 I want to introduce to you a very special circuit called the Wheatstones Bridge. I want to talk about the logic behind this particular circuit. So what's a Wheatstones Bridge you ask? Well, here's an example. Wheatstones Bridge is a circuit that contains four resistors like these and we call them the four arms of the bridge and there is a resistor in between like this. So this is a Wheatstones Bridge. Now before I continue let me ask you, what if I asked you to calculate the current through this circuit? How would you do that? Well, if you ask me, the first thing I always like to do or I always like to think about is, can I reduce this circuit into a single resistance by using my series and parallel formula? Can I do that? Well, I don't see any resistors in series. These are not in series, there is a resistance in between. I don't see any of them in parallel. Again, they are not in parallel because they're not directly connected across each other. So I can't reduce this using series and parallel. And when I can't do that, the next thing I resort to, the general method is Kirchhoff's Laws. But of course, Kirchhoff's Law is tedious. You have to think about currents and then I have to use the loop rule and blah. But here we won't require Kirchhoff's Laws because this is not any Wheatstones Bridge. This is a very special Wheatstones Bridge. What's the speciality you ask? Well, you can see that the resistors here are having the same ratio as the resistors here. One is to two, one is to two. Or you can also say resistors here are in the same ratio as resistors here. One is to five, one is to five. And whenever you have a Wheatstone network like this where the ratios are the same, then we say such a network is balanced, such a bridge. We call it a Wheatstones network or a bridge or a circuit, same thing. We call that a balanced network. You may ask, okay, that's a nice thing but why should I care about a balanced Wheatstone network? Why should I even care about a Wheatstones network? We'll come to that slowly. We'll see the practical application. But the speciality of a balanced network is that it turns out and we will see how, that's the whole idea behind the video to logically see how it turns out and whenever this network is balanced there will be no current flowing through this middle resistor. So in this case the current over here becomes zero. And if the current here becomes zero that means the current here should be exactly the same as the current here. All the current from here flows here because nothing flows down or nothing flows up. And so that means these two come in series. Similarly these two also come in series. Same current flowing through them. So you see in the balanced condition it's as if this resistor was not even connected. So I can completely neglect that resistor and my circuit will not be affected. And the beauty now is that means I can now solve this series, series parallel. And I can solve this. And so often in your problems you'll be given Wheatstones network like this and you may have to identify whether it's balanced or not. If it's balanced then it's your lucky day because then you can solve it immediately. So before we go forward let's do a quick check of our understanding. Here are two networks. I want you to pause the video and see if they are balanced or not. Alright let's look at the first one. This is a Wheatstone network very similar to this. I've just arranged them a little differently and this is usually how the Wheatstone network is shown. And let's see if it's balanced. Well this is in the ratio 1 is to 3. This is in the ratio 3 is to 1. You see that? 1 is to 3 is to 1. So this is not a balanced network. So here current will flow and so if this network was given to you bad day for you. If this was 10 and this was 30 then it would be balanced. What about this one? Well it looks like it's a balanced network. 1 is to 4. 1 is to 4. But if you look carefully you will notice something is wrong here. This is not even a Wheatstone network. Notice Wheatstone network is connected between the red resistor is connected between these two points. This and this point is here now. Can you see that? Because this is where the battery will be connected. I've just rotated this. So don't get confused. This is not even a Wheatstone network and you can directly solve this. These are in series, these are in series the three are in parallel. So don't get confused. Definitely current here won't be 0 because it's not a Wheatstone's bridge. On the other hand if this resistor was connected here then it would have been a Wheatstone's bridge. So you can check the ratios 1 is to 3, 1 is to 3. Now the current over here would be 0. So now let's get back to the point. Why is the current over here 0? How can we logically understand this? You see in the books they derive it using Kirchhoff's laws and everything but I always wondered why if it's a balanced condition why if the ratios are the same why would the current be 0? Well, here's how I like to think about it. First, I would like to make a copy of this circuit without that red resistor and see how the voltages get distributed everywhere. That will give us the clue. So in this particular case current is running and everything over here notice now these two resistors are in series. Similarly these two resistors are also in series. And you know in series current is exactly the same. From Ohm's law, because I want to talk about voltage I'm going to go to Ohm's law Ohm's law tells me that V equals IR V equals IR and so if you have two resistors which have the same current then notice the voltage must be proportional to the resistance. In other words, if the ratio of resistance here is 1 is to 2 the ratio of voltage here must also be 1 is to 2. Voltage here is X the voltage here must be 2X Same as the case here, the voltage here is X, the voltage over here must also be 2X and that means I can now immediately calculate what the voltage is because I know 9 has