 Myself, Mrs. Veena Sunil Patki, Assistant Professor, Department of Electronics Engineering, Vulturen Institute of Technology, Solapur, welcome you for this session. At the end of this session, students can solve the numerical on RLC series circuit. So, let us see the first numerical. A pure resistance of 50 ohms is in series with pure capacitance of 100 microfarads. The series combination is connected across 100 volt 50 hertz supply. Find A, the impedance, B, current, C, power factor, D, phase angle, E, voltage across resistor, F, voltage across capacitor and draw vector diagram. So, first we are going to write down the given data. So, resistance is 50 ohm, capacitance is 100 microfarad, voltage is 100 volt and frequency 50 hertz. So, as the C equal to 100 microfarad. So, here capacitive reactance we can calculate as xc equal to 1 by 2 pi fc equal to 1 by 2 pi into 50 into 100 into 10 raise to minus 6 as capacitance is given in microfarad. Here we are going to multiply that capacitance by 10 raise to minus 6. So, after calculating here we will get the capacitive reactance xc equal to 32 ohms. So, you can see here resistance and capacitance both are connected in series and the V equal to Vm sin omega t is the applied voltage. So, let us calculate the impedance first z equal to under root r square plus xc square equal to under root 50 square plus 32 square equal to 59.4 ohm. Our current calculation i equal to V by z equal to 100 by 59.4 equal to 1.684 ampere and power factor we can calculate by using the formula r by z equal to 50 by 59.4. So, power factor equal to 0.842 leading. So, circuit is capacitive so power factor is leading and phase angle we can calculate as phi equal to cos inverse of 0.842 equal to 32.3. So, here we will get the phase angle as 32.3 degrees. So, voltage across resistor we can calculate as V r equal to i into r that is 50 into 1.684 equal to 84.2 volt and voltage across capacitor is given by i into xc. So, here 32 into 1.684 equal to 53.9 volt is the voltage across capacitor. So, for phasor diagram we are going to use V r equal to 84.2 volt Vc equal to 53.9 volt and phi equal to 32.3 degrees. So, first you draw the current vector V r is in phase with the current and Vc lags with the current as the current leads to the capacitive voltage and the total voltage for resistive and capacitive series circuit for that draw the parallelogram and here we will get the total voltage V. So, theta is the angle between total voltage V and i and here you can see the current leads voltage by 32.3 degrees. Now, we are going to solve the second numerical that is a equal. So, second numerical is a 120 volt 60 watt lamp is to be operated on 220 volt 50 hertz supply. Calculate what value of a non inductive resistance and b pure inductance would be required in order that lamp is run on correct voltage which method which method is preferable and why means the rating of lamp is given and we have to connect that lamp for 220 volt 50 hertz supply and we have to calculate the resistance and inductance. So, first we are going to consider if the resistance is connected in series with the lamp that is non inductive resistance is nothing, but the pure resistance is connected in series with the lamp. So, here you can see the diagram. So, bulb having 60 watt lamp and 120 volt rating is connected in series with the resistance. So, V R is the voltage across that resistance. Now, the total voltage V equal to 220 volt and frequency is 50 hertz. So, the voltage across register we can calculate as V R equal to 220 minus 120 equal to 100 volt because both voltage and resistance both bulb and resistance having the same phase because lamp is nothing, but the resistive and the resistance is connected in series with the lamp. So, both voltages are in phase. So, lamp rating is 120 volt 60 watt. So, current through the lamp we can calculate as P equal to V i or i equal to P divided by V. So, here current through the lamp we can calculate as 60 by 120 that is 0.5 ampere and voltage across resistance is in phase with applied voltage. So, R equal to V R by i equal to 100 by 0.5 equal to 200 ohm. Now, for second condition here if inductance is connected in series with the lamp. So, potential difference across bulb is 120 volt and V L is the voltage across inductor. We can calculate that V L by using this formula. So, the total voltage for series resistive and inductive circuit is given by under root V R square plus V L square. So, after putting the values here the V L equal to under root 220 square minus 120 square equal to 184.4 volt that is the voltage across inductive that is the voltage across inductance. Now, here V L equal to i into X L. So, by putting the values of V L and i we can calculate the X L. So, V L equal to 184.4 equal to 0.5 into L into 2 pi into 50. So, here directly L we can calculate as because X L equal to 2 pi f L directly we have used that formula here and L is equal to 184.4 divided by 0.5 into 3.14 equal to 1.17 Henry. So, here R equal to 200 ohm and L equal to 1.17 Henry. So, method B is preferable that is second method is preferable because in method B there is no loss of power. Ohmic resistance of 200 ohm itself dissipates the large power that is 100 into 0.5 equal to 50 watt. But in inductive circuit pure inductance no power consumption is there power consumption in the inductance is 0. So, here the second method is preferable. So, now pause the video and give the answer for this question the power factor of an RC series circuit is A often 0, B between 0 and 1, C always unity and between 0 and minus 1. So, what is the answer? So, here answer is B between 0 and 1. So, you can refer the book Electrical Technology by B. L. Thareja. Thank you.