 We have been given this combination of blocks and we need to find the equivalent thermal conductivity. Now before I work out this problem, why don't you pause the video and give this one a try first? All right, hopefully you gave this a shot. Now, how do we go about this? Well, we can look at the combination. What do we see when we look at it? So we see that these two blocks, the blue and the yellow block, they are in series with each other and the combination of the blue and the yellow block is in parallel with the green block. So in order to find the equivalent thermal conductivity, what we really need to find is the equivalent thermal conductivity of this combination, which is, well firstly, we can find the equivalent conductivity of the blocks in series and then we can use, then we can find the equivalent conductivity of that combination with the green block that is in parallel. So how did we figure out the equivalent conductivity when the blocks are in series and parallel? Let's write that down. For series, it is R effective. This is the equivalent thermal resistance R1 plus R2 and for parallel, the equivalent thermal resistance would be 1 by R effective. This is 1 by R1 plus 1 by R2. And thermal resistance, how did we write thermal resistance in terms of length, thermal conductivity K and area? Let's write that on the corner. So this is R that is equal to L length of the block or whatever the rod divided by thermal conductivity K into the area A. So first let's try and find the equivalent thermal resistance of the blocks in series. So R effective, R effective really is, we can write this as L divided by K effective. That is the equivalent thermal conductivity into area which is A by 3. The area is A by 3 over here, right? And L is not just single, so we have L plus L, this becomes 2L really. This is equal to R1. So R1 is L divided by 3K into A by 3 plus R2 which is L divided by 2K into A by 3. Now one thing that we notice, A by 3 just gets cancelled, right? We can also cancel this, the length L. So what remains, what remains is 2 divided by K effective. This is equal to, this is equal to 1 divided by 3K plus 1 divided by 2K. This is cancelled. Now we can try and work this out. So what we can do is, we can write this as 2 divided by K effective. This is, we can take the LCM that is 6K and on top will be 2 plus 3. So this will be 5 divided by 6K. So K effective really comes out to be equal to 12K divided by 5. This is the equivalent thermal conductivity of these two blocks in series. Now that we know what's the equivalent thermal conductivity of the combination, we can replace them with a single block and that can look somewhat like this. So now we have one block and these two are in parallel. Now we can use this relation to find the equivalent thermal conductivity. So let's do that. Let us find some space here. Okay, now we have this combination. So for this, we have 1 divided by R, this is the thermal resistance relation 1 plus R1 plus 1 plus R2. R effective or just equivalent thermal resistance is L divided by KA. So this will be 1 divided by L divided by K effective into area effective. So what really is effective area? Well that would be the addition of A plus 3 and 2A plus 3. So A plus 3 plus 2A plus 3 really just comes out to be equal to A. So we can write this as A. This is equal to 1 divided by L divided by, so let's take the pink block 12 by 5 K into A by 3 plus 1 by L. This is K by 2 into 2A by 3. So some things get cancelled. Let's make life simpler. So we can cancel L all the way through. We can cancel this area all the way through. Now this is a little tricky because you can see there's like a denominator, often denominator. So what this, we can just take this up and we can write it in a simplified manner which will be, this will be K effective, K effective. This goes right to the top. This is equal to 12 by 5 K into 1 by 3. You have this factor of 3 right. Plus K by 2 into 2 by 3. So 2 over here gets cancelled and this comes out to be equal to 12 K divided by 15 plus K divided by 3. So over here LCM is 15 and this will be 12 K plus 5 K. So this comes out to be 17 K by 15. So the equivalent thermal conductivity of the combination is 17 K by 15.