Engineering Circuit problem -- Finding potential

[Bptashi]

Homework Statement

We wish to find Va-Vb given that R1 = 1000 ohm, R2 = 2000 ohm, R3 = 3000 ohm and emf = 10v.

Homework Equations

??

The Attempt at a Solution

I started by trying to simply the circuit but couldn't do so because of R3.
then i decided to add potential differences and currents.

Let the current from
DA = I1
AC = I2
AB = I3
CB = I4
CD = I5
BD = I6

loop: ACDA
-I2R2 - I3R1 + emf = 0
loop: ACBDA
-I2R2 - I4R3 -I6R2 +emf =0
loop: ABDA
-I3R1-I6R2+emf =0

current sum
I1 = I2+I3
I2 = I4+I5
I6 = I4+I3
I1 = I5+I6

And i am not sure how to go from here?
Do i put them in a matrix and get rref then backtrack?

 

[cnh1995]

I started by trying to simply the circuit but couldn't do so because of R3.

Have you studied delta-wye transformation?

 

[Jon B]

Since there are no series or parallel resistances use the formula for converting delta to wye networks so you will have series and parallel resistances to work with.

 

[Bptashi]

No we have not learned delta-wye transformation. This is for 1st year college physics(calc based).

 

[cnh1995]

No we have not learned delta-wye transformation. This is for 1st year college physics(calc based).

Delta-wye transform will quickly simplify the network and you'll have your answer in a couple of steps. It would be helpful if you looked it up.
In your attempt, you have assumed 6 currents. Instead, assume only three currents in the beginning and write the other three as the addition or subtraction of any two of these three. For example, current through AB is i1, through AC is i2 and through BC is i3 (from B to C) . Current through BD is simply i1-i3 and current through CD is i2+i3. Write proper KVL equations and solve for 3 variables i1, i2 and i3.

 

[gneill]

Have you covered nodal analysis or just basic KVL and KCL?

 

[Bptashi]

okay i found the 3 equations but instead of writing the actual numbers imma just use letters.

e - xi1-yi2 = 0; we set i2 = something * i1;
e-xi1-yi2-zi3 = 0; replace i2 with i1 which turns the equation to something like e-xi1-yi3 = 0;
e-xi1-yi2-zi3 = 0; same thing ^
then
we have two equations with 2 unknowns and i just solve them?
Thats it?

 

[Bptashi]

Have you covered nodal analysis or just basic KVL and KCL?

If KVL,KCL stands for k something voltage law, k something current law then yes
but nothing like nodal analysis.

 

[Jon B]

Kirchoff's Voltage and current law

 

[gneill]

If KVL,KCL stands for k something voltage law, k something current law then yes
but nothing like nodal analysis.

Okay, then you're going to have to work through the simultaneous equations. Write KVL loop equations for three loops using the currents that you've identified. I suggest the obvious loops:

although you are free to choose any three independent loops so long as every component is included in at least one of the loops.

Then work the KVL and KCL equations to solve for current I3 (the one you've identified as flowing from A to B).

 

[Bptashi]

sorry for replying so late(was studying for a test)... i found the textbook and it says something about the currents going through the resistors are the same because they are symmetrical. So that the current from ab is equal to the current cd but I have not read anywhere that states this. Could you explain? I mean i know that if the R3 was not there then the resistance would be equal(R1+R2= R2+R1) and the current would just be I/2 for each of the path.

 

[gneill]

The most convincing way to show that those currents are equal would be to solve the circuit. But there are arguments based on symmetry that can be used to convince you that it is so.

Suppose you were to reverse the polarity of the voltage source. Since no other changes are made to the circuit you can be sure that the only change in the currents would be to their directions (reversed). By symmetry path DC now has the same role that path AB had previously, the circuit configuration being topologically identical (just rotate the image 180°). So the current magnitude must be the same as before. Flip the voltage source polarity again and the current directions are restored and their magnitudes are kept the same. So the current AB must equal the current CD.

 

[Jon B]

sorry for replying so late(was studying for a test)... i found the textbook and it says something about the currents going through the resistors are the same because they are symmetrical. So that the current from ab is equal to the current cd but I have not read anywhere that states this. Could you explain? I mean i know that if the R3 was not there then the resistance would be equal(R1+R2= R2+R1) and the current would just be I/2 for each of the path.

