 Let's solve a couple of questions on Transformers. Here we have a 1200W AC source which is connected to the primary coil of an ideal transformer and that powers an appliance rated at 100V through the secondary coil. The question is to figure out the current drawn at the secondary coil, IS. Now before we get into this example, why don't you pause the video and first attempt this one on your own. Alright. Now let's try and break down this question. So we have a 1200W AC source. So let us bring an AC source. This is 1200W and it is connected to the primary coil of an ideal transformer. This can be a primary coil of an ideal transformer. And we will talk about this ideal transformer in a bit. First let us bring in the secondary coil. So here we have our secondary coil and this is connected to an appliance. We can represent an appliance by our resistance. Every appliance has a resistance of its own. And this is rated at 100V. So we can write VS this is equal to 100V. Now because the primary coil is connected to an AC source, there will be current in the circuit and the current will continuously change its direction because it is connected to an AC source. At one instant the current is flowing clockwise and at the other instant it might be flowing anticlockwise. Since there is a current flowing through the turns of the primary coil, there will be a magnetic field produced which can look like this. We also know that the direction of the current is changing as it is connected to an AC source. So even the direction of these magnetic field lines changes according to that. We can see that the field lines are passing through these turns of the primary coil. So we can say that there is a flux attached with each turn of the coil. And the same flux passes through each and every turn of the secondary coil. Now when we mention ideal transformer, it means that all the flux that is generated in the primary coil links or passes through the secondary coil. There is no flux leakage and therefore there is no energy loss. Now you might say it doesn't look that way. The secondary coil is at a distance and there are some field lines which are just going to the left. So how can all the flux be passing through the secondary coil? Now it turns out there are ways to make the transformer ideal. So we can use a ferromagnetic core, a ferromagnetic core which kind of pulls or sucks all the magnetic field lines and makes them pass through the secondary coil. We will not go into what that really looks like in this video but let's assume that this is an ideal transformer and all the magnetic field lines, they pass through the secondary coils and the strength of the magnetic field lines that passes through the secondary coil remains the same. Now that we know what an ideal transformer is, why don't you pause the video at this point and think about why is ideal transformer mentioned here? What could be its importance over here? Alright, since this is an ideal transformer, there is no flux leakage and therefore there is no energy loss. So all the energy from the primary coil is being transferred to the secondary coil and that means that this question is just on law of conservation of energy. It has nothing to do with how many number of turns there are in the primary coil or in the secondary coil. So if we can figure out the energy or power which is just energy per second, if we can figure out that in the primary coil and equate it to the secondary coil, we should be able to figure out the current. Let's do that. So from the law of conservation of energy, we can say that the power across both the coils is the same. So that means the power across the primary coil that is VP into IP that is the current that is flowing in the primary coil. This is equal to the power across the secondary coil that is VS into IS and we already know this value that is VP into IP. This is 1200 watts. We also know the voltage across the appliance that should be 100 volts. Now using this we can figure out the value of the current that is drawn at the secondary coil and when we do that this would be 1200 divided by 100 and this comes out to be equal to 12 amperes. So there is a current of 12 amperes that is flowing in the secondary circuit and now we can put in 12 amperes over here and check our answers. You can try similar questions from this exercise in this lesson and if you are watching on YouTube, I've added the link to the exercise in the description. Let's move on to our next question. Here we have a refrigerator which needs 5 amperes current to operate and the primary coil of a transformer with 2750 turns is connected to an AC source that draws a 6 ampere current and the secondary coil powers the refrigerator. Question is to figure out the number of turns in the secondary coil. Again before we get into this, attempt this one entirely on your own first. Alright, so let's bring in our primary and secondary coils. So here we have the AC source and this is connected to the primary coil and here we have the secondary coil that is connected to the refrigerator which we are representing by just a resistance. Now I do not really know how many number of turns are there in the secondary coils. There might be more than 2750, there might be less. Let's just make it this way for now and see the number that we land on. Now for any questions related to transformers, I like to think from the law of conservation of energy. That is all the power in the primary coil is transferred to the secondary coil. And in reality that doesn't really happen but there are ways to reduce power losses and mostly we deal with transformers that are ideal in nature which assume that there is no power loss, there is no flux leakage. So I can write power in the primary coil to be as VP into IP. IP is the current that is flowing in the primary coil and this is equal to the power transferred to the secondary coil that is VS into Is. And Is is the current that is flowing in the secondary coil. Alright this can be a first equation but we are not dealing with number of turns for now so how do we get that factor in? Now we know there is current flowing in the primary coil so there will be magnetic field lines produced and because the current is changing these magnetic field lines are also continuously changing. That means that the flux through the primary coil is always changing and from Faraday's law we know that wherever there is changing flux there will be an EMF induced, there will be an EMF induced whenever there is changing flux. And this EMF is induced in each and every turn of this coil and the primary coil has 2,750 turns. Why don't you pause the video at this point and think about the total voltage across the primary coil. Alright now there is an EMF induced in each and every turn of this coil and there are 2,750 turns. So for total voltage we should be adding the EMF 2,750 times. The total voltage across the primary coil that is VP this is equal to 2,750 into the EMF into the EMF that is induced in just one turn of the coil. And because we are assuming that there is no flux leakage or any energy loss we are assuming that all the magnetic field lines all the magnetic field lines are passing through the secondary coil and also that the strength doesn't change. We can say that the EMF that is induced in each and every turn of the secondary coil is exactly the same as the EMF induced in each and every turn of the primary coil. So the total voltage across the secondary coil that comes out to be that comes out to be equal to that is VS equal to the number of turns in the secondary coil into the EMF. And now we can compare these two equations we see that the EMF is the same if we just divide them if you directly divide them this will come out to be VP divided by VS that is equal to 2,750 divided by the number of turns in the secondary coil that is NS. So now if we try and manipulate the first equation a bit we can write VP divided by VS that is equal to IS divided by IP that is the current in the secondary coil divided by the current in the primary coil. And now we can equate the right hand sides of these two equations and when we do that there's no space over here so let me write this over here this is 2,750 divided by the number of turns in the secondary coil and that is equal to IS, IS is 5 amperes that is given in the question divided by 6 amperes that is again provided in the question the primary coil draws a current of 6 amperes. And now we can solve for NS and it turns out that NS the number of turns in the secondary coil this comes out to be equal to 3,300 again you can try similar questions from this exercise in this lesson and if you're watching on YouTube I've added the link to the exercise in the description.