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Explain briefly, with the help of a labelled diagram, the basic principle of working of an a.c. generator. In an a.c. generator, coil of N turns and area A is rotated at upsilonrevolutions per second in a uniform magnetic field B. Write the expression for the emf produced. A 100-turn coil of area 0.1m^(2) rotates at half a revolution per second.It is placed in a magnetic field 0.01 T perpendicular to the axis of rotation of the coil. Calculate themaximum voltage generated in the coil. |
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Answer» Solution :A.C. GENERATOR Principle :Whenever a closed coil is rotated in a uniform magnetic field about an AXIS perpendicular to the field, the magnetic flux linked with coil changes and an induced emf is set up across its ends. The essential parts of an a.c. generator are shown in the above figure. Initially the armature coil ABCD is horizontal . As the coil is rotated clockwise, the arm AB moves up and CD moves down By Fleming'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=NB A cos omega t` Induced emf will be `E=-(d phi)/(dt)=-(d)/(dt) (NBA cos omega t)= NBA omega sinomega t` or `E=E_(0) sin omega t` where `E_(0)=NBA omega = ` peak value of induced emf Numerical : `N=100, A = 0.1 m^(2), B=0.01T` `V=(1)/(2)` revolution per sec =0.5 r.p.s `:. ` Maximum voltage generated `e_(0) = NBA omega = NBA (2 PI upsilon)` `e_(0) = 100 xx 0.01 xx 0.1xx2xx(22)/(7) xx 0.5` `=(2.2)/(7) = 0.314 `Volt `e_(RMS)=(e_(0))/(sqrt(2))=(0.314)/(1.414)=0.22 V`. |
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