Node Voltage Method, find the unknown nodes.

Thread Starter

r-persaud5000

Joined Nov 22, 2016
3
Hi,

I'm stuck on this node voltage analysis circuit, I understand Node A is equal to 100V because it is coming out of the voltage source. However I can't seem to see the KCL equation for solving the nodes of this circuit, can anyone help me understand this better? My understanding of node voltage analysis is basic at best. I have uploaded a jpeg of the circuit in question.

Node Voltage Question.png

Thanks,

Randy
Humber School of Applied Technology


Moderators note: shown image full size.
 

wayneh

Joined Sep 9, 2010
16,114
Treat A to C and C back to ground as if they were single resistors. In each case you have 2 in series, in parallel with a single resistor. You can calculate an equivalent resistance. This will give you the voltage at C. The rest should be straightforward after that.
 

WBahn

Joined Mar 31, 2012
24,854
I'm stuck on this node voltage analysis circuit, I understand Node A is equal to 100V because it is coming out of the voltage source.
That's actually not why Node A is equal to 100 V. It is 100 V because the 100 V source establishes a 100 V difference between Node A and the bottom-left node AND because the bottom-left node has arbitrarily been designated as having a voltage of 0 V. I can't tell if you grasp this very important distinction or not. A very common mistake that people make is assigning a voltage to the node attached to the positive side of a voltage source as simply being equal to the voltage of that source regardless of what the other side of the source is connected to.

However I can't seem to see the KCL equation for solving the nodes of this circuit, can anyone help me understand this better? My understanding of node voltage analysis is basic at best. I have uploaded a jpeg of the circuit in question.
View attachment 115768
First, forget about Nodal Analysis and see if you can figure out the answer using what you already know. Can you reduce the resistors to a single resistance in order to find the total current? Can you reduce the resistors to an equivalent network of two resistors in series (connected at Node C)? to find the voltage at Node C? Given the voltage at Node C can you determine the voltages at Nodes B and D? If so, do that and put it aside to use as a check on your work later.

How you go from there depends a bit on what you have covered so far and how you have been shown to approach problems. If you have not covered the notion of "essential nodes" (other names are used) yet then you are probably expected to apply nodal analysis at each node directly on the circuit as given. That way you can get practice working with a simple circuit that you can easily analyze as above yet still get practice coming up with and solving multiple node equations. You will then learn that this actually results in more equations and more unknowns than are really needed.

To do Nodal Analysis simply go through each unknown node (B, C, D) and write the KCL equation at that node using the convention that currents flow outward from each node into each connected branch. So if you have a resistor, Rbc, connected between Node B and Node C, then the current flowing outward from Node B is (Vb-Vc)/Rbc. When you write the equation for Node C, the current flowing outward from Node C through that same resistor is (Vc-Vb)/Rbc. As per KCL, the total current flowing outward from each node must be equal to zero.
 

DGElder

Joined Apr 3, 2016
351
First to make it easier to see what is going on redraw the circuit so that the right hand point of the triangle shaped branches is joined with node C - these are actually one node. B and D are not essential nodes so you can ignore them in your KCL equation. The ground works well as a reference node so you can go ahead and call that zero volts. You know that all the current that flows in the circuit comes from the voltage source. And you can see that all of that current must flow in and out of node C. So you only need one KCL equation, with one unknown variable Vc. Once you calculate Vc then the other nodes are easily determined one at a time with ohms law or resistor voltage divider equation.
 

Thread Starter

r-persaud5000

Joined Nov 22, 2016
3
To do Nodal Analysis simply go through each unknown node (B, C, D) and write the KCL equation at that node using the convention that currents flow outward from each node into each connected branch. So if you have a resistor, Rbc, connected between Node B and Node C, then the current flowing outward from Node B is (Vb-Vc)/Rbc. When you write the equation for Node C, the current flowing outward from Node C through that same resistor is (Vc-Vb)/Rbc. As per KCL, the total current flowing outward from each node must be equal to zero.
Thank You for taking your time to help me, in your KCL equations are you writing it as (Vb-Vc)/RbRc or (Vb-Vc)/Rb,c. In other words are you simplifying the circuit by combining the resistor values 5.6kohms and 1.0kohms? or are you multiplying the resistor values?

MOD NOTE: Edited to make quoted material appear quoted.
 
Last edited by a moderator:

MrAl

Joined Jun 17, 2014
6,646
Hi,

Real quick...

In any circuit the first thing you should look for is series connections and parallel connections, and this can reduce the work by a lot.

You should first combine the series resistors, then combine the parallel resistors, but sometimes you have to do that in the other order and sometimes apply that several times. In this circuit for example once you combine two series resistors you end up with two parallel resistors so then you can combine them. You'll see the circuit get more and more simple and then you'll be able to solve it easily.

Once you combine all resistors as much as possible, then you can go from there with your chosen method of analyzing the circuit.
 

WBahn

Joined Mar 31, 2012
24,854
Thank You for taking your time to help me, in your KCL equations are you writing it as (Vb-Vc)/RbRc or (Vb-Vc)/Rb,c. In other words are you simplifying the circuit by combining the resistor values 5.6kohms and 1.0kohms? or are you multiplying the resistor values?

MOD NOTE: Edited to make quoted material appear quoted.
If you multiply the resistances together and then divide a voltage difference by the result, you do NOT end up with a current, you end up with a current divided by a resistance, which is basically meaningless.

Rbc refers to the resistance between nodes B and C. In your circuit it is 5.6 kΩ.

If you leverage knowledge of essential nodes, then you use the voltage difference between Vc and Va and divide by the resistance Rca, which is the resistance between Nodes C and A which is the sum (not the product) of the 5.6 kΩ and the 4.7 kΩ resistors.
 

shteii01

Joined Feb 19, 2010
4,647
The hilarious thing about this thread is that OP has not posted a single node equation.

The reason I point this out is because I, personally, prefer Mesh-Current Method (KVL?). But! Even I can write out node equations. I would not like them. I would not want to solve them. But! I still know how to set them up. Oh! And OP claimed that they have basic understanding of node voltage analysis. LOL.
 

WBahn

Joined Mar 31, 2012
24,854
Let's shift gears a bit (while we wait for the TS to show some effort) and do some estimating.

Between A and C we have two parallel paths of 10 kΩ and 10.3 kΩ. That puts the equivalent resistance of this path at no less than 5.0 kΩ and no more than 5.15 kΩ.

Between C and ground we have two parallel paths of 1.8 kΩ and 2.0 kΩ. That puts the equivalent resistance of this path at no less than 0.9 kΩ and no more than 1.0 kΩ.

Using the easy bounds, that means that we have a total resistance of about 6 kΩ and the voltage at C will be about 1/6 of the applied voltage, or about 17 V (16.7 V) while the total current will be about 17 mA. The current in each individual branch will be pretty close to half of this. The voltage at D will be one-half the voltage at C, or about 8.5 V. The voltage at B will be about 55% of the way from B to A. Since the voltage between B and A is about 83 V, half would be 42.5 V and another 5% would add about 4.3 V taking it to 46.8 V. Adding 16.7 V to that and we have 63.5 V.

If the actual answers aren't pretty close to these, then we know we have done something wrong (and it's possible that it's the estimate that is wrong).
 
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