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Suppose I have a 100 mAh battery at 20V. I connect a 1000 kohm resistor across it. How much heat will be generated and how can I find the temperature rise in the resistor? As the battery operates I think that the current flow will reduce over time but am not sure about the voltage for a real battery. Perhaps I am not giving sufficient information here, I am sorry for that.

I just wish to know, what information is needed to make such calculation? Have you ever done it? In the ideal case (taking only the most significant factors into consideration) what factors are considered to make an estimate of the heat dissipation and temperature rise and why would the real heat dissipation and temperature in the actual practical experiment be different?

I know this question looks hard, but I will be very happy if I can finally have this mystery resolved.

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  • \$\begingroup\$ Power = Current * Voltage (P=I*V). V across resistor here is 20V, I thru 1M (1,000k - typo?) resistor is .02mA. P = .4mW \$\endgroup\$ – dext0rb Apr 10 '13 at 22:14
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    \$\begingroup\$ Please read this previously asked question, and let us know if you still have questions: electronics.stackexchange.com/questions/32996/… \$\endgroup\$ – The Photon Apr 10 '13 at 22:15
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    \$\begingroup\$ So is it 1,000K or not, OP? \$\endgroup\$ – dext0rb Apr 11 '13 at 2:54
  • \$\begingroup\$ wow thanks, the value of the resistor is not that important, its the actual steps that matter. \$\endgroup\$ – quantum231 Apr 20 '13 at 20:39
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The power delivered to a resistor, all of which it converts to heat, is the voltage accross it times the current thru it:

    P = IV

Where P is power, I is current, and V is voltage. The current thru a resistor is related to the voltage accross it and the resistance:

    I = V/R

where R is the resistance. With this additional relation, you can rearrange the above equations to make power as a direct function of voltage or current:

    P = V2/R

    P = I2R

It so happens that if you stick to units of Volts, Amps, Watts, and Ohms, no additional conversion constants are required.

In your case you have 20 V accross a 1 kΩ resistor:

    (20 V)2/(1 kΩ) = 400 mW

That's how much power the resistor will be dissipating.

The first step to dealing with this is to make sure the resistor is rated for that much power in the first place. Obviously, a "¼ Watt" resistor won't do. The next common size is is "½ Watt", which can take that power in theory with all appropriate conditions met. Read the datasheet carefully to see under what conditions your ½ Watt resistor can actually dissipate a ½ Watt. It might specify that ambient has to be 20 °C or less with a certain amount of ventillation. If this resistor is on a board that is in a box with something else that dissipates power, like a power supply, the ambient temperature could be significantly more than 20 °C. In that case, the "½ Watt" resistor can't really handle ½ Watt, unless perhaps there is air from a fan actively blowing accross its top.

To know how much the resistor's temperature will rise above ambient you will need one more figure, which is the thermal resistance of the resistor to ambient. This will be roughly the same for the same package types, but the true answer is available only from the resistor datasheet.

Let's say just to pick a number (out of thin air, I didn't look anything up, example only) that the resistor with suitable copper pads has a thermal resistance of 200 °C/W. The resistor is dissipating 400 mW, so its temperature rise will be about (400 mW)(200 °C/W) = 80 °C. If it's on a open board on your desk, you can probably figure 25 °C maximum ambient, so the resistor could get to 105 °C. Note that's hot enough to boil water, but most resistors will be fine at this temperature. Just keep your finger away. If this is on a board in a box with a power supply that raises the temperature in the box 30 °C from ambient, then the resistor temp could reach (25 °C) + (30 °C) + (80 °C) = 135 °C. Is that OK? Don't ask me, check the datasheet.

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  • \$\begingroup\$ Is there a reason why mankind opted to numbers like 1/4w 1/2w and so on? why not 1/5w or instead? I thought we might have to know about the "specific heat capacity" of the resistor and talk of Joules (unit for energy) but that is not important it seems. We are talking about power here and not energy. \$\endgroup\$ – quantum231 Apr 10 '13 at 23:06
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    \$\begingroup\$ @quantum: 1/5 Watt resistors would be silly since 1/4 Watt are so cheap :-) \$\endgroup\$ – Olin Lathrop Apr 10 '13 at 23:14
  • \$\begingroup\$ @quantum231, the manufacturer has already accounted for specific heat, etc when they specify thermal resistance and endurance in the datasheet - either by some calculation or experimentally. \$\endgroup\$ – bhillam Apr 11 '13 at 1:18
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    \$\begingroup\$ @quantum231: Specific heat capacity isn't relevant other than if you multiply it by the mass of the resistor you can calculate the rate the temperature will rise or fall when power is applied and removed. It's the ability of the resistor to dissipate the heat that determines its operating temperature and that, as the answer says, is determined by thermal resistance to ambient. Rate of temperature rise may be very important in other applications such as impulse heat sealing (like butcher's bag sealer), thermal transfer printheads or even your cooker hob but that's another question. \$\endgroup\$ – Transistor Feb 16 '16 at 18:46
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    \$\begingroup\$ @quantum231 The specific heat capacity will only tell you how quickly the resistor will get hot, which isn't usually important. How hot it gets in the long run depends on how well heat is conducted away, which is a lot more complicated. \$\endgroup\$ – Simon B Mar 21 '18 at 14:53
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Dissipation is just comes from the power law.

The temperature rise is impossible to predict without knowing how well the given resistor dissipates heat. It depends on what it is in contact with (heat sink or not?), what is the air flow, and what is the ambient temperature. The less well the resistor can actually eliminate heat, the higher its temperature will have to rise so that it can dissipate the wattage implied by the power law. We cannot predict this simply from voltage and resistance.

Furthermore, resistors have a temperature-dependent resistance. If the temperature rise is significant, and the coefficient is significant, it may need to be considered.

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    \$\begingroup\$ this is getting interesting. \$\endgroup\$ – quantum231 Apr 10 '13 at 23:05

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