 Let's solve a question on voltage in a series LCR circuit. Here we have the RMS voltage drops across the elements in a series LCR circuit and we can see them. This is across inductor, across capacitor and across resistor. The question is to figure out the RMS value of the supply voltage that is VS RMS in the circuit. Alright, now before I get into this, why don't you pause the video and attempt this one on your own. Alright, hopefully you have given it a try. Now where do we begin to think for this one? Let's see, we have a series LCR circuit and there will be a current flowing in the circuit. This will also be an alternating current. We are asked to figure out the RMS value for the supply voltage. Now from what we know about series circuits, the source voltage that is always equal to the resultant of or the sum of the voltage across all of the components in that circuit. Right? Here we have an inductor, a capacitor and a resistor. So to find the supply voltage, what we can do is we can find the resultant of the voltage across LC and R. Now to find the resultant, to find the resultant of the voltage, we could use the search of law that is the second law KVL but it turns out that the maths for that becomes very complicated. We have differential equations and they are also second order differential equations. An easier way to find the resultant voltage, the resultant source voltage in this case in an LCR circuit is to use phasor diagrams. To find the resultant of the voltages across these three components, we can draw phasor diagrams and the resultant phasor, the resultant phasor can give us a supply voltage. So how do we draw one? Well, let's think about a reference to draw the voltage phasors. So let me draw the phasor for the current, for the current at any random angle. We can draw it at any random angle and so for this one I've just drawn it horizontally on the x-axis. Here is the current phasor. Now using our knowledge of whether voltage leads, lags or is in phase with the current, we can draw the voltage phasor of each element and the resultant of these voltages will give us a supply voltage. Since the voltage across resistor that is VR that is always in phase with the current, so we have the phasor for voltage across resistor in the same direction, in the same direction as that of current. So we have already drawn VR. Now let's think about drawing the voltage across inductor. The voltage across an inductor that leads current by 90 degrees because an inductor resists any change in current. So voltage leads the current in an inductor and how do we draw that over here? We know that it leads it by 90 degrees. So the voltage across inductor, it could look like this. It is 90 degrees ahead, it is 90 degrees ahead from the current. And voltage across capacitor, it lags the current by 90 degrees. So you can imagine that the vector for that would still be vertical, but it will be pointing in the downwards direction because this is lagging the current, lagging the current by 90 degrees. And the length of these phasors that represents the peak values that is the V0 across L, the V0 across C and V0 across R that represents the peak values of voltage drops. Now let us find the vector sum of these three vectors to find the peak value of the supply voltage. We can see that VL, the voltage across inductor and the voltage across the capacitor, they are in the opposite direction, they are opposite to each other. So we can combine them and that would be VL minus VC because VL is more than VC in this one. So the resulted vector for these two, the addition of these two vectors that will be in the up first direction. So it could look somewhat like this. This is VL, this is VL minus VC. Now we have a right angle triangle with us. So we can use the Pythagoras theorem to figure out the resultant and the resultant could be in this direction. Okay, at this point, why don't you pause the video and write an equation, write an equation for the resultant supply voltage. All right, hopefully you have tried it. So the equation would look like this. So let's go through this one. This is the source voltage, the peak value of the source voltage. This square equals VL naught minus VC naught whole square plus the peak value of the voltage across the resistor and square of that. This is using Pythagoras theorem for this triangle. Now since we are provided with the RMS value in this problem, let us try and relate the peak values in this equation to the RMS values and then solve for the RMS supply voltage. We can assume one thing, we can assume that the variation of voltage is sinusoidal. So that means we can use V peak, that is V naught. This is equal to under root 2 into VRMS, okay? And this will be true for each and every voltage drop. So for the voltage drop across the inductor, this will be VL naught that is equal to under root 2 and VRMS for the inductor and similarly for capacitor and resistor as well. So when we substitute this in this main equation, we will get something of this nature. In place of source peak voltage, we have under root 2 into the RMS source voltage whole square and similarly for inductor capacitor and even for the resistor. Now this under root 2 is the common factor over here and we can right away cancel it. When this is square, it just becomes 2 and that comes out from the bracket in each of these terms. So under root 2 just, it just gets cancelled right away. And now what remains is just substituting the values that we have provided in the question. So let's do that. Now this is equal to VL RMS minus VC RMS. So this is 220 minus 211 and whole square of this one plus VRMS whole square. So that is 215 whole square. This will come out to be equal to 46306. But this is VRMS whole square. So when we remove the square, the answer would be the under root of, that would be the under root of this number and that comes out to be equal to 215.2 volts. So this is 215.2. Now you can try more questions from the exercise in this lesson and if you are watching this on YouTube, then you will find the exercise link added in the description.