Dear Sir, Could you please tell me how we can consider no load emf as 238V as at no load back emf would be 250V right? if 1200is no load speed emf corresponding to it should be 250V. Please reply
Sir , in previous lecture , you told that for E(a) vs I(f) graph , speed is constant and I(f) is proportional to flux , but at 6:28 you took the ratio of E(a)/E*(a)=N/N* . Here , due to armature reaction , flux also decreases and as E=k(flux)N and if flux is proportional to I(f) , E=kI(f)N . Then real equation should be E*(a)/E(a)=I*(f)N*/I(f)N
I have also the same doubt someone already asked you that how the no load back emf would be 238.3V instead of 250V as you determined n* with respect to no load speed (n•)
In DC motors, shaft speed is inversely proportional to the flux due to field winding and shaft speed is also directly proportional to armature voltage. But then armature voltage is induced due to the field flux itself. How do you explain this? in the problem u solved while taking no load speed we have to consider back emf as terminal voltage ,then why you took emf at 200A?
this is a good questions, actually it has something to deal with the practical working of dc motor, and the flux weakening at one side of the armature slot teeth and strenghthening on the other side of the same teeth, I will try to put it in a different lecture.
+Ashir Malik the non linear analysis section for every machine is followed by a big numerical..It's a very good numerical also..If u understand them all other numericals will be very easy..
@@ElectricalisEasy Sir the point you mentioned at 7:05 that Speed(new) = (previous speed)*(new back emf/previous back emf) is only applicable when flux is constant so we cann't write that equation because at those emf flux are different.
I think answer is 1330rpm,because high current 195amp means high armature reaction,thereby flux decreases enormously,but we need to meet load torque = induced torque so emf gets constant.thats why the speed increases greater than rated speed
Sir no load speed is equal to 1200rpm corresponding to that Ea = 250V know sir. But while solving you have taken it as 238.3V. And Ea is proportional to flux and speed in this problem . So I think we cannot write this equation n*=no.Ea*/Ea
thank u very much sir..
100% understandable lectures of urs..
thanks thanks
Dear Sir, Could you please tell me how we can consider no load emf as 238V as at no load back emf would be 250V right? if 1200is no load speed emf corresponding to it should be 250V. Please reply
With this the required speed comes as 1118.4 rpm
Same doubt.. please clarify.
Sir , in previous lecture , you told that for E(a) vs I(f) graph , speed is constant and I(f) is proportional to flux , but at 6:28 you took the ratio of E(a)/E*(a)=N/N* . Here , due to armature reaction , flux also decreases and as E=k(flux)N and if flux is proportional to I(f) , E=kI(f)N . Then real equation should be E*(a)/E(a)=I*(f)N*/I(f)N
I am also thinking the same. In that case, n becomes 1364.31 rpm, way bigger then no load 1200 rpm
I have also the same doubt someone already asked you that how the no load back emf would be 238.3V instead of 250V as you determined n* with respect to no load speed (n•)
In DC motors, shaft speed is inversely proportional to the flux due to field winding and shaft speed is also directly proportional to armature voltage. But then armature voltage is induced due to the field flux itself. How do you explain this? in the problem u solved while taking no load speed we have to consider back emf as terminal voltage ,then why you took emf at 200A?
this is a good questions, actually it has something to deal with the practical working of dc motor, and the flux weakening at one side of the armature slot teeth and strenghthening on the other side of the same teeth, I will try to put it in a different lecture.
sir plz solve almost one example of every lecture..
thank u
+Ashir Malik the non linear analysis section for every machine is followed by a big numerical..It's a very good numerical also..If u understand them all other numericals will be very easy..
Sir, N1/N2 = E1/E2 is valid for constant flux right? How can be that valid for reduced flux?
Here with nonlinearity corresponds to the behaviour of terminal characteristics due to the armature reaction.Am I right?
yes armature reaction and saturation etc...
@@ElectricalisEasy Sir the point you mentioned at 7:05 that Speed(new) = (previous speed)*(new back emf/previous back emf) is only applicable when flux is constant so we cann't write that equation because at those emf flux are different.
I think answer is 1330rpm,because high current 195amp means high armature reaction,thereby flux decreases enormously,but we need to meet load torque = induced torque so emf gets constant.thats why the speed increases greater than rated speed
As like DC series motor
Goated class
Sir no load speed is equal to 1200rpm corresponding to that Ea = 250V know sir. But while solving you have taken it as 238.3V. And Ea is proportional to flux and speed in this problem . So I think we cannot write this equation n*=no.Ea*/Ea
Plse increase the voice level ,,in some videos voice is low ....
Yes these are videos of the initial days of the channel, kindly use a headphone for some these videos..