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HACTEHA [7]
3 years ago
10

Given: an 802.11 wlan transmitter that emits a 50 mw signal is connected to a cable with 3 db loss. the cable is connected to an

antenna with 16 dbi gain. what is the eirp power output?
Physics
1 answer:
CaHeK987 [17]3 years ago
8 0

-- A transmitter has 50 mW output power.  

50 mW is equivalent to +17 dBm.

(+17 is a magic number.  It tells us that the transmitter could very well be based on a single modulated Gunn diode oscillator, which, after resonating and filtering to remove the unwanted puree, hash, and garbage, typically delivers right around +17 dBm at the output.  

-- The power passes through a piece of lossy cable, where it loses 3 dB.

+17 dBm went into the cable.   +14 dBm came out of the other end.  

(The lost 3 dBm warmed the cable.)  

-- The power was then coupled (losslessly) to an antenna with +16 dB "gain".

+14 dBm went into the antenna.  It was shaped and focused so that coming out of the antenna in a certain direction, it sounded as loud as a source that's radiating (+14 + 16) = <em>+30 dBm = 1 watt</em> .

This is NOT 1 watt of real power output.  The antenna has no batteries, it isn't plugged into a wall outlet, and it has no actual 'gain'.  

That 1 watt is "eirp" . . . <em>"</em><em>E</em><em>ffective </em><em>I</em><em>sotropic </em><em>R</em><em>adiated </em><em>P</em><em>ower"</em>.  The antenna focuses most of its power in one certain narrow direction, and then, <u><em>in that direction</em></u>, it sounds as loud as an antenna would that took 1 watt and spread it equally in <em>all</em> directions.

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An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
8 0
3 years ago
Read 2 more answers
A typical nuclear fission power plant produces about 1.00 GW of electrical power. Assume the plant has an overall efficiency of
mamaluj [8]

Answer:

mass consumed by 235U each day = 2 kg

Explanation:

electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV

                                            = 6.24151× 10²¹ MeV/s

thermal energy =  0.420 * 250 = 105 MeV

\dfrac{1 GW}{150 MeV}= \dfrac{6.24151\times 10^{21}}{105}

                                      = 5.94 × 10¹⁹ fission/second

                                       =5.94 × 10¹⁹× 24 × 60 ×60)

                                      =  5.13 × 10²⁴ fission/day

mu = 235.04393 ×  1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg

M = mu ×5.13 × 10²⁴

   = 390.1729× 10⁻²⁷ ×5.13 × 10²⁴

M   =  2 kg(approx.)

mass consumed by 235U each day = 2 kg

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True. Electromagnetic energy does not require a medium.

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