Electrical Circuits - Parallel vs. Series

[BuddyBoy]

Homework Statement

Are two copper bars on top of each other parallel or series?

Relevant Equations

R_S = R1 + R2
R_P = (1/R1 + 1/R2)^(-1)

This is a THEORETICAL thought exercise ONLY to help in understanding the concept of parallel vs series.

If I were to have a copper bar of uniform density as a conductor. One end of the bar is at potential A, the other end of the bar is potential B.

Lets say the copper bar has a resistance of 10 ohms, just a randomly chosen number that is easy to do math with.

If I were to remove the power source from the circuit, remove the copper bar and cut it down the middle with some magical tool, such that one piece of the bar sits directly on top of the other piece. One piece is 5 ohms, and so is the other piece. The copper bar is placed back into the circuit in the same spot, such that one piece sits directly on top of the other, aligned perfectly so that one does not overhang the other.

Now are the two pieces considered to be in series or parallel?

My initial thought that the two pieces are identical to the one whole piece, same shape, so I should consider the two copper pieces as a whole has 10 ohms of resistance. The act of simply cutting it in the middle makes the two pieces in parallel and the resistance of the two pieces is now 5/2 ohms?

We think of parallel circuits as this:

But what if R1 and R2 is not connected at only two points, but infinitely many points? Were in this thought exercise R1 is not a resistor but a copper bar with a resistance of 5 ohms, and R2 is also a copper bar of 5 ohms? The two bars are placed directly on top of each other, lined up perfectly with each other.

Not sure if it matters in this thought experiment, but what if the two bars are not 100 % lined up with each other? One still lays on top of the of the other, but one bar overhangs the other by a few millimeters because a human placed the two bars on top of each other and "eye balls it" to align the two pieces together.

 

[phinds]

Your scenario is unclear. Draw a before and after picture. Simply cutting a conductor in the middle results in an open circuit, so I'm not sure what you are talking about.

 

[BuddyBoy]

Your scenario is unclear. Draw a before and after picture. Simply cutting a conductor in the middle results in an open circuit, so I'm not sure what you are talking about.

Thanks for the reply. Sorry for not explaining this well.

The two rectangles that are 3 inches by 7 inches, are some nickel plates that go to some other part of the circuit. One of these rectangles is at potential A while the other is at potential B. Potential A is so ever slightly higher than the other, because of the resistance of the copper bar.

In between these two rectangles exists my copper bar, that is 11 inches long and 3 inches tall. The two nickel plates have full contact with the two faces of the copper bar.

The Nickel Plate that is at electrical potential A is not visible from this view point sketched above. Only the nickel plate at electrical potential B.

The copper bar has a uniform density and is 10 ohms.
I remove the power source from the circuit and remove the copper bar. Leaving the two nickel plates in place with the rest of the circuit.
I cut the copper bar down the middle perfectly, resulting in two copper bars, each with 5 ohms resistance.
I reassemble the two copper bars so that they are in the same spot in the circuit, one is directly on top of each other, aligned perfectly.

Not shown above because it's blocked out of view, is the nickel plate at potential A.
The two copper bars are in full contact with the two nickel plates.

Are the two copper bars in Series, Parallel or Neither, and how would I determine the resistance? I had assumed that the total resistance would still be 10 ohms for the two copper bars between the two nickel plates. As opposed to the two copper bars are now in parallel so the equivalent resistance is now 5/2 ohms.

Please excuse my poor artwork.

I then had the follow on question that what if Copper Bar A overhangs Copper Bar B by a few millimeters because a human places them on top of each other, and they are not perfectly aligned.

Thanks for help in understanding this, and I hope this is easier to understand and I explained it better. Let me know if I didn't.

 

[SammyS]

Homework Statement: Are two copper bars on top of each other parallel or series?
Relevant Equations: R_S = R1 + R2
R_P = (1/R1 + 1/R2)^(-1)

This is a THEORETICAL thought exercise ONLY to help in understanding the concept of parallel vs series.

If I were to have a copper bar of uniform density as a conductor. One end of the bar is at potential A, the other end of the bar is potential B.

