Power dissipated by a resistor on a coaxial cable

In summary, the conversation is discussing a problem involving the calculation of R, which represents the resistance of a resistor connected between two cylinders. The question asks for a demonstration that, for any value of R, the power dissipated in the resistor is equal to the rate at which electromagnetic field energy is propagating along the cable. The Poynting vector from part (b) is used to find this rate. The conversation also mentions the need to satisfy V=IR for steady-state conditions and the potential for adjusting the values of lambda and/or I to achieve this. There is some confusion about the problem and its origins.
  • #1
gausswell
Homework Statement
Find the power dissipated by the resistor.
Relevant Equations
P=IV, P=V^2/R
I need help with part c.
8a975be3d5d22daa109677a5638ef173.png

My solution:
0eadcf5e76b91bbe5c2e9d8b32705d81.png

Is there an other way to do this other than dimensional analysis?
P.S "dr an infinitesimal radius", it ofcourse should be dz.
 
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  • #2
If possible, please type your work rather than post images of your work. It makes it easier for us to quote specific parts of your work.

I'm not following your calculation of R. In part (c), R represents the resistance of a resistor that is connected between the two cylinders. R can have an arbitrary value. The question asks you to show that for any value of R, the power dissipated in the resistor equals the rate at which electromagnetic field energy is propagating along the cable in the space between the two cylinders (as found using the Poynting vector from part (b)).

EDIT: For steady-state conditions with the resistor in place, we need to satisfy ##V = IR##. ##V## is determined by ##\lambda## (and ##a## and ##b##). So, if the values of ##\lambda## and ##I## are specified, then ##R## would need to have the value determined by ##V = IR##. Or, if ##R## is chosen arbitrarily, then ##I## and/or ##\lambda## would need to be adjusted so that ##V = IR##.
 
Last edited:
  • #3
This is a very bad problem (or at least badly stated). May I inquire from the OP the exact origin?
What is the answer, please, for the Poynting vector in part b?
 

FAQ: Power dissipated by a resistor on a coaxial cable

What is the formula for power dissipation in a resistor on a coaxial cable?

The power dissipation in a resistor on a coaxial cable can be calculated using the formula \( P = I^2 R \), where \( P \) is the power dissipated, \( I \) is the current flowing through the resistor, and \( R \) is the resistance of the resistor.

How does the frequency of the signal affect power dissipation in a coaxial cable?

The frequency of the signal affects power dissipation primarily through the skin effect and dielectric losses. At higher frequencies, the skin effect causes the current to be distributed unevenly, increasing the effective resistance and thus the power dissipation. Dielectric losses also increase with frequency, leading to higher power dissipation.

What role does the characteristic impedance of the coaxial cable play in power dissipation?

The characteristic impedance of the coaxial cable affects how efficiently power is transmitted along the cable. Mismatches between the characteristic impedance of the cable and the load (resistor) can lead to reflections, standing waves, and additional power dissipation. Properly matching the impedance minimizes these effects and reduces unnecessary power loss.

Can power dissipation in a coaxial cable be minimized, and if so, how?

Yes, power dissipation in a coaxial cable can be minimized by ensuring proper impedance matching, using cables with low-loss materials, and operating at frequencies that minimize skin effect and dielectric losses. Additionally, keeping the cable length as short as possible and using high-quality connectors can help reduce power dissipation.

How is power dissipation measured in a coaxial cable setup?

Power dissipation in a coaxial cable setup can be measured using a combination of techniques. One common method is to measure the current through and the voltage across the resistor using an oscilloscope or multimeter, then calculate the power using \( P = I^2 R \) or \( P = V^2 / R \). Another method is to use a network analyzer to measure the insertion loss of the cable and calculate the power dissipated based on the known input power.

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