So P = I*V because V is the amount of energy per coulomb and I is the amount of electrons going past a cross sectional area in one second.
So if we do the math, since V = J/C and I = C/sec, if we multiply those values we get J/sec which is the amount of energy turned to other energy from electric potential energy and the equation seemingly works out because the Coulombs disappear from equation and I can choose not to overthink about it . (But i want to really understand the whole thing)
But the thing confusing part is that it doesn't make sense intuitively because no matter how hard I try to come up with something meaningful,just multiplying the rate of flow which is the amount of coulombs passing through a cross sectional area by the potential difference seems very difficult to connect. (because for example lets say it takes an electron to move through a potential difference of 1V in 2 seconds, and if the current is 1A, the electron behind it will enter the resistor(light bulb) when it is halfway through the resistor)
It only makes sense if let's say, 3 coulomb of charge moves through a potential difference of 1V in 3 seconds, which means it converted 3 joule of electrical energy in 3 seconds, therefore it has a power of 1J/sec = 1 watt. Does charge instantly get converted to other energy and the 2 points that has a potential difference is just a cross section?
-
$\begingroup$ As indicated here, we encourage users to search and research before asking questions, and this feels like the kind of query that can be found elsewhere. $\endgroup$– secavaraCommented Apr 24, 2022 at 16:44
-
1$\begingroup$ "...because V is the amount of energy per coulomb". The potential difference, $V$, between two points is the work required per unit charge (work per coulomb of charge) to move the charge between the two points. Then the rate of work (power) is the voltage times the amount of coulombs per second (current). $\endgroup$– Bob DCommented Apr 24, 2022 at 16:46
-
$\begingroup$ @secavara I actually did a lot of research and was reading that wikipedia page exactly hours ago but if you actually see the contents of the page it doesn't really answer my question. $\endgroup$– ikigai20Commented Apr 24, 2022 at 16:54
-
$\begingroup$ I isn't the amount of electrons going past a point but the amount if charge. Clearly since the basic unit of charge is that on the electron there is a relationship between amount of charge and number of electrons. $\endgroup$– jimCommented Apr 24, 2022 at 18:05
10 Answers
multiplying the rate of flow which is the amount of coulombs passing through a cross sectional area by the potential difference seems very difficult to connect
Sorry, but there isn't much more to it than that.
It might help to consider that if we're calculating power from current and potential difference we're typically not just talking about a current through a cross-sectional area, but along a path with non-zero length. So there is a separation between the two points (or equipotential regions) that charge is moving between, and we assume that all the current that leaves the first point reaches the second point, so that we can talk about the current from one place to the other, rather than just across a surface.
But with that said, the potential difference tells you the energy transferred by each element of charge, so if you know the rate of charge being transferred you know the rate of energy being transferred. And "power" is the name we use for a rate of energy being transferred.
You describe it perfectly, but then say you don't understand it, so I think you really are just over-thinking it.
-
1$\begingroup$ Yes but i thought like, potential difference is something constant(for example some has 12V, some has 1.5V, and if the current is going through a light bulb, the amount of work done per coulomb(in this case Electrical PE to light and thermal energy), it must move some distance since deltaPE= qEd.If 1coulomb of charge goes through a 1V of potential difference(which is some distance) it dissipated 1 joule of energy, but what if the charge takes more than 1 second to go that distance, that must mean another 1 coulomb of charge will start to do work before the first wave of electrons do work? $\endgroup$– ikigai20Commented Apr 24, 2022 at 17:05
-
$\begingroup$ If the 1 coulomb takes 2 seconds to go through the same potential difference then the power is just one half the previous value. This also means that the current is just one half the pevious value. And no matter how long it takes for a specific pack of charge to go through the 1 volt, other packs will enter and exit during the same time. At any given time the amount of charge moving between two points in a DC circuit is the same. $\endgroup$– nasuCommented Apr 25, 2022 at 1:11
-
$\begingroup$ @ikigai20, Typically the charge flows continuously through the device, l like water through a hose. If I set the flow rate of my garden hose to deliver 1 liter per minute of water, it doesn't pop out 1 liter of water, then wait a minute with nothing coming out before another liter appears, it just flows in a slow trickle continuously, and if I consider any 1 cm segment of the hose, there will be a small amount of water in it at any time I choose to look (if I were to use a higher flow rate, like 1 liter per second, I'd find the hose entirely full of water at any moment I look). $\endgroup$ Commented Apr 25, 2022 at 15:12
-
1$\begingroup$ @ikigai20, because that's the definition of "1 ampere". If it was more or less than 1 C/s, then the current would be more or less than 1 A. $\endgroup$ Commented Apr 25, 2022 at 17:32
-
1$\begingroup$ Also we know that charge isn't stored in an (ideal) resistor. But charge is a conserved quantity. So the exact same current that goes in one terminal has to come out the other terminal. This is a form of Kirchhoff's current law (or, KCL is an application of the principle of conservation of charge). $\endgroup$ Commented Apr 25, 2022 at 17:34
Consider a charge $q$ moving with speed $v$. In $t$ seconds it will move a distance $d=vt$, so the current (charge per second passing by that cross section area) is $I = q/t = qv/d$.
