Power Transmitted in a Coaxial Cable

In summary: To do step 1 you will need one item of data given to you.The outer radius of the space between the conductors can be found by using Gauss's theorem. Gauss's theorem states that the E field is related to the potential difference between the conductors by V = E(r) . So, in order to solve for V, you need to know the radius of the space between the conductors.
  • #1
Marcus95
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2

Homework Statement


A coaxial transmission line consists of an inner cylindrical conductor of radius a = 1 mm and a
cylindrical outer conductor chosen to make the characteristic impedance 75 ohm. The space
between the conductors is lled with a gas which can stand a maximum eld of 105 V/m
without dielectric breakdown. Estimate the maximum mean radio-frequency power that
can be transmitted along this line into a matching load.

Homework Equations


On transmission line
## Z = \sqrt{L/C} ##
For coaxial with radius a and b:
##C = \frac{2\pi \epsilon \epsilon_0}{ln(b/a)} ##
##L = \frac{\mu_0 ln(b/a)}{2\pi} ##

The Attempt at a Solution


The voltage between the two conductors is by Gauss's law:
## V = \int \boldsymbol{E} \cdot d\boldsymbol{r} = E_0 ln(b/a) ##
hence:
## V_{max} = E_{max} ln(b/a) ##
also:
##Z = \sqrt{\mu / \epsilon \epsilon_0} \frac{ln(b/a)}{2\pi} ##
Using ##P = VI = V^2/Z ## and asuuming ##epsilon \approx 1## gives:
##P_{max} = 0.21 GW##, but the answer is supposed to be 104 W.

What am I doing wrong? Many thanks!
 
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  • #2
Marcus95 said:
The voltage between the two conductors is by Gauss's law:$$
V = \int \boldsymbol{E} \cdot d\boldsymbol{r} = E_0 ln(b/a)$$
This equation doesn't make sense. The left side is in volts, the right side is in volts/meter. Check your derivation.
 
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  • #3
kuruman said:
This equation doesn't make sense. The left side is in volts, the right side is in volts/meter. Check your derivation.
Thank you, I got it!
 
  • #4
kuruman said:
This equation doesn't make sense. The left side is in volts, the right side is in volts/meter. Check your derivation.
I had my doubts about this step but for another reason. There is vector potential ##A## inside coaxial cable, so it is
##E=\nabla V+\frac{\partial A}{\partial t}##

so that
##V=\int Edr## doesn't hold (because it is equivalent to ##E=\nabla V##.
Do we make the approximation ##A\approx 0 ##?
 
  • #5
It's not necessary to invoke the magnetic potential.

You want to use Gauss's theorem to relate E field to potential difference between the conductors. So first you want to determine the geometry of the cable, i.e. the inner and outer radii. Is there a relation between these radii and Z0?

Then: use Gauss's theorem relating the E field to the potential difference V and the geometry and compute the max. allowable V for the given max. E.

Don't forget they want the rms power figure (I think - wording is not clear to me). Also assume an ideal matched line.
 
  • #6
rude man said:
It's not necessary to invoke the magnetic potential. Then: use Gauss's theorem relating the E field to the potential difference V and the geometry and compute the max. allowable V for the given max. E.

Don't forget they want the rms power figure (I think - wording is not clear to me). Also assume an ideal matched line.
Can you show how you find V using Gauss's theorem(have to solve Poisson's equation?) without using that ##V=\int Edr## because that last integral becomes path dependent when there is vector potential.
 
  • #7
Delta² said:
Can you show how you find V using Gauss's theorem(have to solve Poisson's equation?) without using that ##V=\int Edr## because that last integral becomes path dependent when there is vector potential.
Delta² said:
Can you show how you find V using Gauss's theorem(have to solve Poisson's equation?) without using that ##V=\int Edr## because that last integral becomes path dependent when there is vector potential.

We are dealing solely with electric, not magnetic, fields. I have never run into a vector potential such that E = ∇ x Ae the way we write H = ∇ x A (or sometimes B = ∇ x A). The former is in fact impossible if there is charge present since ∇⋅E = ρ, not zero, so no vector potential can exist. In any case, forget all vector potentials.

You can relate V to E by solving Laplace's equation for the space between the conductors but you'd still need to do step 1 below. It would in fact be good to do so as a check if you went with my suggestion which is:

(1) Find the outer radius of said space; (2) relate E(r) to V, the potential between the conductors using Gauss; (3) find where E is the largest, then you can determine the max. allowable V given the max allowable E.

To do step 1 you will need one item of data given to you.
 
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  • #8
Actually, solving Laplace's equation is the more direct route. Use the cylindrical coordinate expression ∇2V = (1/r) d/dr (r dV/dr) = 0 since there are no derivatives with respect to θ or z.

But again, your first task is to find the outer radius of the space between the conductors.
 
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What is power transmitted in a coaxial cable?

Power transmitted in a coaxial cable refers to the electrical energy that is sent through the cable to power a device or transmit signals.

How is power transmitted in a coaxial cable?

Power is transmitted through a coaxial cable using a central conductor, usually made of copper, which carries the electrical current. Surrounding the conductor is a layer of insulating material, followed by an outer conductor or shield, and then another layer of insulation. This design helps to minimize interference and signal loss.

What factors affect the power transmitted in a coaxial cable?

The power transmitted in a coaxial cable can be affected by various factors such as the length and diameter of the cable, the type and quality of the materials used, and the frequency of the signal being transmitted. Additionally, external factors such as temperature and electromagnetic interference can also impact the power transmission.

How can I calculate the power transmitted in a coaxial cable?

The power transmitted in a coaxial cable can be calculated using the formula P = I^2 * R, where P is power, I is current, and R is the resistance of the cable. The resistance of a coaxial cable can be determined by its length, diameter, and material. It is also important to take into account any losses from impedance mismatch or external interference.

What are the advantages of using a coaxial cable for power transmission?

Coaxial cables have several advantages for power transmission, including low signal loss, high bandwidth capabilities, and improved signal quality. They also have a strong shielding design that makes them less susceptible to external interference. Additionally, coaxial cables are relatively inexpensive and easy to install, making them a popular choice for power transmission in various applications.

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