 Kaniga buys two bulbs of the following power ratings to decorate her lawn during Diwali. So we can see there are two bulbs, A, B, we have the power rating. She connects them in series across a 6V battery as shown. So we have the two bulbs connected in series and it is connected across a 6V battery. We need to figure out the power dissipated in each of these bulbs. Power dissipated in A and power dissipated in B. Now these kind of questions can get a little tricky. So why don't you pause the video first, try to think about strategy, how will you approach this question and then we will solve it together. Alright, hopefully you have given this a shot. So let's see what all do we have. We have the power ratings of these two bulbs, A and B. We have the battery across which they are connected 6V and we know that the bulbs are connected in series. So we need to figure out the power dissipated, right? Since they are connected in series, one thing that we can be sure of is that the same current will pass through each of the bulbs. The current I will pass through A and the current I will pass through B. If we were able to figure out this current, then maybe we can figure out the power dissipated, right? What else do we know? We know the power rating of A and B. Well, from here, from this data, we can try and figure out the resistance of these two bulbs because we know that power is V square by R. So we can figure out R, this is V square by P. We already know what P is individually for these bulbs. You also know volt, like this is the 10 volt 25 watt power rating of A. This is 10 volt 50 watt power rating of B. So we can figure out R, A and R, B. We can do that. We can figure out R, A and R, B. Then what if we added them? We found the equivalent resistance of the circuit and using that, using Ohm's law, then we can figure out the current. Okay, so we have some strategies. So let's go step by step. Let's figure out the equivalent resistance of the circuit. Before that, we will have to figure out the individual resistances of these two bulbs. So that becomes the first step. R, A is really R, A. This is V, A square divided by P, A. And R, B, this is V, B square divided by P, B. So R, A, this is V, A square. This is 100. I'm just writing 100, 10 square divided by 25. This comes out to be equal to 4 Ohms. And R, B, R, B, let's write that over here. R, B is V, B square, again 100 divided by P, B. So that is 50. This is 2 Ohms. Right? All right. That is going crazy. So, all right. Let's find the equivalent resistance now. We have 4 Ohms and 2 Ohms. So when you find the equivalent resistance, this is R, A plus R, B. This is really 4 plus 2 and this is 6 Ohms. We know the equivalent resistance. We know the voltage of the cell. Using Ohm's law, using Ohm's law, what we can do is we can find the current that is passing through this circuit. So that is the second step using Ohm's law. Ohm's law is V that is equals to IR. If you need to figure out the current, I is equal to V by R. So I, this is equal to V by R and this is 6 divided by 6, 1 ampere. So 1 ampere of current is passing through A and 1 ampere of current is passing through B. Now we know the individual resistances. We know the current passing through it. Let's try and find the power dissipated across, power dissipated in each part. So that will be our third step. Third step is power dissipated in A. This is equal to I square into RA and power dissipated in B. This is I square into RB. So I is just 1 into RA which is 4 Ohms and for power dissipated in B, this is I square again 1 into 2 Ohms. So this is really 4 Watt. Power dissipated in A is 4 Watt and power dissipated in B is 2 Ohms. So there you go. These are the power dissipated in each of these two volumes.