Node voltage Problem? - Simple

[fouadalnoor]

Hello guys,

I am trying to solve this circuit by finding out what the voltage across the R1 resistor is...

I cant seem to get the right answer (21.7mV)

Any help!

PS: we are supposed to use Nodal voltage Analysis

 

[MikeMl]

The right answer is 0.0217391 V, see below:

Use Kirchoff's current law (Summation of currents into any node is zero) to write a system of n equations in n unknowns, and the solution will come out of that. I could do it using algebra, but I am lazy, so I let LTSpice do it for me

 

[fouadalnoor]

Well, I get as far as this:

1. I know that The current splits through the two 500ohm resistors is -0.5mA

2. I want the current going through R1 to get the voltage. I know the current the current going into the node N1 will split between R3 and R2.

Here I get stuck.

I know that I can take 0.5mA from the current going through R3 and I can take 0.5mA from the current through R2 to get the two currents going into R1.

I can then take them away from each other and multiply by R1 to get the voltage across R1.

BUT, what's the current's flowing through R3 and R2? Do I use current divider? ... I keep getting it wrong.

Thanks for the help again Mike! (even though this is separate from the Logic stuff!)

 

[Ratchit]

fouadalnoor,

Well, I get as far as this:

1. I know that The current splits through the two 500ohm resistors is -0.5mA

2. I want the current going through R1 to get the voltage. I know the current the current going into the node N1 will split between R3 and R2.

Here I get stuck.

I know that I can take 0.5mA from the current going through R3 and I can take 0.5mA from the current through R2 to get the two currents going into R1.

I can then take them away from each other and multiply by R1 to get the voltage across R1.

BUT, what's the current's flowing through R3 and R2? Do I use current divider? ... I keep getting it wrong.

Thanks for the help again Mike! (even though this is separate from the Logic stuff!)
​

All that verbiage and musing above is not giving you the node equations you said are wanted. Can you make an attempt to write the node equations?

Ratch

 

[MikeMl]

Here is some guidance:

ps: I just noticed that V(N3) = 0.

 

[MrAl]

Hello there,


It always helps to look for symmetricalness in a circuit and in this circuit we find just that with R4 being equal to R5. This leads to quick simplifications. Since these simplifications are best shown graphically, i've included a diagram.

Referring to the diagram, in Fig 1 we have the original circuit. If we look carefully we can see that R1 is in parallel with two resistors R4 and R5 which are both the same values (500 ohms each). This means that if we were to divide R1 into two pieces R1a and R1b that were exactly one half the original value then the center tap would have the same potential as the center tap of the two 500 ohm resistors, and that means we can short the two center taps together and end up with the same circuit. This is the circuit shown in Fig 2.
Note that with this seemingly small transformation we have now greatly simplified the circuit, because now R1a is in parallel with R4 and R1b is in parallel with R5 and this means we can calculate a new value for both of these and reduce the circuit to four resistors This is shown in Fig 3.
Now we note that both sides of the circuit each have two resistors in series, so we can again simplify by calculating their combined series resistance on each side. This circuit is shown in Fig 4.
Finally we see that this circuit has been reduced to two branches where we can use simple current division to calculate the two branch currents.
Now knowing both branch currents we can refer back to the original circuit (Fig 5 with blue arrowheads showing the branch currents) and use simple Ohms Law to calculate the voltage across R3 and across R2, then subtract those two to get the final voltage across R1.
Note that if R4 and R5 where not the same we would have had to divide R1 up differently.

 

[fouadalnoor]

Thanks a LOT! I do find it a little hard to see how R1 is in parallel with R4 and R5, though I think if I look at it a little longer I shall get it.

 

[The Electrician]

Thanks a LOT! I do find it a little hard to see how R1 is in parallel with R4 and R5, though I think if I look at it a little longer I shall get it.

The technique MrAL has shown here doesn't work in general. Just because R4 and R5 are equal in value, and their series combination is in parallel with R1 doesn't mean that you can split R1 into two equal value resistors (each equal to R1/2) and connect their junction to the junction of R4 and R5, and expect ALL the circuit properties to remain unchanged. Whether or not this is true depends on the rest of the circuit.

Hello there,


It always helps to look for symmetricalness in a circuit and in this circuit we find just that with R4 being equal to R5. This leads to quick simplifications. Since these simplifications are best shown graphically, i've included a diagram.

