My 14 yr old son is studying electricity. I thought an interesting demonstration would be to place a penny across the terminals of a nine-volt battery (at a dead short). With its copper content, I thought it would be a good conductor. Then I’d show how a dime would not conduct very well and build up resistance (as evidenced by heat). We did this three times, using a new battery for each coin each time. We monitored the temperature of both coins with a thermometer. To our chagrin, the penny heated up 20 degrees hotter! Why?
Conductors get hot because of current. The more current, the hotter it gets. For a given voltage, the lower resistance passes more current. This is why the penny got hotter: it conducts better, as you surmised, and therefore passes more current at 9 V than a dime does. You had the right idea, you just had the concept backwards.
Doesn’t a poor conductor (such as a stove element) build up heat when current is applied? In my experience with welding cables, heat is always an indicator of loose connections (poor conduction). Thanks for the reply.
A few thoughts:
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Your sample size and (presumably) the number of measurements taken are statistically small. It’s therefore difficult to make any kind of judgment on the data.
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Nowadays, pennies are not pure copper; they’re copper plated. When you calculate DC resistance, you must use the volume resistivity.
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The temperature will be very closely related to the power dissipated by the coin, i.e. P[sub]c[/sub] = V[sub]c[/sub][sup]2[/sup]/R[sub]c[/sub], where V[sub]c[/sub] is the voltage across the coin and R[sub]c[/sub] is the resistance of the coin. V[sub]c[/sub] is not 9 V; instead, V[sub]c[/sub] ≈ 9 * R[sub]c[/sub] / (R[sub]s[/sub] + R[sub]c[/sub]), where R[sub]s[/sub] is the battery’s source resistance. This means that the power dissipated by the coin will not keep getting higher as the coin resistance goes down. Instead, maximum power will be attained when R[sub]c[/sub] = R[sub]s[/sub]. Note that contact resistance also dissipated heat.
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The battery will get hot (due to internal resistance), as will the coin. If the battery gets hotter than the coin, heat will conduct from the battery to the coin. If the coin gets hotter than the battery, heat will conduct from the coin to the battery.
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How are you measuring the coin’s temperature? Measuring surface temperature is notoriously difficult due to stem loss and thermal contact resistance.
Consider this to prove the fallacy of your thinking. The battery doesn’t discharge when there is nothing across the terminals. Why? Because the resistance of the air gap between them is extremely high. Your logic says that the battery will superheat even more with nothing across the terminals.
Resistance doesn’t cause heat - resistance plus current flow causes heat. But, typically, high resistance will limit current flow, so it’s not a simple equation.
Stove elements are made of high-resistance nichrome (or similar) wire because this allows the lelments to be be made more compact. That is, you need a shorter length of wire to get a suitable resistance whih will meet the current draw requirements of the design. If you used copper wire, you’d need a LOT more of it to limit the current appropriately, disregarding the fact that copper would probably melt at those temperatures.
All conductors have resistance, and thus all conductors dissipate heat due to current flow. (Superconductors would be the only exception.) If you attach an ideal voltage source to a variable resistor, the heat power dissipated by the resistor will increase as you lower the resistance. If you attach an ideal current source to a variable resistor, the heat power dissipated by the resistor will increase as you increase the resistance. For a stove element, we can assume the 240 VAC approximates an ideal voltage source. Therefore, as the resistance of the stove element is decreased, the power dissipated by the stove element increases.
But as noted in my first post, there’s no such thing as an ideal voltage source; all real and unregulated voltage sources have a certain amount of source resistance. Therefore, lowering the resistance will allow the resistor to dissipate more power up to a point. When you lower the resistance to the point where it is equal with the source resistance, maximum power will be dissipated by the resistor. If you continue to lower the resistance, the amount of power will decrease. Note, however, that in the real world it gets a bit more complicated than this, since the source voltage and the source resistance are both functions of current.
Could then, resistance + flow = friction? Or are those two totally different functions that both result in heat?
True. But the battery’s case is also shunting the terminals. A little bit of leakage does occur through the case.
Sometimes high resistance means high heat. Sometimes high resistance means low heat. It all depends on how it is being driven. None-the-less, the equations (to a first-order approximation) are pretty simple.
Let’s look at the equations. We have V = IR (voltage equals current times resistance), and P = VI (power (dissipated as heat, in this case) equals voltage times current). Combining these, we can say P = I[sup]2[/sup]R, or P = V[sup]2[/sup]/R. So now we have to ask, what’s being held constant? We’re using the same battery in both cases, and a battery produces a constant voltage, not a constant current (this is actually only an approximation, and not a very good one in this case, but it’s better than assuming constant current). So V = 9 volts, and we use P = V[sup]2[/sup]/R. So decreasing the resistance will increase the power put out as heat, and thus increase the temperature.
Another thing to be careful of in this experiment, though: The penny and the dime will have different specific heats, due to being made from different materials, and they’re also different sizes. Which means that the one with more heat won’t necessarily be the one at higher temperature. One could get around this problem by dropping both into the same amount of water, and measuring the temperature change of the water, but that would make the temperature changes rather small, and perhaps difficult to measure.
I know you’ve gotten lots of good answers already, but I’m going to try to sum things up for you a bit.
If a pair of current-source electrodes were shorted with your coins, your expectations would be correct. At a set current, a low resistence would heat up less than a larger resistance.
However, 9-volt batteries are voltage sources, which will provide variable current based on load. So, the current is higher in low resistences, so heat can build more (though that still depends on the materials, connections and cross-section of the coins).
By the way, after you ruin a 9-volt or two, pull one of them apart. Show your son that 9-volt batteries are properly called that, as they are made up of a battery of 6 individual 1.5-volt cells. Same with 6-volt lantern batteries (only it’s 4 cells, natch). AAA, AA, C, D and such ‘batteries’ are not actually batteries, and are properly called cells.
This is certainly true from a theoretical standpoint. But there’s no way in hell a 9V battery can maintain 9 V across a coin. (If we assume the coin has a resistance of 0.1 Ω, the current would be a whopping 90 A. Ain’t gonna happen.) In this case, we would have to model it as a resistance divider using the battery’s source resistance and the resistance of the coin. (See my equations in my first post.) And when you do that, you’ll probably find that the actual voltage across the coin is less than 0.5 V. (I bet it’s closer to 0.1 V).
Also, wouldn’t the better conductor of heat pass the energy along more efficiently to the thermometer?
Daniel
Yes, Daniel, so you keep the thermometer on the coin until it settles down to a fixed reading. As long as there’s temperature differential, the thermometer will carry on rising, poor heat conductivity will just make it happen more slowly.
I had to think about this for a moment…
Current’s constant throughout a circuit. Therefore, if one part of a series circuit (e.g., your poor connection, with a good conductor either side of it) has much higher resistance than the rest, the power being dissipated there (I[sup]2[/sup]R) is much higher than elsewhere in the circuit. There’s your disproportionate heating. I think.
Like I said, it’s a bad approximation here. But I’m not sure what the source resistance in a 9-volt battery, and it’s generally assumed to be small. Then again, the resistance across the coin is going to be small, too…