Question

A generator has a coil with 1260 turns that are each 0.090 m in diameter. The coil rotates at 3600 rpm in a 0.830 T magnetic field. If a 20.0 Ω resistive load is connected in parallel with the generator, what is the average power used by the load?

Answer 1

Step-by-step explanation:

Average power in AC is given as

`P = i_(rms) V_(rms) cos\phi`

As we know that when AC is connected across the Resistor then we will have

`cos\phi = 1`

now we have

`P = (i_oV_o)/(2)`

here we have

`V_o = NBA\omega`

`V_o = 1260(0.830)(\pi* 0.045^2)(2\pi (3600)/(60))`

`V_o = 2508.15 Volts`

now we have

`P = (2508.15 * 2508.15)/(2 * 20)`

`P = 1.57 * 10^5 Watt`