 Next we are going to study circuits with capacitors and inductor, when it is a DC supply inductor is zero resistance path and capacitor is an open circuit but with AC they are different. So there is one easy way to analyze the alternating current circuit that uses a mathematical concept called phasors. So this is just a brief introduction, we are not going to too much of detail of phasor like what it is and all that, we just talk about what all things are useful with respect to phasor and that is it. So phasors are nothing but rotating vectors, they are rotating vectors. One thing is finding the average value and other is to get the exact value as a function of time. So now when you draw a voltage on x and y axis you get a wave sort of thing. So you cannot just draw waves and it becomes very easy to convey the meaning of the wave. So phasor is the easier way to convey what do we mean by Vm sin omega t or Im sin omega t plus pi and all that. So if there is a vector that starts from this axis, if it rotates with angular velocity omega this angle will be what? Omega n to t, fine. This vector just keeps on rotating, alright. You will see that after time t if you take its projection on the y axis you will get the vector magnitude times sin of omega t, fine. This vector could be voltage, could be current. Let us say this is voltage whose peak value is V0. So this vector is denoting the peak value, the length of the vector is the peak value of voltage or current, getting it? So if this is voltage this length will become how much? V0 sin omega t, getting it? So it becomes easy to represent. We will say that the rotating vectors projection on the y axis is the voltage that is going on, fine. Suppose current is I0 sin of omega t minus pi by 2 then how will you represent here in the phasor? It will be 90 degree behind. It will be like this 90 degree behind this so like that you will have an idea. So this is current, right now current is behind the voltage by 90 degree, okay. So we are going to, I mean that is it, okay. We are not going to too much of detail of phasors. So we are going to use this particular knowledge to analyze the circuits further. Any doubt on phasors? Write down, inductor in AC voltage, inductor with alternating voltage, okay. So if you have inductor, inductance is L, V is equal to V0 sin omega t, like this. You need to find same thing, like VRMS, IRMS, average power V is given I, these five things. This thing is already given, this is V0 sin omega t. You need to find other four. Try to write Kertschoff's equation, the potential difference across inductor is LDI very deep. Then you will have to integrate and all. Let us see, all of you got this V0 sin omega t minus LDI by dt is equal to 0. What is this? Then you will get LDI by dt, this DI is V0 by L sin omega t dt, okay. This is DI, this is not I, okay. So I will become what? V0 by omega L sin omega t plus a constant, we do not know the limits, okay. But one thing is pretty clear, average of voltage is 0, so average of current should also be 0, fine. So when you take average of current, you will get what? This becomes 0, so this becomes constant, so that should be 0, constant should be 0. Oh yeah, yeah, yeah, yeah, yeah. And minus actually, but that minus should only be direct, okay. Now first back down this, the average of current should be 0, so constant is 0, alright. So I get current as minus V0 by omega L plus of omega t, fine. This thing, if you compare with voltage, voltage is what? V0 sin omega t, if you write down in terms of sin, we get V0 by omega L sin of omega t minus pi by 2, yes or no? So current is behind the voltage by pi by 2 or voltage is ahead of current. So current you can write down here as, and one more thing I will tell you, this is sin of omega t minus pi by 2. This thing, omega into L, is it behaving like a resistance? If omega, if omega L is large, current becomes small, so it behave like a resistance but we cannot call it resistance. So some rhyming term we call it, we call it reactance, fine. So reactance are two types, capacitor and for inductor. This is capacitive reactance, that is equal to omega L, yeah, inductive reactance. And Vrms will be what? V0 by root 2 and Irms will be what? This is again sin only, if you square it, it will become V0 by root 2 times omega L, isn't it? What about power? Average power is how much? How much? 0. 0 is what? Instantly, this power is V into I, V is V0 sin omega t, I is minus V0 by omega L cos omega t. You will get minus V0 square by 2 omega L sin 2 omega t, sin omega t cos omega t divided by, I mean, that into 2 is sin 2 omega t, okay? When you take average of this, it becomes 0 or not? So if it is pure inductive circuit, it will not consume any power, right? Now can you represent in phasor voltage and current? If you say this is voltage, where will be the current? 