What is the resistance between two resistors?

[ImpCat]

Homework Statement

Homework Equations

R=V/I

The Attempt at a Solution

Would it be 10ohms? Since PQ is between two 10ohms resistor.

 

[berkeman]

Homework Statement

View attachment 109814

Homework Equations

R=V/I

The Attempt at a Solution

Would it be 10ohms? Since PQ is between two 10ohms resistor.

Are the two resistors in parallel or series or some other configuration?

Hint -- what do parallel resistors share?

 

[ImpCat]

Are the two resistors in parallel or series or some other configuration?

Hint -- what do parallel resistors share?

They have the same voltage across both resistors? But the voltage across the two resistors is after the voltage has been reduced by the first resistor right? I still am not seeing how I can solve the equation with just these information. :<

 

[berkeman]

They are on purpose trying to make the question confusing. one terminal goes to one side of each resistor, and the other terminal goes to the other side of each resistor. Are the resistors in parallel?

 

[CWatters]

Try colouring all wires connected to "P" one colour and all wires connected to "Q" another colour.

 

[ImpCat]

Try colouring all wires connected to "P" one colour and all wires connected to "Q" another colour.

But how would I know the resistance just from the resistance of the other two resistors?

 

[gneill]

But how would I know the resistance just from the resistance of the other two resistors?

What does your textbook or course notes say about how resistors combine to create a net resistance?

 

[ImpCat]

What does your textbook or course notes say about how resistors combine to create a net resistance?

In parallel, the resistors combine by the rule 1/r1 +1/r2 = 1/rTotal, in series the resistors combine by r1+r2=rTotal. So the net resistance in this circuit would be rTotal= r1+ 1/(1/r1 +1/r2)

 

[gneill]

So the net resistance in this circuit would be rTotal= r1+ 1/(1/r1 +1/r2)

How do you justify that? Can you explain your reasoning?

 

[ImpCat]

How do you justify that? Can you explain your reasoning?

As in a series, the total resistances are simply added up, and in parallel, the resistances are added up by 1/r1 +1/r2 = 1/rTotal. So if I added up the parallel resistors and pretend it is just 1 resistor with a resistance of 1/(1/r1 +1/r2)=rX, then the total resistance in the circuit would be rX+r1(Where r1 is the 10ohms resistor in the series)=rTotal.

 

[berkeman]

As in a series, the total resistances are simply added up, and in parallel, the resistances are added up by 1/r1 +1/r2 = 1/rTotal. So if I added up the parallel resistors and pretend it is just 1 resistor with a resistance of 1/(1/r1 +1/r2)=rX, then the total resistance in the circuit would be rX+r1(Where r1 is the 10ohms resistor in the series)=rTotal.

Very good. So in the end, are these two resistors in series or parallel? What is your final answer?

 

[ImpCat]

Very good. So in the end, are these two resistors in series or parallel? What is your final answer?

The unknown resistor and 1 of the known resistor are in parallel, but on the whole both are in series with another known resistor. Yet I still don't understand how to convert all of these into the value of the unknown resistor. :(

 

[NascentOxygen]

So if I added up the parallel resistors and pretend it is just 1 resistor with a resistance of 1/(1/r1 +1/r2)=rX, then the total resistance in the circuit would be rX+r1(Where r1 is the 10ohms resistor in the series)=rTotal.

No.

Once you have accounted for a resistor, you can't then include it in a second accounting.

Has anyone told you to redraw the circuit, using as many trials as is necessary, until you can draw them unequivocally as a pair of resistors in either (a) a series arrangement, or (b) a parallel arangement. It must be one or the other.

 

[ImpCat]

No.

Once you have accounted for a resistor, you can't then include it in a second accounting.

Has anyone told you to redraw the circuit, using as many trials as is necessary, until you can draw them unequivocally as a pair of resistors in either (a) a series arrangement, or (b) a parallel arangement. It must be one or the other.

Actually, I used v1 both times as the two resistors provided are of the same ohms, and I was too lazy to change its symbol. So you can treat one of the v1 as vb or vy. Sorry for the confusing equations xD. Also, I did convert the circuit into a series, by adding the parallel resistors together and treat it as a series along with the 10ohms resistor.

 

[NascentOxygen]

Actually, I used v1 both times as the two resistors provided are of the same ohms, and I was too lazy to change its symbol. So you can treat one of the v1 as vb or vy. Sorry for the confusing equations xD. Also, I did convert the circuit into a series, by adding the parallel resistors together and treat it as a series along with the 10ohms resistor.

I have no idea what all that is about.

►► So, is your answer to the question that started this thread now A, B, C or D?

 

[ImpCat]

I have no idea what all that is about.

►► So, is your answer to the question that started this thread now A, B, C or D?

I still have no idea, I tried totaling the resistance on the parallel resistors (Rtotal) and treat it as a series with the 10ohms resistor. I then assumed that the current that passes out of the 10ohms resistor is equivalent to the current into the parallel resistors:

Current out of 10ohms resistor: I=V/10

Current into parallel resistors: V/10 = I1+I2
V/10 = V1/10 + V1/X (As the voltage is the same between parallel resistors where V1 is the voltage)

but I still cannot solve for x.

