Total current through two wires

[tenshi01]

Homework Statement

A resistor is made out of a low wire having a length L. Each end of the wire is attached to a terminal of a battery having a constant voltage Vo. A current I flows through the wire. If the wire were cut in half, making two wires of length L/2, and both wires were attached to the battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal), what would be the total current flowing through the two wires?

Options are:
a.) 4I
b.) I
c.) I/2
d.) 2I

Homework Equations

The Attempt at a Solution

I was thinking it would be 2I based on R=p*L/A. So if the resistor is halved than the current is is doubled? (I=V/R)

 

[rude man]

Homework Statement

A resistor is made out of a low wire having a length L. Each end of the wire is attached to a terminal of a battery having a constant voltage Vo. A current I flows through the wire. If the wire were cut in half, making two wires of length L/2, and both wires were attached to the battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal), what would be the total current flowing through the two wires?

Options are:
a.) 4I
b.) I
c.) I/2
d.) 2I

Homework Equations

The Attempt at a Solution

I was thinking it would be 2I based on R=p*L/A. So if the resistor is halved than the current is is doubled? (I=V/R)

Sorry, no.
If the wire of length L had resistance R, what is resistance of length L/2?

The what happens to total R if you put the two half-lengths in parallel?

 

[tenshi01]

I know in parallel, 1/R = 1/R₁ + 1/R₂
Would the resistor not change then?
and current would be halved?

 

[rude man]

I know in parallel, 1/R = 1/R₁ + 1/R₂
Would the resistor not change then?
and current would be halved?

No.

If I halve the wire what happens to the resistance of each half?

 

[tenshi01]

R/2² so R = 1/4 ?

 

[rude man]

R/2² so R = 1/4 ?

Why the power of two?

Again: what is the resistance of L/2 if the resistance of L is R?

 

[tenshi01]

I was going off the equation R_total=R_long/n²

 

[lendav_rott]

The way I see it is let s say you have a water pipe that has the "I" to set a value of litres/s. No matter what the change you do to the pipe, cut it in half, the amount of water flowing through will still be the same - there is no change in the source that gives you the water.

Or..if your voltage is V0, it means there is a potential difference, so you can have the pipe set on a slope with a certain degree and a flow of water coming from above flowing through the pipe, if you cut the same pipe in half getting 2 pipes with the same diameter, you should think that you will get more water, which would be true only if the intensity of the flow changes, but you are set on a fixed slope (V0), so the intensity remains the same, thus you won't get more water in the same time.

This is only a theory at the moment, I am not sure of the actual answer, but I would say the current remains the same - I

 

[rude man]

Sorry, tenshi and lenday, you're both wrong. Lenday, I would be cautious about drawing analogies that may not be appurtenant.

Until tenshi answers my question I cannot be of further help. BTW the wire is cut perpendicular to its axis, as you would normally cut a wire to shorten it.

 

[lendav_rott]

Well the wire's resistance is determined by the length, diameter and material.
R = material_constant * length / diameter
so let s say R = M*L/D , where M is the constant and D is diameter.
If you cut the thing in half perpendicularly, which you would do anyway, it would be R = M*L/2D so essentially we have 2 wires with equal resistances L/2D and for the sake of simplicity we assume M and D as 1 so 2 wires have R = L/2 for both of them.

Now here is something that is reallly confusing - it says the wires' ends are connected to the battery's terminals. In an ideal case our "I" will go through the roof - shortcircuited.

By imagining this crazy scenario, the wires ought to be parallel:
In the original case we have V0/L = I
2 resistors L/2 in parallel - R = L/4
Now we have:
4V0/L = 4I so 2I in either wire

Makes little sense though, If you grab on the 1st uncut wire, you will live - If you grab on the cut wire, you die. How?

 

[CWatters]

You got the right answer 4I.

The resistance changes from R to "R/2 in parallel with R/2" which is R/4.

You can see that in R=p*L/A because the change halves the length and doubles the area. the equation becomes..

R^'=p*(L/2)/(2*A)

Makes little sense though, If you grab on the 1st uncut wire, you will live - If you grab on the cut wire, you die. How?

What kills you is the current flowing through your body. The voltage on the wire is the same so I'd expect no difference to the current flowing through your body if you grabbed hold of it. The resistance of your body is likely to be << than that of the wire.

 

[lendav_rott]

Oh yes, how stupid of me - voltage stays the same all the time. But now I am trying to think how my initial theory wouldn't work, what is the flaw there? By this "I to 4I" I should be getting 2 times more water through either of the pipes, I thought for sure they should act in a similar manner, both the current flow and the water flow :/ Well it was worth a shot atleast, even though I am wrong about it :D

 

[CWatters]

Your water pipes need to model resistors not ideal wires...

Imagine your water pipes are very small bore so they restrict the flow of water and there is a pressure drop when water flows through them. You need to apply a pressure difference between the ends to cause water to flow.

If you keep the pressure constant but halve the length of pipe you might expect the flow rate to double.

Putting two pipes in parallel will also double the flow rate.

Net result is four times the flow rate.