90 degree behind, like this. Both will rotate with same angle of velocity, omega. Can you doubt? See, Vrms into Irms is incidental, it is not a formula that Vrms into Irms should be power. It happens to be that when it is pure ray and resistance. And one thing is this, that this inductive reactance can change with omega. That is something weird because it is like changing resistance with frequency. It is a function of frequency. You can change the resistance with frequency and it makes sense. What is it? Omega means what? Rapidly the current is changing or rapidly the voltage is changing. So inductors should resist it more. That is why its reactance is increasing. Now write down capacitor with AC. For C, try to find out all of those things. So if charge is Q on 8, the potential difference will be Q by C. So the loop rule will say V0 sin omega t minus Q by C is equal to 0. So I will get Q equals to C V0 sin omega t. How much is the current? DQ by dt. So current will be DQ by dt. That will be equal to C times omega V0 sin omega t. This you can write it out as you can write this down as 1 by omega C of omega t. So 1 by omega C behaves like a resistance. This time it is capacitive reactance 1 by omega C. Just to show that it is like a resistance. This is capacitive reactance. You can write this down in terms of sin as V0 by XC into sin of omega t plus pi by 2. So now you will see that current is ahead of voltage by 90 degrees. Earlier current was behind. So this is I and this is V. So we will have I RMS as V0 by root 2 times XC. V RMS as V0 by root 2. What is the average power? So it will be this with current into voltage. Again sin 2 omega t will come. Average power will be 0. So if you draw the phase diagram and you say this is voltage, where will be the current? It will be ahead of voltage. This is current. Both of them are working with a measure. Can I doubt? Yes. So these are the basic circuit elements. The basic circuits. Only resistance, only capacitor, only resistance, only the inner things. Now we will be combining them. But before we do that, let us solve few questions. No doubts on this. Write down a pure inductor of inductance 25 millihenry. Inductance is 25 millihenry. It is connected to a source of voltage 220 volts. Find out inductive reactance XL and RMS value of current if frequency of source is 50 hertz. If I do not say anything, it means RMS. Every time in a question, if it is not mentioned, what it is, it is RMS. How is it? How is it answered? XL. What is the formula for XL? Omega L. Omega L, how much is it? 2.2. 2.2? How much is that? 2.2. Omega is how much? 7.8. 7.8. Unit? Omega is 7.8. Not omega, XL. It forms. Because V naught by that is current, right? Yeah. So V naught by anything that is current means it has to 2.2 volts. You got it? 7.8. Omega is 2 pi into frequency. This into L. This is 7.85 ohms. So frequency into 2 pi is omega. Do not directly use frequency as omega. What is IRMS? 2.2 by 8. So 2.2 by 8. 7.8 is approximate. Approximate. So 2.2 by 8. Approximate, huh? 2.8. Is it 2.2 by 8? Approximate. Approximate. This is exact number. This is 28m. This is the RMS. Okay. This is a theoretical question. Tell me. A lamp is connected in a series with a capacitor. Suppose you have an inductor like this, draw this. You have an inductor and there is a lamp that is connected like this. There is an AC supply. So lamp will glow. Okay. Now what you are doing is you are putting an iron rod inside, inside the coil. What will happen to the glow of the bulb? Increase, decrease, remain same. Iron. Iron is not touching anything. What will happen to the glow of the bulb? bulb is glowing because of what? Why it is glowing? Because it is consuming power. I square R. What will happen to I square R? R will not change. So what will happen to current? If current increases, glow will increase, current decreases, glow will increase. What will happen to that? This is what is the problem. You are not. Why? Because reactants will increase. Reactants will increase. The L value will increase or not? L will increase. So omega L will increase. Omega L increases, reactants increases. So current decreases. Global decreases. Next question. A 15 micro farad capacitor is connected to 220 volt 50 hertz supply. Find out capacitive reactants and the current. RMS and the P. This is like the basic question so that you know you are able to. These are the questions which will help you to get familiar with the concept. Once you get familiar, then you have to solve those XC vermin stuff. How many XC? 222. Who will calculate? After writing. Sir XC 222.21. 212. Yes 222. So I have the answer. 1.0 is I have the 1.0. Omega has got 2 pi into F. Yeah. Into C 15 into 10 is power minus 6. Yeah. You know again I am getting the 222. This is 10 raised to power 6 divided by 2 into 15 is 100. So this becomes 10 raised to power 4 divided by 15 pi. So it will be 100 divided by 15. Into 100 divided by pi. So this is like 33. something and this is how much? It is like 7. So 7 into 33.3. I mean this is approximate okay. 33 into 7. 7 is the 21, 21, 22. It is a rough calculation. The answer is 222. What is IRMS? IRMS is nothing but 220 divided by XC. That is 1. 220 divided by 212. This is rough. The exact value is 212. So it is 220 divided by 212. Okay. 1.0 at Konbola. If I got 298. I see. You have evaluated. Okay. Rough what you are doing. Don't do anything like that. Answer is 1.04. Peak current is how much? Peak current is root 2 times of this which is 1.47. Now tell me what will happen if you increase the frequency. If you double the frequency what all things will change? XC will become how much? XC will become what? If it is right now 212. What will become if you make omega 2 omega? XC will be 212 divided by 2. Yes 1 of 6. Right. And current will double. Yeah double. Fine. Just by changing frequency you are able to increase the current. But then at our home there is a fixed frequency of current that is coming in. That is 50 hertz. You don't usually change the frequency.