Referring to the diagram, in Fig 1 we have the original circuit. If we look carefully we can see that R1 is in parallel with two resistors R4 and R5 which are both the same values (500 ohms each). This means that if we were to divide R1 into two pieces R1a and R1b that were exactly one half the original value then the center tap would have the same potential as the center tap of the two 500 ohm resistors, and that means we can short the two center taps together and end up with the same circuit. This is the circuit shown in Fig 2.
Note that with this seemingly small transformation we have now greatly simplified the circuit, because now R1a is in parallel with R4 and R1b is in parallel with R5 and this means we can calculate a new value for both of these and reduce the circuit to four resistors This is shown in Fig 3.
Now we note that both sides of the circuit each have two resistors in series, so we can again simplify by calculating their combined series resistance on each side. This circuit is shown in Fig 4.
Finally we see that this circuit has been reduced to two branches where we can use simple current division to calculate the two branch currents.
Now knowing both branch currents we can refer back to the original circuit (Fig 5 with blue arrowheads showing the branch currents) and use simple Ohms Law to calculate the voltage across R3 and across R2, then subtract those two to get the final voltage across R1.
Note that if R4 and R5 where not the same we would have had to divide R1 up differently.

This only works because of a couple of special circumstances in this circuit:

1) The problem asks for the DIFFERENCE between V(N2) and V(N3); that DIFFERENCE is unchanged by the change you made going from your Fig 1 to your Fig 2. The individual voltages at node 2 and node 3 are different in the two cases, however.

2) It's only because of one additional special circumstance that the DIFFERENCE is unchanged by the connection you made from the junction of R1a and R1b to the junction of R4 and R5. That circumstance is the fact that node 1 is energized by a current source (which has theoretically an infinite source impedance). If node 1 is driven by a voltage source, or if there were another resistance from node 1 to node 4 causing the equivalent source to have less than infinite output resistance, then the DIFFERENCE between V(N2) and V(N3) would not be the same in Fig 1 and Fig 2.

Generally, making the connection between the junction of R1a and R1b and the junction of R4 and R5 is only permissible (causes no change in ANY of the circuit behavior) if the voltage at those two junctions is identical BEFORE you connect them.

 

[mbarazeen]

see the attached file, this is how you have to work to solve it using nodle method.

 

[Grossel]

Convert to from triangle into star structure

Hi.

Also an easy method to solve this is to convert R1, R2 and R3 into an equivalent star resistor network.

 

[MrAl]

The technique MrAL has shown here doesn't work in general. Just because R4 and R5 are equal in value, and their series combination is in parallel with R1 doesn't mean that you can split R1 into two equal value resistors (each equal to R1/2) and connect their junction to the junction of R4 and R5, and expect ALL the circuit properties to remain unchanged. Whether or not this is true depends on the rest of the circuit.



This only works because of a couple of special circumstances in this circuit:

1) The problem asks for the DIFFERENCE between V(N2) and V(N3); that DIFFERENCE is unchanged by the change you made going from your Fig 1 to your Fig 2. The individual voltages at node 2 and node 3 are different in the two cases, however.

2) It's only because of one additional special circumstance that the DIFFERENCE is unchanged by the connection you made from the junction of R1a and R1b to the junction of R4 and R5. That circumstance is the fact that node 1 is energized by a current source (which has theoretically an infinite source impedance). If node 1 is driven by a voltage source, or if there were another resistance from node 1 to node 4 causing the equivalent source to have less than infinite output resistance, then the DIFFERENCE between V(N2) and V(N3) would not be the same in Fig 1 and Fig 2.

3)Generally, making the connection between the junction of R1a and R1b and the junction of R4 and R5 is only permissible (causes no change in ANY of the circuit behavior) if the voltage at those two junctions is identical BEFORE you connect them.

Hello Electrician,

Actually, this is a general topology transformation that works in any circuit and the reason i showed this example was because i wanted to show how we can sometimes simplify circuits by doing these kind of transformations. Note how with this circuit we never had to do any real network analysis, just simple calculations and simple graphical transformations. That's the beauty of it. There's another example right here on this forum where we did another network using ideas like this (topology transformations with equivalent impedances) but im not sure where it is now.

1) The WHOLE idea was to find that voltage and we of course keep this end goal in mind when we start out which means we know before hand that what we are doing is going in the right direction. That's part of what network simplification is all about. We dont simplify if it's not going to produce the required result, that's obvious, but sometimes we can find an intermediate result which aids in the determination of the end result. For example, sometimes we calculate the parallel resistance of two resistors even though in the end we need to know the current through only one of those resistors. Two resistors in parallel with another resistor in series with those two would mean we would first want to calculate the parallel resistance in order to find the total current. Once we find the total current we could then use current division to calculate the individual currents. There we had to modify the topology first too (three resistors into two resistors) although we dont usually think of it that way. Once done, we get information that leads to the answer we really are seeking.
2)Sometimes it's possible to allow a change just to get some intermediate result which helps in the end result as in #1 above with the parallel resistors.
3)Actually, if you note my last sentence in my previous post, you'll see that i noted that we might have to choose the two resistors differently according to the ratio of the other two resistors. In other words, we dont depend on the happenstance of the voltage between the two new resistors, we FORCE that voltage to ALWAYS be the same as the other node so that we can ALWAYS short the two. That's part of the method. In other words we have to impose the constraints:
A) R1=R1a+R1b (to satisfy the impedance requirement)
and
B) R4/R1a=R5/R1b, the ratios of the two pairs of resistors have to be the same (to satisfy the node voltage requirement).
With the resistors shown, we satisfy both of these conditions so the new network is equivalent and thus simplifies the math greatly. For another example, if R1 was 1800 ohms and R4 was 500 ohms and R5 was 400 ohms we would have had to spit R1 up into 1000 ohms and 800 ohms instead.
Taking both constraints into account we end up with the following formula for R1a:
R1a=(R1*R4)/(R5+R4)
and then R1b=R1-R1a of course.

