AC generator: The coil in a 50-Hz ac generator has 283 turns, each having an area of 3.0 × 10⁻² m² and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator? A) 120 V B) 320 V C) 200 V D) 240 V E) 220 V

Question

Quizplus · Accepted Answer

The peak output voltage of an AC generator is given by:V = N * A * B * w * sin(wt)where:N is the number of turns in the coilA is the area of each turnB is the magnetic field strengthw is the angular frequency (2 * pi * f)t is the timeIn this problem, we have:N = 283A = 3.0 × 10^-2 m^2B = 0.12 Tf = 50 Hzw = 2 * pi * f = 314 rad/sPlugging these values into the equation, we get:V = 283 * 3.0 × 10^-2 m^2 * 0.12 T * 314 rad/s * sin(314t)The peak output voltage is the maximum value of V, which occurs when sin(314t) = 1. Therefore, the peak output voltage is:V = 283 * 3.0 × 10^-2 m^2 * 0.12 T * 314 rad/s = 120 VTherefore, the correct answer is A) 120 V.

AC Generator: the Coil in a 50-Hz Ac Generator Has