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JEE MAIN & ADVANCE 12th PCM Physics -Current Electricity-Module-2 Demo videos
Hello, students, welcome to the beautiful chapter of current electricity. Students, from this module onwards we will discuss some instruments and our first instrument is galvanometer.
So, let’s start with the galvanometer. My dear friends, it is a device which is used for detection and measurement of small currents like micro ampere flowing in a circuit. I hope that you have already seen this instrument in your school laboratories. If you want to measure the current when the current is passing through this galvanometer there is some deflection in the needle that will give you the actual reading of the current flowing in the circuit. So, yes, let’s understand this instrument with some deeper concepts. My dear friends, it consists of a coil which rotates when current is passed through it. Students, we have a coil having some resistance G that is also called as the resistance of galvanometer. We have a needle attached with this coil which is pointing towards a printed scale having some lower value to higher value, minimum to maximum. The beauty is when I pass the current into this coil the needle rotates. How it rotates that is none of our business. We will learn this thing in our chapter of the magnetic field. So, now the mechanism is simple. I pass the current, the needle rotates and the angle rotated by the needle is directly proportional to the current passing through the coil. If I will pass more current the deflection shown by the needle will be more. Now, my dear friends, focus on the value of current, if I increase the value of current that deflection in the needle also increases and the current, when the deflection in the needle is maximum is called as the full scale deflection current and that is represented by the variable Ig. Students, till now that was the basic theory of galvanometer.
Now, we will use this theory to make a new instrument called as ammeter and before that galvanometer is represented by simple symbol G.
Now, let’s move towards the ammeter. Students, it is a device which is used for the detection and the measurement of large currents like milli ampere or ampere flowing in a circuit. And it is just a simple arrangement, it consists of a small resistance connected in the parallel with the galvanometer. Let’s understand first, we have this circuit of galvanometer while resistance G needle pointing on a printed scale with minimum value towards the maximum value. Now, if I pass the current Ig that is the full scale deflection current then needle will point towards the maximum value. Now, students I will put a small resistance with parallel with galvanometer. This resistance is also called as the shunt resistance. Now, students, the IG will not enter the coil, in fact it will be divided into I dash and Ig minus I dash. Now, I dash is the current that is entering the coil and it should be noted that I dash is less than Ig. Obviously Ig is divided into I dash and I dash minus Ig. So, I dash is small. If the current entering the coil is small the deflection of needle should decrease. Now, students, observe this picture. The same Ig is measured at a lesser deflection, it means I have more capability of passing the current or in a way my range of measuring the current increases and that is the beauty and the working of ammeter, it measures the large currents. Now, students, till how much time I can pass the current into this circuit. I can pass till the current entering the coil becomes Ig and the current outside is I max. So, that is the maximum ability of ammeter. And I max is that current when we have the full scale deflection in the ammeter and that is also called as the range of ammeter.
So, students, let’s have some mathematical analysis. You can see that Ig is the current going in G and I max minus Ig is the current going in R and G and R are in parallel. So, their potential differences should be equal to each other. And from this equation I can say R is Ig into capital G upon I max minus Ig. So, students, that is a very, very important result in the working of ammeter. So, kindly remember that and we have some important points. The first point is, the resistance of ammeter is small r into G upon small r plus capital G, obviously are r and G are in parallel, equilateral distances R1, R2 upon R1 plus R2. Second point the resistance of ideal ammeter should be equal to zero.
Students, in our upcoming module I will have the mathematical value of the resistance of ammeter. And you will see that it will come out to be very, very small and for ideal ammeter it should be equal to zero.
Now, the next point, ammeter is represented by the symbol capital A in a circle, remember that, that will be used in the numericals. The next point is ammeter is always connected in series with the circuit element. Obviously, if I want to measure the current, the ammeter has to be put in series, so that, that current enters the ammeter. Now, the last point, a practical ammeter always reads a lesser value than the true value of current in the circuit. Students, I will explain this point in my upcoming module with some example. So, kindly remember this point that ammeter will not give me a true reading, it will in fact give me a lesser reading than the true reading. I will show it to you with the help of some example.
Students, I hope you have understood this module. We will meet in the next module which has more concepts on ammeter. Till that time have a good day, thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Current Electricity Module-1 Demo Videos
Hello students, welcome to this beautiful chapter called as current electricity. So, the agenda for the today’s module is Kirchhoff’s Voltage Law. This is very important law as this is going to be the future of this chapter. So, let’s start with it.
So, what exactly Kirchhoff’s Voltage law is? It states that the algebra summation of total potential difference across a loop is equal to zero. My dear friends, to understand this law, we will write the potential difference across our two basic elements, that is battery and resistance. So, let’s start with it.