Lets say the copper bar has a resistance of 10 ohms, just a randomly chosen number that is easy to do math with.

If I were to remove the power source from the circuit, remove the copper bar and cut it down the middle with some magical tool, such that one piece of the bar sits directly on top of the other piece. One piece is 5 ohms, and so is the other piece. The copper bar is placed back into the circuit in the same spot, such that one piece sits directly on top of the other, aligned perfectly so that one does not overhang the other.

Now are the two pieces considered to be in series or parallel?

My initial thought that the two pieces are identical to the one whole piece, same shape, so I should consider the two copper pieces as a whole has 10 ohms of resistance. The act of simply cutting it in the middle makes the two pieces in parallel and the resistance of the two pieces is now 5/2 ohms?

Nice thought Exercise !

Now look at the sentence that I have emphasized in bold font. You are absolutely correct there.

However, that means that each of the two identical pieces carries 1/2 of the current for the given voltage source. What does that tell you about the resistance of each piece ?

 

[BuddyBoy]

Nice thought Exercise !

Now look at the sentence that I have emphasized in bold font. You are absolutely correct there.

However, that means that each of the two identical pieces carries 1/2 of the current for the given voltage source. What does that tell you about the resistance of each piece ?

Thanks for confirming. I just wanted to be sure.

I'm not sure what you mean. The resistance of each copper bar is independent of the voltage and current going through it, right? So each piece would still be 5 ohms?

 

[SammyS]

Thanks for confirming. I just wanted to be sure.

I'm not sure what you mean. The resistance of each copper bar is independent of the voltage and current going through it, right? So each piece would still be 5 ohms?

Whatever the voltage is, only 1/2 the current goes through each of the two pieces. Therefore, each piece has more resistance than the initial bar. Right?

 

[BuddyBoy]

Whatever the voltage is, only 1/2 the current goes through each of the two pieces. Therefore, each piece has more resistance than the initial bar. Right?

The potential difference across each half would remain the same as the initial bar A - B.

The current through each half piece would be half the initial current.

If you cut the bar in half each piece would have half the resistance of the initial piece.

I don't understand why each half piece would have more resistance than the initial piece.

I'm thinking of the cross sectional area times the resistivity of copper times the length. The cross sectional area is reduced by half the initial piece when cut down the middle, so the resistance would be half of the initial piece? This is independent of the voltage across it or the current going through it.

So if the initial piece was 10 ohm, each half piece would be 5 ohm.

I'm not really sure why the resistance would go up for each half piece, more than the initial piece.

 

[SammyS]

The potential difference across each half would remain the same as the initial bar A - B.

The current through each half piece would be half the initial current.

That pretty much answers your question.

For the same potential difference, either half only allows half of the current to pass, compared to the current through the initial piece. Therefore, the resistance of each half is 20 ohms, not 5 ohms. Your Thought Exercise proves it.

 

[BuddyBoy]

That pretty much answers your question.

For the same potential difference, either half only allows half of the current to pass, compared to the current through the initial piece. Therefore, the resistance of each half is 20 ohms, not 5 ohms. Your Thought Exercise proves it.

I see what your saying. In order to maintain the same potential difference, and half the current going through each piece, the resistance of each half piece would have to be double the initial piece, so 20 ohms.

This is very interesting to me. If I cut my 10 ohm bar in half, each half piece would be 5 ohms, if measured outside of the circuit? Upon inserting it into the circuit each half piece would be 20 ohms?

This is very interesting to me. You would think with half the amount of material as the initial piece, the resistance would be less, but it actually increases.

I remember learning this equation in school. Cross sectional area of each half piece would be half, everything else would remain the same, so resistance would be half or 5 ohm. A bit confusing.

 

[Tom.G]

I'm not really sure why the resistance would go up for each half piece, more than the initial piece.

Only because there is a misunderstanding of the phrase "cut it down the middle".

This was clarified in the next sentence: "One piece is 5 ohms, and so is the other piece."

I believe that @SammyS interpreted the cut to be lengthwise rather than crosswise.

