In an LC circuit in which C = 4.0 mu F the maximum potential difference across t

Question

In an LC circuit in which C = 4.0 mu F the maximum potential difference across the capacitor during the oscillations is 1.7 V and the maximum current through the inductor is 40 mA. (a) What is the inductance L? (b) What is the frequency of the oscillations? (c) How much time does it take for the charge on the capacitor to rise from zero to the half of its maximum value? One type of hologram consists of bright and dark fringes produced on photographic film by interfering laser beams. If the hologram is illuminated with white light, the image will be reproduced multiple times, in different pure colors at different sizes. A) explain why: b) which color correspond to the largest and smallest images, and why? A diffraction grating has 300 rulings/mm, and a strong diffracted beam is noted at theta = 25 degree. (a) What are the possible wavelengths of the incident light? (b) What colors are they? An oven with an inside temperature T_0 = 317 degree C is in a room having a temperature T_r = 27 degree C. There is a small opening of area 8.0 cm^2 in one side of the oven. How much net power is transferred from the oven to the room? The remove an electron from the surface of sodium is 2.2 eV. Find the work needed to maximum wavelength of light that will cause photoelectrons to be from sodium. How maximal velocity will have electrons if the surface of sodium will be illuminated with ultraviolet light (wavelength 425 mm? What is more dangerous, a radioactive material with a short half-life time or a long one?

Explanation / Answer

Ans.(1)

Given capacitance C = 4.0 uC

maxmium potential difference during the oscillation is 1.7 V

and the maximum current through the inductor is 40 ma

(a)
Since capacitance C = q / V

Hence the maximum charge on the capacitor is

q0 = C*V

= (4.0 uF)*(1.7 V)

= 6.8 uC

And the maximum current in the inductor is given by
I0 = q0*omega

hence the angular friquency
omega = (I0/q0)

= (40/6.8)

= 5.88*10^3 rad/sec

As omega = (1/root(L*C))

Hence the inductor l is given by

L = 1/(omega^2)*(C)

L = (1/(5.88*10^3)^2*(3.7*10^-6))

L = 4.59 mH

(b)
friquency of the oscillation is

v = (omega/2*pi)

= (5.88*10^3/2*3.14)

v = 936.30 Hz

(c)
Hence the time period is

T = 1/v

= 1/936.30

T = 1.06*10^-3

Hence the time for the charge to increase from
zero to maximum value is

= T/4

= 1.06*10^-3/4

= 2.65*10^-4 s

= 0.265 ms

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