Final answer:
The peak voltage of the generator can be calculated using Faraday's law of electromagnetic induction. With an angular velocity of 336.3 rad/s, a coil area of 0.0283 m^2, in a 0.75 T magnetic field, and 200 turns, the peak voltage is approximately 1421.2 V.
Step-by-step explanation:
To calculate the peak voltage of a generator, we can use Faraday's law of electromagnetic induction, which states that the induced emf (electromotive force) in a coil is proportional to the rate of change of the magnetic flux. The formula for the peak induced emf (or peak voltage) in a coil is given by:
E = NABωsin(ωt),
where E is the peak emf, N is the number of turns in the coil, A is the area of the coil, B is the magnetic field strength, and ω is the angular velocity of the coil.
First, we need to calculate the angular velocity (ω) in radians per second (rad/s) from revolutions per minute (rpm). Since 1 revolution is 2π radians and there are 60 seconds in a minute, we can use:
ω = (3200 rpm)(2π rad/rev)(1 min/60 s) = 336.3 rad/s
Next, we calculate the area A of the coil:
A = πr^2, where r is the radius of the coil.
Since the diameter is 0.19 m, the radius r = 0.095 m.
So, A = π(0.095 m)^2 = 0.0283 m^2
Given that N = 200 turns and B = 0.75 T, we can calculate the peak voltage:
E = NABω = (200)(0.0283 m^2)(0.75 T)(336.3 rad/s) = 1421.2 V
Therefore, the peak voltage of the generator is approximately 1421.2 volts.