 Let's solve a couple of questions from phase angles in series LCR circuits. For the first one we have RMS voltages across the elements in a series LCR circuit. So the RMS voltage across inductor is 70, across capacitor is 215 and across resistor is 160. The question is to figure out whether the current is leading or lagging the supply voltage, vs is the supply voltage and also to figure out with what angle with what phase angle is it leading or lagging. Alright before I get into this may pause the video and see if you can attempt this on your own. Alright hopefully you have given it a try. So how do we start thinking about this question? Now it is asking me whether the current is leading or lagging the supply voltage but nowhere in the question do I know what the supply voltage is. So maybe that's one thing that I need to figure out and when I know what the supply voltage is I should be able to tell whether the current is leading or lagging it. Now the question is how do we figure out the supply voltage? We know that this is a series LCR circuit so the supply voltage would be the resultant of voltages across the inductor capacitor and the resistor but for LCR circuits using a second rule becomes very difficult it becomes very complicated mathematically. So the approach that we adopt is drawing a phasor diagram. So maybe let's draw that containing the net supply voltage and the current in the circuit so that we can look at the angle between these phasors and then we will be able to tell which of them leads and by how much. So for phasor diagrams we we bringing the coordinate axis and we start off by drawing the current vector because then we will have a difference for all the other voltages. So when we draw the current vector then we can we can take any direction let's say the current is along the x-axis it could be any direction. Now using a knowledge of whether the voltage leads or lags the current we can try and draw vectors for all the other voltages. We know that voltage across the resistor is in phase with the current. So the resistor vector that will be along the current direction because it's in phase and you see this is v0r because the vector lens over here they represent the peak value of these voltages. So v0 it represents the peak value v0r represents the peak value across the resistor and voltage across the inductor that leads the current by 90 degrees. So if it leads the current it will be it will be in a vertical direction and it will be pointing upwards because that's how you show that's how you show that the voltage is ahead of the current by 90 degrees. Similarly for capacitor for capacitor the voltage lags behind the current by 90 degrees. So for that this is how this is how it can look like. You can see that the length of the peak voltage across the capacitor is slightly more than v0l because I'm taking some hints from the numbers that are given to me in the question. If the vc rms that's more than vlrms then the peak value of voltage across the capacitor should be more than the peak value of voltage across the inductor. Alright now let's try to figure out the supply the resultant supply voltage. So we know that v0l and v0c they are in opposite directions. So in order to find the resultant of those two vectors we can write v0c minus v0l and the resultant vector is pointing downwards and we already have one vector that is pointing to the right that is the voltage across the resistor. Now we have a right angle triangle on our hands and the resultant would be in this direction right here. We can find the magnitude even of this resultant supply voltage using the Pythagoras theorem but for this one maybe we don't really need to do that because we only need to tell whether the current is leading or lagging the voltage and we need to figure out the phase angle. So in order to tell whether the current is leading or lagging we should bring in the current vector also right. So here it is this is the current vector and now we see that the resultant the supply the supply voltage and the current they are of course not in phase with each other. There is a difference of angle and just from the look of it we can see that the current is leading the voltage the supply voltage by the phase angle of phi it is leading the supply voltage by phi. Now let's try and draw a triangle to figure out this angle of phi. So we have one voltage across the resistor the peak voltage across the resistor is v0r the length is slightly more in this I'm drawing it in a slightly more exaggerated manner and then we have v0c minus v0l. Okay now the resultant the resultant is in this direction right here this is the resultant supply voltage and the angle that we are interested in the angle that we are interested in that is this angle this angle that is phi. Now how do we figure out this angle why don't you pause the video at this point and maybe maybe try to establish a relation of phi with the peak voltages that you see in the triangle. All right one thing that we can do is we can translate this vector right over here and now this right angle triangle we can write we can write that tan phi tan phi this is equal to perpendicular upon base and in this case perpendicular is the red vector v0c minus v0l. So let us write that this is v0c minus v0m divided by v0r that is the base v0r. All right now one thing that we can assume is that this is a sinusoidal AC source so we know how the peak voltage and the rms voltage will be related to each other and those two they are related by this expression v0 is equal to under root 2 v rms. So for each peak voltage we can substitute the rms value so when we do that this becomes equal to under root 2 vc rms minus under root 2 vl rms divided by under root 2 vr rms. Okay all the under root 2 gets cancelled right away and we know the values of all the rms voltages across all the components so when we substitute that this becomes this becomes tan phi equal to 215 minus 70 divided by 160 this comes out to be equal to 0.91. Now phi, phi would be