 I, Mrs. V.S. Patki, Assistant Professor, Department of Electronics Engineering, Valkan Institute of Technology, Solapur, welcome you for this session. At the end of this session, students can simplify the purely inductive circuit for AC supply. This purely inductive circuit means if we consider inductance, this is the symbol of inductance, if AC voltage is connected to this inductance, inductance L is measured in Henry, I current flows through this circuit, V equal to Vm, sign omega t is the AC voltage connected across this pure inductive circuit. Pure inductance means here we are going to consider the resistance for this inductor is zero. To discuss about the relation between this voltage and the voltage drop across this inductor, we are going to indicate that voltage drop across this inductor as VL. So, this VL is the inductive voltage drop across this pure inductance. This inductance is the property which opposes change in current. Due to this AC voltage here the current changes through this coil and the self induced EMF, will be developed across this inductor and that self induced EMF is given by, that is VL equal to L di by dt. So, only one inductor is connected across this AC voltage V equal to Vm sign omega t. So, we can equate that VL as Vm sign omega t. So, we can write down this equation as L di by dt equal to Vm sign omega t. So, rewrite down this equation as Di equal to Vm by L sign omega t. So, integrating on both side we will get I equal to Vm by L integration of sign omega t dt. So, we will get here Vm by L integration of sign omega t is minus cos omega t divided by omega. So, I equal to Vm by omega L minus cos omega t. Now, we can write down this minus cos omega t as minus cos omega t equal to sign omega t minus pi by 2. So, if we write down this minus cos omega t equal to sign omega t minus pi by 2, in this equation we will get the current equation as I equal to Vm by omega L sign omega t minus pi by 2. But here omega L is equal to omega is 2 pi f into L equal to XL. This is the inductive reactance. This term XL is called as inductive reactance that is the opposition of the inductor for change in current which is measured in terms of ohms. So, we can write down this Vm by XL as I m maximum current. So, this I m equal to Vm by XL. If we put that here we will get the current equation as I equal to I m sin omega t minus pi by 2. So, this is the current equation for this purely inductive circuit. Now, we have to compare the current equation and this voltage equation. Where in voltage equation theta is omega t and in current equation theta is omega t minus pi by 2. So, minus pi by 2 is the phase difference for this voltage and current. So, here we can say that current is lagging to this voltage by 90 degree. So, this negative sign indicates that the current is lagging to this voltage. So, what is the meaning of that lagging? So, you can see here we are going to draw the cycles for this voltage and current here. So, if we draw the voltage cycle here first. So, voltage cycle is the reference for this because here omega t is the angle t. Voltage or current we can consider here on y axis current cycle starts after 90 degree from this voltage cycle. So, here you can draw the current cycle like this. So, current cycle starts after pi by 2 radians or 90 degree. That is why we can say the current is lagging to this voltage by 90 degree. Now, we can draw the phasor diagram here. So, positive x axis is the reference on reference we are going to consider the voltage. So, this is the voltage vector and here the current is lagging to this voltage by 90 degree. So, here you can draw the current vector here. This is the lagging direction. So, we can say that the current is lagging to voltage by 90 degree. So, you can see here current is lagging to voltage by 90 degree. So, theta equal to 90 degree. So, power factor for this circuit is again cos theta equal to cos 90 equal to 0. So, instantaneous power P is given by V into I. Voltage equation is Vm sin omega t. Current equation is I m sin omega t minus pi by 2. So, we will get Vm I m sin omega t and here minus cos omega t. So, minus Vm I m. So, we can use here one formula that is sin 2 omega t by 2. For this sin omega t and cos omega t we can use this mathematical formula. So, we will get the equation P equal to minus Vm I m by 2 sin 2 omega t. So, here only one term is there. So, we can calculate the average power. So, if we calculate the average of this sin 2 omega t that is equal to 0. So, the average power for this inductive circuit is 0. Means we can say that in purely inductive circuit power consumption is 0. Again we can compare this power equation with this voltage equation. You can see here the omega t is the angle for this voltage equation. But here in this power cycle the frequency is double to omega t. Means double frequency is there than this voltage cycle. We can indicate that graphically also see here. So, this is the voltage cycle. So, current is lagging to this voltage. So, current cycle starts after this voltage cycle by 90 degree. So, voltage or current you can take on this y axis. So, here the current cycle is there. So, here the current is negative. So, if we want to draw the power cycles on this diagram, make the four parts of this cycle. So, here one voltage cycle make four parts. Voltage is positive here, current is negative. So, product of voltage and current is here negative. Voltage and current is positive here. So, product is positive. Again here the current is positive, voltage is negative. So, product is negative here. Here both voltage and current is negative. So, product is positive. So, we will get the positive cycle here. So, two positive cycles and two negative cycles. Means one voltage cycle has the two power cycles. You can see here the frequency of the power cycle is double that of the voltage cycle. But the average of this both cycle is zero. We can again indicate that graphically also. So, average power for this purely inductive circuit is zero. Now pause the video and think about this question. What is the phase difference between voltage and current for purely inductive circuit? Yes, what is the answer? For purely inductive circuit phase difference is ninety degree. You can refer the book Fundamentals of Electrical Engineering and Electronics by B.L. Thareja. Thank you.