The absolute thermal resistance can be expressed in the following formula:

$$R = \frac{\Delta T}{P} \left[\frac{K}{W}\right]$$

It shows how big is the temperature difference between two points in the object per watt of applied energy. So, for known R = 3 K/W and P = 2 W we will get 6 K drop.

What happens with this power loss (temperature)? In case of electric analogy the lost current voltage drop (energy loss) goes into heat, but it is not clear for me where this energy goes to. Or is it only mathematical model which is used only in calculations?


1 Answer 1


I'd say it's a bit more than a mathematical model, but you are taking the analogy to a place where it can't be applied.

First to clear up some terminology in your question about the analogy. The temperature loss is equivalent to a voltage loss when going through a resistor, not the power lost from the circuit. In the electrical analogy for a circuit, there is no lost current; the current has to be constant across for a single loop/line.

Current is also analogous to heat transfer rate, and that's where the analogy starts to fall apart. Heat transfer rate already measures where the energy is going. That's the point, it measures the thermal energy transfer per unit time. You're already measuring where the energy goes (and it's rate) when you measure $\dot Q$ (or $P$ in your equation).

So it basically falls apart because an electric circuit is only telling you where the charge goes; which doesn't directly measure energy. But the heat transfer equations tell you where the heat goes, which is directly telling you about the energy.

  • $\begingroup$ oh, yes of course voltage not current =) So, therefore, I can say that if heat transfer rate say how much and WHERE energy goes the thermal resistance tells me how much and where energy doesn't go. So, in simple words, thermal resistance doesn't show me energy which was lost somehow but the energy which didn't go to where it suppose to go. Is that closer to the truth? $\endgroup$
    – alberand
    Nov 16, 2019 at 19:28
  • $\begingroup$ @A.Albershteyn It doesn't really tell you where the energy doesn't go, just like resistance doesn't really tell you where the current doesn't go. It just tells you how hard it is for the energy to go there; which then relates it to the temperature difference. Just like how telling you how hard is it for the current to flow allows you to relate it to the voltage difference. $\endgroup$
    – JMac
    Nov 16, 2019 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.