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xenn [34]
3 years ago
13

Determine the dimensions (a, b) of an air-filled rectangular waveguide that providessingle mode operation over the frequency ran

ge from 9 GHz to 14 GHz.
Physics
1 answer:
Dima020 [189]3 years ago
8 0

Answer: the required dimensions (a, b) are; ( a= 1.667 cm, b = 1.0714 cm )

Explanation:

Given the data in the question;

air-filled rectangular waveguide  

μ₀ = C

m = 1 --- { for single mode operation}

f₀ = c/za        {we have equation}

denote the f₀ = 9 GHz  

f₀  = Carrier frequency

f₁₀ = 14 Ghz        Range or cut off frequency

so we have to find;

a = c/2f₀ and b = c/2f₁₀

so

f₀ = 9 GHz

a = c/2f₀  

a = (3 × 10⁸) / ( 2× 9 × 10⁹)

a = 0.01666 cm

a = 1.667 cm

f₁₀ = 14 GHz

b = c/2f₀  

b = (3 × 10⁸) / ( 2× 14 × 10⁹)

b = 0.01071 m

b = 1.0714 cm

Therefore the required dimensions (a, b) are; ( a= 1.667 cm, b = 1.0714 cm )

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Explanation:

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n=\dfrac{q}{e}

n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}

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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
Nezavi [6.7K]

Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0  x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12    (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))  

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR)  --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides  

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x 10^{-9  

t = 23.9ns

Thus, it took 23.9ns  for the potential difference between the deflection plates to reach 75 volts

6 0
3 years ago
Read 2 more answers
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