tan inverse of 0.91 and this is equal to 42 degrees. Okay so the current, current is leading the voltage and it is leading with a phase angle of 42 degrees. Let's move on to our second example now. Now for the second one we have reactants of an LCR circuit xc the capacitive reactants minus the inductive reactants xl that is equal to 40 ohms and the net impedance z is 60 ohms. The question is to figure out whether current is leading or lagging the supply voltage and again with what phase angle. The question is asking the same thing as the previous one but the information given is different. Now we know the difference between the capacitive and the inductive reactants and also the net impedance. All right again before I get into this first try this one on your own. All right now these questions which ask for a phase difference can be solved using a shortcut as well and that shortcut is called an impedance triangle. We will see what an impedance triangle looks like and also we will arrive at one going through the underlying physics and maths of it. So in this question we see that xc minus xl is a positive number. So we see that xc minus xl this is greater than zero that means that xc is greater than xl the capacitive reactants is greater than the inductive reactants and now if you multiply both of the sides by let's say the peak value of the current if you multiply both the sides by i0 if we do this then i0c this is nothing but the voltage the peak voltage across the capacitor. So this is v0c which is greater than which is greater than v0l. i0xl is the peak value of voltage across the inductor. So we see that the voltage across the capacitor is more than the voltage across the inductor just for this case for this particular series helcia circuit. That means if we have to draw a phasor diagram to find the net supply voltage the diagram would look somewhat like this. It is similar to the one that we got in the previous question where the voltage across the capacitor was more than the voltage across the inductor. We can also draw we can also draw a vector for the current and that will be that will be in the same direction as the voltage as the voltage across c resistor because voltage across the resistor is in phase with current. This right here is let's say this is i0. Now over here in place of v0c in place of v0c we can write just like this we can write i0xc and in place of v0l we can write i0xl. Also in place of v0r in place of v0r we can write i0r. When we do that then the phase difference the phase angle tan phi or like just five when we do that when we write tan phi this would become this would become equal to v0c-v0l divided by v0r but in place of v0c we are writing i0xc and in place of v0l we are writing i0xl. Similarly in place of v0r we are writing i0r. One thing that would just get cancelled right away you can see is i0. So let's see that this is i0xc-i0xl divided by i0r. So i0 just gets cancelled and now what you get is tan phi which is a phase angle this is equal to xc-xl divided by r. So this means if if there's a relation like this then this means a triangle there are triangle can look can look like this over here. In this case a phase angle this is your phi and tan phi is equal to xc-xl divided by r. In place of v0s in place of supply voltage you see z because v0s this is equal to this is equal to i0 into the total opposition that is provided by the circuit and the total opposition is the net impedance so this is i0z. This right here is called this right here is called an impedance triangle. Now for such questions where we know the net impedance or let's say the resistance and also the capacitive and inductive reactances we can directly use this triangle to figure out the phase angle between the current and the supply voltage. In this question we had xc greater than xl but if the inductive reactance that is xl if that is greater than the capacitive reactance then the triangle could look like this and the phase angle would be this angle right here phi. Now you can use these triangles right away to figure out the phase angle that's a shortcut but it is important for you to know the underlying physics and maths behind it so that even if you forget this triangle you can still arrive at it using the maths in the physics of it and we did that all in the previous question in the previous example and also you can see how we arrived at an impedance triangle using the voltage phasor diagram and finally arriving at a triangle which has reactances and resistance and also the net impedance. All right now using this triangle why don't you pause the video use this triangle and try to figure out the phase angle what is the phase angle between the current and the voltage. All right hopefully we have given this a try so in this case now we do not know what the resistance is we do not really know what the resistance is so we will try to avoid the base of this right angle triangle and in doing so we should write we should write sine phi this is equal to perpendicular that is xc minus xl divided by z the net impedance in this circuit and xc minus xl is 40 this is divided by 60 so this becomes this becomes 2 by 3 and phi would be sine inverse of 2 by 3 so this would be finally 42 degrees this would be 42 degrees and also we can see that the current in this case is again leading it's leading the supply voltage as you can see from this diagram the current is leading the supply voltage and this phase angle is 42 degrees so whenever the capacitive reactance is more than the inductive reactance in a circuit current will always lead the voltage and when the inductive reactance if this is more if the inductive reactance is more than the capacitive reactance if this is a positive number then you can see that the current in this case would lag the current would be along along the resistance it would lag the net supply voltage all right you can try more questions from this exercise in this lesson and if you're watching on youtube the exercise link is added in the description