NODAL VOLTAGE ANALYSIS

In electric circuits analysis, nodal analysis, node-voltage analysis, or the branch current method is a method of determining the voltage (potential difference) between "nodes" (points where elements or branches connect) in an electrical circuit in terms of the branch currents.

STEPS INVOLVED

1. Select a reference node. Label it "REF". If there is a ground node in the circuit you may choose that as the reference, but it is not required. It is often advantageous to select the node with the most wires coming into it.
2. Label the other nodes in the circuit (V1, V2, etc.).
3. Write KCL equations for each node (in terms of the node voltages). Each term in the equation should be a current, but the term is written using ohm's law (I=V/R) so that they contain the node voltages you labeled in step 2.
• If you write all the currents entering a node, then the node you are focusing on will show up negative in that equation and all other node voltages will show up positive.
• If you write all the currents leaving a node, then the node you are focusing on will show up positive in that equation and all other node voltages will show up negative.
• The sum of all currents leaving equals zero (don't forget to write the =0).
4. Solve the N equations in N unknowns. This will give you all the node voltage values. If you are asked to find a current, just use Ohm's law to find the current in one more step.

What are some difficulties I might run into?

A voltage source in a node-voltage analysis can be a headache. If there is a voltage source connecting two nodes (with no resistance), then you must add an extra variable (call it Ix, the current through the source). Then you will need an extra equation as well. The extra equation is simply that the difference between the two node voltages is the value of the source.
A dependent source can be a headache. Simply write the currents as you normally would for an independent source (but writing down the dependency for the value). Then write an extra equation to describe the dependency in terms of the node voltages.
A capacitor or inductor can be a headache when writing the time-dependent node voltage equations. If the element is in series with another element in a branch, this is a good time to use an auxilary node.

QUES1

Solve for the current through the 5 ohm resistor and the current throughthe 18V source using Node-Voltage Analysis. The reference node was selected and the three other nodes of the circuit were labeled V1, V2, and V3. Note that the node voltages are all with respect to the reference. So for example, V1 is the voltage with the "+" at the label V1 and the "-" at the reference node. The following equations can then be written: Current leaving V1 (v1 - v4)/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0 Current leaving V2 (V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0 Current leaving V3 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: 8*V1 -2*V2 -3*V3 = 36 -20*V1 +37*V2 = 12*V3 = 0 -5*V1 -2*V2 + 15*V3 = 120 Continue with substitution until you have it solved.

QUES2

Solve for the current through the 5 ohm resistor and the current through the 4V source using Node-Voltage Analysis.

First, select a reference node and label the other nodes. Since each node has the same number of connected branches (4), we'll simply choose the bottom node as the reference. There are only two other nodes, which we will label V1 and V2.

Now write KCL at each node (except the reference):
KCL at V1:
-5A + V1/5 + (V1-V2)/10 + [V1-(V2+4)]/10 = 0
Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up.
KCL at V2:
(V2-V1)/10 + V2/2 - 2A + [V2-(V1-4)]/10 = 0
Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up.
Now gather terms (multiplying through by 10 to clear up the fractions):
4V1 - 2V2 = 54
-2V1 + 7V2 = 16
Now solve the set of 2 equations with 2 unknowns.
V1 = 17.08V
V2 = 7.17V
We can now determine the current through the 5 ohm by Ohm's law:
I = V1/5 = 3.41A
The current through the 4V source can be found as:
I = [V1-(V2+4)]/10 = 0.59A