Question

A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms current of 60.0 A.

Required:
a. What is the peak current produced?
b. What is the area of each turn of the coil?

Answer 1

Peak current= 84.86 A

Area of each turn = 0.029 m^2

Step-by-step explanation:

The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.

Hence;

Irms= 60.0A

Io= Irms/0.707

Io = 60.0/0.707

Io= 84.86 A

Vrms= 0.707Vo

Vo= Vrms/0.707= 170/0.707 = 240.45 V

From;

V0 = NABω

Where;

Vo= peak voltage

N= number of turns

B= magnetic field

A= area of each coil

ω= angular velocity

But ω= 2πf = 2×π×95= 596.9 rads-1

Substituting values;

A= Vo/NBω

A= 240.45/550×0.025×596.9

A= 0.029 m^2