Question

A 500-loop circular armature coil with a diameter of 6.0 cm rotates at 190 rev/s in a uniform magnetic field of strength 0.60 T . What is the rms voltage output of the generator?

What would you do to the rotation frequency in order to double the rms voltage output?

To double the output voltage, you must decrease by two times the rotation frequency.

To double the output voltage, you must increase by four times the rotation frequency.

To double the output voltage, you must treble the rotation frequency.

To double the output voltage, you must double the rotation frequency.

Answer 1

a.   Peak Voltage   V₀   =   ω* N * B * A      -----------(1)

Here angular frequency ω   =   190 rev/s

                                                =   190* 2π      rad/s

                                                =   1193.2 rad/s

   

   No. of turns in the armature   N   =   500

   Area of one loop   A   =   π *r²

                                       =   3.14* (6.0 * 10^-2 / 2)²

                                       =   0.2826* 10^-2   m²

   Magnetic field   B   =   0.6 T

   V₀   =   1193.2* 500 * 0.6 * 0.2826 * 10^-2

         = 1011.59 V

   V_RMS   =   V₀/ √2

               =   715.31 V --> Answer

   b.   From equation(1)      V_RMS   α   V₀   α   ω

   Hence to double V_RMS,   ω must also be doubled.

   ω'   =   2ω

         =   2* 190

         =   380   rev/s

   We would double the rotation frequency to 380 .

To double the output voltage, you must double the rotation frequency.