Anonymous
Anonymous asked in 科學及數學其他 - 科學 · 6 years ago

# phy energy

1. The filament in a lamb is sometimes coiled to increase its resistance.

--> is it true that increase in resistance=increase in power in this case? Could you please explain?

2. why is the radiation power intercepted by the Earth = (solar constant)(pi)(square of radius of the Earth) but not (solar constant)(surface area of the Earth/2)?

Thanks!

Update:

1. So why coiled filament is used in lamps?

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• 天同
Lv 7
6 years ago

1. For a fixed voltage V, the power P consumed by the lamp, of resistance R, is given by,

P = V^2/R

An increase in resistance R of the lamp leads to a decrease of power P, and vice versa.

2. It is because the earth is NOT a flat surface, it is spherical. Hence, half of the earth is exposed to the incoming solar energy. Due to the spherical nature, not all incoming solar rays are incident normal to the earth surface. The term pi.R^2 is the "effective catchment area" for these solar rays.

Your use of the term (surface area/2) is NOT correct, because you have assumed that all rays are incident normally to the earth surface.

The mathematical derivation is as follows:

Let R be the radius of the earth. Consider a ring on earth surface with radius r and width dr. Hence, we have the folllowing relations:

r/R = sin(a)

Area of ring, dA= 2.(pi)r.(dr) = 2.(pi)R.sin(a)(dr) = 2.(pi).R^2.sin(a)(da)

Let I be the solar constant in unit of w/m^2. Hence, power caught by the ring

dP = (I.cos(a)).[ 2(pi).R^2.sin(a).(da)] = I(pi).R^2.sin(2a).(da)

Power P caught by the earth

P = integral {I.(pi).R^2.sin(2a).(da)} with angle a from 0 to pi/2

i.e. P = I.(pi).R^2.{[-cos(pi) + cos(0)]/2} = I.(pi).R^2[(1+1)/2]

P = I.(pi)R^2

2014-05-31 22:00:39 補充：