today we are going to talk about capacitors and these are used in radio circuits along with resistors and inductance now first of all I need your help to answer this question I have a capacitor which is connected to a DC source and I find that no current flows through this so if I have a little lamp over here the lamp does not glow which means no current is flowing through this capacitor this seems to make sense because you know that it is an insulating medium present between the plates of the capacitor so current can't flow through it now if I look at this AC source which is connected to the capacitor I find that a current is flowing through it so say I put a little lamp over here in the circuit this lamp closed which means current is flowing in this now how is it that current is flowing in this case but it is not flowing in the case where a DC source is connected see in case of the DC source let's call these plates let's say a and V now the plates a and B acquire a positive charge and a negative charge respectively now during that very short while when the capacitor is getting charged to a voltage which is given by Q over C during this very short interval while the capacitor is getting charged which means current is flowing in the wires during that time this lamp will glow but then very soon it goes off the current stops flowing now let's see how current is flowing in this case say we have an alternating voltage which continuously charges and then discharges the capacitor while while charging the capacitor the voltage across the plates of the capacitor rises and the charge also builds up let's say this is the charge now the charge on the plates of the capacitor is going to increase when the voltage across the plates of the capacitor decreases the charge on the plates of the capacitor will also decrease then when the voltage increases in the opposite direction the capacitor gets charged in the reverse direction which means that this will become positive and this will be negative so this is when the voltage reverses and you can see that this is going to be the charge is going to be in phase with the voltage it's in step with it now as the voltage decreases and once again comes back to zero the charge on the plates of the capacitor also come back comes back to zero now this is true because Q is given by C UB but if you take a look at the current the current which is flowing in the circuit is given by DQ over DT now we need to understand this a little more in detail so that we understand the relation between the voltage and the current when an alternating voltage is applied to a capacitor the alternating voltage which is applied to the capacitor is a time varying voltage which can be represented by a sine function V is equal to VM sine Omega T where VM is the amplitude of the voltage and Omega is the angular frequency now this voltage is applied across the capacitor so the charge Q on the capacitor is C times V which is C times VM sine Omega T and we just saw that the charge is also in step with the voltage which means that there is no phase difference between the voltage and the charge next we can write I as DQ over DT since the current is given by the variation of the charge so differentiating the charge we get C and VM are constants and differentiation gives us Omega times cos Omega T and now we are going to write this as V M over 1 by Omega C times sine Omega T plus PI by 2 writing the current I equals I M sine Omega T plus PI by 2 we can see that the current also varies sinusoidally though there is a phase difference of Pi by 2 we'll just look at this a little while later now coming back to this term over here I M is equal to VM over and we write this as X C so XC is 1 by Omega C and this is the capacitive reactance so over here view using the term reactance rather than resistance for instance we have AI is equal to V over R so R is the resistor but over here we are using the term reactance now the SI unit of capacitive reactance ohms same as that of resistance and if you plot a graph between the frequency of the alternating voltage which is applied and XC now you can see XC is 1 by Omega C so this will vary inversely as the frequency for low frequencies the reactance is extremely high and for high frequencies the reactance decreases this also explains what we saw in the beginning that when frequency is 0 then the capacitive reactance is infinite since this is 1 by Omega C so a capacitor does not allow DC current to flow through it because the capacitive reactance is in finite will show the phase difference between the current and the voltage on the graph and then using a phasor diagram so we know that V is VM sine Omega T and for this case I was given as given by I M sine Omega T plus PI by 2 so first if you draw the current sorry first if you draw the voltage and say with we show the voltage as a sine curve now you can see that let's say this is Omega T over here and this is V so this is 0 this will be PI and this is 2 pi now we will just write the current as a cos curve because that makes it easier for us to plot it now you know that cos 0 is 1 so the current will have its maximum value which is IM cos pi by 2 is 0 cos pi is negative 1 so the current will be minus i am at this instant cos 3 pi by 2 is again 0 and cos 2 pi would again be 1 now joining these points shows us the variation of the current in case of the capacitor so you can draw a continuous curve for this and this is the variation of current and the black curve shows the variation of voltage and you can see that there is a phase difference of 90 degrees or PI by 2 between them with the current leading the voltage by 90 degrees which means that the current passes its maximum of water cycle ahead of the voltage next we'll use faizal's to show the variation of voltage and current we have the voltage V which is shown over here and this is the amplitude of the voltage see if you recall a projection on the y-axis which would be this quantity over here this gives the instantaneous value of the voltage cv is VM sine Omega T V is the instantaneous value and VM is the amplitude so the length of the phasor over here gives us the amplitude and we are using the amplitudes to show the phase difference between the voltage and the current similarly we can draw a current phaser and the current phaser will need the voltage phasor by 90 degrees so let's say this is the current phaser and the difference over here between these two is 90 degrees and this would be Omega T the angular speed with which these phases rotate is omega which is the angular frequency of the alternating quantities so you can just think of this for instance you can think that you have an imaginary clock where you have two needles and these keep rotating at an angular speed of Omega and they are at all times at a difference of 90 degrees from each other try these questions to test your understanding the answers are given a short while later SI unit of capacitance is farad SI unit of capacitive reactance is ohms when the voltage across the capacitor is maximum the charge on its plates is also maximum but the current is minimum the capacitor oppose lower frequencies more not higher lower frequencies more than it opposes higher frequencies voltage phasor lags the current phaser current is ahead of the voltage for an applied voltage whose peak value is V not the peak value of the current I naught will be equal to V naught over the capacitive reactance X C so the RMS value of the current is I naught by root 2 and in this case it will be V naught over Omega C which is 1 by Omega C times 1 over root 2 if you've got all the questions right don't forget to give yourself a pat on your back and do tell me your name you can write it in the comment box below if you haven't got them right don't bother go back to the lesson go through it once again and then try the question thank you
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