Electric field is the gradient of the potential, so the electric field that exists within that distance $d$ is $E=V/d$, so $V=Ed$ (imagine a short distance or a constant field so no derivative needed for this hand-waving argument).
Now multiply: $$VI = (Ed)(\frac{qv}{d}) = qEv = Fv$$ Force times velocity: power.
Hope that helps!
-
$\begingroup$ Thanks for the answer but i do have one question. The V that you got from E=V/d must be some arbitrary number depending on the distance but in circuit problems, the voltage drop is already set before(for example 1A current going through 1V so the power must be 1W) but how do we know if that 1A of current goes through the distance of 1V in exactly 1 seconds? $\endgroup$– ikigai20Commented Apr 25, 2022 at 13:21
-
$\begingroup$ Do you mean 1 C of charge rather than 1 A of current? $\endgroup$– garypCommented Apr 25, 2022 at 22:35
-
$\begingroup$ oh my bad i meant to say 1C of charge $\endgroup$– ikigai20Commented Apr 26, 2022 at 15:46
-
$\begingroup$ Slightly abusing notation we know that the definition of power is $P = E/t$ or in the case of an electrostatic problem $P=qV/t$. Turn that around $t=qV/P$. You told me that $q=1$C, $V=1$V, and $P=1$W. So we get $t=1$s. $\endgroup$– garypCommented Apr 27, 2022 at 17:32
Force on a charge $\lambda dl$
$f dl = \lambda \vec{E} dl$
Amount of work done by $\vec{E}$ moving a charge $\lambda dl$ with a velocity $\vec{v}$ in time dt
$dw = \lambda \vec{E} \cdot \vec{v} dl dt$
Rate at which this happens (power)
$dp = \lambda \vec{E} \cdot \vec{v} dl$
Substitute in: $\vec{I} = \lambda \vec{v}$
$dp = \vec{E} \cdot \vec{I} dl$
Integrate to find the total amount of work an electric field $\vec{E}$ does on a distribution
$ P = \int \vec{E} \cdot \vec{I} dl$
Negative charges move opposite the field, so $\vec{I}$ and dl are in the opppsite direction so can be swapped *** read bottom ***
$P= -\int I \vec{E} \cdot \vec{dl} $
When current is constant, it is independant of the integral and can be factored out.
$P= -I \int \vec{E} \cdot \vec{dl}$
Sub in $v = -\int \vec{E} \cdot \vec{dl}$
$$P=IV$$
This makes sense , as a potential difference between A and B being Postive , would mean positive work is done on negative charge.
*** the reason I have said that I and dl are in opposite directions is a technical detail of my proof, in the beginning, I assumed that dl = v dt, meaning I'm finding the work done given the electrons move in the direction I'm evaluating my potential ( from negative to positive), which creates a current opposing my dl vector***
To get a basic intuition for electrical equations I tend to think of a hydrological analogue. When water is moving through waterways, the difference in elevation corresponds to a potential difference, and the rate of water moving though a canal corresponds to the current. Water likes to move from high places to low places, like positive charges like to move from high electric potential to low potential; the fact that in most situations it is the negative charges (electrons) that actually move to produce current is an unfortunate detail that can be ignored (for electrical equations there is no distinction between current produced by positive charges moving one way or negative charges moving in the opposite direction, so no intuition is lost by thinking of positive charges moving).
