Nodal Analysis with Voltage Source

Nodal Analysis with Voltage Source

There are 2 cases:

1st case:  If a voltage source is connected between the reference
node and a non reference node, we simply set the voltage at the non reference
node equal to the voltage of the voltage source.

2nd case: If the voltage source (dependent or independent) is connected between two non reference nodes, the two non reference nodes form a generalized node or super node; we apply both KCL and KVL to determine the node voltages.

To know about super node is a formed by enclosing a (dependent or independent)
voltage source connected between two non reference nodes and any
elements connected in parallel with it.

Example:

Question:
     Solve for the current through the 5 ohm resistor and the current throughthe 18V source using Node-Voltage Analysis.

We don't know the current through the 12V nor through the 18V, and we cannot use Ohm's law to find it, since they are not resistors. So add unknown current labels to the circuit, as shown:

Now we can write the KCL equations:
iy + (V3 - V2)/5 - ix = 0

Current leaving V4

-iy + (V4 - V1)/2 = 0

Current leaving V5

ix + (V5 - 0)/1 = 0

Extra Equations

We now have five equations, but we have seven unknowns (the 5 unknown voltages plus 2 more unknown currents). We can get two extra equations -- just look at the voltage sources that caused the problems in the first place, where we had to add ix and iy. We'll write the voltage relationships between the node voltages:

V4 - V3 = 12V
V3 - V5 = 18V


Now we have 7 equations and 7 unknowns. To solve these equations, let's simplify it down to fewer equations. Solve for V4 and solve for V5 from those last two we added:

V4 = V3 + 12V
V5 = V3 - 18V


Substitute these two equations into the previous five, so that we eliminate V4 and V5. We now have 5 equations and 5 unknowns:

(v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
iy + (V3 - V2)/5 - ix = 0
-iy + ([V3 + 12] - V1)/2 = 0
ix + ([V3 - 18] - 0)/1 = 0


Continue using substitution (Gaussian elimination) to narrow it down. Next, solve for ix in the last equation:
ix = 18 - V3 and substitute back to get 4 equations and 4 unknowns:

(v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
iy + (V3 - V2)/5 - [18 - V3] = 0
-iy + ([v3 + 12] - V1)/2 = 0

Next, solve for iy in the last equation:
iy = ([v3 + 12] - V1)/2 and substitute back to get 3 equations and 3 unknowns:

(V1 - [V3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
([V3 + 12] - V1)/2 + (V3 - V2)/5 - [18 - V3] = 0

Now multiply each equation by a constant to clear out fractions. That is multiply by the Least Common
Denominator (multiply first by 6, second by 60, third by 10).

3 * (V1 - [V3 + 12]) + 2 * (V1 - V2) + 3 * (V1 - 0) = 0
12 * (V2 - V3) + 15 * (V2 - 0) + 20 * (V2 - V1) = 0
5 * ([V3 + 12] - V1) + 2 * (V3 - V2) - 10 * [18 - V3] = 0

Next, multiply through

3*V1 - 3*V3 - 36 + 2*V1 - 2*V2 + 3*V1 = 0
12*V2 - 12*V3 + 15*V2 + 20*V2 - 20*V1 = 0
5*V3 + 60 - 5*V1 + 2*V3 - 2*V2 -180 + 10*V3 = 0

Now gather like terms or use matrix (basket) to solve:
8*V1 -2*V2 -3*V3 = 36
-20*V1 +47*V2 = 12*V3 = 0
-5*V1 -2*V2 + 17*V3 = 120