Homework Statement: Are two copper bars on top of each other parallel or series?
Relevant Equations: R_S = R1 + R2
R_P = (1/R1 + 1/R2)^(-1)

The act of simply cutting it in the middle makes the two pieces in parallel and the resistance of the two pieces is now 5/2 ohms?

I agree.

I then had the follow on question that what if Copper Bar A overhangs Copper Bar B by a few millimeters because a human places them on top of each other, and they are not perfectly aligned.

For your purposes, and the usual introductory approach, it makes no difference, the resistance is still 5/2 Ohms.

Note: If the power source was Radio Frequency (high frequency), there could be a difference, depending on the geometry, the relative proximity and orientation of the pieces, and the frequency. That howeever leads to another whole can of worms. Wait another year or three before you dive in to that!

Cheers,
Tom

 

[BuddyBoy]

Only because there is a misunderstanding of the phrase "cut it down the middle".

This was clarified in the next sentence: "One piece is 5 ohms, and so is the other piece."

I believe that @SammyS interpreted the cut to be lengthwise rather than crosswise.

I agree.

For your purposes, and the usual introductory approach, it makes no difference, the resistance is still 5/2 Ohms.

Note: If the power source was Radio Frequency (high frequency), there could be a difference, depending on the geometry, the relative proximity and orientation of the pieces, and the frequency. That howeever leads to another whole can of worms. Wait another year or three before you dive in to that!

Cheers,
Tom

Thanks for the clarification.

I did mean cutting the bar into two identical pieces in half. Not sure which dimension we are calling length. But I meant this way.

So why are the two pieces considered to be in parallel? I guess I never really got a good definition of parallel.

I always pictured parallel to be this model/picture. Two physically separate paths for current to travel with only two common points.

However this representation/picture is misleading via this thought exercise. The two copper bars act as two separate paths for current to flow. The two copper bars are in contact with each other as they are on top of each other. There is countably infinitely many points in common between the two paths in this thought exercise.

So these two resistors are in parallel

But so are these two. If you imagine the resistor symbol to be slightly thicker, representing one on top of the other, to represent the two copper bars, one placed on top of each other. One surface of one copper bar is in full contact with the other surface of the other copper bar.

Interesting. So the act of simply cutting the bar in half reduces the overall resistance between the two nickel plates. This is a bit perplexing to me. The act of cutting them in half created two separate paths of current to flow. I guess I just thought placing them on top of each other into the original shape would be equivalent to one path as I would have countably infinite common points between the two paths

 

[SammyS]

@Tom.G :

Yes, I interpreted OP to mean that the bars were cut length-wise and then slapped back together as if they hadn't been cut. (That's why he says the resulting resistance is 10 ohms.) See the final figure in Post #3 showing Copper Bar A and Copper Bar B.
I think he simply assumed that cutting the bar in half would mean that the resulting resistance would also be cut in half .

 

[BuddyBoy]

@Tom.G :

Yes, I interpreted OP to mean that the bars were cut length-wise and then slapped back together as if they hadn't been cut. (That's why he says the resulting resistance is 10 ohms.) See the final figure in Post #3 showing Copper Bar A and Copper Bar B.
I think he simply assumed that cutting the bar in half would mean that the resulting resistance would also be cut in half .

Yea this is what I meant, and what I assumed.

Cutting the bars in half results in half the resistance = Wrong

But why?

I guess I had just assumed from this equation.

If my cross sectional area is half, than so is my resistance. Why is this wrong?

But I see what your saying, if I have the same voltage A - B.
The current is cut in half in each piece.
So therefore the resistance of each half must be double the initial piece or 20 ohms.
Cutting the original 10 ohm bar in half resulted in each piece having double the resistance of the initial piece.
A bit perplexing to me, as I have half the material.
Ohhhhh... it's because you get
2 * resistivity * (length/(original cross sectional area))

Lastly, with the two halves on top of each, assembled in the circuit, each half having 20 ohms, would the overall resistance between the two nickel plates be 40 ohms (ignoring the resistance of the two nickel plates)?