This isnt the only kind of topology transformation that comes up in network analysis though, see also the more commonly known delta to wye and wye to delta transformations here:
https://en.wikipedia.org/wiki/Y-%CE%94_transform
Note that we end up with some pretty drastic changes sometimes but the resulting sub circuit still acts the same way.

 

[The Electrician]

Hello Electrician,

Actually, this is a general topology transformation that works in any circuit...

In the original circuit, replace the 1 mA current source with a 1 volt voltage source. Now the voltage across R1 is 64.51613 millivolts.

What do you get using your transformations? The current in R3a is now 4.44444 mA, which, if the same current flows in the original R3 means that there is a voltage drop of 125/225 = .55555 volts. The current in R2a is now 2.8571 mA, which, if the same current flows in the original R2 means that the drop across R2 is .7142857 volts. The difference is 158.73 millivolts.

This is not the same as in the pre-transformation circuit.

Apparently, this transformation doesn't work (get the correct answer) for just any circuit.

 

[MrAl]

In the original circuit, replace the 1 mA current source with a 1 volt voltage source. Now the voltage across R1 is 64.51613 millivolts.

What do you get using your transformations? The current in R3a is now 4.44444 mA, which, if the same current flows in the original R3 means that there is a voltage drop of 125/225 = .55555 volts. The current in R2a is now 2.8571 mA, which, if the same current flows in the original R2 means that the drop across R2 is .7142857 volts. The difference is 158.73 millivolts.

This is not the same as in the pre-transformation circuit.

Apparently, this transformation doesn't work (get the correct answer) for just any circuit.

Hello again,


Ok i see what you are getting at now. The source has to be a current source or else we can not meet the requirements for the node. The question was for a current source though, so we did get it right.
However, given a voltage source then we would have an entirely different circuit which we could approach using different transformations that do lead to a valid solution (see attached diagram).

Transforming the circuit from Fig1 to Fig4 in the diagram, again simply combining various resistances we end up with a simple voltage divider circuit where we can easily calculate the partial voltage at node 2. Now knowing this voltage, we can backtrack and calculate the partial voltage for node 3. Then, transforming the same circuit from Fig 5 to Fig 8 we can again end up with a simple voltage divider to calculate the other partial voltage for node 3 and again backtrack and calculate the other partial voltage for node 2. We then have two partial voltages for each of the nodes 2 and 3.
The voltage for node 2 is then the sum of partials for node 2, and the voltage for node 3 is then the sum of partials for node 3. Knowing these two again we can subtract and get the voltage across R1.

I guess i find these kinds of circuit transformation very interesting because they simplify the math. It's not usually too hard to transform the circuit but i guess it could get tricky.

 

[The Electrician]

The circuit under analysis here is, of course, a loaded unbalanced Wheatstone bridge.

You've used the superposition method after splitting the source into two sources. This splitting technique is well known:

**broken link removed**

and superposition is a method that is generally applicable to any circuit. However, the method you used in post #6 is not generally applicable because it connects two nodes that are not at the same voltage before the connection is made. It worked in that case (where the difference between V2 and V2 was wanted; the individual values of V2 and V3 changed after your transformation) because the source was a current source, but it won't always work for all circuits whereas superposition will.

However, with respect to simplifying math, it's true that some transformations simplify math, but it is a well-known deficiency of the superposition method that it increases the required math because the circuit has to be solved repeatedly for each source. The source splitting technique you used in post #13 doubles the number of sources and the amount of math.

Using minimum math, this circuit can be solved with two simple equations in two unknowns, no transformations needed:

7*V2 - 2*V3 = 4
-2*v2 + 5*V3 = 2

 

[MrAl]

Hello again,


I assumed we already thought about the more conventional ways of analyzing circuits, I like to stay open to different ways of doing it. If you are trying to tell me that superposition is not a good thing you are going to have a very hard time convincing me.

Note that the transformations in the first circuit almost allow us to solve the entire circuit in our heads without using simultaneous equations, which some people would prefer not to use.

Also note that i am not trying to force you to do it this way or that way, do it however you feel more comfortable with.

