 Ω resistor, we will know the current through the 9 V battery. Now it should be clear why it was convenient to label the "node" at the upper right as VB + 9. Using i = v/R, the current through 8 Ω resistor is
[(VB + 9) - 5]/8.
We have two equations and two unknowns. Solve by any method you wish. Once way is shown below.
2 = (VA - VB)/3 + (VA - VB)/6 +
(VA - 5)/7 + VA/4
2 + 5/7 = VA(1/3 + 1/6 + 1/7 + 1/4) - VB(1/3 + 1/6)
2.714 = .893VA - .5VB
0 = (VB - VA)/3 + (VB - VA)/6 + (VB + 9 - 5)/8
-4/8 = -VA(1/3 + 1/6) + VB(1/3 + 1/6 + 1/8)
-.5 = -.5VA + .625VB
The result is
VA = 4.69 V
VB = 2.95 V
Take care to avoid round-off error when working with decimals. Some people prefer to work with fractions to avoid this problem.
Before going on to the homework, you should complete Tutorial 4 on the node voltage method.