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## Class 12 Physics (India)

### Course: Class 12 Physics (India) > Unit 3

Lesson 11: Wheatstone bridge# Wheatstone bridge & its logic

Wheatstone bridge is a special circuit consisting of 5 resistors. When the resistances in the adjacent arms have the same ratio, no current flows through the middle resistor! This is called a balanced Wheatstone bridge.
It's used in calculating unknown resistances using a meter-bridge set up. Created by Mahesh Shenoy.

## Want to join the conversation?

- What would the practical applications be for this type of circuit?

I just can't see how it would be useful in real life

I understand that it makes life easier for everyone with less math to do, but I don't think anyone will be wasting resistors to make wheatstone's bridges(2 votes)- You use this in cars for air pressure calculations in the air manifold system, for example.(1 vote)

- I have a not-so-good doubt- can't the 5 ohm and the 1 ohm resistor be parallel?(2 votes)
- No, because of the resistor in the middle of them. One of the properties of parallel-connected resistors is that the voltage on both of them is the same but here it is not the case because of the extra resistor in the middle.(1 vote)

- Can we apply the properties of a balanced wheatstone bridge even if the resistors in the diagram (original) are replaced with capacitors? If so, why?(1 vote)
- At4:30this is Wheatstone bridge....(0 votes)
- That's not a Wheatstone bridge. Notice that in the example given to the top-right, in the same format, the middle 'red' wire, is connected only to the arms of the Wheatstone bridge, and not externally.

This arrangement you're talking about would be placing all three wires (and thus resistances) in**parallel**, since they all have the same start and ends points.(2 votes)

## Video transcript

i want to introduce to you a very special circuit called the v stones bridge i want to talk about the logic behind this particular circuit so what's the wheatstone's bridge you ask well here's an example wheatstone's 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 v stones 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 in parallel and when i can't do that the next thing i resort to the general method is kirchhoff's loss but of course kisha's law is tedious you know i have to think about currents and then i have to use the loop rule and but here we won't require keshav's loss because this is not any wheatstone's bridge this is a very special wheatstone's bridge what's the specialty 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 victim's 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 witchtones 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 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 wheatstone's 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 all right 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 which turns network is shown and let's see if it's balanced well this is in the ratio one is to three this is in the ratio three is to one you see that one issue three threes to one 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 okay 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 one is to four one is to four but if you look carefully you will notice something is wrong here this is not even a wheatstone's network notice which store network is connected between the the 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 like rotated this so don't get confused this is not even a which turns network and you can directly solve this you know these are in series these are in series three are in parallel so don't get confused definitely register uh current here won't be zero 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 and now we can check the ratios one is two three one issue three now the current over here would be zero so now let's get back to the point why is the current over here zero how can we logically understand this you see in the books they derive it using keshav'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 zero 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 all right so in this particular case uh 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 so 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 one is to two the ratio of voltage here must also be one is to 2. so the voltage here is x the voltage here must be 2x same is the case here if the voltage here is x the voltage over here must also be 2x and that means i can now immediately calculate what the voltages 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 a 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 three volt drop meaning these two points must also have the same voltage so voltage at point a must exactly equal voltage at point p 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 it 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 zero 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 isn't it 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 zero and now hopefully you'll 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 this network is used in calculating practically values of unknown resistors for example let's say there is a wire over here there's 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's a current here or not and then we can take any of one of these resistors as a variable resistor or 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 uh the value becomes zero when you hit i don't know maybe seven ohms when this goes to seven ohms let's say the current over here goes to zero now i know aha this wheat stone network is balanced therefore i the the resistance here must be exactly the the ratio here must be exactly the ratio here and since this is one is to seven this must be one is to seven so the resistance over here is seven 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 sir 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 b also zero can you pause and think a little bit about this all right 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 the the ratios of the resistances here is exactly the same as over here one is to two is to five one is to two is to five which means the voltages will also get divided similarly right one is two is to five minus two is to five and therefore the voltage drops will be the same just like 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 jee 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 again 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 all right 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 one is your two one into two one is to do and therefore the voltages 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 which turns network you can solve it so long story short a basic wheatstone network is four resistances connected this way with a 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 with network