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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?

Respuesta :

Answer:

Explanation:

Average power in AC is given as

[tex]P = i_{rms} V_{rms} cos\phi[/tex]

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

[tex]cos\phi = 1[/tex]

now we have

[tex]P = \frac{i_oV_o}{2}[/tex]

here we have

[tex]V_o = NBA\omega[/tex]

[tex]V_o = 1260(0.830)(\pi\times 0.045^2)(2\pi \frac{3600}{60})[/tex]

[tex]V_o = 2508.15 Volts[/tex]

now we have

[tex]P = \frac{2508.15 \times 2508.15}{2 \times 20}[/tex]

[tex]P = 1.57 \times 10^5 Watt[/tex]

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