PURDUE UNIVERSITY NORTH CENTRAL
Electrical & Computer Engineering Technology Department
ECET 107 (Prof. Smith)
Fall, 2008
Solving for Voltages and Currents with Nodal Analysis
Let’s solve the same circuit that we used in the Mesh Analysis handout, but using Nodal Analysis.
STEP 0 – NO VOLTAGE SOURCES ALLOWED!
Nodal analysis will only work if there are no voltage sources in the circuit. Since this circuit has three
voltage sources, we must convert them all to current sources, as shown below.
NOTE: Both circuits always have the same resistor value!
In this circuit, we would make the following substitutions:
E1 → I1 = E1 / R1 = 2V / 1Ω = 2 A
E2 → I2 = E2 / R2 = 4V / 1Ω = 4 A
E3 → I3 = E3 / R5 = 2V / 4Ω = 0.5A
ECET 107
Nodal Analysis Handout
Fall, 2008 – Page 2
STEP 1 – DRAW & LABEL NODAL VOLTAGES.
Number each node, except the ground or reference node at the bottom. Label each node’s voltage as
V1, V2, etc. NOTE: This V1 and V2 are not in any way related to E1 and E2 from the original circuit
Here’s the circuit after making the voltage substitutions and labeling the nodes:
STEP 2 – ADD ALL CONDUCTANCES ATTACHED TO EACH NODE.
At each node, add up the conductance of all the resistors attached to it. Remember, a resistor’s
conductance is one over the resistance: G = 1 / R. (The units used for conductance is the Siemans (S),
named after Ernst Werner von Siemens, who founded the German electrical firm of Siemens AG.) This
gives us the following equations:
Node 1 → G1 + G2 + G3 = 1/R1 + 1/R2 + 1/R3 = 1/1 + 1/1 + 1/2 = 1 + 1 + 0.5 = 2.5S
Node 2 → G3 + G4 + G5 = 1/R3 + 1/R4 + 1/R5 = 1/2 + 1/3 + 1/4 = 0.5 + 0.333 + 0.25 = 1.083S
STEP 3 – SUBTRACT THE CONDUCTANCES THAT ARE BETWEEN NODES.
Subtract any shared conductances. This gives us the following equation:
Between Nodes 1 & 2 → G3 = - 1/R3 = -1/ 2 = -0.5S
STEP 4 – ADD CURRENT SOURCES INTO EACH NODE AND SUBTRACT THOSE FLOWING OUT.
At each node, add the sources flowing into the node and subtract those that are flowing out. This will
give us the total current flowing into the node. This yields the following equations:
Node 1 → I1 + I2 = 2A + 4A = 6A
Node 2 → I3 = -0.5A (Note that this is negative since it is flowing out of the node.)
SIDE NOTE: In Mesh Analysis, we solved the equation R * I = V in matrix form, to find I .
With Nodal Analysis, we will solve the equation I = (1 / R) * V = G * V in matrix form, to find V .
ECET 107
Nodal Analysis Handout
Fall, 2008 – Page 3
STEP 5 – WRITE THE EQUATION MATRICES.
Fill in the matrix G with (a) Step 2’s conductance sums on the diagonal and (b) Step 3’s shared
conductances in the other cells.
Fill in the matrix I matrix, using the current sums from Step 4.
Here’s a table showing all the values from Steps 2, 3 and 4. With a large circuit, or if you are not
comfortable with matrices, filling in this table first makes it easier.
Node 1 G’s
Node 2 G’s
Net Current Into Node
Node 1 G’s
Sum = 2.5S
(-) Shared = -0.5S
Sum = 6A
Node 2 G’s (-) Shared = -0.5S
Sum = 1.083S
Sum = -0.5A
Now, write the matrix equations for G and I :
G =
2.5
----0.5
-0.5
---1.083
STEP 6 – USE MATLAB TO SOLVE THE MATRIX EQUATION.
Now simply solve Ohm’s Law for V as follows:
G * V = I → G-1 * G * V = G-1 * I
I =
→
6
---0.5
V = G-1 * I
The Matlab screen is shown below. Remember (1) to begin and end a matrix the square brackets - [
and ], (2) leave a space between values and (3) separate each row with a semi-colon - ;.
EDU» G = [ 2.5 -0.5 ; -0.5 1.0833 ]
G =
2.5000
-0.5000
-0.5000
1.0833
EDU» I = [ 6 ; -0.5 ]
I =
6.0000
-0.5000
EDU» V = G^(-1)*I
V =
In other words,
2.5424
0.7119
Å Let’s write this as V2 = 712mV
V1 = 2.542V
V2 = 712mV
ECET 107
Nodal Analysis Handout
Fall, 2008 – Page 4
Let’s check our answers with PSPICE. (I labeled the current directions and voltages +/- from PSPICE.)
Looks like all our answers for V1 and V2 check out!
STEP 7 – USE THE MATLAB ANSWERS TO FIND ALL V’s AND I’s IN THE CIRCUIT.
Use these values to calculate all the circuit’s voltages and currents. As in the Mesh example, the
voltage polarities (+/-) and current directions from the original circuit are used as the reference. Thus,
a negative voltage or current means that it is actually the reverse of the direction labeled in the original
circuit. Also, we have to convert the current sources back into the original voltage sources. This makes
it easier to make a mistake, so be careful!
E1:
V = E1 = 2V
I = IR1 (Calculated below as 542mA, DOWNWARD.)
E2:
V = E2 = 4V
I = IR2 (Calculated below as 1.458A, UPWARD.)
V = E3 = 2V
E3:
I = IR5 (Calculated below as 678mA, DOWNWARD.)
R1:
V = V1 - E1 = 2.542V – 2V = 0.542V (The + side is on the TOP.)
I = V / R1 = 0.542V / 1Ω = 0.542A = 542mA (Direction is DOWN.)
V = V1 – E2 = 2.542V – 4V = -1.458V (The + side is on the BOTTOM.)
R2:
I = V / R2 = -1.458A / 1Ω = -1.458A (Direction is UP.)
V = V1 – V2 = 2.542V – 0.712V = 1.83V (The + side is on the LEFT.)
R3:
I = I2 = 915mA (Direction is LEFT to RIGHT.)
R4:
V = V2 = 712mV (The + side is on the TOP.)
I = V / R4 = 0.712V / 3Ω = 0.237A = 237mA (Direction is DOWNWARD.)
R5:
V = V2 + E3 = 0.712V + 2V = 2.712V (The + side is on the TOP.)
I = V / R5 = 2.712V / 4Ω = 0.678A = 678mA (Direction is DOWNWARD.)
Checking these answers with the Mesh Analysis results, we can verify that they are the same!