If total power dissipated in a DC circuit is given by $P = VI$ and power dissipated as heat is given by $P=I^2R$, since these are numerically equal, wouldn’t that mean all the power is dissipated as heat? But that is of course not the case- then how do we calculate the power dissipated as electrical energy to different components (e.g. a filament bulb) in a circuit?
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Qmechanic
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nerdingout
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1The formulae, $P=VI$, $P=I^2R$, and $P=\frac{V^2}{R}$ all give the same value for $P$ when there is a well-defined, $R$. Which one you'd choose depends on which two of $I$, $V$, or $R$ you know. (Sometimes it is less easy to know one of them than, to know the others.) Also, for some "two-terminal devices," $R$ is not well defined. $R$ is not defined for diodes (including LEDs), for electric motors, for inductors and capacitors, etc. For such non-Ohm's Law components, $P=VI$ will tell you how much power the component consumes, but the formulae involving $R$ are unhelpful. – Solomon Slow Mar 20 '23 at 12:22
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The power delivered to a two-terminal circuit element is $P = VI$, where $V$ is the voltage across it and $I$ is the current through it. Circuit theory says nothing about what kind of energy conversion takes place in the element. Depending on the element, there might be conversion to heat (as in a resistor), light (as in an LED), mechanical energy (as in a motor), sound (as in a loudspeaker), and so on.
In an incandescent light bulb, the electrical energy is first converted to heat, and then partially radiated as light. The efficiency of this conversion will depend on the light bulb.
Puk
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I understand. But why do textbooks (while explaining the heating effect or current) describe undesirable power loss as heat as equal to I^2 R and V^2/R, which are in fact equal to the total power dissipated? What am I misunderstanding here? – nerdingout Mar 20 '23 at 06:23
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In ordinary resistors (which those textbooks are likely referring to since those equations include $R$), the energy is dissipated as heat. This is sometimes undesirable and sometimes not. The total power and power dissipated as heat are the same in this case. Incandescent bulbs are specifically designed so that the filament reaches very high temperatures in response and radiates lights. In principle there is still an intermediate energy conversion into heat. – Puk Mar 20 '23 at 06:27
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The two equations $P=VI$ and $P=I^2R$ describe the same power; which equation you use depends on the information you have available.
When we think of a simple circuit as a battery connected by wires to a light globe (say), we assume the battery and wires are perfect, so all the resistance in the circuit comes from the light globe. In reality this is not the case; there will be some resistance in the wires, and also in the battery.
If we measure the voltage across the light globe and the current flowing through it we can work out the power it puts out (which equals the electrical power it uses) using $P=VI$. Some of this will be heat, but some will not be.
The current is the same flowing all through the circuit, so if we know the resistances of the two wires and the battery we can use $P=I^2R$ (using these three components' values of $R$) to find the electrical power loss in them. These power losses are almost always heat.
If we add the four amounts for $P$, we get the electrical power being used/lost by the system, which will be the same as the power produced by the chemical reactions in the battery.
Peter
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Your misunderstanding may come from the fact, that $R$ is varying, e.g.
- open battery, $I=0, R=\infty, P=V/\infty = 0$
- battery short circuit, $ I=V/R_{short}, P \gt 0$, battery heating up
Electric components are idealized thingies. E.g. resistors develop heat, while sources, capacitors and inductors don‘t. If real ones do, as they do, you need to make them less ideal, i.e. add resistors to them, to account for loss of energy.
To add, you can model a real life battery by and internal ideal voltage source $V_{ideal}$ and an internal series resistancs $R_{internal} > 0$. If you short-circuit that one, current is limited to $I_{max} = \frac{V_{ideal}}{R_{internal}}$ and hence $P_{max}$, i.e. you can't draw more current from said battery. Just compare the big battery with the smaller one, both supplying $1.5$ V.
For real life batteries even $R_{internal}{(t)}$ changes over time, reflecting the batteries discharge or loss of energy over time. Again, compare the run time when supplying the same device with a small and a big battery of $1.5$ V.
MS-SPO
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