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I have a very simple circuit.

12V/1A -> 10 ohm resistor - > ground.

My question is why once I turn the power supply on the resistor catches fire? I Googled and did bump into "wattage rating" but I can't figure out why such small resistor (I mean 10 ohms) is a problem while when I substitute that one for 10k resistor everything seems fine (no fire/smoke).

My "explanation" is that 10 ohms "gives" smaller resistance, thus it should be okay while 10k gives much more resistance thus it should be a much more strained component. Well it seems it works exactly opposite.

If you can give me a formula + explanation like I'm 10 years old you will make me a very happy man.

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  • \$\begingroup\$ The clarification everyone failed to ask is where you get 1A from? Is this your desired current or is it the current you think you get if you put 12V across 10ohm (it is not), or is it the current limit of your supply? \$\endgroup\$
    – Lundin
    Jul 6, 2023 at 6:51
  • \$\begingroup\$ Formally, with 10 ohm resistor you formally overload the power supply as well. \$\endgroup\$
    – fraxinus
    Jul 6, 2023 at 7:28

7 Answers 7

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A 12V/1A supply means it will supply 12 V whatever the current (as long as the current doesn't exceed 1 A).

Ohms law for resistors is U=R*I, or if we put I on the left size, I=U/R (I=current, U=voltage, R=resistance). So the bigger the resistor, the LESS current you get flowing through it.

U=12 V, R=10 Ω -> I=12/10=1.2 A (which is a lot, and even a bit more than allowed by your power supply)

U=12 V, R=10 kΩ = 10000 Ω -> I=12/10000=0.0012 A =1.2 mA (which is far less).

If you want the power dissipated by a resistor (or any other component), you just multiply the voltage by the current.

R=10 Ω:P = U * I = 12 * 1.2 = 14.4 W

R=10 kΩ: P = U * I = 12*0.0012 = 0.0144 W

A typical resistor (the through hole type you get in most hobby kits) is rated 0.25 W max: so if it is a 10 Ω one, 14 W is far too much, and it will burn. With 10 kΩ, you are fine.

Another way to formulate it: if you have a constant voltage, then the higher the resistance, the less current will pass through the resistor, and therefore the less power will be dissipated as heat.

NB: If you have a constant current source, then it is the other way around - the higher the resistance, the more power you get in the resistor.

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    \$\begingroup\$ Note also that the voltage component in Ohm's Law is expressed as V (voltage) or E (electromotive force) outside of Europe. electronics.stackexchange.com/q/99584/2028 \$\endgroup\$
    – JYelton
    Jul 5, 2023 at 16:44
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    \$\begingroup\$ yeah . . . I was so confused having "V=IR" burned in my head 😄 \$\endgroup\$
    – Mike M
    Jul 6, 2023 at 9:56
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    \$\begingroup\$ As a brit I would certainly say V is far more common than U here. \$\endgroup\$ Jul 6, 2023 at 11:54
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    \$\begingroup\$ As an Italian I am too. U is sometimes used because of it being a potential energy, but V for Voltage is much more common (it also honors an Italian scientist, so...). \$\endgroup\$
    – Simone
    Jul 6, 2023 at 14:38
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First of all, welcome to the world of electronics! Second, I hope this experience will not discourage you as I guarantee that your situation has been experienced by almost everyone who tried their first basic circuit exercise.

Your resistor burned primarily because your circuit exceeded the rated power designed for your resistor. A standard small resistor is usually rated for 0.25 Watts. But to explain how this works I will introduce to you two fundamental equations that will guide you throughout your electronics journey.

My "explanation" is that 10 ohms "gives" smaller resistance, thus it should be okay while 10k gives much more resistance thus it should be a much more strained component. Well it seems it works exactly opposite.

Based on this, it seems like you are not familiar with the fundamental law of electronics called the Ohm's Law.


Ohm's Law

$$ V = I * R$$

where V is voltage, I is current, R is resistance.

Based on this, it tells you that the current and resistance are inversely proportional, while V is directly proportional to both I and R.

In math form, if your resistance goes up, the current must go down:

$$ I_{10} = \frac{12V}{10\Omega} = 1.2A $$ $$ I_{10k} = \frac{12V}{10k\Omega} = 0.0012A $$

As you can see, the 10 ohm resistor will draw 1.2 amps, while the 10k ohm resistor will draw 1.2 milliamps. It's an inverse relationship.

After this, the next formula that you should learn is the Power Law.


Power Law (derived from Joule's First Law)

$$ P = I * V $$

where P is power, I is current, V is voltage.

This formula was derived from Joule's Law of Heating and it shows how electrical power is simply a product of both current and voltage. Electrical power is directly proportional to both current and voltage.

Now if we calculate the power drawn by each resistor:

$$ P_{10} = 12V * 1.2A = 14.4 W \implies \text{greater than 0.25 Watt rating} $$ $$ P_{10k} = 12V * 0.0012A = 0.0144 W \implies \text{lesser than 0.25 Watt rating}$$

Based on this, it tells you that a 10-ohm resistor would draw much more power than a 10k-ohm resistor and thus will exceed the power (heat) rating of that resistor causing it to blow up.

