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

Answer:

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

$P = \frac{i_oV_o}{2}$

here we have

$V_o = NBA\omega$

$V_o = 1260(0.830)(\pi\times 0.045^2)(2\pi \frac{3600}{60})$

$V_o = 2508.15 Volts$

now we have

$P = \frac{2508.15 \times 2508.15}{2 \times 20}$

$P = 1.57 \times 10^5 Watt$