Is the Capacitance of a Twin Lead TV Antenna Cable Within Specifications?

In summary, the conversation discusses the use of a TV antenna cable with specific specifications for capacitance and length. The cable has two 0.50mm-diameter wires spaced 12mm apart, and the total capacitance should not exceed 1000 pF. The potential difference between the conductors is given by V = lambda / (pi*epsilon_0) * ln((b-a)/a). Using the general definition for capacitance, C=Q/V, it is concluded that the cable does not meet the specifications as its calculated capacitance is greater than 1000 pF. Other attempts to calculate the capacitance were also discussed but deemed incorrect.
  • #1
tnutty
326
1

Homework Statement



Your home video studio will use a TV antenna cable often known as "twin lead." Two 0.50mm--diameter wires are spaced 12mm apart. Your equipment requires that the total capacitance of the cable not exceed 1000 pF. The cable is 30m long. A technical paper on the cable gives the potential difference between the two conductors as

V = lamda / (pi*epsilon_0 ) * ln ( (b-a) / a )

where a is the radius of the wire and b is the separation.

Is your cable within the specifications? ____ (my answer is NO), see below

Attempt :

C = epsilon_0 * A / D

A = 2piRL
A = 2 * pi * ( 0.5/2 * 10^-3 ) * (30 m)
D = 12mm * 10 ^ -3

C = epsilon_0 * 2 * pi * (0.25 * 10^-3) * (30) / (12 * 10^-3)

= 3.47 * 10 ^ -11

which is greater than 1000pF = 10^-9, thus it should be NO, I think. Is this correct?
 
Last edited:
Physics news on Phys.org
  • #2
Any one have any Ideas ,

I also tried this :

C = Q/V

V is given , and Q is lamda * L, but I don't think its right
 
  • #3
Use the general definition for capacitance, C=Q/V.
 
  • #4
I did that as well.

C = Q/V

V is given, and I use Q = q*Lamda

But I think its not correct though.
 

FAQ: Is the Capacitance of a Twin Lead TV Antenna Cable Within Specifications?

What is capacitance?

Capacitance is a measure of an object's ability to hold an electrical charge. It is represented by the symbol C and is measured in units of farads (F).

How do you calculate capacitance?

The capacitance of an object can be calculated by dividing the charge stored on the object (Q) by the potential difference across the object (V). This can be written as C = Q/V.

What factors affect capacitance?

The capacitance of an object is affected by the surface area of the object's plates, the distance between the plates, and the type of material between the plates. In general, larger surface area and smaller distance between plates results in a higher capacitance.

Why is capacitance important?

Capacitance is important because it allows for the storage and release of electrical energy. This is useful in many applications such as storing energy in batteries, smoothing out electrical signals, and filtering out unwanted frequencies in a circuit.

How is capacitance checked or measured?

Capacitance can be checked or measured using a multimeter or a capacitance meter. These devices use different methods to measure capacitance, but both provide an accurate reading of the capacitance of an object.

Similar threads

Capacitance Calculation for Dielectric-Filled Coaxial Cable
Replies
32
Views
5K
RadioInstigator: Building a SIGINT
Replies
5
Views
3K
Optimizing TV Antennas with Yagis: A DIY Guide for Better Reception
Replies
20
Views
7K

Hot Threads

Recent Insights

Back
Top