An electric generator consisting of copper wire with 32 turns and a radius of 2
ID: 1465922 • Letter: A
Question
An electric generator consisting of copper wire with 32 turns and a radius of 2 m. The armature rotates at 60 times per second (60 Hz) in a 0.124 T constant magnetic field.
a. Calculate the peak value of the voltage produced be this generator.
b. Calculate the root-mean-square value of the voltage.
c. If the average power produced by this generator is 25 Megawatts, determine the RMS current supplied by this generator.
d. We wish to transport this energy to a small city. The total resistance in the 10 km of power lines is about 2. Calculate the power loss when the generated current is sent through the 2- ohms of resistance.
e. What can be done to minimize the power loss? Explain with physics.
Explanation / Answer
given data
no of turns, N = 32
r = 2 m
area of loop, A = pi*r^2
= pi*2^2
= 12.57 m^2
frequency, f = 60 Hz
b = 0.124 T
angular frequency, w = 2*pi*f
= 2*pi*60
= 377 rad/s
a) Vmax = N*A*B*w
= 32*12.57*0.124*377
= 18804 volts
b) Vrms = Vmax/sqrt(2)
= 13296.4 volts
c) P_average = Irms*Vrms
==> Irms = P_average/Vrms
= 25*10^6/(13296.4)
= 1880 A
d) Power dissipated at resistor, P = I^2*R
= 1880^2*2
= 7.07*10^6 Watts
e) By increasing ouput voltage and decreasing output current, we can decrease power loss.
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