Without R3 the current through A - C would be the same as the current through A - B. However the voltage across A - C (6 2/3V) would be twice the voltage across A - B (3 1/3V). The voltage difference between C and B is 3 1/3V. This source of current when adding R3 3000 Ohms will increase the current through A - B and decrease the current A - C. You still have to solve the mystery.

 

[NascentOxygen]

i found the textbook and it says something about the currents going through the resistors are the same because they are symmetrical. So that the current from ab is equal to the current cd but I have not read anywhere that states this.

I recommend that you stay with this exercise until you can see why the two R1 resistors must inevitably carry identical currents, and can use that fact to determine its value. Before you complete your course of study there is sure to be another question arise where symmetry affords a quick solution. This exercise will be good practice. [emoji57]

 

[Jon B]

I recommend that you stay with this exercise until you can see why the two R1 resistors must inevitably carry identical currents, and can use that fact to determine its value. Before you complete your course of study there is sure to be another question arise where symmetry affords a quick solution. This exercise will be good practice. [emoji57]

R3 adds to the imbalance in the current of the left and right branches and the two R1 resistors do not carry identical currents with R3 present. Op said "We wish to find Va-b."

 

[gneill]

R3 adds to the imbalance in the current of the left and right branches and the two R1 resistors do not carry identical currents with R3 present. Op said "We wish to find Va-b."

An "imbalance" in the potentials at nodes B and C does not preclude the currents through the two R1 resistors having the same magnitude (and vice versa).

For example, suppose that the battery EMF was 12V and that the identical currents through the R1's was such that they drop 4 V, while the currents through the R2's both drop 8 V. For both sides of the bridge the total potential drop is 8 + 4 = 12 V, and there will be a potential difference of 8 - 4 = 4 V across the bridge resistor.

 

[NascentOxygen]

An "imbalance" in the potentials at nodes B and C does not preclude the currents through the two R1 resistors having the same magnitude (and vice versa).

For example, suppose that the battery EMF was 12V and that the identical currents through the R1's was such that they drop 4 V, while the currents through the R2's both drop 8 V.

With these figures I see a problem, given that R₂ = 2R₁.

 

[gneill]

With these figures I see a problem, given that R₂ = 2R₁.

My example was meant to be arbitrary, not necessarily conforming to the given problem. Sorry, I should probably have stated this.

 

[NascentOxygen]

R3 adds to the imbalance in the current of the left and right branches and the two R1 resistors do not carry identical currents with R3 present. Op said "We wish to find Va-b."

If you would like to finish solving this relying on just Kirchoff's Laws, then we'll be able to see how close the R₁ currents are to being equal.

 

[Jon B]

VA to VB will not be close to VC to VD... Discovering that is the purpose of this problem.

 

[NascentOxygen]

VA to VB will not be close to VC to VD... Discovering that is the purpose of this problem.

Right. As for you working on the calculations to confirm what you're saying, it's a double thumbs-up, is it?

 

[Jon B]

Don't need calculations after working with Wheatstone Bridges selecting precision resistors and temperature cycling to select them for opposing temperature coefficients to use in pairs in devices that are used in variable temperature environments.

 

[The Electrician]

VA to VB will not be close to VC to VD... Discovering that is the purpose of this problem.

It appears that Bptachi has gone on his way. I don't think it would be giving away much if Jon B would tell us his result for the voltages Vab and Vcd. I'm rather curious myself.

 

[Jon B]

It appears that Bptachi has gone on his way. I don't think it would be giving away much if Jon B would tell us his result for the voltages Vab and Vcd. I'm rather curious myself.

I don't have a Wheatstone bridge. But if you put a short across R3 you can see that the upper R1 will provide the lion's share of current to the lower half of the circuit 1000 Ohm in parallel with 2000 Ohm and increase the voltage across the upper R1. What would be the best method to find the answer with a resistance in R3 position? The recommendation of cnh1995 sounded good.

 

[NascentOxygen]

What would be the best method to find the answer with a resistance in R3 position? The recommendation of cnh1995 sounded good.

The method you proposed in #3 will do fine.

 

[The Electrician]

I'm puzzled by your post #22, Jon B. I got the impression that perhaps you didn't need calculations to determine the voltages, because of your experience with selecting precision resistors for Wheatstone Bridges. Did I misunderstand what you said?