to be divided in the ratio 1 is to 2 and when you divide that into 1 is to 2 you get 3 and 6 and that's why I took 9 because it's easy to divide that but you couldn't take, you could have taken any values of the voltage and the same will be over here. Here also this would be divided in the 1 is to 3 ratio, 3 volts and 6 volts. So voltages get divided equally because the ratios are the same so what? What does that do? Well now let's look at the voltages at these two points to do that remember whatever is the voltage here same is the voltage here so the voltage here and here is exactly the same isn't it? If you want to call that as 9 and you want to call this as 0 then this would be 9 and this would be 9 as well Now notice when I go from here to here there's a 3 volt drop but when I go from here to here there's again a 3 volt drop meaning these two points must also have the same voltage so voltage at point A must exactly equal voltage at point B that's the consequence of having the same ratios of the resistance so what you ask? Well now notice because the voltages are exactly the same over here if I were to go ahead and now in this circuit when it's running if I were to put a resistor in between or a capacitor in between doesn't matter what you put in between notice the voltage across it is the same therefore I mean the yeah these two voltages are the same therefore the potential difference is 0 there is no difference in voltage there is no potential difference and therefore there will be no current flowing over here and that's why a balanced Wheatstone's network has no current does that make sense? is that wonderful? the whole idea is when the ratio of the resistances are the same the voltage also gets divided in the same resistance and therefore these two points will end up with the same voltage making sure there is no potential difference that's why the current goes to 0 and now hopefully you will agree that it doesn't matter what voltage you put over here what are the values of the resistances over here or over here all that matters is that the ratio here must be exactly the same as the ratio over here and then you will find no current flowing in between now you might ask okay so what's the application of this why should I care about this well this is a this network is used in calculating practically values of unknown resistors for example let's say there is a wire over here there is some object over here whose resistance we don't know and we want to calculate it what we can now do is we can set up a Wheatstone's network like this and instead of a red resistor over here we can put a galvanometer a galvanometer basically detects current it will tell you if there is a current here or not and then we can take any of one of these resistors as a variable resistor something that's resistance can be changed and now all you have to do is keep changing this resistance and look at the value of the galvanometer so as you keep changing this resistance let's say the galvanometer becomes the value becomes 0 when you hit I don't know maybe 7 ohms when this goes to 7 ohms let's say the current over here goes to 0 now I know aha this Wheatstone network is balanced therefore I the resistance here must be exactly the ratio here must be exactly the ratio here and then this is 1 is to 7 this must be 1 is to 7 so the resistance over here is 7 times 5 35 ohms amazing right so you can use this and we will use this in a future video called Meter Bridge to explore how to use Wheatstone's network in practice to calculate unknown resistance now we could stop over here but I want to go one step forward and generalize this something that might help you in your competitive exams so let me bring back the original circuit my question to you is do you think that this circuit can be extended let's say I add more resistances over here something like this what do you think about this one would this also be balanced it's a more complicated circuit now but would it be balanced with the current here and here be also 0 can you pause and think a little bit about this alright let's see we can apply the same logic the first step is to first neglect the resistances and then see how the voltage gets divided now notice that the ratios of the resistances here is exactly the same as over here 1 is to 2 is to 5 1 is to 2 is to 5 which means the voltages will also get divided similarly right 1 is to 2 is to 5 and therefore the voltage drops will be the same as before and that means the voltages here must be the same as here voltages here must be the same as over here in other words there will be no current flowing over here which means you can solve the same circuit here as well beautiful right extended Wheatstone's network something you might see in je let me show you another way in which we can extend this circuit what if we extend the circuit this way what do you think about it now pause the video I know it looks very complicated but pause the video and think a little bit about it the batteries are connected across this and this alright again notice we can use the same logic if you forget about these resistors in between then the voltage should get divided equally because again they have the same ratio 1 is to 2 1 is to 2 and therefore the voltage is here here and here must be the same so no current will flow over here and so you can neglect this and you can solve this circuit if I hadn't given you the color coding just look at how complex this looks like typical you know competitive exams problem looks very complex but by using the same logic of Wheatstone's network you can solve it so long story short a basic Wheatstone network is four resistances connected this way with the resistance in between it's balanced when the resistances are in the same ratio when that happens the voltages get divided equally and therefore the voltage across this red resistor becomes the same no potential difference no current flows through it balanced Wheatstone network