 Hello, everyone. I am Mrs. Veena Sunil Patki, Assistant Professor, Department of Electronics Engineering, Valchan Institute of Technology, Solapur. So, at the end of this session, students will be able to derive the EMF equation of transformer and can solve the numerical based on it. First, we will discuss about the working principle of transformer. When the source of alternating current is applied to the primary winding of the transformer, magnetizing current flows and it produces alternating flux in the core of transformer. The produced alternating flux gets linked with the secondary winding of the transformer by mutual induction as it is alternating flux in nature. So, according to Faraday's law of electromagnetic induction, first law states that if any conductor or coil links with any changing flux, there must be an induced EMF in it. And second law states that the magnitude of induced EMF is directly proportional to rate of cutoff fluxes. So, magnitude of mutually induced EMF in the transformer can be found by EMF equation of transformer. So, if we consider the n number of turns in the winding, phi m is the maximum flux in the core in Weber. As per the Faraday's law of electromagnetic induction, e is given by minus n d phi by dt, where phi is the instantaneous alternating flux and represented as phi is equal to phi m sin 2 pi f t. As the alternating voltage is connected to the primary of the winding, so phi has the nature of AC. So, we can write down the e equal to minus n d phi m sin 2 pi f t by dt. So, after finding out the derivative of this equation, so we will get the equation e equal to minus n phi m cos 2 pi f t into 2 pi f. As the maximum value of cos 2 pi f t is 1, the maximum value of induced EMF e is given by em equal to n phi m into 2 pi f. To obtain the rms value of induced EMF, divide this maximum value by root 2. So, e rms is given by em by root 2. So, for this equation e rms equal to n phi m into 2 pi f divided by root 2. And after calculating 2 pi divided by root 2, we will get the value 4.44. So, e rms equal to 4.44 phi m f into n volts. So, this is the EMF equation of transformer. Now, we will consider this diagram where the primary and secondary windings are connected on the core. So, n 1 and n 2 are the primary and secondary turns, e 1 and e 2 are the primary and secondary EMFs and v 1 and v 2 are the primary and secondary voltages, i 1 and i 2 are the primary and secondary currents. So, we can write down the e 1 equal to 4.44 phi m f n 1 and e 2 equal to 4.44 phi m f n 2. So, voltage ratio of transformer, it is expressed by the ratio of primary and secondary voltage. So, voltage ratio is given by e 1 by e 2 and the turns ratio of transformer is given by n 1 by n 2. That is also equal to e 1 by e 2 because the induced EMF in the transformer is directly proportional to the number of turns of the primary and secondary. So, we can write down the turns ratio and voltage ratio as e 1 by e 2 equal to n 1 by n 2. So, if we consider the ideal transformer in that transformer will be 0. So, we can write down the input power equal to output power that is v 1 i 1 equal to v 2 i 2 and from this equation we can write down the v 2 by v 1 equal to i 1 by i 2 and if we equate all the ratio with one equation then we can write down that ratio as a transformation ratio k equal to v 2 by v 1 equal to v 2 by n 1 n 2 by n 1. So, transformation ratio k equal to v 2 by v 1 equal to n 2 by n 1 equal to i 1 by i 2 where k is the constant and that is the transformation ratio of transformer and if n 2 is greater than n 1. So, k is greater than 1 then that transformer is called as the step up transformer. If n 2 is less than n 1 then k is less than 1. So, the transformer is called as step down transformer. Now pause the video and write the Faraday's second law of electromagnetic induction. So, answer is induced EMF in the coil is directly proportional to rate of cutoff fluxes. So, e is equal to minus n d phi by dt. Now, we will solve the numerical based on this 80 kva 3200 oblique 400 volts single phase 50 hertz transformer has 111 turns on the secondary winding. So, calculate 1 number of turns of primary winding then second primary and secondary full load current then third cross sectional area of the core if the maximum flux density is 1.2 tesla. So, first you write down the given data that is the kva rating of transformer equal to 80. So, v 1 equal to 3200 v 2 equal to 400 volt and the frequency F equal to 50 hertz and number of turns on the secondary n 2 equal to 111 that is 111. Now, we have to calculate the value of n 1 number of turns of the primary then i 1 the current flowing through the primary winding then i 2 current flowing through the secondary winding and a is the nothing but the cross sectional area of the core. So, we know the transformation ratio that is k equal to v 2 by v 1 equal to n 2 by n 1 equal to i 1 by i 2. So, put the values of v 1 and v 2 and n 1. So, we will get the equation as a 400 divided by 3200 equal to 111 by n 1. So, after calculation we will get the value of n 1 as 888. Now, we can write down the equation that is kva that is kilo volt ampere into 10 raise to 3 equal to v 1 i 1 equal to v 2 i 2. So, we will get 80 into 10 raise to 3 equal to v 1 i 1 if we put the value of v 1 in this equation. So, after calculation you will get the value of i 1 that is 25 ampere and if we use the kva into 10 raise to 3 equal to v 2 i 2 put the value of v 2 in this equation. So, 80 into 10 raise to 3 equal to 400 into i 2. So, after calculation we will get the i 2 equal to 200 ampere. Now, maximum flux density is given by b m equal to phi m by a where b m is the maximum flux density phi m is the maximum fluxes in the core a is the cross sectional area. So, from this equation we can write down phi m equal to b m into a. So, if we put this phi m equal to b m into a in voltage equation of the transformer we will get v 1 equal to 4.44 b m a f into n 1. So, put the values of e 1 b m f n 1 in the above equation where e 1 is nothing, but the v 1 that is the primary voltage b m is the maximum flux density in the core f is the supply frequency and n 1 is the number of turns for the primary winding. So, we will get the equation a equal to v 1 divided by 4.44 b m f n 1. So, after putting the values of b m f n 1 in this equation. So, we will get the equation a equal to 3200 divided by 4.44 into 1 into 50 into 888. So, after calculation we will get the value of a equal to 1.35 into 10 raise to minus 2 that is in meter square. So, cross sectional area is given by a equal to 1.35 into 10 raise to minus 2 meter square. So, if we see this numerical we can calculate the cross sectional area or the fluxes or the e m f induced in the primary or secondary or we can calculate the number of turns by using these equations. So, we will get the equation a equal to 1.35 into 10 raise to minus 2 meter square.