Physics, 30.08.2019 05:30 sammybrain

An l-c circuit containing an 82.0-mh inductor and a 1.60-nf capacitor oscillates with a maximum current of 0.800 a . assuming the capacitor had its maximum charge at time t= 0, calculate the energy stored in the inductor after 2.40 ms of oscillation.

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Answer from: Diegosolorzano50

19.5 mJ

Explanation:

Assuming perfect components without resistance or losses, the circuit should oscillate indefinetly.

The circuit will have a natural pulsation of

$w = \frac{1}{\sqrt{L * C}} = \frac{1}{\sqrt{82e-3 * 1.6e-9}} = 87304 rad/s$

$f = \frac{w}{2\pi} = \frac{87304}{2\pi} = 13895 Hz$

$T = \frac{1}{f} = \frac{1}{13895} = 7.2 \mu s$

So, by the time t = 2.4 ms, 333.33 cycles would have passed

$\frac{2.4 ms}{7.2 \mu s} = \frac{2400 \mu s}{7.2 \mu s} = 333.33$

Therefore it would be at one third after the beginning of the cycle. The circuit would be in an equivalent state as t = (7.2 us)/3 = 2.4us

At t=0 the capacitor is fully charged, so the voltage is maximum and the current is 0. The current will increase towards a maximum of 800 mA at t=T/4, then decreas to 0 at t=T/2, decrease to -800 mA at 3T/4 and go back to 0 at t=T following a sine wave.

The equation of this sine wave would be

I(t) = I0 * sin(w * t)

I(T/3) = 0.8 * sin(w * T/3)

Since w = 2π/T

w * T = 2π

$I(T/3) = 0.8 * sin(2 \pi /3) = 0.8 * 0.866 = 0.69 A$

The current stored in an inductor is

$E = \frac{1}{2} * L * I^2$

$E = \frac{1}{2} * 82e-3 * 0.69^2 = 0.0195 J = 19.5 mJ$

Answer from: Quest

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Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

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