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denpristay [2]
3 years ago
7

Consider a 5 m long, air-filled section of a coaxial transmission line, given that the radius of the inner conductor is 10 cm an

d the inner radius of the outer conductor is 20 cm? If the current is 100 mA, how much magnetic energy is strored in the air medium between the conductors. This problem is similar to Example 2 in the Time Varying Fields folder in the Example Bank.
a) 200.89 nj
b) 3.47 nj
c) 10.45nj
d) 80.9nj
Engineering
1 answer:
Helga [31]3 years ago
4 0

Answer: b) 3.47 nj

Explanation:

Given that;

length l = 5m

radius of inner conductor r = 10cm = 0.1m

radius of outer conductor D = 20cm = 0.2m

current I = 100A = 100×10⁻³ = 0.1

medium between conductor in air u₀ = 4π × 10⁻⁷

Energy in a coaxial cable transmission line is

w = u₀ /2π I² en(b/a)

we substitute

L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)

L =3.4657 × 10⁻⁹ J

L = 3.4657 nJ ≈ 3.47 nJ

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B×π×r^2=I×2.1×e^-3π×0.025^2

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U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

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3 years ago
Find the current Lx in the figure
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Explanation:

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\frac{9}{9 + 3}  \times 4 = 3\\   \frac{2}{2 + 8} \times 3 = 0.6a \\  = 0.6 \: milli \: amper

6 0
3 years ago
What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

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Work done by a system during a process can be considered as a property of the system. a)True b) False
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Air in a piston-cylinder assembly is compressed isentropically from state 1, where T1 = 35°C, to state 2, where the specific vol
ivanzaharov [21]

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(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg

Explanation:

Solution

T₁ = 35°C = 308

the first step to take is to Use the Table A-17: Ideal gas  properties for air:

Now,

At T₁ = 308 K

V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)

So,

V₁ =219.818 kJ/kg

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Applying Table A-17, at Vr₂ = 58.178

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Hence,

(b) q - w = v₂ -v₁

= 0 -w  = 549.996- 219.818

w = 330.178 kJ/kg

4 0
3 years ago
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