 I, 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 numerical of DC generator. So, let us see the first numerical. The output of DC shunt generator is 450 ampere at 250 volt, if armature resistance is 0.04 ohm and shunt field resistance is 50 ohm, what is the generated EMF? So, first let us write down the given data. So, shunt generator current is 450 ampere is nothing, but the load current I equal to 450 ampere, terminal voltage V is given as 250 volt, armature resistance R A equal to 0.04 ohm and shunt field resistance R S H equal to 50 ohm. So, here you can see in this diagram shunt field winding resistance is given and the armature winding resistance is given and the generated EMF is given as V plus I A R A and the I A is given as I S H plus I L, but I S H equal to V by R S H because the total voltage V will drop across the shunt field winding. So, we can calculate the shunt field current equal to 250 divided by 50 equal to 5 ampere that is I S H. The armature current as I A equal to I S H plus I L already I L is given as 450. So, armature current is calculated as 5 plus 450 that is 455 ampere and E G is given for shunt motor as V plus I A R A equal to 250 plus 455 into 0.04. So, after calculation here we will get E G equal to 250 plus 18.2 equal to 268.2 volt. Now let us see the second numerical a DC machine develops an EMF of 250 volt when driven at 800 rpm with flux per pole of 0.03 Weber. It is desired that this EMF be increased to 300 volt at 1000 rpm. What should be the value of the flux per pole under the new circumstances? So, let us see first given data here the generated EMF is 250 volt that is given as E G 1. Generated speed is N 1 equal to 800 rpm and the flux per pole phi 1 equal to 0.03 Weber means when the generator rotates with 800 rpm the generated EMF is 250 volt and flux per pole is 0.03 Weber. New generated EMF E G 2 is 300 volt at N 2 equal to 1000 rpm and we have to calculate the new flux per pole at this circumstances. So, for existing values we are going to write down the generated EMF as E G equal to phi ZN by 60 P by A. This is the formula for generated EMF. After putting the values here we will get E G is nothing but the E G 1 that is 250 volt equal to phi is 0.03 into Z into N is 800 divided by 60 into P by A. So, this is the first equation. So, here we have the first equation like this now for new values we can write down the second equation as E G equal to 300 equal to phi Z into 1000 divided by 60 P by A and by dividing these two equations here we will get 250 by 300 equal to 0.03 into 800 divided by phi into 1000. So, after calculation here we will get phi equal to 0.03 into 800 into 300 divided by 250 into 1000 equal to 0.0288 Weber. So, the new phi is equal to 0.0288 Weber. This is the answer for flux per pole at new circumstances. Now pause the video and think about this question what is the use of commutator in DC generator? What is the answer for this question? Generally, the commutator is used to convert AC into DC. So, that commutator work as a rectifier, mechanical rectifier to convert alternating current to DC current in generator that is the answer for this question. Now, let us see the next example a four pole DC machine with wave wound armature having 400 conductors and with flux per pole is 20 milli Weber and the armature and shunt field resistance are 0.042 Ohm and 75 Ohm respectively. The machine has to supply a load of 600 lamps of rated 100 watt each at 250 volt. The line drop, voltage drop in connecting wire 6, 10 volt calculate the speed at which the generator should be driven. The number of poles P equal to 4, number of parallel paths A equal to 2 and number of conductors Z equal to 400 then flux per pole phi equal to 20 milli Weber equal to 20 into 10 raise to minus 3 Weber. And armature resistance is given as 0.042 Ohm and the shunt field resistance that is given as RSH equal to 75 Ohm and the terminal voltage V equal to 250 volt. And here the load on the generator is given as PL equal to 600 lamp of 100 watt so we can calculate that as a 600 into 100 watt that is 60000 watt. And line drop is given as 10 volt and we have to calculate the speed of armature. Generated EMF is given as phi Zn by 60 P by A and EG equal to V plus IARA plus VLD. So P equal to V into IL so from that we can calculate the line current equal to so total power is 60000. So here we are going to consider the 60000 divided by this voltage V equal to 250 equal to 240 ampere is the line current and shunt current equal to V by RSH 250 divided by 75 that is equal to 3.33 ampere. The armature current equal to ISH plus IL equal to 240 plus 3.33 that is 243.33 ampere. So we can calculate the generated EMF equal to 250 plus 243.33 into 0.04 plus line voltage drop that is 10 equal to 269.73 volt that is the generated EMF. Now the generated EMF is given as phi Zn by 60 P by A from this we can write down the speed n equal to 60 into EG into A divided by phi Zp. So after putting the values in this formula here we will get 60 into 269.73 into 2 divided by 20 into 10 raise to minus 3 into 400 into 4 equal to 1011.48 rpm that is the speed of the DC generator. You can refer the book Electrical Engineering by B. L. Thareja, Principles of Electrical Machines, V. K. Mehta and Rohit Mehta. Thank you.