 Assistant Professor, Department of Electronics Engineering, Walchand Institute of Technology in Sallapur, welcome you for this session. At the end of this session, students can simplify the purely capacitive circuit for AC supply. Now, so purely capacitive circuit means if we consider a one capacitor connected to AC supply, so AC supply is connected to this capacitor, C is the capacitance in Farad, this AC voltage V equal to Vm sin omega t is connected across this capacitor, where V is the instantaneous voltage, Vm is the maximum voltage, omega is the angular frequency and t is the time. This AC voltage is connected across this capacitor in this session we are going to discuss about the current relation with this voltage. If we connect this AC voltage how this capacitor works, so before going to that first let us discuss about this capacitor. This capacitor is nothing but the two conducting plates connected together in between that the dielectric material is provided, so two conducting plates are there, if we connect here the DC voltage the capacitor charges to this voltage V. So, due to this positive of the battery all the positive charges accumulate on this plate and here the negative charges accumulates. Here we can draw the charging process like this, so voltage across this capacitor develops like this. So, at the starting maximum current flows through this and when the full voltage developed across this capacitor current stops through this circuit, so we can say that in this capacitive circuit this DC blocks because when this capacitor charges to this voltage no current flows through the circuit. So, we can say this capacitor blocks DC, this is the AC cycle if we connect this AC cycle to this capacitor, this capacitor fully charges to this maximum voltage then again the discharging process is there due to this decrease in voltage is there. So, again here charging and discharging process is there, so continuous current flows through the circuit, so we can say that capacitor passes AC that is why we can use these capacitors in filter circuits. Now, if we connect this AC voltage V equal to Vm sin omega t to this capacitor, this Q charges move through the circuit which is given by Q equal to C into V. So, this Vc is the voltage developed across this capacitor, this Vc is nothing but this applied voltage Vm sin omega t. So, we can write down this equation as C into this V is nothing but Vm sin omega t Vm sin omega t. So, here we will get the equation Q equal to C Vm sin omega t. Now, differentiate this equation with respect to time, so by differentiating this equation so differentiate this equation with respect to time. So, we will get the equation dQ by dt equal to C Vm derivative of this sin omega t is omega cos omega t. So, we can rewrite down this equation as dQ by dt is rate of flow of charges is given by current I Vm divided by this omega c we can write down as 1 by omega c here and this cos omega t we can write down as sin omega t plus pi by 2. So, 1 by omega c is nothing but 1 by 2 pi fc is xc that is called as capacitive reactance. This capacitive reactance is the opposition of the capacitor for change in voltage, so that is measured in terms of ohms and this xc if we use here Vm by xc Vm by xc is given by maximum current that is I m. So, if we replace this term by I m we will get the current equation as I equal to I m sin omega t plus pi by 2. So, this is the current equation this is the voltage equation compare that both here you will get the angle omega t plus pi by 2 where pi by 2 is the phase difference for this voltage and current and this plus sign indicates that current is leading to this voltage by 90 degree. Now, we can draw the cycles and then also we can draw the phasor diagram for this. So, here time voltage or current we are going to consider on this y axis. So, voltage is the reference cycle current leads to the current leads the voltage by 90 degree. So, current cycle starts its cycle before this voltage by 90 degree. So, this is the angle pi by 2 radians or 90 degree. So, you can draw this current cycle like this. So, current leads the voltage by 90 degree. We can also draw the phasor diagram. So, if we consider here the voltage on the reference x axis then current leads here this is the direction of the leading. So, this current leads the voltage by 90 degree. Now, you can calculate the instantaneous power for this P equal to V into I, V is given by V m sin omega t, I is given by I m sin omega t plus pi by 2. So, V m I m sin omega t into cos omega t. This sin omega t plus pi by 2 is nothing but cos omega t. Again by using formula we can write down here V m I m by 2 sin 2 omega t. So, this is the power equation for this circuit. So, this 2 omega t indicates that power cycle has the double frequency than the voltage cycle. We can also prove this by graphically. So, see here this is the voltage cycle, this is the current cycle, current leads the voltage by 90 degree, current leads the voltage by 90 degree. Now, if we want to draw the power cycles make the 4 parts of this voltage cycles. So, these are the 4 parts of the voltage cycle. Now, see here product of voltage and current is the power. So, voltage and current both are positive. So, here we will get the product is positive here, voltage is positive, current is negative. So, product is negative here. Again both voltage and current is negative. So, product is positive here, here this current is positive, but voltage is negative. So, product is again negative here. So, 1 voltage cycle has the 2 power cycles. So, here 2 positive cycles and 2 negative cycles are there. So, average power for this we can calculate this average power equal to 0. Sin omega t average of that is also 0. Graphically you can calculate that average power for this is also 0. Now, pause the video and think why capacitor is used in filter circuit. So, what is the answer? Capacitor blocks DC and passes AC that is why capacitor is used in filter circuit. You can refer the book fundamentals of electrical engineering and electronics by B. L. Thareja. Thank you.