Question

A generator has a coil with 1260 turns that are each 0.050 m in diameter. The coil rotates at 3600 rpm in a 0.310 T magnetic field. If a 50.0 Ω resistive load is connected in parallel with the generator, what is the average power used by the load?

Answer 1

Answer:

Average power dissipated in the coil is 833 W

Explanation:

As we know that the amplitude of induced EMF in the coil is given as

$E_o = NBA\omega$

so we have

N = 1260 turns

B = 0.310 T

$A = \pi (0.025)^2$

$A = 1.96 \times 10^{-3} m^2$

$\omega = 2\pi(3600/60)$

$\omega = 377 rad/s$

Now EMF is given as

$E_o = 288.6 V$

now Average power is given as

$P = \frac{E_o^2}{2R}$

$P = \frac{288.6^2}{2(50)}$

$P = 833 W$

Answer 2

The average power used by the load of the generator coil is 833W.

How to calculate average power?

The average power of a coil can be calculated by using the following formula:

Eo = N × B × Aw

Where;

• N = number of turns
• B = magnetic field (0.310T)
• A = ?

A = 22/7 × 0.025²

A = 1.96 × 10-³

W = 2 × 22/7 × 3600/60

W = 377rad/s

Eo = 1260 × 0.310 × 1.96 × 10-³ × 377

Eo = 288.6V

P = E²/2R

P = 288.6²/2(50)

P = 833W

Therefore, the average power used by the load of the generator coil is 833W.

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