Doubt discussion 17:00 to 31:25 Here driving equation for phase-a Ankit sir have rotated conducter clockwise and angle get was Wt, but for phase-b and for phase c sir have rotated the conductor anticlock wise. So emf we got is not at 120 degree phase displcement. I am rectifying it by rotating the conductor of phase-a anticlockwise So angle we get wil be -Wt . And also rotating the conductor of phase-b and phase-c anticlockwise So we will get correct 120 degree phase displacement.
Anyone reading this can confirm this....I think the equation derived for emf induced with AC excitation (25:40sec) is wrong .. there is some error in derivative
Here We assume coil rotating & rotor stationary...but actually always rotor rotated....means if we assume rotor is rotated in clockwise direction then we saw the coil is rotated in opposite directions means reverse direction
Just for simplification... If assumed the same direction, then sequence would have been reveresd... Instead of b, c would come meaning wt - 120 in c and wt+120 in b...
Field wdg ko jab ac supply diye to derivative kuchh samajh me hi nhi aaya...loved all ur lectures but this one repeating since morning but still not clear..
Sir ji muje samj nhi aaya derivation syn. Machine ka ...filed wdg ko AC supply dege to sin2wt kese aaya & -N $m w cost wt ki term kaha gai sir ji????? Please ans me
Dekho I understand your problem. What I have did ***Rotation rule Only keeping B(magnetic field axis) reference, take the rotation For phase A Take -wt(negative value of wt) For phase b Take (120-wt) For phase C Take(240-wt) Keep it mind while doing phase b sir hesitately change the direction of rotation that means it not only change for b ,it also change for a,b,c So you have to check it again for phase a back. Don't do the anticlockwise angle is pogetive rule. Just do from reference B axis how much angle is "decreased" at time t from t=0 **Because Follow carefully for phase b & c it is decreased then how for phase a it is increased? it should also decreased.(I think) Again one point Take only angle between B--a B--b B--c Never take B--a B--b' B--c' X(wrong) ***If sir not change the direction while calculating b phase then also we get the balance EMF But in this time sequence would be change from reference phase sequence. i .e our reference phase sequence is abc 120digree phase difference but It would come in case of EMF induced Ea,Ec,Eb 120 digree sequence I did by this above rule I may be wrong🤗🤗🤗but my problem is solved And *****ALL THE NEGATIVE SIGN BECOME POGETIVE.
respected ankit sir thanks sir for your amazing content i have a query sir i always got confused in wave form of mmf and flux in different machine how can i get cleared
Go through the privious lectures from DC machine as well as transformer don't see from middle.& Never find the shortcut if you really want to understand 🤗
According to the orientation of conductors in stator which are placed in sequence abc with 120 deg apart rotor should rotates in cw so that it could cuts conductors in sequence of abc.. Actually sir has taken acw by mistake.. So assume it cw even in "a" phase
Magnetic Field and Area in same direction gives you the flux = 0, hence, angle between B and dA is 90 degree. And in EMF, sin theta is there, so sin 90 = 1, which gives you the maximum EMF generated along the magnetic field
Sir, iska matlab ki rotor me magnetic field tab tak varying nhi hota jab tak rotor move nhi kr rha ho. And jab rotor move krega to magnetic field sinusoidal varying hoga due to chamfered pole in salient pole and stepped number of turn in cylindrical rotor.
@@aseemprakash9971 current carrying conductor produces magnetic field..the rotor is rotating by prime mover(turbine) until it is not excited by dc supply. after excitation the magnetic field of rotor rotates with the rotor and get synchronous speed.. now apply concept of chamfered pole
Horizontal component becomes A cos theta, and Vertical component becomes A sin theta, here theta = 120 - wt. Hence, simply it becomes A sin (120 - wt). Only we have to consider the vertical component of the area, otherwise for horizontal component in same direction of B, Area Vector dA will be perpendicular to B, that will give the flux = 0 (B.dA)
