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I'm trying to calculate the resistance of a copper bar in a DC current, but I'm not sure how to do it. The only things I've seen online calculate the resistance of a wire and ask for its gauge, but a bar is not a wire and it doesn't have a gauge. It's a rectangular prism shape and its dimensions are 1.3cm x 2.5cm x 30cm. You don't have to do the math for me I just would like a formula for getting the resistance.

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While you can look up the resistivity of copper on wikipedia or Kaye & Laby, you'll find that the answer is given in units of 10^-8/m. I don't know about you, but I always miscount powers of 10 when using that sort of unit. Instead I commit the following to memory

1 metre of \$1mm^2\$ copper wire has a resistance of 17 \$m\Omega\$ (roughly)

If you can't remember that resistance is proportional to length, and inversely proportional to area, then just think series resistors, or parallel resistors.

10m of wire would have 10x the resistance, it's like ten 1m lengths in series.

If the wire is 4mm2, it will have 1/4 the resistance, because it's like four 1m lengths in parallel.

Copper, like all pure metals, has a fairly strong temperature coefficient of resistance, which is why I've only given the resistivity to 2 significant figures. The tempco is 0.4% per degree C, or a whopping 10% change for 25 degree change in temperature. It can be useful to remember that when you want to estimate the temperature rise of (say) transformer windings, just measure their resistance cold, and again when hot.

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  • \$\begingroup\$ +1 For the transformer temperature measurement trick. \$\endgroup\$ – KalleMP Apr 26 '18 at 15:55
  • \$\begingroup\$ @KalleMP does that +1 mean you're going to give me an upvote ;-) \$\endgroup\$ – Neil_UK Apr 26 '18 at 16:01
  • \$\begingroup\$ Sure, I forgot or missed the button. Your mnemonic is also something handy to keep in mind when out in the field, it gives a way to sanity test calculation or table results. \$\endgroup\$ – KalleMP Apr 26 '18 at 16:28
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R= \$\frac{\rho L}{A}\$ where L is length, A is cross-sectional area and \$\rho\$ is resistivity.

Take care to use compatible units(!).

\$\rho\$ is a function of temperature. Over a reasonable range, \$\rho\$ of most elemental metals is approximately proportional to absolute temperature. At very cold temperatures, weird things happen, some metals (lead, for example) see their resistivity suddenly drop to zero and other metals such as copper level off at a value that is strongly indicative of their purity.. the ratio of that to the room temperature resistivity is called RRR = Residual Resistance Ratio.

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  • \$\begingroup\$ I would also add that the formula above assumes that the point of contacts at the extremes of the bar are 'ideal', so that the current density is uniform across the bar section. If this is not true, the becomes higher. \$\endgroup\$ – Vladimir Cravero Apr 26 '18 at 15:25
  • \$\begingroup\$ Does the work history of the material affect \$\rho\$ for bar stock like it does rolled sheets? (I mean, of course it does, but maybe bar stock is all sold nicely annealed) \$\endgroup\$ – The Photon Apr 26 '18 at 15:42
  • \$\begingroup\$ @ThePhoton I think it does, I have slightly different numbers for annealed copper and regular copper. But only slightly. \$\endgroup\$ – Spehro Pefhany Apr 26 '18 at 16:25
  • \$\begingroup\$ Little tip that I recently learned... use \displaystyle THEN YOUR EQUATION and it makes your fractions much larger. You can see the results on my answer. \$\endgroup\$ – KingDuken Apr 26 '18 at 16:25
  • \$\begingroup\$ @SpehroPefhany You're welcome! :) works for integrals too... Practically anything that you want larger, you can use \displaystyle \$\endgroup\$ – KingDuken Apr 26 '18 at 16:27
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There are variants of Ohm's Law that are not your typical \$V=IR\$. Henceforth, Georg Ohm has also made some important discoveries of how to measure the resistance of a particular object. For instance, your copper bar (bar is a little ambiguous but I'm assuming you're talking about an ingot or something).

There's a concept called "surface current" that brings your typical line current that travels in 2-D space into a 3-D space. This is noted as:

\$J=I/S\$, where \$J\$ is your surface current, \$I\$ is your line current, and \$S\$ is the surface area of your object. But if you were to bring the current into 3-D space, you also have to bring the voltage into 3-D space. Therefore, we have to utilize the electric field noted as \$E\$. We can make a relationship between the surface current and electric field as:

\$J=\sigma E\$ where, \$\sigma\$ is the conductivity of copper.

You may also find \$E\$ by another relationship known as "Poisson's Equation".

\$E=-\nabla V\$, where \$\nabla\$ is the gradient operator and \$V\$ is your electric potential.

Once you find your electric potential, you are ready to solve the problem... and now it's pretty straightforward. Now you just have to solve for \$R\$... which is easy. Sphero gave you equation already...

\$ \displaystyle R=\frac{\rho \ell}{A}\$

But.... if you weren't given \$\rho\$:

\$\displaystyle V= R\iint_{S}I\cdot dS \rightarrow \boxed{R=\frac{V}{\iint_{S}I\cdot dS}}\$

So KingDuken? Why did you give me all of these equations if all I was asking for was just the resistance?

Good question! I wanted to give you the 3-D insight of resistance :)

EDIT: You know, I just saw your profile and I saw that you were 16 years old. So perhaps using an equation that involves an understanding of multivariate calculus is perhaps not the best way to answer your question... Not saying that a 16-year-old wouldn't understand but there's not a lot of teenagers who not have taken their third semester of Calculus. Let me give you a more acceptable formula:

\$ \displaystyle R=\frac{V}{I\times WHL}\$, where \$W,H,L\$ are the width, height, and length of your copper prism.

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