Node-Voltage or Nodal Analysis Our next set of techniques commonly used in circuit analysis are Nodal and Loop Analysis. Both of these techniques give you a systematic way of solving a problem. In nodal analysis, the following 4 steps should be followed.

1) Label all nodes (label one of them as the reference) 2) Define currents in each resistive branch 3) Write KCL at non-reference nodes 4) Express currents in terms of Voltage (I = V/R)

The goal is to find all node voltages: the voltage from a node (+) to reference (-) Example:

Nodes are labeled (1, 2, 3). Let us call node 3 the reference node. Currents were also defined in each resistive branch (I1, I2, and I3) At node 1, I notice that I don’t need to write KCL because that node voltage is known. The voltage between node 1 and reference, V1, is equal to 50 V.

Go to node 2 and write KCL: I1 + 3A = I2 + I3

Now, express each current as V/R starting with I1 I1 is attached to nodes 1 and 2. The direction as drawn shows it leaving node 1 so express this current as the difference between node voltages → (V1-V2)/5 If you don’t see that, write KVL in the inner loop and call the voltage across the 5 Ω resistor V5:

KVL: -50 + V5 + V2 = 0 V5 = 50-V2 or V1 – V2

I1 = V5/5 or (V1-V2)/5

Now go back to the KCL equation:

I2 + I3 = I1 + 3A

And express I1, I2, I3 as V/R. Note the 5 Ω resistor is the only one expressed as a difference between voltages divided by R. The 10 and 40 Ω resistors are both connected to the reference node so V2 is the voltage across both of them.

Multiply both sides by 40 to clear the fraction

4V2 + V2 = 8(V1-V2) + 3(40)

5V2 + 8V2 = 8(50) + 3(40) because V1 = 50 V.

13V2 = 520 or V2 = 40 V.

Once you know all node voltages, you know all currents…

Things to keep in mind when doing nodal analysis:

• When a Voltage source is connected between a pair of nodes, you know the value of that node voltage with respect to the reference node. Typically, the minus sign is connected to the reference node, otherwise the node voltage is negative.

• When you have a resistor connected to the reference node, draw the current flowing toward the reference node.

• When you have a resistor that is not connected to the reference node, the current flowing through it is the difference in node voltages divided by the value of resistance (see below)

Example: Use Nodal Analysis to find Io and Vo

Step 1. Label Nodes. This circuit has 3 nodes (2 + 1 reference node) so the unknown node voltages will be Va and Vb. Va is the voltage from node a to reference and Vb is the voltage from node b to reference.

Step 2. Define a current in every resistive branch.

Io is already defined.

I chose to make I1 the current flowing through the 6 kΩ resistor; direction doesn’t matter.

I chose to make I2 the current flowing through the 12 kΩ resistor; direction is chosen to be downward since this R is connected to a reference node.

Step 3. Write KCL at nodes a and b.

Step 4. Express unknown currents as a difference in voltages divided by the resistance. Notice that the 3 kΩ and 12 kΩ resistors are connected to the reference node so they are simply the node voltage divided by the resistance.

Now you have 2 equations, 2 unknowns (Va and Vb). Multiply the top equation by 6×103 and the bottom equation by 12×103

Simplify:

Multiply the second equation by 3 and subtract the second equation from the first equation:

36 = 9Va – 3Vb

- 48 = 2Va – 3 Vb

-12 = 7 Va

Or Va = -12/7 V. Substitute that information to find Vb.

12 = 3 (-12/7) – Vb

Vb = -36/7 – 12

7Vb = -36 – 84

Vb = -120/7 V.

Now that you know the node voltages, both Vo and Io can be found.

From the circuit, Vo = Va - Vb = -12/7 – (-120/7) = +108/7 V.

And Io = Va = -12/7 = -12/21 mA 3kΩ 3×103

Supernode

A special situation called a “supernode” exists when you have a voltage source between two non-reference nodes.

Example:

Label nodes and define currents:

Identify the supernode (between nodes 1 and 2). The difference between node voltages = 12 V giving you an equation:

V1-V2 = 12. (V1 comes first to satisfy KVL). KVL: -V1 + 12 + V2 = 0 or V1-V2 = 12

Write KCL at the supernode (every current entering the supernode = every current leaving)

KCL at supernode: 8 A = I1 + I2

Now, express each current (except 8A because it is a known current) in terms of node voltages

8 A = V1/3 + V2/6

Multiply both sides by 6

(6*8) = 2V1 + V2

OR 2V1 + V2 = 48 Combined with the equation above gives you 2 equations V1 - V2 = 12 If these 2 equations are added together, you get… 3V1 = 60 or V1 = 20 V. AND, V2 = V1 – 12 = 8 V.

Once you know the node voltages, it is easy to find each current:

I1 = 20/3 A. and I2 = 8/6 = 4/3 A. There is a technique called Cramer’s Rule that allows you to solve for unknowns by using only determinants of the matrices formed from the equations. For example, the two equations above written in matrix form look like:

V1 and V2 can be expressed as the determinants of the following matrices:

Notice the determinant of the first matrix is the denominator in each case. That same matrix is used for the numerator except replace the first column (in the case of V1) with the coefficient matrix (last matrix). For V2, replace the second column with the coefficient matrix.

Another supernode example:

Label nodes and define currents (Io defined for you). You know V1 and V4 already (V1 = 6 V, V4 = -4 V) You also know the supernode voltage

(V3-V2 = 12)

Write KCL at the supernode:

I1 = I2 + Io + I3

Express current in terms of node voltages:

Clear the fraction

V1-V2 = 2V2 + V3 + V3-V4

Substitute known node voltages:

These equations written in matrix form:

Using Cramer’s rule:

Io = V3 = 38/5 = 38/10 mA = 3.8 mA 2k 2k