Coaxial cable-physics problem

Question

Please help with this problem.

Problem:

Power P is transmitted over a distance D on a coaxial cable. The radius of the cable's inner conductor with negligible resistance is Ri, and the radius of the thin-walled, similarly resistance-free outer tube is R0. There is a vacuum both outside the cable and between the inner wire and the outer tube, direct current flows through the cable.

(the current flows down the surface of the outer cylinder, radius R0, and back along the inner cylinder, radius Ri)

Question:

At which end of the coaxial cable - right or left - is the generator and the consumer located?

What is the transmission line's current if the outer tube's wall is not stretched either inward or outward?

I appreciate any help you can provide.

Answer 1

I will answer as best possible your question(s), however several statements in your problem statement are troublesome to me.

First of all, coaxial cable is rarely used for direct current, i.e. DC. Coax is used for RF (radio frequency) or microwave signals. A coaxial cable has a Z_o characteristic impedance that is determined by the dimensions of the inner wire running down the center of the cable and the diameter of the outer shield, you call a tube.

Any cable is a reciprocal device, in other words, it does not matter what end of the cable you insert the signal, in this case the RF, and where you connect the load, the receiver of the power being conveyed through the cable.

The formula for the characteristic impedance of a coaxial cable is ( and I wish I could use a true equation editor here)

Z_o = 138/sqrt(e_r) x log₁₀(D/d) where e_r is the relative dielectric constant

D is the diameter of the outer shield

d is the diameter of the wire going down the center

Using your definition of terms, Z_o = 138/sqrt(e_r) log₁₀(R_o/R_i).

Because you specify there is no resistance (nor any other deleterious effects such as capacitance or inductance), the P_out is the same as P_in. Also, the current is directly impacted by the applied voltage which in the problem statement is not defined.

All RF sources have an internal source impedance which is inescapable, in other words, it is always there. Most RF sources are 50 Ohms which is assumed here. Maximum power is transferred from a source of Z Ohms to a load of Z Ohms. Therefore, in this example, the source impedance is 50 ohms and I am assuming your load impedance is also 50 ohms so that your load receives the maximum amount of power.

E_r = 1.0 since the dielectric of the cable is air.

Using the formula above for cable Z_o, the current that flows in the cable is V/(R_Source + R_Load) = V/100. At DC Z_o does not enter in whatsoever. In real world applications R_Load is likely different than the source 50 Ohms or load and with that occurs transmission losses. The transmission coefficient for current is precisely

t_i = 2 Z_o/(Z_o+Z_Load)

Note: a perfectly lossless coaxial cable terminated "exactly" in the characteristic impedance of the cable, has no impedance mismatch loss, no standing waves, and behaves the same, theoretically, whether it is 1 inch in length or 100 feet in length.

This answer has bounced around somewhat because the actual problem statement is more of a DC current question than RF. More information is needed to work this problem in the light that it was likely intended.