6.10 Power Supplied, Power Dissipated from Office of Academic Technologies on Vimeo.

**6.10- Power Supplied Power Dissipated**

Alright, let’s now consider power in electric circuits. Now, let’s consider a simple circuit which consists of a power supply (like a battery), and the terminals of this battery is connected to a component. Let’s throw that component in the form of a rectangle, something like this. Therefore, as we turn the switch on, we will generate a potential difference between the ends of this component.

Let’s say this point is *a*, this point is *b*, and that potential difference is going to be equal to as much as the potential difference generated by this power supply, by this battery. So once we turn the switch on, therefore, a current will emerge from the positive end and it will flow through this circuit to the negative end of this power supply.

Now, obviously, as soon as we turn the switch in on position, once we have a complete path for the charges to move, and again, we will assume that the moving charges are positively charged carriers), so that they will move along this path and this is the end that they have higher potential. Because as you recall, potential is associated with the positive charges greater than the potential associated with the negative charge and therefore, the potential energy. So the potential energy is greater at this end, in comparing to the other end, the negative terminal of the power supply.

So these charges, in that way, are all done from high electrical potential energy region, towards the low electrical potential energy region, like a ball is rolling down along an incline, from the top of the incline towards the bottom of the incline. So as these charges move along this path and come to this component, which we represented with a box over here, can be a resistor for example, that it will have a certain resistance.

So as they go from *a* to *b*, they are going to transfer some of their energy to the function of this box. In other words, they will lose some of their energy as they go through this box. And once they emerge, they are not going to have the same electrical potential energy that they had before. Therefore, now, once they come to the negative terminal of the power supply, they will not have this energy that they had before they enter into this component.

So if you look at this process for the movement of the charges from point *a* to point *b*, and let’s say some *dq* amount of charge is going through from *a* to *b*. The energy of this charge will be reduced — and let’s make a note of that over here — the energy of incremental charge *dq* will be reduced by the charge times the potential difference between these two points.

Now as you recall, the potential difference between two points were described as the electrical potential energy per unit charge. Therefore, if we look at the energy, then that’s going to be simply charge times the potential difference. Now as *dq* amount of incremental charge moves from point *a* to point *b*, therefore, their energy is going to be reduced by *dq* times the potential difference between those two points. In other words, they will transfer that much of their electrical potential energy for the function of that component over here. So, we can express then this reduced electrical potential energy, *dU*, as *dq* times the potential difference between those two points.

Well, we recall that the current was *dq* over *dt*. So from there, we can express the incremental charge *dq* in terms of the current, as *i* times *dt*, duration of time, such that, the charges are moving from point *a* to point *b*. Therefore, we can express *dU*, then, the incremental amount of electrical potential energy decrease, as charge *dq* goes from point *a* to point *b* through the potential difference of *V**ab* as *i* *dt* times *V**ab*. From here, if we divide both sides by *dt*, we will have *dU* over *dt* is equal to *i* times *V**ab*.

But the *dU* over *dt* is the rate of change of energy, rate of change of electrical potential energy. And that is also equal to rate of electrical energy transfer to this box. So, that is by definition equal to power. Therefore, this term represents rate of electrical energy transfer. Of course, this energy is then going to be consumed by this component, whatever the function that is going to do or convert this energy into the other form of energy.

Well, if we look at the units over here before we go further, the current times the potential difference. The unit of current is in amps and the unit of potential difference is in volts. So we have current times volt. And if we write down these quantities in explicit form, the current is coulomb per second, charge per unit time. And the potential difference of volt, is basically, energy per unit charge, and that is joule per coulomb. From here, the coulombs will cancel, we’re going to end up with joule per second. Joule per second is the energy per unit time, and we have a special name for that, which we call it “watt”.

Well, from Ohm’s law, we know that *V*, the potential difference, is equal to current times the resistance. In other words, the potential difference between two points in a given electric circuit is equal to amount of current passing through those points times the resistance between those two points. And if we use this expression in our power, then we will have — let’s denote power with *P* — *P* is equal to *i* times *V.* And instead of *V*, we can write down *i* times *R*. Then we end up with *i*2*r*.

Of course, we can also express this in terms of the potential difference. In other words, we can also write this quantity as for *i*. We can write down as *V* over *R* times *V*. That’s going to give us *V**2* over *R*. And these quantities for power: *i*2*R* or *P* is equal to *V**2* over *R* represents the resistant dissipation.

**Therefore, if the current is flowing through resistance, a resistor let’s say, with a resistance of R, this much of electrical potential energy will be converted into heat in every second. Or, if, again, we have a resistance of R between two points in the conducting medium, and if the potential difference between those two points is V, again, when the current flows through that region, we’re going to end up in terms of the potential difference and resistance, V2 over R of electrical potential energy being converted into heat in every second.**