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From my book:

"A length of wire is cut into five equal pieces. The five pieces are then connected in parallel, with the resulting resistance being 2.00 Ω. What was the resistance of the original length of wire before it was cut up?"

I don't understand what exactly I'm looking for. What resisters? This isn't like any other problem or example in my book. I don't even know what a parallel with five pieces looks like. I tried using the parallel equation, assuming there are five resistors:

$\ \frac{1}{2}=\frac{1}{x}\cdot 5$

Then each resister would have would be 0.40Ω, but that can't be right.

I've only been given the equations for $\ R_{eq}$ in both series and parallels.

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up vote 2 down vote accepted
  • Resistor is anything that pose resistance to flow of charges in a circuit. Here is how it looks like: enter image description here $$R=\rho \dfrac lA$$

Where $\rho$ is material property, $l$ is length and $A$ is cross sectional area.

Here is how 5 resistors in parallel look like:

enter image description here

and the circuit diagram showing the same:

enter image description here

In parallel configuration voltage drop across each resistor is same and an equivalent resistance can be calculated using this:


Now your problem.

Let's say that original Wire had $R$ resistance, and so 5 equal lengths of it's would have $r=R/5$ resistance each.($R=\rho l/A$; $r=\rho [l/5]/A$)

Which when in parallel gives equivalent of $r/5=R/25$ resistance across $x$(see the circuit diagram, each $R_1, R_2=R/5$ ) Hence we get the original resistance as:

$$R/25=2\Omega\iff R=50\Omega$$

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Seems like you're not making the connection between the actual physical setup and the equations. So, here's a translation:

1) A length of wire is a resistor. By resistor what we mean is that when we apply a potential difference $V$ between the two ends (like from a battery) the resulting current $I$ is given by $V = IR$ where $R$ is the resistance.

2) The length of wire in the question is cut into five equal pieces. I hope you have no issues with this statement.

3) By 'in parallel' what that means is that you join one end of all five pieces together (A), and you join the other end of all five pieces together (B). Then you apply a potential difference between A and B.

Why 'parallel'? Because, well, the wires are now in parallel. Like in a 400m track there are 8 parallel lanes because 8 runners can run at the same time.

In the question, the resistance of this setup is 2 Ohms.

4) The other configuration would be 'in series'. Then you join the five pieces end by end, like - - - - -, so you get back the original wire.

Why 'series'? Because well, the wires follow one another and so they are in series. Like in a 400m track in the 4x100m, the 4 runners follow one after another. not at the same time.

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Let the resistance of the original wire be R.

R = ρ (L/A)

Now l = L/5

R’ = ρ (L/5A)

Or R’ = R/5

Now 5 resistors of R’ are connected in parallel

1/R(net) = 5/R + 5/R + 5/R + 5/R + 5/R


1/2 = 25/R


R = 50 Ω.

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The wires have resistance. You had 5 resistors in series. If connected in parallel, they give a known parallel resistance.

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This answer is a little too brief. You havn't really explained anything further than what's given in the question. – Nic May 6 '13 at 12:23
You ask where are the resisters. I answer. And, after that you tell that I do not answer anything more than is given in the question? What? – Val May 6 '13 at 12:51
I deleted some inappropriate comments. – David Z May 8 '13 at 17:26

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