 We have a scary looking circuit with many many resistors connected across a battery. Our goal is to find the current through one of these resistors, so 20 ohms. How do we do that? My go-to method is always to try and reduce this circuit to see if I can find some series resistors, parallel resistors, keep reducing it until I can find voltage or current across those resistors. Once I can find them, I'll stop and I'll find all of those values and then I'll come back. Okay, so let me show you what I mean. So let's first step is to try and reduce this circuit. Can we find some resistors that are in series or in parallel? Well, I see these two in parallel because they're connected across the same points. Think about it, this point has the same voltage as this point, right? Because there's nothing in between, no energy loss in between. This point is the same as this point. So these two are connected in parallel, so that's great. I can reduce that. Anything else do I find? For example, are these two in series? Looks like that, but it's not really because for series, see if the current through here and here is the same and it may not be. Whatever current flows here, some of it can flow here and it might split and some of it might flow here. So they're not in series. These are not in series. What about these two? Are they in series? No, because the current from here, well, it can split. Remember for series, the current needs to be the same. Not in series. What about these two? They're also not in series. So I find these two to be in parallel, so that's great. What about these two? What do you mean these two, you might say? Like there's nothing over here, right? Yes, there's nothing over here. That means there is zero resistance here and that's important and we will see why that's important. So let me just draw that over here. Sorry, not 10, zero, zero ohms because I have a zero ohms in parallel with 15 ohms. You see that these two are in parallel. So let me reduce these two. Let me reduce these two and see what happens. So let me just do that. So for parallel, I have one over RP equals one over R1 plus one over R2. So for these two, it becomes one by 20 plus one by five. And if I take the common denominator 20, I get one plus four. That is five. So let me just write that five here. Five by 20 becomes one over four. But that's one over RP. So RP in this case is four, which means these two resistors, I can replace it with one single resistor connected to this point of four ohms. What about these two? I want you to pause the video and think about what is the parallel combination of these two. Alright, let's see if you use the formula, I might end up getting something weird. So we'll get one over zero plus one over 15. Which what do you get? Well, let's see if I cross multiply, I get 15 plus zero. So that's 15 divide by zero. That's one over RP. So what is RP reciprocal of this zero by 15? So RP in this case just gives me zero. So these two give me zero resistance. What does that mean? What is it going on over here? Well, think of it this way. If there is no resistance over here, whatever current flows here, all of the current from here will flow here. There'll be none of the current flowing from this branch. Because remember, current likes the path of the least resistance. This has the minimum resistance. And so all the current flows here. Whenever you have something like this, just a wire with zero resistance, we call that as a short, a short circuit. So since I short circuited these two points, there'll be no current flowing through anything else, all the current flows here. And therefore these two can be replaced with a zero ohm resistor, which is basically another short. So if I do that, so let me just show you what it looks like. There you go. So here is our 10 ohm. Here is that four ohm resistor replacing these two, which I got over here. And this I have replaced over here with a zero. Remember, zero means a wire which has no resistance. Does that make sense? Okay. Now that I have this, I ask myself, okay, can I find voltage and current through all these resistors? If I can, I'll stop and I'll go back. Otherwise I'll continue. I can't find anything because I don't know the voltages here, right? The 10 volt gets divided, but I do see that these are all in series. And so let me go one step further. So series, I know I have to add them so I can do it mentally. Six plus four 10, 10 plus 10 is 20. So I'll replace these three with one single resistor of 20 ohms. And if I do that, there you go. I get this. And now I know that this entire 10 volt must be coming across this. So I know the voltage here is 10 volt. So what I do now is since I know the voltage, I'll find the current over here. What's the current? Well, remember V equals IR ohms law. So I equal to V divided by R. V is 10. R is 20. V divided by R is 10 by 20. That's one over two. So the current and what direction is the current? It's going to be positive to negative. It's going to be this way. So the current here is one by two amp. And now what I do is I keep going back because my eventual goal is to come over here. So let me go back and I go back from here to here. Look, this wire is the same as this wire. This wire is the same as this wire. Remember all these three together became the 20 ohm. So that stays the same, which means the current stays the same here. So the current here, here, here is the same. It's half ampere, half ampere. The current through this register is also half ampere. What do I do next? Well, if I know the current, I'll find the voltage. That's the game. If you know the voltage, you find the current. If you know the current, you find the voltage. And I try to do that for all the resistors because it just makes it simpler to go back. So what is the voltage here? Well, V equals IR. So I is half, R is 10. So 10 into half five volt. So here voltage must be five volt. What about voltage here? IR, half into four, two volt. IR, three volt. And let's just check three plus two plus five is 10. And that's exactly that 10 volt that got divided. See, I found voltage and current everywhere. Go back. And I go back. Well, I know the voltage here. That's the same three volt. I know the current here. So let me just draw that. Let me just do that. So I know this current is half ampere. I know this voltage is three volt. Okay, what about here? Well, I know that there is absolutely no current flowing here. All of that half ampere is flowing only here. So this half ampere flowing from here to here. Okay, now comes the main part. Again, here also I have half ampere. And here also I have the voltage of five volts. So that's five volt. Okay, what happens here? Remember, this four ohm came from the combination of these two. What do I do here? Can you pause and think about this? All right. This is a parallel combination, right? In parallel, the voltage stays the same. I know the voltage here is two ohms, sorry, two volt. So the voltage here and the voltage here must be the same, two volt. Does that make sense? So I know this voltage is two volt. I know this voltage is two volt. And then you can now find the current. I can now find the current here and I can find the current here. V equals IR. So over here current should be I is equal to V divided by R. So it's one over 10. And in fact, it's not going down. It should be going up actually. So it's one over 10 amps. And what is the current here? It's even if it's not asked, I can just calculate it. It is two over five amps. And if you check it, you'll find that these two add up to give you half amp. That's the half amp that we got over here. Now I know this looks like there's a lot of steps involved, but I promise you if you practice this, you'll get it. The whole game is to first reduce the circuit from the voltage find the current, come back. If it splits as a series resistor, the current stays the same, you find the voltage. And if you come back and it splits as a parallel resistor, the voltage stays the same and you find the current and you keep doing that. Keep practicing that, you'll get it.