Derive an expression for the induced emf in an AC generator. An a.c. generator consists of a coil of 50 turns and area 2.5 m^2 rotating at an angular speed of 60 rad/s in a uniform magnetic field B= 0.3 T between two fixed pole pieces. The resistance of the circuit including that of coil is 50 Ω. What is the maximum current drawn from the generator?

Dear Student
In an AC generator as the armature coil rotates in a magnetic field. Flux Associated to it changes due to rotation which in turn induces emf and current. Derivation as follows:

Solution of the second part of your question is as follows:-

 Emf from the AC generator source is given as E= NBAsin(ωt) where ω = frequency of source
Now Induced EMF will be maximum when sin(ωt)=1 i.e. ωt=90o
Here, N=50  A=2.5m2  ω=60rad/s  B=0.3T
Putting these values we get, 
Emax= 50*0.3*60*2.5= 2250V
Hence Maximum Current Drawn i.e. Imax= Emax/R = 2250/50= 45 A
Hence Maximum Current Drawn is 45 A
Regards

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