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 v_{ab} which
is v_{a} - v_{b}, so

i = (v_{a} - v_{b})/R.

Stated in words, this is: "The current flowing from *a *to *b* is
the voltage at *a *minus the voltage at*b* divided by the
resistance". Similarly, we can say the current that flows from *b* to
*a* is

(v_{b} - v_{a})/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

i_{ab} = -i_{ba}, just as

v_{ab} = -v_{ba}.

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 "V_{B} + 9" is a non-essential
node. It is labeled for convenience. We see that the voltage at this point is 9
V higher than V_{B}.

We now write KCL at the remaining two nodes:

-2 + (V_{A} - V_{B})/3 + (V_{A} - V_{B})/6 +
(V_{A} - 5)/7 + V_{A}/4 = 0

(V_{B} - V_{A})/3 + (V_{B} - V_{A})/6 +
(V_{B} + 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:

(V_{A} - V_{B})/3,

and in the second, we have:

(V_{B} - V_{A})/3.

This is not inconsistent, because

(V_{A} - V_{B})/3 = -(V_{B} - V_{A})/3.

The term

(V_{B} + 9 - 5)/8

will take a little explanation. We want a term for the current between the
nodes labeled V_{B} 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 V_{B} + 9. Using i = v/R, the current
through 8 Ω resistor is

[(V_{B} + 9) - 5]/8.

We have two equations and two unknowns. Solve by any method you wish. Once way is shown below.

2 = (V_{A} - V_{B})/3 + (V_{A} - V_{B})/6 +
(V_{A} - 5)/7 + V_{A}/4

2 + 5/7 = V_{A}(1/3 + 1/6 + 1/7 + 1/4) - V_{B}(1/3 + 1/6)

2.714 = .893V_{A} - .5V_{B}

0 = (V_{B} - V_{A})/3 + (V_{B} - V_{A})/6 +
(V_{B} + 9 - 5)/8

-4/8 = -V_{A}(1/3 + 1/6) + V_{B}(1/3 + 1/6 + 1/8)

-.5 = -.5V_{A} + .625V_{B}

The result is

V_{A} = 4.69 V

V_{B} = 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.

- Identify (mark) the essential nodes.
- Choose a reference node (usually the node with the most branches connected to it).
- For each voltage source, write an appropriate equation, or define the voltage at a point on the circuit. It may be necessary to define a current in some voltage sources.
- Assign a variable name to each other essential node.
- Write Kirchhoff's Current Law for every node (except the reference node and nodes where the voltage is known) using Ohm's Law.
- Count your equations and unknowns (they should be equal).
- Solve.
- Is your answer reasonable in terms of magnitude, sign, and units?

Before going on to the homework, you should complete Tutorial 4 on **the node
voltage method**.

- Write the node voltage equations for this circuit. Do not solve.

- E = 200 volts, Is = 11 A, R1 = 15 Ω, R2 = 40 Ω, R3 = 10
Ω, R4 = 20 Ω, R5 = 30 Ω. Find V using the node voltage
method. The answer is an integer.

- E = 12I volts, Is = 2.5 A, R1 = 16 Ω, R2 = 6 Ω. Find I using
the node voltage method. The answer is an integer.

- E1 = 150 V, E2 = 120 V, E3 = 70 V, Is = 4I amps, R1 = 600 Ω, R2 =
17 Ω, R3 = 10 Ω, R4 = 60 Ω, R5 = 5 Ω, R6 = 100
Ω, R7 = 8 Ω. Find I using the node voltage method. The answer
is an integer.