# Power loss in cables

1. Apr 16, 2010

### xunxine

I found in a book the equation for the power loss caused by the heating effect of current in the transmission lines as below:
Ploss = Iloss2R = (Pout/V)2R

A friend of mine commented that the last bit "(Pout/V)2R" is unnecessary, confusing and wrong (but won't tell me why).
Is my friend correct? Is there a way to calculate power loss using voltage?
Someone pls help explain.

Thanks!

2. Apr 16, 2010

### Redbelly98

Staff Emeritus
Ploss = V2/R​
where V is the voltage across the length of the cable (i.e. not the voltage between two cables).

3. Apr 16, 2010

### xunxine

Does this mean we can't substitute I = P/V into the equation to get:
Ploss = (Pout/V)2R,
which is wrong,
but instead use I = V/R to get:
Ploss = V2/R?

How are these different?

4. Apr 16, 2010

### Redbelly98

Staff Emeritus
What do you mean by Pout, how does it differ from Ploss?

5. Apr 16, 2010

### Antiphon

Your friend unfortuately is the one who is confusing and wrong. The equation says (the power delivered by the line / V) is the current in the line which is correct. The rest of the formula is the same as the first part. (Xunxine has it right.)

6. Apr 16, 2010

### Redbelly98

Staff Emeritus
Now that I realize (I think) what the terms mean, I agree with xunxine. Xunxine is correct, but it is also confusing. V refers to the output of the power source, not the voltage across the length of cable. R refers to the resistance of the length of cable, not the resistance connected across the power source.

7. Apr 17, 2010

### xunxine

Hi Redbelly 98 & Antiphon, thanks for your replies! Pardon me cos i think i'm confused again.

I thought Pout refers to the power at the receiving end, let's say, homes. It would be smaller than the power produced in the power station. So the amount of power lost (due to heating effect) would be given by the equation in question, ie. Ploss. Am i right to say this?

Then V refers to ...?

Ok, so which is the least confusing way to find the power loss in transmission cables? By not using the equation with V? That is, using just
Ploss = Iloss2/R ?

8. Apr 17, 2010

### Redbelly98

Staff Emeritus
Yes, I think that is least confusing, since I is the same for the cables, load, and power source (thus you can just refer to current as I without any specifying subscript). Also, it's probably clearer to use Rcable instead of just R.

9. Apr 17, 2010

### Bob S

The ac power loss (at MHz frequencies) in coaxial cables is mostly skin-effect (eddy-current) losses in the center conductor, and is measured by attenuation (decibel loss (dB) loss) per 100 feet or meters. See table:

http://www.hamuniverse.com/coaxdata.html

The dc resistance of the center conductor, or the characteristic impedance Z of the coax has very little to do with the power loss. For a matched cable and termination, the forward power is V2/Z, where Z is the characteristic impedance, but this formula is the forward signal power, and not the power loss.

Bob S

10. Apr 18, 2010

### xunxine

Thanks Bob S, but i think the info & reference website is a bit too technical for me. i barely understood it.

I found another book that explains this equation. I scanned a portion (attached). Here it relates power loss to current in the cable, not current loss, ie.
Ploss = I2R
and not
Ploss = Iloss2R

So we can just take the values of current & resistance in the cables to get power loss? This matches Redbelly98's explanation then! And it greatly simplifies things...

But then, to get the value of current, the value of voltage of source is still needed?

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11. Apr 19, 2010

### Redbelly98

Staff Emeritus
Yes, and also the power of the source.

12. Apr 19, 2010

### Bob S

The power loss in transmission lines (coaxial cables) is usually due to skin-effect (eddy current) losses in the copper conductors, as I pointed out in post #9. The losses are NOT directly related to the DC resistance or the characteristic impedance of the transmission line. The losses are usually quoted in dB or decibels, which are in turn based on the base-10 logarithm of the output to input voltage or power ratio. See

http://en.wikipedia.org/wiki/Decibel

For power loss, dB = 10Log10(Pout/Pin)

or Pout=10(dB/10)Pin

For many coax cables, the attenuation (which is frequency dependent) is listed in

http://www.hamuniverse.com/coaxdata.html

Specifically for RG-59, the attenuation for 100 MHz is listed as 3.4 dB per 100 feet (equal to 0.034 dB per foot). Note that the attenuation due to skin-effect losses scales as the square root of frequency. So for x=300 feet of RG-59 coax at 100 MHz, the power loss can be calculated as

Pout/Pin= 10(-0.034·x/10) = 10(-10.2/10) = 10(-1.02) = 0.095

or Pout = 0.095·Pin

This assumes that the coax cable is terminated in its characteristic impedance, e.g., 73 ohms for RG-59. For comparison, the impedance of free space is about 377 ohms.

Bob S

13. Apr 19, 2010

### ruko

How is characteristic impedance of a transmission line determined? Disregarding shielding and interference possibilities which line is better for carrying a TV signal, 73 ohm coax or 300 ohm twin lead?

14. Apr 19, 2010

### Bob S

For an air-filled coaxial cable (without dielectric), the characteristic impedance Z is simply related to the impedance of free space:

Z = (377/2π)Ln(D/d) ohms

where D is the (inner) diameter of the outer conductor, d is the diameter of the inner conductor, and Ln is the natural logarithm.

Z=300-ohm twin line closely matches the impedance of a folded (magnetic) half wave antenna, and is relatively lossless because of the high voltage to current ratio.

In general, the characteristic impedance of a transmission line is

Z = sqrt(L/C)

where L and C are the inductance and capacitance per unit length.

Bob S

15. Apr 19, 2010

### Redbelly98

Staff Emeritus
True. I was treating this as an introductory-level physics question, it seemed from Post #1 that was an appropriate level for this thread. Possibly xunxine is more versed in electronics than I originally thought.

16. Apr 20, 2010

### xunxine

Thanks all for the explanations and extra info, which is actually beyond my level. Maybe I'll understand them better one day. Original post is for GCE O Level. We do cover eddy current but not in detail.