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    \$\begingroup\$ Their "first basic circuit exercise"? I've been doing electronics on and off for decades. I set a diode on fire last week. \$\endgroup\$
    – Sneftel
    Jul 6, 2023 at 9:06
  • \$\begingroup\$ @Sneftel Setting diodes on fire is advanced electronics usually quite outside of linear behavior. A resistor will behave pretty linearly right to the point of stopping to be a resistor. \$\endgroup\$
    – user107063
    Jul 6, 2023 at 11:03
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    \$\begingroup\$ @Sneftel haha touche. But congrats you advanced from burning resistors to diodes ;D \$\endgroup\$
    – micropyre
    Jul 7, 2023 at 2:31
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Ohm's law is U=R * I, so I=U/R with the current (assuming a given voltage) being larger when the resistance is smaller. The power P=U * I, or P=U * U/R, so your power here is 12 V²/ R. If R = 10 \$\Omega\$ and the power source is good for the 1.2 A of current you'll be drawing, you are burning off 14.4 W of power.

Typical resistors may be rated for 0.25 W which is way way too weak. A 10 \$\Omega\$ resistor rated for 25 W will work fine but get pretty warm. But it will be pretty large with cooling fins.

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If you can give me a formula + explanation like I'm 10 years old you will make me a very happy man.

There are already two comprehensive answers but I'll add to try and speak to this part of the question, since I don't personally build intuition from formulas. You write:

My "explanation" is that 10 ohms "gives" smaller resistance, thus it should be okay while 10k gives much more resistance thus it should be a much more strained component.

Put aside the resistor for a moment and consider what voltage is. Voltage is measured in joules per coulomb, or "energy per electron". No matter what resistor you put in that circuit, every single electron that flows through will dissipate the same amount of energy in doing so. The resistor just controls how many electrons flow through per unit of time. High value resistors reduce the amount of current (rate of electron flow) more than low value ones do, so the resistor doesn't "give" anything to the circuit, it would be more correct to say that it takes away (which is why it ends up in the denominator of many equations).

On strain:

This isn't really a meaningful measure of what's happening to the resistor, since parts can fail in lots of different ways. In specific, your resistor is overheating. This is because like we said, the energy per electron is constant, and once you pick your resistor, the energy per time is too. All the energy the electrons are losing has to go somewhere, and in this case, it's into your resistor. As the resistor gains energy, it heats up. Hot objects conduct heat energy to their environment faster the warmer they are, so if energy flows into the resistor at a constant rate, it will rise in temperature until it is flowing out as heat at the same rate. This is called Newton's Law of Cooling.

$$ \dot{Q} = hA\Delta T $$

It states that the rate energy is transferred (the Q) equals the difference in temperature between it and the environment, multiplied by the area for transfer, multiplied by a magic constant h. From this its clear that raising the temperature difference increases the rate energy flows out. In your case, the temperature the resistor would need to be to dissipate heat that quickly is above the combustion temperature of the material.

You can also increase the area instead - this is what radiator fins do and you can bolt them onto many parts (for example, power transistors).

You can also adjust the h constant. In practice, h is based on a whole bunch of things like the materials concerned, the flow of fluids over the hot surface, and so on. One easy way to increase h is to surround the hot object in a liquid. You can immerse a resistor in water and drive much more current through it than it can usually handle, because the water will conduct heat away a lot faster than air will (do not though, since water can electrolyze, there are shorting concerns, etc).

Manufacturers give wattage ratings based on specific assumptions about how their resistors will be used, and therefore how quickly energy will flow out.

Finally, as you've noted, you can adjust the amount of power being dissipated by changing the value of the resistor.

Bonus off-topic answer: What happens when the resistor can't combust because there's no oxygen? In this case, the temperature can get quite high - high enough that significant amounts of energy start to flow out as blackbody radiation in addition to the regular conduction / convection to the surroundings, and the resistor will begin to glow 💡

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It's true that a 10k resistor will heat up more than the 10 ohm resistor if the current is the same. This is because for a given current the heat is proportional to the power dissipated by the resistor, which is in turn proportional to the resistance.

However in your case, the current depends on the resistance, and you need to take that into account.

Start by calculating the current as voltage divided by resistance. Then since the power is the voltage times the current, you end up with power being the voltage squared divided by the resistance. You have the same voltage for either resistor, so the heat is going to be inversely proportional to the resistance.

Since the 10 ohm resistor has 1/1000 the resistance of the 10k resistor, the heat will be 1,000 times greater.

By the way, the current rating of your power supply is the maximum that the power supply can handle and doesn't determine the amount of current that flows

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Here's an answer I'd give a 10 year old -- purposefully short and imprecise, meant mostly to give intuition.

The more electrons that flow through your circuit, the more heat will be generated, and the more your resistors will catch fire. The thing about resistors is that they resist or oppose the flow of electrons. And a resistor with a higher resistance (more ohms) will resist the flow of electrons more. So your 10k (10000) ohm resistor really slows those electrons down, and that means not much heat is generated. But the 10 ohm resistor lets the electrons run a bit wild, and then you get the magic blue smoke.

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It seems that your power source is constant voltage, but not constant current, and 12V is not enough to push firestarting current through a 10 kOHhm resistor.

Set up your circuit again, but with a long gap left in the place of resistors. In simplified terms, this gap could be regarded as an infinite resistance. However, you will notice that the gap doesn't seem to be strained at all! The voltage is constant (12V), so the current is zero. That means no heat generation.

Now, it should be obvious why decreasing the resistance increases heat generation. It allows more current to pass through, which equals more heat generation.

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