A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.

Physics

1 answer:

Pachacha [2.7K]3 years ago

6 0

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

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(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$