 I, Mrs. Veena Sunil Patki, Assistant Professor, Department of Electronics Engineering, Valchan Institute of Technology, Solapur, welcome you for this session. At the end of this session, students can analyze RL series circuit through AC. So, register and inductor both are connected in series, VR is the voltage across register and VL is the voltage across inductor, R is measured in terms of Ohms, L inductance is measured in terms of Henry, V equal to Vm sin omega t, AC voltage is connected across RL series circuit. So, VR is given by I into R, VL is given by I into XL, where XL equal to 2 pi FL that is measured in terms of Ohm, where F is the frequency and L is the inductance in Henry. So, total voltage V is the vector sum of VR and VL is calculated as vector addition as V equal to VR plus VC. For resistive circuit voltage and current both are in phase and for inductive circuit current is lagging to inductive voltage by 90 degree, steps to draw the phasor diagram. So, first you draw the current vector, current I is taken as a reference because current is same for both, register and inductor and voltage across register is in phase with the current and for inductive voltage current is lagging to the voltage. So, VL we are going to draw as a leading to current vector. So, here vector sum of 2 voltages VR and VL for calculation of that draw the parallel diagram VL and VR and here this is the total voltage V. Now, this is the theta in between voltage and current and so this is the phasor diagram for this series circuit. Now, you can see here the current is lagging to voltage by theta degree. From this vector diagram we can draw the voltage triangle here VR equal to IR, VL equal to IXL and V equal to IZ and theta is the angle between voltage and current. So, voltage triangle this is the voltage triangle and from this voltage triangle we can calculate the total voltage V equal to under root VR square plus VL square. So, by putting VR and VL as IR and IXL we will get the total voltage as under root IR square plus IXL bracket square. So, if we take that current as a common we will get Z equal to under root R square plus XL square and we can write down the total voltage as V equal to I into Z where Z is the impedance measured in terms of ohms and Z is also calculated by V by I or we can also write down that I equal to V by Z. So, we can also draw the impedance triangle from this voltage triangle. So, simply eliminate the current from that IZ IXL and IR will get the impedance triangle and from that impedance triangle we can write down that Z equal to under root R square plus XL square and in rectangular form Z equal to R plus J XL. In polar form Z equal to magnitude of Z angle of theta. So, phase angle from this voltage triangle and impedance triangle in RL circuit current lags voltage by theta angle is called as phase angle. So, we can calculate that tan theta from voltage triangle and impedance triangle as VL by VR or XL by R and from that we can write down that theta equal to tan inverse of XL by R and we can write down the voltage equation that is V equal to Vm sin omega t and from this phasor diagram we can write down the current equation as I equal to Im sin omega t minus theta because the current is lagging to the voltage by theta degree. So, from voltage triangle we can draw the power triangle only by multiplying that V equal to IZ, VL equal to IXL and VR equal to IR by current that is nothing but I square Z or the I square XL, I square R or we can also write down that as a I into V, I into VR, I into VL. So, here we will get the three powers as active power P equal to I into VR or we can also write down that as a Vi cos theta that active power is measured in terms of watts. We can also write down the reactive power as Q equal to I into VL or Q equal to Vi sin theta. This reactive power is measured in terms of VAR that is volt ampere reactive and the apparent power S is given by I into V or we can also write down that as a V into I that apparent power is measured in terms of VA or volt ampere. Now pause the video and think about this question. What is the relation between voltage and current for purely resistive and inductive circuit and for RL series circuit? So, think about this. So, what is the answer? For purely resistive circuit voltage and current both are in phase. For purely inductive circuit current is lagging to VL by 90 degree and for RL series circuit current is lagging to voltage by theta degree. So, this is the answer for this question. Now, we are going to discuss about the power factor that is very important for electrical circuits. From this voltage triangle, from this impedance triangle and from this power triangle we can write down the power factor that is equal to cos theta equal to VR by V equal to R by Z equal to P by S. So, cosine of the angle between voltage and current is nothing but the power factor. So, actually this name indicates the factor of the power which is consumed in the circuit that power factor indicates the consumption of the power in the circuit. So, we are going to discuss about the power in RL series circuit. So, voltage is given by V equal to Vm sin omega t and the current equation is I equal to Im sin omega t minus theta for RL series circuit and instantaneous power is given by P equal to V into I. So, by using this above voltage and current equation, we can write down the equation as P equal to Vm sin omega t into Im sin omega t minus theta. Now, by using trigonometric, so we can rewrite the above equation as P equal to Vm Im by 2 cos theta minus cos 2 omega t minus theta. So, here by using trigonometric formula, we can write down this equation and here we will get the equation as P equal to Vm by root 2 Im by root 2 in bracket cos theta minus cos 2 omega t minus theta. These two we are going to split as a root 2 into root 2. So, here we will get the equation as P equal to Vm by root 2 Im by root 2 cos theta minus Vm by root 2 Im by root 2 cos 2 omega t minus theta. Now, if we calculate the average of this power, so here we will get the P equal to average of that first term and average of that second term. So, the P equal to Vrms into Irms cos theta because here average of cos 2 omega t minus theta is 0. So, here we will get the average power as P equal to Vrms Irms cos theta where cos theta is nothing but the power factor of the circuit and Vm by root 2 and Im by root 2 is nothing but the Vrms and Irms. So, we can draw the power cycles voltage voltage and current cycles like this you can see here the current is lagging to the voltage by theta degree voltage and current cycles are given and power cycles we can draw here. So, the positive power cycle indicates that continuous power consumption is there in the circuit. You can refer the book Electrical Technology by B.L. Thareja. Thank you.