How Much Power Can a Coaxial Cable Transmit Before Breakdown?

[SiberianS;pth]

Homework Statement

A coaxial transmission line consists of an inner cylindrical conductor of radius 1mm and a cylindrical outer conductor chosen to make the characteristic impedance 75 ohms. The space between the conductors is filled with a gas that can stand a maximum field of 10^5 Vm^-1 without dielectric breakdown. Estimate the maximum mean radio-frequency power that can be transmitted along this line into a matching load.

Homework Equations

The impedance of a co axial cable is Z=Zo*ln(b/a)/n*2*pi where b is outer radius, a inner radius and n is sqrt(permittivity).

The Attempt at a Solution

So for the empty coaxial cable I had Z=Zo*ln(b/a)/2*pi

I rearranged this to work out the value of the outer radius which I got as 3.49mm.

I used the distance between inner and outer radius and the max electric field to work out potential difference. I then tried to the use the standard P=Vi formula to work out power but my answer is about 8 times too big. I assumed all the power is transmitted due to impedance matching.

 

[berkeman]

Homework Statement

A coaxial transmission line consists of an inner cylindrical conductor of radius 1mm and a cylindrical outer conductor chosen to make the characteristic impedance 75 ohms. The space between the conductors is filled with a gas that can stand a maximum field of 10^5 Vm^-1 without dielectric breakdown. Estimate the maximum mean radio-frequency power that can be transmitted along this line into a matching load.

Homework Equations

The impedance of a co axial cable is Z=Zo*ln(b/a)/n*2*pi where b is outer radius, a inner radius and n is sqrt(permittivity).

The Attempt at a Solution

So for the empty coaxial cable I had Z=Zo*ln(b/a)/2*pi

I rearranged this to work out the value of the outer radius which I got as 3.49mm.

I used the distance between inner and outer radius and the max electric field to work out potential difference. I then tried to the use the standard P=Vi formula to work out power but my answer is about 8 times too big. I assumed all the power is transmitted due to impedance matching.

Welcome to the PF.

The breakdown voltage is based on the peak voltage, while the power is based on the RMS voltage. Still, that should only give you a difference of √2, not 8. Can you post your work so we can check the numbers?