 Let's solve a question on impedance in a series LCR circuit. Here we have an AC source rated 200 volts and 140 pi hertz frequency which powers a series LCR circuit. An RMS current of 10 amperes flows in a circuit and the resistance of the resistor is given to be as 8 ohms. The inductance is 12 millihenry and the question is to figure out the capacitance of this capacitor right here. Alright again before I get into this why don't you pause the video and first attempt this one on your own. Alright so we need to figure out the capacitance of this capacitor. Now where do we begin in this question? I know from Ohm's law that current and voltage are related to each other and in this case if the voltage is an AC source and you have an RMS value of a current in a circuit and also an RMS value of the voltage that is given to be as 200 volts. So we can relate these two with this expression. The I RMS that is equal to that is equal to V RMS divided by divided by some opposition right and the opposition in this one in an LCR circuit is denoted by the symbol the letter Z. This is the total opposition to the current in this circuit. This collective opposition is called the impedance the impedance of the circuit. So if you want to figure out the value of the capacitance maybe we need to figure out the value of the impedance the total impedance in the circuit because we already know IRMS that is 10 amperes and VRMS that is 200 volts. So if you figure out Z we should be able to figure out C considering we know the resistance and also the inductance of this inductor. So now the question is how do we figure out this value of Z? Is there a way to do that? Now we can find this expression using a phasor diagram. We can draw a phasor diagram for all the voltages in the circuit and maybe try to find the resultant of the voltages across LCR to find the supply voltage. So let's see that let's do that. Let's draw a coordinate axis and let's start off with the phasor of current so that we have a reference for all the other voltages. Here we have the current vector. Now using a knowledge of whether the voltage leads or lacks the current we can try and draw the vectors for all the other components that is L, C and R. So let's begin with the one let's begin with the resistor one because we know that the voltage across the resistor is in phase with the current. So VR lies on the same direction as that of current. Then we know that voltage across inductor it leads the current by 90 degrees. It leads the current by 90 degrees. So we can draw it in this manner it is vertically up and the angle between the voltage and the current is 90 degrees and you can see how the voltage is 90 degrees ahead of the current. Similarly for the voltage across capacitor we have VC and this one is lagging the current. The voltage lags the current by 90 degrees. So we still have a vertical vector but it is 90 degrees before or lagging the current. The length of all of these vectors that represents the peak value of the voltages and okay let's try and find the resultant now. So we know that VL and VC are in opposite direction. One thing that we are assuming over here is that the voltage across the capacitor is more than the voltage across the inductor. So when we do VC minus VL what we can do is we can add a magnitude sign so that we are only interested in the magnitude of the difference the negative sign is taken care of. The other voltage that is left is the voltage across the resistance there it is. So now we can try and find the resultant of these two. Now we have a right angle triangle on our hands and the resultant will be in this direction and if we try and write that if we try and write this this would be V0 square equal to this is equal to VC minus VL whole square plus VR square and these are the peak values. These are the peak values so we can add we can add a knot in the beginning just to show that these are these are your peak values. All right so we have some expression on our hands. We have the expression for the peak supply voltage and it has the peak voltage across capacitance, inductance and the resistance and you can see how they are being operated. Now one thing that I can do with this is I know that the peak voltage across the resistance I can express that I can write V0R I can write V0R in this manner. I know that this is equal to the current that is flowing the peak value of the current that is flowing through the resistor and let me let me write that as I0 I0 into the value of this resistance that is just R and this is not changing R is what it is so I can write the peak value of voltage across the resistance in this manner. Similarly what I can do is I need to bring in these factors right I need to bring in R, L and C so to be able to do that maybe I should express the peak voltage across the capacitance also in this manner. So how do we do that? V0C this can be equal to I0 the maximum the peak value of the current flowing through the capacitor I0 into the capacitive reactance. This is the capacitive reactance. This is the opposition to the current provided by the capacitor. Similarly for inductor V0L V0L this will be equal to I0 into X into XL and this is the inductive reactance the opposition provided by the inductor to the to the current. So now when we replace all of these in this expression right here let's see what do we get we still have V0 square so let's just keep it at it is V0 square this is equal to I0XC minus I0XL whole square plus I0R square. Okay I can further manipulate this expression what I can do is I can take I0 outside this bracket that will be I0 square and I will have an I0 square from here also. So I can take I0 square to be common across this entire this entire term. So when we do that this will be this will be V0 square that is equal to this is equal to I0 square and a bigger bracket now and then we have XC minus XL whole square plus R square. Alright I can remove the squares and take this entire phrase on the left hand side. So I'm I'm approaching something that is kind of resembling this this expression right here not with the RMS values but we know that RMS and peak are related to each other with with with an under root somewhere in between we'll come to that but let's take this phrase to the left hand side first. So when we do that I can write this as I0 this is equal to V0 divided by this whole thing right here and we will also have an under root over here over this entire thing. Okay lastly we know that this is an AC source and let's assume that this is sinusoidal in nature. So there is a relation between the peak and the RMS voltage and that that is given by that is given by V0 equal to under root 2 VRMS. Similarly for current you have I0 equal to under root 2 IRMS. Now what we can do is we can substitute in place of V0 we can write under root 2 VRMS. So when we do that this becomes under root 2 VRMS and base of current we can write under root 2 IRMS. Now we can see that the under root 2 part just gets cancelled right away and finally if we compare this with the very first expression we can see that we have finally figured out the expression of the impedance. Impedance is under root XC minus XL whole square plus R square. Okay now let's try and figure out the value of the capacitance. So for that let me make some space. All right now we know IRMS and VRMS. So IRMS is just 10 and VRMS is 200. So Z would be equal to 200 VRMS divided by 10 IRMS. So this is just 20. Okay so 20 this 20 that is equal to this entire thing. Now XC is nothing but the capacitive reactance and that is given by 1 upon omega C and XL XL is the inductive reactance. This is just omega L and what is omega over here? Omega is the angular frequency but in the question we are only provided a frequency not the angular one and in fact we can convert one into other. There is an expression for that that is omega equals to 2 by F. So we can figure out the angular frequency this will be 2 pi into 140 divided by pi and this comes out to be equal to 280 radians per second. So that is the value of omega and we already know the value of the inductance that is 12 millihenry and we already know the value of the resistance that is 8. So now what remains is only one unknown that is C. So let's try and figure that out. So this 20 that is equal to XC minus XL whole square plus R square and when we remove the under root if we remove the under root you will get 400 on the left hand side because that is what that is what 20 square is. So if we remove this we will have 400. How did we get 400? So we remove the square root we remove the square root and when we do that we have to square the we have to square both the sides. So that is 400 equals to this entire thing. Okay and R square would be 8 square that is 64. So we can take R square on the left hand side and when we do that when we do that this will be this will be 400 minus 64 that is 336 336 equal to XC minus XL whole square and instead of XC now we can write 1 upon omega C so with a bracket 1 upon omega C minus omega L and whole square of this expression. Now we know the value of omega that is 280 and we know the value of L that is 12 into 10 to the bar minus 3 because this is in milli millihenry. So what we can do is we can substitute those values and figure out the value of the capacitance and I strongly encourage you to do all those steps math just do the maths and what you should land on the final answer the final value of the capacitance this comes out to be equal to 0.16 milli Farads. All right you can try more questions again from this lesson and the exercise link is in the description if you are watching this on YouTube so I do encourage you to click on it and try more types of questions from this unit.