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Consider a generator that rotates its 200-turn, 0.15 m diameter coil at 3700 rpm in a 0.95 T field.

Randomized variables:
- Diameter (d) = 0.15 m
- Frequency (f) = 3700 rpm
- Magnetic field (B) = 0.95 T

Calculate the peak voltage of the generator.

Answer :

Final answer:

To calculate the peak voltage of a generator, we use the formula Emax = NABw, applying values for the coil's turns, rotation speed, coil diameter, and magnetic field strength, yielding a peak voltage of approximately 1292.3 V.

Explanation:

To calculate the peak voltage of a generator, we need to understand the formula that relates the speed of rotation, the number of turns in the coil, the magnetic field strength, and the area of the coil. The formula for the peak voltage (Emax) generated by an electrical generator is given by

Emax = NABw,

where N is the number of turns in the coil, A is the area of the coil in square meters, B is the magnetic field strength in teslas, and w is the angular velocity in radians per second.

In this scenario, the coil has 200 turns and rotates at 3700 rpm in a 0.95 T magnetic field, with a coil diameter of 0.15 meters. First, we convert the rotation speed to radians per second:

3700 rpm = 3700 * (2π/60) rad/s.

Then, we calculate the area of the coil

(A = π*(d/2)^2).

With these values, we can find the peak voltage using the formula.

Let's go through the calculation step-by-step:

  • Convert rpm to rad/s: 3700 rpm = 387.6 rad/s
  • Calculate the area of the coil: A = π*(0.15/2)^2 = 0.0177 m2
  • Apply the values to the formula: Emax = 200 * 0.95 * 0.0177 * 387.6
  • This gives a peak voltage of approximately 1292.3 V.

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Rewritten by : Barada