# Why isn't a paper clip shocking me when I apply a 1.5V battery to it

I am trying to wrap my head around basic circuit theory, and I don't understand this simple experiment I am trying to do. I have a $1.5\,{\rm V}$ AA alkaline battery, with an internal resistance of about $0.14\,\Omega$ (measured that using a voltmeter). I have simply taken a paper clip, unfolded $\sim 6\,{\rm cm}$, and tried to calculate the theoretical resistance of the paper clip using the following formula. I will note that the paper clip is pure metal, it doesn't have a plastic coating like some do.

$$\Omega = \frac{\rho L}{\pi r^2}$$

A basic search on Google got me a $\sim 5 \times 10^{-7}$ resistivity for the steel clip. The length is $0.06\,{\rm m}$ and the radius of such a small paper clip is about $0.0015\,{\rm m}$. Plugging that all in I get a theoretical internal resistance of the wire of about $0.00424\,\Omega$.

Using the ideal Ohm's Law, $$I = \frac{\epsilon}{(R+r)} = \frac{1.36}{0.14+0.00424} = 9.4\,{\rm A}$$

So why isn't a paper clip shocking the crud out of me when I hook it up to the positive and negative terminals on the AA battery? Or has my thinking skewed me somewhere?

• Why would current flowing through a paper clip hurt you? The current isn't flowing through you, it's flowing through the clip. Second, batteries have internal resistance, so your current calculation is incorrect. A 9V battery cannot source hundreds of amps. Mar 28 '17 at 20:43
• @DanielSank The paper clip is bare metal, it isn't insulated. The calculation includes both internal resistance of the battery (hence the terminal voltage of 1.36V) and the theortical resistance of the wire, yet mathematically I get a super high Amp. So where and what am I doing wrong? Mar 28 '17 at 20:46
• I thought it was a 9V. For a AAA your numbers are fine. Anyway, so what about the insulation? The current is not flowing through your body so you don't feel a shock. Where's the problem? Mar 28 '17 at 20:48
• I would assume if I touch a live electrical power line it would shock me, if I grounded myself. I see the paperclip as a miniature power line concept. Its bare metal carrying amps. Mar 28 '17 at 20:49
• The power line would shock you because it's at hundreds or thousands of volts, so it puts an appreciable current through you. The AAA battery doesn't put much current through you. Mar 28 '17 at 20:51

The problem is that the internal resistance of the battery is not Ohmic for currents this large. What this means is that, as the current drawn from the battery increases, the internal resistance of the battery will also increase. For a material in which resistance is due to scattering (like a metal) this is true for a large range of currents, but in a battery where charge is carried by a chemical reaction the current and voltage are not linearly related. So, when you put a tiny resistance across the battery, the internal resistance should increase greatly. Alkaline AA batteries can't be operated too far from $1\,{\rm A}$ without substantial voltage drop I believe. The $0.14\,\Omega$ number you are using should only hold for currents smaller than this. The paper clip is definitely not carrying $9.4\,{\rm A}$. You would realize it. (Here's one at $20\,{\rm A}$).

You can use this fact for a back of the envelop estimate of the internal resistance for large currents. The battery is capable of supplying $0.5\,{\rm A}$ with a very small resistance across it. We can neglect this resistance as it will be small compared to the internal resistance which supplies nearly all of the voltage drop. Thus $$r_{\rm int} = \frac{1.5\,{\rm V}}{1\,{\rm A}} = 1.5\,\Omega$$ The internal resistance should be on the order of $10\times$ larger than the small current specification. If you get your hands on a multimeter you can test this! (Be sure to use the high current settings. Otherwise the fuse usually burns out at $200\,{\rm mA}$ and you will make your equipment manager angry.)

Now, if you were to hook yourself up to the battery in parallel, your body has a resistance of around $10^4\,\Omega$. Only a very small current of approximately $1.5\times 10^{-4}\,{\rm A}$ would pass through your body.

Lastly, in your calculation am I correct in assuming you got the $1.36$ by subtracting off the internal resistance? You should not do that. The total voltage of the battery should be used and including an internal resistance will supply the voltage drop.

I've also heard that NiMH batteries can discharge very fast. Bridging one of those might produce currents like this, but I wouldn't try it unless you want to be short a battery and up some burns.

• Thank you for clearing that up! I will research more about the "non-ohmic" nature of a ideal emf source. However, I would like to add another question. When I short circuit the 1.5V battery with the paper clip, and put the two prongs of the voltmeter on the positive and negative terminal I get a voltage reading of about ~0.35V. What has happened to the 1V that was recorded when I used just the prongs on the battery itself, with no paper clip? Mar 29 '17 at 13:12
• @macas Non-ohmic behavior is quite common. We use small lightbulbs to demonstrate it in labs. When you use the two prongs on the battery, you are connecting a large resistance across it so only a small current flows. This leads the internal voltage drop of the battery Ir_{int} to be small and you get a large voltage reading . When you hook it up in parallel to the paperclip, the current through the battery is large and the internal resistance causes a larger voltage drop. Mar 29 '17 at 18:53

I'm gonna use the data I got from Wikipedia (I simply googled 'Resistance offered by the human body'.) It cites NIOSH as the source of the data.

It says that under dry conditions the resistance offered by the human body is around 100,000 $\Omega$.
Wet or broken skin drops the resistance of the body to around a 1,000 $\Omega$.
And that at high voltage, electrical energy quickly breaks down human skin to reduce its resistance to around 500 $\Omega$.

Given the conditions under which you are performing your experiment, I assume you aren't wet and that you sustain no injury which has broken your skin. Moreover the voltage you are dealing with is by no means 'high'. Given that, your body offers a resistance of 100,000 $\Omega$. Plugging in the value of that resistance in the equation for the current (the resistance of the naked paper clip can he neglected now, it's nothing compared to what your body offer), the current comes out to be about $9×10^{-5}$ Amperes. That's like 0.00009 Amperes. And such a negligible current poses no hazard to you.

Furthermore the internal resistance is not ohmic as mentioned by @user2860104. Hence the internal resistance which you state isn't a constant. It varies.

But the dominant factor here is the resistance offered by your body.

Please would amateur experimenters stop doing silly things like short circuiting batteries with paper clips ! This kind of stupidity is only a couple of steps removed from the lad at school who tried to measure the internal resistance of the local substation by connecting an AVO on the low ohms range to the 240V mains outlet. Needless to say that could have been fatal and/or burnt down the school. VERY luckily it only resulted in a £200 bill for a new AVO-meter and some electrical repairs. I also heard of someone attempting the same type idea with a 12V lead acid car battery. The battery exploded. Acid is liquid. Use your imagination.

Experimenting with ANY source of electrical power is potentially dangerous. Always run your idea past someone with suitable knowledge and experience BEFORE you do the experiment. It very silly to risk possible permanent injury (or even death) just to try to learn something like the fine points of Ohm's Law !

• This is 1.5v battery - not a power substation. Relax there daddy-o. Mar 31 '17 at 13:40