Question

An ac electrical generator is made by turning a flat coil in a uniform constant magnetic field of 0.225 T. The coil consists of 33 square windings and each winding is 15.0 cm on each side. It rotates at a steady rate of 745 rpm about an axis perpendicular to the magnetic field passing through the middle of the coil and parallel to two of its opposite sides. An 8.50-Ω light bulb is connected across the generator.

(a) Find the voltage and current amplitudes for the light bulb.

Answer 1

The voltage amplitude for the light bulb is 550.54 V, and the current amplitude is 64.83 A.

Step-by-step explanation:

The voltage and current amplitudes for the light bulb can be found using Faraday's law of electromagnetic induction and Ohm's law. First, let's calculate the peak voltage. The peak output voltage of the generator can be determined using the formula: V = NABω, where N is the number of windings, A is the area of each winding, B is the magnetic field strength, and ω is the angular velocity. Plugging in the given values, we have: V = (33)(0.15^2)(0.225 T)(2π(745/60) rad/s) = 777.91 V.

Next, to find the voltage amplitude for the light bulb, we divide the peak voltage by √2, since the voltage varies sinusoidally to give a maximum value of V and a minimum value of -V. V_amplitude = 777.91 V / √2 = 550.54 V.

To find the current amplitude, we can use Ohm's law. The current amplitude can be calculated using the formula I = V_amplitude / R, where R is the resistance of the light bulb. Plugging in the values, we have: I = 550.54 V / 8.50 Ω = 64.83 A.