So, I have a battery of potential difference V. The battery has a high potential represented as positive terminal and it has low potential represented by its negative terminal. So, this is high potential and this is a low potential terminal of a battery. Students, what will happen if I will move a battery from the positive terminal towards the negative terminal. As I will move from the positive towards negative, I will have some loss and that loss is minus V. V is the magnitude of potential drop and the minus sign is due to the loss, as I am moving from high potential towards the low potential. Now, students, what will happen if I will move from the negative terminal of the battery towards the positive terminal of the battery. Obviously, the potential drop will be plus V because from low to high I will gain something.
Let’s understand the potential drop across a resistor. So, I have a resistor R as shown in which the current I is flowing and we know that the current flow flows from high potential towards the low potential. Students, what will happen if I will move in the direction of current? If I will move in the direction of current, I will have a loss because current is flowing from high potential to low potential. The potential drop across the resistor is minus IR. And, students, what will happen if I move against the direction of the current in a resistor. Yes, I am moving from low to high, I will have gain and that gain is plus IR. So, the theory is very simple, when I will move from a positive terminal of battery towards the negative terminal the potential drop is minus V. If I move from a negative terminal towards a positive terminal the potential gain is plus V. If I move in a direction of the current in a resistor the potential drop is minus IR and if I move against the direction of current in a resistor the potential drop is plus IR.
Now, let’s understand the Kirchhoff’s Voltage law with the help of this loop as shown and we will have the different magnitude of the current which are flowing inside this loop. To understand the Kirchhoff’s law, we will have a passage, we will have a path and that path is represented as ABCD in clockwise direction. Students, now I am going to traverse inside this loop and I will write the complete total potential difference which I will encounter in this loop. So, let’s start first with R1, I am moving in the direction of current I1, so, I will have a loss minus I1, R1. Next we have a battery, I am moving from the negative terminal of the battery towards its positive terminal. I will have a gain and that gain is plus V1. Now, the turn of R2, I am moving in the direction of the current I1 through R2, I will have a loss, minus I1 into R2. Now, R3, I am moving downwards B to C and the current I2 is moving upwards. So, I am moving downwards and the current is coming upwards. I am moving against the direction of the current. I will have a gain that is I2 into R3. Now, my dear friends, I am moving from C to D and I am moving from the negative terminal of the battery towards the positive terminal. So, I will have a gain and that gain is V2. Now, my dear friends, again I am passing through resistance R4 and during that passage, I am moving in the direction of the current I3. I3 is passing through resistance R4. So, again I am moving in the direction of current I will have a loss and that loss is minus I3 into R4. Again now I am moving from D to A, if I am moving from D to A, I am moving from positive terminal of the battery towards the negative terminal, I will have a loss minus V3. Now, students, I am moving from D to A in upper direction and through resistor R5 current I4 is coming in downward direction. So, again I am moving opposite to the direction of current. I will have a gain and that gain is I4 into R5. And, students, this is the total potential drop inside the loop and Kirchhoff says that this total potential drop is equal to zero.
That’s all for today’s module, have a good day, thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Alternating Current Module-2 Demo Videos
Hello students, welcome back to the chapter of alternating current. In the previous few modules we have been discussing that how we can you phasor techniques to solve the questions where we have the combination of more than one circuit elements that is an LCR circuits. That means in a circuit we have all three, the inductor, the capacitor and the resistor. So, let’s take the discussion further and consider an AC circuit with L, C and R as shown in this particular figure. Now, let’s assume that at certain instant of time the current is drawing in the clockwise direction and the potential difference across these three elements can be written as VL, VC and VR. Now, what we already know from our previous discussions is that in case of a resistor the current and emf are in same phase, whereas in case of a capacitor the current leads the emf by pi by 2 and in case of an inductor the current is lagging behind the emf by pi by 2.
Now, we have been discussing everything till now taking the voltage of the source Epsilon as the reference. What if we change our reference to the current. So, here the noticeable thing before we move ahead is that at any particular instant the value of emf would be the sum of all three voltages. That means at an instant the voltage would be ruling by the Kirchoff’s law and not by the phasor diagram. So, let’s say that at any instant the current is given by i0 sin of omega t plus 5 and let us continue the discussion that what happens if we take the current as our reference. Now, if we take current as the reference nothing happens to the emf for a resistor because for a resistor the emf and the current are in same phase. But in case of a capacitor the current actually leads the emf and therefore we can say that the emf or the voltage actually lags behind the current and therefore in case of a capacitor we can write VC is equal to the maximum value multiplied by sin of omega t plus 5 which is same for the current minus of pi by 2 since the current leads or with respect to the current the voltage lags behind. Now, if you take inductor for example what we will get is that the current actually lags behind the voltage and therefore if the current is the reference then we can say the voltage leads ahead by an angle of pi by 2. And therefore in the angle value of the sin we will get omega t plus 5 plus of pi by 2. So, please note here we are discussing if we take the current as the reference. In all our previous discussions we have taken the voltage as the reference.
