Question

What is the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field?

a) 96.0 V
b) 192 V
c) 288 V
d) 384 V

Answer 1

To calculate the peak voltage of the generator, we use the formula V_peak = N * A * B * w, where N is the number of turns, A is the area of the coil, B is the magnetic field strength, and w is the angular velocity. By plugging in the given values and calculations, we find that the peak voltage is 472.44 V, which is closest to option c) 288 V.

Step-by-step explanation:

To calculate the peak voltage of the generator, we can use the formula:

V_peak = N * A * B * w

Where:

N is the number of turns in the coil (200)

A is the area of the coil (π * r^2, where r is the radius)

B is the magnetic field strength (0.800 T)

w is the angular velocity (3600 rpm)

First, we need to find the radius of the coil. The diameter is given as 0.100 m, so the radius is 0.100/2 = 0.050 m.

Now we can calculate the area of the coil: A = π * (0.050)^2 = 0.00785 m^2.

Next, we convert the angular velocity from rpm to rad/s: w = 2π * 3600/60 = 376.99 rad/s.

Finally, we can calculate the peak voltage:

V_peak = 200 * 0.00785 * 0.800 * 376.99 = 472.44 V.

Therefore, the correct answer is option c) 288 V.