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Conversion of galvanometer into ammeter

To convert a moving coil galvanometer to an ammeter, we add a low shunt resistance, but why? The shunt resistance carries the excess current, ensuring the right amount of current through the galvanometer. Created by Mahesh Shenoy.

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  • hopper jumping style avatar for user Aditya Chauhan
    okay I understood
    but how is it conversion from galvanometer to ammeter.
    (1 vote)
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  • blobby green style avatar for user emaanfatima009988
    I still didn't understand. firstly how can we control what amount of current should pass through the galvanometer and shunt resistor? secondly, the current which will pass through the shunt, how will it go through the galvanometer? and from where are we supposed to calculate the actual current value as the galvanometer here only measures up to 10 milli ampere?
    (1 vote)
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    • leaf yellow style avatar for user St0dent
      "how can we control what amount of current should pass through the galvanometer and shunt resistor"
      By changing their resistances, making one path easier for the current to flow.

      "the current which will pass through the shunt, how will it go through the galvanometer"
      Same reason as above.

      "where are we supposed to calculate the actual current value as the galvanometer here only measures up to 10 milli ampere?"
      This is all relative, the 10 micro amperes is something we have labeled, we can just change the labels, just like how he removed the "sticker" of the micro in the beginning. We labeled it 10 micro amperes by looking at when it reached equilibrium when a known current was passed through it.
      (1 vote)
  • blobby green style avatar for user minuet4anna
    Okay but I don't understand how, if you send in 5 amps of current through the circuit, only 5 micrometers of current flow through the galvanometer. Shouldn't it be 10 micro amps, resulting in 5 Amps minus 10 micro amps flowing through the shunt?
    (1 vote)
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    • leaf yellow style avatar for user St0dent
      Both the current through the galvanometer and the shunt resistance would decrease. So in your case, the shunt would have 5 - 5microamps of current and the galvanomter would have 5microamps. This is because the resistance is fixed, so if the voltage is changed, the current would and vice versa.
      (1 vote)
  • blobby green style avatar for user Ema Smith
    how to convert mircoseconds to one fortnight
    (0 votes)
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Video transcript

your friend owns a galvanometer which you might know detects very tiny amounts of current in this example she owns one which can detect a maximum of 10 micro amperes of current but she wants to work on an experiment where she wants to measure currents up to 10 amperes now she doesn't want to buy a new ammeter so she comes to you and says hey can you somehow make this galvanometer measure up to 10 amperes and being a good friend you say sure i can do that that looks like a lot of work and so maybe you just pay me 200 rupees for that and she says okay cool and so the question we want to try and answer in this video is how do you extend the range of this galvanometer from 10 microamperes to 10 amperes now at first it might seem like we would have to understand the inner workings of a galvanometer and we might have to open things up to actually do this but guess what we don't we can actually do this without understanding anything about the inner workings of the galvanometer all we require is some basic knowledge of electricity so here's how i like to think about it we want to make sure that when we send 10 amps of current 10 amps of current through this galvanometer the reading should show 10. if we can achieve this we are done and of course we can get rid of that micro you know we can change that sticker over there that's not a big deal we are done and the reason we are done is because if we achieve this then if there are lower currents say for example if there is 5 amperes of current that is going through then automatically the deflection will become half right the deflection is proportional to the current so all we have to do is basically achieve this when 10 amperes goes the reflection should show 10. but remember in reality this galvanometer will only show 10 as a deflection when only 10 microamperes goes through this so in other words what we need to do is when 10 amperes is flowing into this galvanometer somehow we need to ensure that only 10 micro ampere actually flows through the galvanometer does that make sense this is what we need to achieve and the question now is how do we do that i mean how do i make sure when i'm sending 10 amperes through a wire only 10 microamperes flows there where will the rest of the current flow [Music] what we can do is we can we can introduce an alternate path we can introduce a path like this in parallel this way such that the rest of the current flows through this does that make sense so in in this particular example what we'll have to do is um 10 microamperes flows over here so the rest of the current which would be 10 amperes minus 10 microamperes that should flow through the alternate path if we achieve this we are done then you know what we could do we could package it and make sure your friend does not see anything inside then your friend will only be able to see this much when you show when you put 10 amperes it will show a deflection of 10. when you put 5 amperes it will show up to deflection of 5 because in reality 5 microamperes will be going through and the rest would be going through over here we would have achieved what we wanted so now that we understand the trick the secret behind this the next question is how do we design it and what i mean is how do we ensure that exactly 10 minus 10 microamperes flows over here how do we ensure that i want you to think a little bit about this this is no longer a galvanometer question this is a question about basic electricity from 10 amperes how do i ensure that only 10 microamperes flows here and the rest of the current flows over here can you think a little bit about how would you approach this problem pause the video and give it a try all right hopefully you've tried but if not don't worry here's a clue the amount of current that comes out of this branch will completely depend upon the resistance of this particular wire think about it if this wire has a lot of resistance then most of the current will just flow through the galvanometer and that thing is going to get blown up your friend will not be happy with that on the other hand if the resistance is incredibly small very small then almost all the current will flow from here and again nothing will happen to the galvanometer but your readings will be not fine and so the question now becomes what should be the value of this resistance such that when 10 ampere flows over here only 10 microampere flows to the galvanometer and the rest flows through this one that is the question we're going to try and answer and again i want you to pause now and think about how would you find the value of the resistance again i'll give you some clue think about the the galvanometer and this resistor they are connected in parallel so somehow can you use that property try and build an equation go ahead pause the video and give it a try all right here's how i like to think about it because these two are in parallel they would have the same voltage and so i would build the equation by saying the voltage across the galvanometer must be exactly equal to the voltage that gets built across this resistor so what's the voltage across my galvanometer of equals ir ohm's law and so the voltage across my galvanometer is going to be the current through the galvanometer so that's going to be 10 microamperes times the resistance of that galvanometer and i don't know what that resistance is let me just call that call that r g and that should equal the voltage across this resistor which is the current through this resistor that's going to be 10 amps minus 10 microamperes times the resistance and i'm done from this the resistance becomes now all you will have to do is plug in the value of the resistance of the galvanometer which can be calculated we know how to calculate resistance of devices practically if you're solving a problem in your exams then the resistance of the galvanometer would be given to you and once you plug in you would have found out the required resistance to connect in parallel to convert this into an ammeter so then what you will do is you will find the required resistance which will be very small number because this number is pretty small and then you will see that all you really require is a wire a copper wire say of some appropriate length you'll have to calculate what that is so you go to a store buy a wire of say i don't know 10 rupees 20 rupees connect it across and we have our ammeter ready i just want to mention a technical note over here this resistance that we add in parallel to extend the range of our galvanometer we give a name to it we call it shunt resistance so this is called shunt resistance and the word shunt basically means an alternate path so i'll write that so it's kind of like saying you see the current is following an alternate path right so to follow a different path to follow an alternate path is what we call as shunting and so if you ask say a physicist or maybe an electronic person you know how do you convert a galvanometer into an ammeter or how do you extend the range of a galvanometer they would just say add an appropriate shunt resistance which basically means at a resistance in parallel and now you understand why we're doing that yeah because we want to make sure that most of the current actually flows through that chart finally if you're wondering why i haven't given you any formula that's because i don't want you to remember any formula whenever a question is asked on this please use your basics please use this concept formula can only be confusing so i don't remember formulae whenever i'm asked any question like this i will go back to my basics and i'll do it this way so once you're done with this you have to cover this up your friend should not see it your friend should not see how cheap this was so you would cover this up in a box and then you would sell it to her and then she will pay you and you would have earned a profit but you and i know the real secret