 

[NascentOxygen]

I tried totaling the resistance on the parallel resistors (Rtotal) and treat it as a series with the 10ohms resistor.

Why do you keep referring to 3 resistors when there are only 2?

Once you have worked out the equivalent resistance of the pair of resistors, then STOP! That is your answer! There is nothing more to be done!

The question being addressed is: with what single resistor could you replace this two resistor arrangement so that the current between P and Q would be unchanged (obviously when some voltage is applied between P and Q).

 

[ImpCat]

Why do you keep referring to 3 resistors when there are only 2?

Once you have worked out the equivalent resistance of the pair of resistors, then STOP! That is your answer! There is nothing more to be done!

So since the known resistors are in a series, do I just add them together to get the unknown resistance? I'm really confused ._.

 

[NascentOxygen]

the resistance on the parallel resistors

the known resistors are in a series

You have not yet determined whether the two resistors are in a parallel arrangement, or are in series?

How would you recognize a parallel arrangement? How would you recognize a series arrangement?

 

[ImpCat]

You have not yet determined whether the two resistors are in a parallel arrangement, or are in series?

How would you recognize a parallel arrangement? How would you recognize a series arrangement?

A series arrangement would be where the resistors are connected next to each other. Parallel would be where the path of the resistors are divided. So the unknown resistor and 1 of the known resistor are in parallel. There is also one known resistor of 10ohms that are in series to the parallel resistors. Right?

 

[NascentOxygen]

A series arrangement would be where the resistors are connected next to each other. Parallel would be where the path of the resistors are divided.

You probably do know, but that is not a clear enough answer to earn any marks in an exam.

So the unknown resistor and 1 of the known resistor are in parallel. There is also one known resistor of 10ohms that are in series to the parallel resistors. Right?

There are no unknown resistors here. There are two known resistors, they are known to be 10Ω each. There are no more resistors; there are just two resistors. There are not 3 resistors, and not 4 resistors. Just two!

 

[ImpCat]

You probably do know, but that is not a clear enough answer to earn any marks in an exam.

There are no unknown resistors here. There are two known resistors, they are known to be 10Ω each. There are no more resistors; there are just two resistors.

I realize that, I just treated PQ as a resistor-which I realize is a big mistake. But if I don't treat it as a resistor, how would I find the resistance between PQ?

 

[NascentOxygen]

There is no additional unshown resistance between P and Q, though it wouldn't make any difference if there were. All we are interested in doing in this exercise is replacing what you can see with its equivalent.

 

[ImpCat]

There is no additional unshown resistance between P and Q, though it wouldn't make any difference if there were. All we are interested in doing in this exercise is replacing what you can see with its equivalent.

I see now, thanks for your help! The answer would then be B. It's just that the phrase "Resistance between PQ" confused me. Like how can you have a resistance between two points, don't you need a sort of conductor?

 

[ImpCat]

Perhaps this picture will help you understand the problem more clearly . It just shows the same circuit drawn in different ways .

ATTACH=full 109845[/ATTACH]

Thanks for the picture, really helped my understanding

 

[NascentOxygen]

I see now, thanks for your help! The answer would then be B.

And to confirm to an examiner that B is not just a guess, how do you justify giving that as the answer?

 

[Nidum]

Perhaps this picture will help you understand the problem more clearly . It just shows the same circuit drawn in different ways .

 

[ImpCat]

And to confirm to an examiner that B is not just a guess, how do you justify giving that as the answer?

By treating it as a parallel circuit so 1/10 + 1/10 = 1/r
r=5

 

[NascentOxygen]

By treating it as a parallel circuit so 1/10 + 1/10 = 1/r
r=5

Well, an examiner could ask how did you recognize it to be a parallel arrangement, and not a series arrangement of two 10 Ω resistors?

 

[ImpCat]

Well, an examiner could ask how did you recognize it to be a parallel arrangement, and not a series arrangement of two 10 Ω resistors?

I would reply that in a series circuit, there would only be one path for electrons to flow. Whereas in this particular circuit diagram, there are two paths for the current to flow toward the resistors-as seen by the wire connecting the sides of the resistor and in between the resistors. As such this is a parallel arrangement, not a series arrangement.

 

[epenguin]

There are two problems that students frequently manifest dealing with these elementary (in this case very elementary) circuit problems. One is they cannot recognise a topology. No longer recognise the same circuit when it is redrawn in a slightly different looking way. What can I say? - if this were an underground railway system how many ways can you go from P to Q? What do you have to go through?

(Students are lucky they rarely meet nonplanar circuits!)

The other point is that they are too formulaic. They have been intimidated and made tense and rigid by formulae. Instead think physically! When you have things in parallel I recommend don't think of resistances, think of conductances. You have a voltage across two resistors, i.e. mediocre conductors, in parallel, in this case of equal conductance. If one for a given voltage across it will carry a certain current, how much current will the other with the same conductance and the same voltage across it carry?

After which what will be the total current?...

Not really a 30+ post problem.

Similar thinking applies in the cases of capacitance.