How to Clean Up Circuit Diagrams Without Contact?

[dimpledur]

Homework Statement

Essentially I am pursuing clues as to what I should do in order to clean up the following circuit:

.


I'm kind of confused regarding the portion where it says "no contact". Help?

 

[gneill]

Points b and c are connected, so they are the same node.

 

[dimpledur]

Does that mean I can merely add up the values of the resistors on lines abd and acd such that R=400+400+600+220=1620, thus Req=[(1/1620)+(1/320)]^{-1}?

 

[gneill]

Does that mean I can merely add up the values of the resistors on lines abd and acd such that R=400+400+600+220=1620, thus Req=[(1/1620)+(1/320)]^{-1}?

No, it means that between a and b there's 400 and 600 Ohms in parallel, and between b and d there's 400 and 200 Ohms in parallel.

Imagine taking hold of point c and swinging it up out of the page and over, placing point c on top of point b. Points b and c are connected, so they're the same node, so this is not altering the underlying topology of the circuit in any way.

 

[dimpledur]

Alright, then it would be as follows:
R1=[(1/400)+(1/600)]^{-1}
R2=[(1/400)+(1/200)]^{-1}

R3=R1+R2, and thus Req=[(1/R3)+(1/320)]^{-1}

 

[cupid.callin]

@ gneill

as bc is just a wire so points b a&c are at same potential so there will be no current through it ... so can't ve just ignore wire bc ?

 

[gneill]

@ gneill

as bc is just a wire so points b a&c are at same potential so there will be no current through it ... so can't ve just ignore wire bc ?

No. Point a is not wired to point c except via a resistor. Point a is not wired to point b except via a resistor. Only points b and c are connected together.

Points b and c are only at the same potential because they are wired together.

 

[cupid.callin]

then of course there is no current in wire bc and and current in ab will continue to bd --- & current in ac will continue in ed ...

so i still think that we can remove wire bc ...

 

[gneill]

then of course there is no current in wire bc and and current in ab will continue to bd --- & current in ac will continue in ed ...

so i still think that we can remove wire bc ...

You can remove the wire only if you connect the resistors at c to point b. Otherwise, there will be current in the wire. A wire may carry current even if the potential at each end is the same (or as close to the same as makes no difference for low resistivity wire). The wire is forcing the potentials to equalize.

In the case of this circuit, if you ignore the two central resistors and solve the remaining loops for the currents, you'll find that there's about 20 mA flowing from b to c through the wire.

 

[dimpledur]

So essentially the effective resistance should be R=174.

 

[gneill]

So essentially the effective resistance should be R=174.

Yes. 174.1Ω (don't forget your units).

If you carried out all the operations using fractions, you'd find 2960/17 Ω.

 

[dimpledur]

Interesting. Thanks for the help! Additionally, is there a trick as to finding the potential at a specific point in a circuit when there is more than one voltage supply?

 

[gneill]

Interesting. Thanks for the help! Additionally, is there a trick as to finding the potential at a specific point in a circuit when there is more than one voltage supply?

It would depend upon what shortcuts are available, if any, for the particular circuit. In general you solve the node equations or the loop equations and "walk a path" to the point of interest, summing up the voltage drops or rises along the way.

 

[dimpledur]

Since there would be two or more voltage supplies, would you execute that as many times as you have voltage supplies, and then just sum the result?

 

[gneill]

Since there would be two or more voltage supplies, would you execute that as many times as you have voltage supplies, and then just sum the result?

I'd probably solve the equations with all the supplies in place using KVL loops (unless some quirk of the circuit topology makes it obvious that another method(s) would be preferable).

I suppose you could use superposition to sum up the 'partial results', but it seems like a lot of extra work to me; you'll be solving as many sets of equation as you have supplies.

 

[dimpledur]

Alright. Sounds good. As an example, if I am trying to find out the voltage at C in the diagram above, would it merely be (32 V)(600 olms)(174.1 olms)^{-1}

 

[gneill]

Nope. You'd make a voltage divider out of the equivalent resistances for the 400 || 600 and 400 || 220, which are 240 and 141.9 Ohms respectively. Then the voltage at b and c (since they are at the same potential) would be 32V x 141.9/(141.9 + 240).

If you'd previously solved for all the currents in the circuit, then you'd just walk voltage potentials from d to b or c.

 

[dimpledur]

@ gneil

I recall you stating there is about 20 mA of current flowing through bc. How did you get that? When I calculated it, I. Got 52 mA. Here's how:

If current flows from abcd then it encounters 2 resistors and I =32/(400+220). Idea?

 

[gneill]

@ gneil

I recall you stating there is about 20 mA of current flowing through bc. How did you get that? When I calculated it, I. Got 52 mA. Here's how:

If current flows from abcd then it encounters 2 resistors and I =32/(400+220). Idea?

It's not as simple as that. There are two voltage dividers involved, so it's what's known as a 'bridge circuit'. They influence each other. Note that you could also have chosen the path acbd and obtained a different result.

One way to solve for the current is to convert each of the voltage dividers and the supply voltage into an equivalent Thevenin model. You end up with a series circuit with two batteries and two resistors. Easy. I've attached a picture.

 

[dimpledur]

Nope. You'd make a voltage divider out of the equivalent resistances for the 400 || 600 and 400 || 220, which are 240 and 141.9 Ohms respectively. Then the voltage at b and c (since they are at the same potential) would be 32V x 141.9/(141.9 + 240).

If you'd previously solved for all the currents in the circuit, then you'd just walk voltage potentials from d to b or c.

One quick question regarding this statement. This formula tells us the voltage difference across the resistors, correct? Therefore, if the question is asking for Vb, I would have to take the magnitude of the voltage source and subtract the result from the voltage divider in order to determine what Vb is, correct?

 

[gneill]

One quick question regarding this statement. This formula tells us the voltage difference across the resistors, correct? Therefore, if the question is asking for Vb, I would have to take the magnitude of the voltage source and subtract the result from the voltage divider in order to determine what Vb is, correct?

Point d is tied to the battery negative, so it's the designated "ground" potential and reference point for the circuit. Vb is determined with respect to that. So no, you don't have to subtract the battery potential.

 

[dimpledur]

Hmm, I think I understand. I would subtract it the question were asking for Vab then, correct?

 

[gneill]

That looks right.

 

[dimpledur]

Thanks for all your help!

 

[dimpledur]

It's not as simple as that. There are two voltage dividers involved, so it's what's known as a 'bridge circuit'. They influence each other. Note that you could also have chosen the path acbd and obtained a different result.

One way to solve for the current is to convert each of the voltage dividers and the supply voltage into an equivalent Thevenin model. You end up with a series circuit with two batteries and two resistors. Easy. I've attached a picture.

How exactly did you come up with the Thevenin equivelant in this picture? Couldnt you just find the current from abcd and then subtract the current from acbd?

 

[gneill]

How exactly did you come up with the Thevenin equivelant in this picture? Couldnt you just find the current from abcd and then subtract the current from acbd?

32V battery and voltage divider abd. Thevenin equivalent between points b-d.

32V battery and voltage divider acd. Thevenin equivalent between points c-d.

Why don't you try your method and see if the results match?

 

[dimpledur]

Yeah, it doesn't match. I ended up with 19.6 mA.

 

[gneill]

Right. That's because the currents split at the b and c junctions. So in your proposed path abcd, the current also has a path via bd that is not being accounted for. A similar situation holds for the path acbd, where the cd path is overlooked.

 

[zgozvrm]

FYI -

You could redraw the circuit as in my attachment.