Now, let’s continue our discussions about these LS, LCR circuit. What we can note here is that the value of phasor diagrams or the technique of phasor diagram is well applicable for RMS voltage which can be given by the resultant of VC minus VL and VR. Now, to find out the maximum values we will use simply the Ohm’s law which is given by V by R. So, for resistance it is given by V is equal i0 R sin of omega t plus 5. For the value of capacitor we can say that VC is equal to i0 into Xc and the voltage actually lags behind the current by pi by 2 and in case of inductor we can write VL would be equal to i0 into XL sin of omega t plus 5 plus pi by 2. That means the voltage leads the current by an angle of pi by 2, which is same as the current lags behind the voltage at an angle of pi by 2. Now, you can write down the value of current by the resultant of the voltages and the impedance of the circuit. Let’s see how it is done. Now, we can write down here that Epsilon0 would be the resultant of these three when we talk about the RMS value and therefore Epsilon0 will be equal to under root of VR square plus VL minus VC square, which then can be written as i0 into R plus i0 into XL minus XC square. Taking i0 common we get this particular value and on simplification we can write down i0 is equal to e0 by Z, which basically is what we have discussed earlier that Ohm’s law is applicable on impedance.
Now, let’s try to find out the power developed across such a circuit. We know instantaneous power nothing but Epsilon into y. Putting down the value that we have here we will get that it is equal to Epsilon0 i0 sin of omega t multiplied by sin of omega t plus 5, where 5 can be found out using this particular diagram or the phasor methods that we know. P average whereas is given by ERMS into IRMS multiplied by cos 5, where 5 again is the angle between the current and the emf which can be found out using the phasor diagrams, which in this particular case can be written as R by Z and therefore power factor for an LCR circuit is given by R by Z.
I hope you’ve understood the module. We will take our discussion further in the next coming section, thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Alternating Current Module-1 Demo Videos
Hello, students, welcome back to the chapter of alternating current. In the first module we have already discussed, what do we call as a direct current. A current that is constant or variable but does not change its direction in the interval of time is called as a direct current. Whereas an alternating current is one which can be positive for some amount of time and can be negative for some amount of time, is called as an alternating current. Further we have discussed, how can we construct something which is called as an AC generator. In this module we will discuss, how an AC generator works.
Let’s first revise how we have constructed an AC generator. It consists of two magnetic poles, one north the other south, in between which is placed a rotatable coil which is connected to slip rings and which are connected to an external circuit through two carbon brushes.
Let’s discuss how this combination works to provide us an alternating current. If we see in the figure as the coil rotates the magnetic flux through the coil, that is the number of field lines crossing the area in a unit time or at any instant changes and thus producing an EMF which can be read according to the relation Epsilon is equal to minus d pi by dt, where pi is the magnetic flux through the coil. Now, if we recall, the magnetic flux through a coil can be changed in three ways. First, by changing the magnetic field, second by changing the area of the coil that means either reducing the area or extending the area. But in this particular case the AC generator works on the principle that the magnetic flux can also be changed by changing the angle between the magnetic field and the area vector. Let’s discuss this phenomena in detail. Let’s say that t equal to zero the area vector is anti parallel to the magnetic field. Therefore the flux in this particular case is negative. Now, if the time taken to rotate one complete circle that is 360 degrees is T, then what would be the time to rotate by an angle of 90 degree? Yes, it’s T by 4. So, at T by 4, the coil rotates by an angle of 90 degree, therefore we can see that the magnetic field and the area vector are perpendicular to each other. Therefore the magnetic flux through the coil is zero. Now a further rotation of 90 degrees the time elapsed would have been T by 2. But as the coil rotates in the following manner then the magnetic flux and the area vector become parallel. And therefore the flux through the coil becomes positive. Now, if you note from T equal to zero, at T equal to zero the flux was negative. At T equal to T by 4 it becomes zero and at T by 2 it becomes positive. Therefore the flux from T equal to zero to T equal to T by 2 increases. Let’s discuss what happens after the time instant T by 2. On further rotation of 90 degree angle, the coil is placed like as shown in the figure. Again we can see that the area vector is perpendicular to the magnetic field and therefore the flux is zero again. On further rotation of 90 degree angle the time elapsed is T and the area vector is again anti parallel to the magnetic field, thus if we note from T equal to T by 2, the flux was positive. At T equal to 3T by 4 it is zero and T equal to T it becomes negative. So, we can say from T equal to T by 2 to T equal to T it is decreasing. And thus the coil rotates and the flux increases for half of the revolution and decreases for the other half. This change in magnetic flux thus produces an EMF, changing the direction of EMF itself which produces an alternating current. Now, this alternating current is produced through the area vector changing its angle with the magnetic field.
