# Nodal Analysis

Discussion in 'Homework Help' started by badCircuitAnalyst, Feb 12, 2015.

Feb 12, 2015
5
0
I'm working on his problem and I know the answer and how it is derived. There is an older thread on this specific problem; however, I do not understand why the answer was derived in such a fashion.

My approach to the problem using nodal analysis was to add the three currents stemming from the top node and set them equal to zero.

I set the node where all three resistors meet to be the reference node. Substituting the currents with voltages and resistance yielded:
(30V/2kΩ) + (20V/5kΩ) + (Vo/4kΩ) = 0
The correct answer is derived from the following equation:
((30V-Vo)/2kΩ) + ((20V-Vo)/5kΩ) + (Vo/4kΩ) = 0

I don't understand why Vo is supposed to be subtracted from each of the voltage sources when it is not between the path from the non-reference node to the node. I find I have problems similar to this in many circuit problems. I do well in subjects from understanding why things work the way that they do, and it seems that in this subject I seldom understand what assumptions can be made or why. Any help is greatly appreciated!

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2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,070
What is the current through R1 in this simple circuit?

Since V1 is in series with R1, then the voltage of V1 added to the voltage across R1 must add up to Vo (10V in this example).

What is the voltage across R1?. Since I named the intervening node to be "tap", and I choose the bottom node to be the 0V reference (Gnd) in this circuit, then the voltage across R1, by Ohms law is (Vtap-0) =IR = 2000*I = Vtap.

So, 30 + Vtap = 10. Rearranging, Vtap = 10-30 = -20, so by substitution in the equation above, I=Vtap/2000 = -20/2000 = -0.01
That means that V1 is supplying 10mA, and the current is actually flowing upwards toward Vo.

If we didn't assume that V0 was known apriori, we could say that Idownwards = Vtap/R1 = (Vo-30)/R1, which means Iupwards = -Idownwards = (30-Vo)/R1.

3. ### mjakov New Member

Feb 13, 2014
20
5
Hi!
Let us denote the uppermost node with a letter A, and the lowermost node with letter B as reference or ground. This means that all other node voltages are in reference to the ground voltage, i.e., node B. Having a common reference is useful, since in this way we can setup our system of equations for larger circuits. Also, it is usual to name all the nodes you analyze.

If you assume that all three currents are flowing out of node A, then their sum needs to be zero - Kirchoffs Current Law. For the current through the rightmost resistor you just need to find the voltage difference between the resistor ends and divide by the resistance - Ohm's law. For the leftmost and middle currents the procedure is the same. However, to obtain the voltage on the upper end of each resistance you need to subtract the value of each voltage source from Va. The voltage sources create a constant voltage difference between node A and the nodes below each source of 30V and 20V respectively. There is a voltage drop towards the resistors, because node A is on the plus side and the upper terminals of the two resistors are on the negative side of the voltage sources.

4. ### WBahn Moderator

Mar 31, 2012
19,154
5,177
The problem is that you don't understand Ohm's Law (don't feel bad, it's a stumbling block for many).

The key is to understand that Ohm's Law relates the resistance of a resistor to the voltage across THAT resistor and the current through THAT resistor. The voltage across the 2Ω resistor is NOT 30V. 30V is the voltage across the 30V supply, which is the voltage between the top node and the node on top of the 2kΩ resistor. The voltage across the 2kΩ resistor is the difference between the voltage on the node on top of the resistor and the voltage on the node on the bottom of the resistor (since you are defining the current to be the current flowing through the resistor from top to bottom). The voltage on top of the resistor is (30V-Vo) and the voltage on the bottom is 0V. So the voltage across the resistor is (30V-Vo).

It can help to label all of the nodes and then work from the node labels. If we label the top node Va, the node on top of the 2kΩ resistor Vb, the node on top of the 5kΩ resistor Vc, and the bottom node Vg (ignoring that you've defined Vg=0V for now), then our node equation for Node A becomes:

((Vb-Vg)/2kΩ) + ((Vc-Vg)/5kΩ) + ((Va-Vg)/4kΩ) = 0

We then have the following constrained imposed by the two supplies:

(Va-Vb) = 30V
(Va-Vc) = 20V

We also know that our desired result, Vo, is given by

Vo = (Va-Vg)

We can then substitute all of this in and get the answer even if we never assign any node as our reference.