The equation $P=V*I$ for the power dissipated in a resistor is then analogous to the hydrostatic power dissipated by a current of water flowing between points of different elevation (and which could be converted to other forms of energy as in the turbines of a hydroelectric plant). This power is the product of (gravitational potential change due to) the difference in elevation times the rate of flow (in mass of water displaced per unit of time). Each unit of water dissipates energy equal to the product of its mass and the change of its gravitational potential (the latter being height change times gravitational acceleration, but you can think of this as the "tension" between the reservoirs between which the water is flowing). Different units of water dissipate energy independently and the total is formed by summing over all units of water displaced. It should not be intuitively difficult to see that the power dissipated is proportional both to the difference in elevation between source and sink (producing pressure for the water to flow), and to the rate of flow that actually takes place.
Along the lines of the answer by @garyp ,
it might be good to note that $VI=Fv$ in that answer
hints of a "mechanical analogy".
Thinking in terms of [generalized-]work (as mechanics or thermodynamics),
$d{\cal W}={\cal F}\ d{\cal x} = \ (\mbox{Electric Potential}\ V)(\mbox{displaced charge}\ dq)=V\ dq $,
the electric potential can be considered a generalized-force as a "work done per unit displaced-charge", thinking of charge as a configuration variable. (It may help to think of "charging up a capacitor" as displacing charge from one plate onto the other plate.)
Then $P=\frac{d{\cal W}}{dt}$.
In a different context, for some steady-state situation,
the associated instantaneous power (assuming that $V$ is constant in time)
is then $P=\frac{d{\cal W}}{dt}= V\frac{dq}{dt}=VI $ (analogous to $P=F\frac{dx}{dt}=Fv$).
Other answers address why power is given by the product of voltage and current, on grounds of physics. Regardless of the physical meaning we associate with this product, the sum of voltage-current products of all branches (or components) of a network (or circuit) is guaranteed to be zero by Tellegen's theorem: $$\sum_k V_k I_k = 0. $$ where the sum is over all branches. In circuits where the voltage-current product corresponds to power, this is a statement of conservation of power (i.e. power supplied is equal to power consumed), which is reassuring. In other words, based on network theory alone, the voltage-current product is a suitable candidate to represent power because it is conserved.
Tellegen's theorem is actually much more general than this. In fact, the components of the circuit need not correspond to physical devices at all: the only conditions are that the branch voltages obey Kirchhoff's voltage law and the branch currents obey Kirchhoff's current law. Under these very general conditions, the total "power" will still be zero.
"Power" as a concept, on its own, doesn't really make a lot of sense unless you think about what is doing the work. That's because that's what it is: it is the rate of doing work or more generally, utilizing energy.
In this case, what is happening is the electric field is doing work on the electric charges, and thus expiring energy in the process. That field, in turn, is being powered by whatever power source is used in the circuit, which is pumping energy into it to fuel its doing the work.
When a charge interacts with that field, it feels a force: $F = qE$. When that charge is pushed by such force over a distance $d$, work $W = Fd$ is done. Depending on the nature of the intervening medium, that push may happen over a long or a short time and as a result, the rate ($\frac{W}{t}$ or, if variable, $\frac{dW}{dt}$) at which the electric force field does work on the charge will be different.
The force over the charge of a small volume in an electric field is: $\delta F = \delta q E = \rho A\delta x E$, where $\rho$ is the local density of movable charges and $A$ the wire cross section. If there is no magnetic induction effects acting, the electric field can be expressed as the derivative of a scalar potential $V$: $$E = \frac{dV}{dx}\implies \delta F = \rho A\delta x \frac{dV}{dx}$$ when $\delta x \to 0$: $$dF = \rho A dV$$ Multiplying by velocity of the charge, the left side is the power by definition: $$dP = \rho Av dV = IdV$$
Integrating over a length of wire : $$P = VI$$
Simplify it. Power is energy in application of delivery to a new or transitioning it's system. The qualities of a engine, motor or conductor dictate how much energy shall be delivered as power in comparison of movement/resistance in a given time frame of that energy's conveyer. Laws of thermodynamics and thermal properties of what's carrying the energy. Waste heat is produced thru out all duration of transit
Let's take water as an analogy. V is the height of fall of the water and I is the volume that falls on the mill wheel.
In electricity, V is the potential difference between the source and the sink and I is the number of electrons that are moved through an electric drive or another sink.