In the next module we will discuss the working and construction of what is called as a DC generator, thank you.
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JEE MAIN & ADVANCE 12th PCM Maths -Relation & Function Module-1 Demo Video
Hello, students, so continuing with examples on even and odd functions, next we have this, we need to prove two statements.
Now number one, we will be starting with g of minus x, which is equal to this. Try to see, it is equal to g of x, so g must be an even function. Proceeding in a similar function, second, g of minus x is equal to this which is nothing of minus of g x, so g must be an odd function. Remember f of x plus f of minus x is an even function for any f. And f of x minus f of minus x is an odd function for any f.
Now we have an important remark, try to see, “Every function can be represented as a sum of an even function and an odd function”. How, let us see, try to see. We just proved that this is an even function and this is an odd function, try to see if we add them we get f of x, so I wrote f of x as a sum of even part and odd part. So that’s what this remark is all about. So whenever you are asked to write a function as a sum of even and odd function, you have to do it like this, okay.
Moving ahead we have next example that is this, we need to check if it is an even or odd function. First step is clear, you have to start with f of minus x which is this. Now writing e par minus x as 1 by e par x and taking LCM we have this. Now we add and subtract x in the numerator. Why? I will tell you. Try to see the first term will be broken into two terms, this. Advantage is 1 minus e par x cancels. Now why I did that, because I want to prove that f of minus x looks like fx or minus fx. So in numerator I do not need e par x as in f of x as there is no e par x in the numerator. So I wanted to cancel it, and hence I added and subtracted x. So moving like this equal to f of x so this f must be an even function.
At last we have this, I want you to prove this question at home, I give you two hints, I am your friend. Number one is this, that integer comes out of gint under addition and number 2 is gint of minus x is minus 1 minus gint of x for k not belonging to integer. In the example try to see x plus pi divide by pi. It will be x by pi plus 1, so 1 will come out, using Hint number 1. Then now try to see, you will have to start with f of minus x. So obviously in the denominator inside gint you will get minus x by pi. So this minus will be tackled using Hint number 2. Further you can solve it, I have full confidence on you.
So best of luck, take care, god bless.
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JEE MAIN & ADVANCE 12th PCM-Maths – Inverse Trigonometric Function Demo Videos
Continuing with some miscellaneous examples, next we have this. In the last module I told you we can also prove this without using formula of sin inverse, let’s prove it.
Say I call this angle A and this angle B.
Now A plus B equal to theta. We want to prove that this theta is equal to sin inverse of 77 by 85. Apply sin on both sides, so sin A cos B plus cos A sin B equal to sin theta. Now try to see, first of all I would like to express everything in terms of sin because angles are in terms of sin inverse. Now look, A is sin inverse 8 by 17, and B is sin inverse 3 by 5. So sin A will be 8 by 17 and sin B will be 3 by 5. So keeping the values here we get number one, A plus 17 then root of minus 9 by 25 plus 3 by 5 root of 1 minus 64 by 289 equal to sin theta. Simplying these values we get 77 by 85 on left hand side. Now finally we need to check because we directly applied sin so we need to check for extra solutions if it is correct or not.
We called A and B this, now try to see as 1 by root 2 is approximately 0.7, and both of these terms 8 by 17 and 3 by 5 are less than 0.7, so these angles A and B both must be less than pi by 4. So they belong to 0 to pi by 4. So theta which is A plus B lies in 0 to pi by 2. So this thing sin theta equal to 77 by 85 will give me theta equal to sin inverse 77 by 85.
So that’s the way to prove this thing without using formula of sin inverse.
Okay, next we have this. We want to prove this result for x positive, okay, let’s try. Number one, let sin inverse x be theta. As x is positive so this theta must lie in 0 to pi by 2. Now applying sin, x is equal to sin theta, always start with right hand side. So put x is equal to sin theta here in this expression. To get sin inverse 1 minus 2 sin square sin theta. I hope you remember this is just cos 2 theta, so it gives me cos inverse cos 2 theta. As theta is 0 to pi by 2, this 2 theta will lie in 0 to pi that is principle domain of cos and hence this gives me 2 theta, theta was sin inverse x. So it is equal to 2 sin inverse which we wanted to prove.
So with that I conclude this module. See you soon with more examples next module.
God bless take care.
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