InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
State the working of a.c. generator with the help of a labelled diagram.The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity omega. Deduce the expression for the alternating e.m.f. generated in the coil. What is the source of energy generation in this device? |
|
Answer» Solution :A.C. Generator: See Q.30 (or) 2008, O.D. Set-1 The ESSENTIAL parts of an a.c. generator are shown in the figure. Initially the armaturecoil ABCD is horizontal. As the coil is rotasted clockwise , the arm AB moves up and CD moves down. By Flemming's right hand rule, the induced current flows along ABCD, In second half rotation, the arm CD moves up and AB moves down. The induced current flows in the opposite direction i.e. along DCBA. Thus an alternating current flows in the CIRCUIT The magnetic flux linked with the coil at any instant is `""phi="NA A "cos omegat` `"Induced emf will be" ""E=-(dphi)/(dt)=-(d)/(dt)("NBA cos "omegat)=NBA omega sin omegat` `or ""E=E_(0) sin omegat` Where, `E_(0)=NBA omega`= peak value of induced emf. (b) To calculate the magnitude of e.m.f induced Suppose N=number of turns in the coil. A=area of enclosed by each turn of coil. `overset(rightarrow)B`=strength of magnetic field. `theta=` ANGLE which NORMAL to the coil makes with `overset(rightarrow)B` at any instant t,  `therefore` Magnetic flux linked with the coil in this position. `""phi=N(overset(rightarrow)B.overset(rightarrow)A)` `""=NBA cos theta` `""=NBA cos omega"".....(i)` Where `omega` be the angular velocity of the coil. At this instant t, if e in the e.m.f induced in the coil, then `e=(-dphi)/(dt)=(-d)/(dt)(NAB cos omegat)=-NAB (d)/(dt)(cos omegat)` `""=-NAB(-sin omegat)omega` `""e=NAB" "omega sin omegat......(ii)` The induced, e.m.f. will be maximum, when `""sin omegat="maximum "=1` `therefore""e_("max")=e_(v)=NAB_(omega)xx1.......(iii)` Put in (ii)`""e=e_(0) sin omegat` (c ) Mechanical energy is converted into electrical energy. ' |
|