 

[The Electrician]

If you are trying to tell me that superposition is not a good thing you are going to have a very hard time convincing me.

Quite the contrary; I said "...superposition is a method that is generally applicable to any circuit." My only criticism of superposition is that it usually involves more math because you have to solve the circuit repeatedly for each source. But, what is good about it is that it always gives correct answers, no matter whether the circuit is driven by current sources or voltage sources.

Also note that i am not trying to force you to do it this way or that way, do it however you feel more comfortable with.

Nor am I trying to force you to use any particular method. My purpose was to point out that you offered a method (in post #6) that isn't generally applicable, and after I described its shortcomings, your later asserted "...this is a general topology transformation that works in any circuit.", an incorrect assertion. The part of the transformation that you described that doesn't "work in any circuit" was the connecting of the junction of R1a and R1b to the junction of R4 and R5.

When one performs various transformations to reduce a circuit, one must be sure that at each step of the way the reduced circuit still has the same behavior in the rest of the circuit that hasn't been transformed. When connecting two circuit nodes together, the rule that allows us to know that the rest of the circuit isn't affected, is that the voltages present on the two nodes must be the same before they are connected.

This was not the case in post #6, and the voltages at node 2 and node 3 changed between your Fig 1 and Fig 2. Their difference didn't change because the source was a current source, but with a voltage source the difference would have changed. Had the problem been to find one or both of the individual voltages at nodes 2 and 3, the method you described fails.

You didn't point out that the transformation you used is not generally applicable, and a reader of the forum who didn't realize that, might try to use the method with another circuit which didn't use a current source and get a wrong result. In fact, the OP commented in post #7, "I do find it a little hard to see how R1 is in parallel with R4 and R5..."

It seemed to me that the general inapplicability of the method of connecting nodes whose voltages are different before the connection is made should be mentioned lest readers take if for a method that always gives correct results. That's why I described the shortcoming of the method; not to force anyone to use a method they don't like, but to suggest that they beware of using a method that may not always give a correct result just because it gives a good result in a special case.

 

[MrAl]

Quite the contrary; I said "...superposition is a method that is generally applicable to any circuit." My only criticism of superposition is that it usually involves more math because you have to solve the circuit repeatedly for each source. But, what is good about it is that it always gives correct answers, no matter whether the circuit is driven by current sources or voltage sources.

That's not entirely true however. If you are going to criticize superposition "because it usually involves more math because you have to solve the circuit repeatedly for each source" that's not a good enough argument because the resulting circuits are often greatly simplified (as in the example i did previously where we reduced the circuit into simple voltage dividers). Thus, if you are going to say there are more circuits to solve then you also have to say that those circuits are simpler too.

You didn't point out that the transformation you used is not generally applicable, and a reader of the forum who didn't realize that, might try to use the method with another circuit which didn't use a current source and get a wrong result. In fact, the OP commented in post #7, "I do find it a little hard to see how R1 is in parallel with R4 and R5..."

Yes, after careful consideration i have to agree, and am in favor of abandoning the idea in order to avoid confusion. I went through the trouble of proving that the method will be more difficult to apply for some circuits, and for some circuits it may not be possible to meet the requirements of the node. It's a bit hard to determine if the requirements of the node are met or not sometimes, so i think it is best avoided pending further investigation.
We still have the other great methods on hand so nothing's lost.

 

[mbarazeen]

Hello there,


Referring to the diagram, in Fig 1 we have the original circuit. If we look carefully we can see that R1 is in parallel with two resistors R4 and R5 which are both the same values (500 ohms each). This means that if we were to divide R1 into two pieces R1a and R1b that were exactly one half the original value then the center tap would have the same potential as the center tap of the two 500 ohm resistors, and that means we can short the two center taps together and end up with the same circuit. .

dear MrAl,

just looked at the way you solved it, as you mentioned dividing R1 into two to make it in parellel to R4 & R5 will not work in this case. after divide it into two & beore connecting, if you note the direction of voltages R1a & R1b will have a voltage drop in the same direction but R4 & R5 will be in opposite. the junction will not have the same potential to connect them to resolve the problem.

if you have doubt work out in both methodss & check the answeres.

 

[MrAl]

dear MrAl,

just looked at the way you solved it, as you mentioned dividing R1 into two to make it in parellel to R4 & R5 will not work in this case. after divide it into two & beore connecting, if you note the direction of voltages R1a & R1b will have a voltage drop in the same direction but R4 & R5 will be in opposite. the junction will not have the same potential to connect them to resolve the problem.

if you have doubt work out in both methodss & check the answeres.

Hello,


Im not sure what the reason behind your post is. I already mentioned yesterday that i was in favor of abandoning the idea even though it worked in the circuit with the current source. I also noted that i had proved that it would not work in some circuits.
Read my previous post.