Now, let's take a look at power in a very simple electrical circuit. We've got this battery with the negative side down here. The positive side up there, and I've indicated the voltage gain when I go from the battery, through the battery from bottom to top. The current is going to circulate around go through the battery, bottom to top and then down through the resistor. Now, you know, a resistor we always indicate the, the voltage drop across the resistor. If the current is going downward in this resistor, then the voltage is higher at this end than that end, it's pushing current through the resistor. So, I've been very careful to indicate the polarities on everything here. So, if I imagine I have a little bundle of charge q and I start down here in the low side of the battery. Well, the battery's going to boost the energy of that charge by raising it through a potential difference V. So, this side adds energy to charge. This side, the charge is moving through the resistor from a high potential to a lower potential. So, it's losing energy. And the work, the energy that it's lost is the work done by the charge. Now, what's happening is the energy that's being lost in the resistor is going into heating it up a little bit. So, it, it produces a little bit of heat, temperature goes up. So, the point, the important point here is that this is a closed system. Whatever work the battery's doing that amount of energy is then in turn dissipated in the resistor. Now, the, remember, work, power is the work, or the energy per unit time. So, the power then is q times V, that's the energy, or the work done on the charge. The energy increase of the charge moving up through the battery. And if I divide by the length of time that takes to happen at least according to the units, this works out to be a current times a voltage. So, q over time, that's the amount of charge I'm moving per unit time, that's current. So, it's, it's coulombs per second, q over P. So, it's coulombs per second. That's current times the voltage. So, that's the formula, the basic formula, for power dissipation. So now, if we look at the resistor and use Ohm's Law, and combine, and we combine that with the I V, I times V expression. So, Ohm's Law says V is I times R, or I is V over R. If we plug that in where we had I, we put V over R. This becomes V squared over R. And I could alternatively have plugged in I R where V was. And I would get I squared R. So, this is probably a familiar formulas. I squared are heating of a resistor. So the, one thing to note here is it doesn't matter what direction the current is flowing through the resistor. Power is always dissipated in the resistor. And that shows up here because of the second power. So, if I have current it doesn't matter if it's a plus or minus, I square it and it always comes out to be a positive number. There's always power dissipated by the resistor, or in the resistor. Now, we have to be a little bit careful about the sign convention, here. This can become very confusing, and this all sounds a little a little dry and pedantic. But it's really important to get this straight to avoid confusion later on. So, we're going to say, when power, we calculate power as a plus number that corresponds to the circuit element delivering power. If it's a negative amount of power, it's absorbing power. Now, a battery. I go from the low side to the high side in the current, if the current is going up through the battery from low to high. This battery is delivering power. I'm taking charge. I'm raising it from a low potential to a higher potential. And if I do that at a certain rate of charge transfer through this battery equal to I, then this battery is delivering power. So, I and V are in the same direction. So, V is increasing upward and I is pointing upward. So, I times V is a positive number in that case. So that corresponds to this battery delivering power, according to our convention. Now, if I had a situation where I were forcing a current through the battery from the high side to the low side. Now, I'm putting charge in the high side and I'm pushing it down to a lower potential so there's energy lost by the charge. So, in that case, the battery is absorbing power, which is perfectly fine. And that happens in your car battery every time you get in and go for a ride. So, the in this case, the I and the V are in opposite directions. And so, I times V, one of these is negative. And the product is then negative. Okay, now here's an example that I want to talk about. And the, to compute the power absorbed or delivered by each element in this simple circuit. So, this is a very simpleminded sort of battery charging circuit in a way. you can think of your car battery is nominally 12 volts and your alternator puts out 14 volts something above 12. And there's a very small internal resistance in the charging part of the circuit. And there may be a larger internal resistance in your battery, especially if it's getting old. so if I take and I represent this situation with this simple circuit. And I just assign, I say, that the current is going around this way I can use Kirchhoff's voltage law and start at the bottom. Go up through this first battery. So, I pick up 14 volts. Then, I drop 1 Ohm times I. So, there's the, the voltage drop across the 1 Ohm resistor. Then, the voltage drop across the 9 Ohm resistor is 9 times I. And then, I'm going through the battery from the high side to the low side, following the current in the defined direction. And so, there's a voltage drop of 12 volts. So, if I solve this the current is 0.2 amperes. So, the circulating current in this circuit is 0.2 amps. So now, let's go back and look at the power delivered or absorbed by every element in this circuit. So, the battery, I have defined the voltage, minus on the bottom, plus on the top. And I have 0.2 amperes going up through the battery. So, it's 14 volts times 0.2 amperes, it's 2.8 watts, plus 2.8 watts. So, that power is being, that's how much power is being delivered by the battery. Now the, the 12 volt battery, I have the current going down through it from plus to minus. So, the voltage is plus on top, minus on the bottom, but the current is pointing downward, so that's negative. And so, this is 12 volts times negative 0.2, so that's negative 2.4 watts. So that's how many, how much power is being absorbed by the battery. Then, if I look at the resistors I have 0.2 amps. So, I squared times R is negative 0.04. So, I'm going from the high to the low side and the current is pointing upwards, so these are opposed. So, this is negative, and its always negative in a resistor. Then, the 9 Ohm resistor, I squared times 9 is negative 0.36. Now, the important thing to note here is that, if I add all of this up, the 14 volt source is delivering 2.8 watts. And then, this battery, the 12 volt battery, is absorbing 2.4 watts and the resistors together are absorbing 0.4 watts. If I add together the total power being absorbed by all of the elements and the power being delivered by this battery, it has to sum up to zero because this is a closed system.
