# inductance losses in wiring

Discussion in 'General Electronics Chat' started by justrob, Apr 25, 2009.

1. ### justrob Thread Starter New Member

Apr 25, 2009
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Hi All,

I am trying to understand if there is a material power loss associated with the inductance & capacitance in a wire that is carrying A/C electricity?

I know I can calculate the losses in the wire due to resistance using P = sq(I)xR but is there a material loss from inductance/capacitance in the wire?

Ultimately I am trying to understand if there will be a (material) difference in losses between between a length of cable carrying say 220V DC or 220V AC?

Thanks in advance for any assistance!

Rob.

2. ### mik3 Senior Member

Feb 4, 2008
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What do you mean by material power loss?

3. ### Wendy Moderator

Mar 24, 2008
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Usually, 50 and 60Hz is such a low frequency that what you are talking about isn't a problem, unless you're talking about a 100 miles (2500 Km) of wire, and even then, it will be low.

What isn't low, however, is resistive losses. They are significant, almost always, and it gets worse with more current. As copper (or any metal) ages, it gets worse, as the metal crystalizes and develops even more internal resistance, sometimes in large amounts. My Dad and I had to replace wires going to our water pump in the country because the line basically ate all our voltage.

How familar are you with Ohm's Law?

The whole point AC exists is it can be boosted with a transformer, and dropped the same way, reducing the losses associated with resistance.

4. ### b.shahvir Active Member

Jan 6, 2009
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I just need to understand one thing! Capacitance and inductance are reactive (wattless) components. Then how can the contribute to power losses in transmission lines?
Thanx

5. ### mik3 Senior Member

Feb 4, 2008
4,846
65
Capacitance and inductance don't cause power losses themselves. The power losses come from the resistance of the wire when current flows to charge this capacitance or inductance. This is for power loss in the wire.

If you consider the load, it will receive less voltage due to voltage drop across the capacitance or inductance in the wires.

6. ### b.shahvir Active Member

Jan 6, 2009
444
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But capacitance and inductance drops a portion of supply voltage across itself resulting in lower voltage (and hence power) on the receiving end. Doesn't it account for power loss?

7. ### mik3 Senior Member

Feb 4, 2008
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I would say it reduces the efficiency of the system.

8. ### b.shahvir Active Member

Jan 6, 2009
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I was looking forward to a more so technically specific answer!

9. ### justrob Thread Starter New Member

Apr 25, 2009
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mik3: When I say material I mean is there a power loss incurred that could affect the practical efficiency of the system

Bill Marsden: Yes I am familiar with the concepts you were asking about

To explain in more detail what I was trying to understand:

Is there any difference in actual wiring losses between an AC system and a DC system when running at the same levels of voltage/current due to the inductance/capacitance induced when using AC.

If I give an example of a 10 m wire with 1ohm resistance per metre. Thus, total resistance 10 ohms. If you run 1 amp (@ 220V say) through this wire to a load your power loss in the wire would be P = sq(1) x 10 = 10 Watts.

If I run this in a DC circuit my wiring loss is 10 Watts.

If I run this in an AC circuit will there be additional loss due to the inductance/capacitance generated on the wire? Also, just to confuse things some more is it this characteristic of AC that causes 'skinning' or is that another loss that needs to be considered for AC (which from my understanding would reduce the effective cross-section and hence increase the resistance of the wire)?

Lastly when a circuit is Open would an AC circuit continuously experience losses that an equivalent DC circuit wouldn't?

Thanks for your help all, apologies if I'm not that clear.

10. ### mik3 Senior Member

Feb 4, 2008
4,846
65
In an AC circuit more power will be lost in the wires due to the current needed to charge the capacitance and inductance if the wires.

Skin effect is due to eddy currents created in the wire by the changing magnetic field around the wire.

If an AC circuit is open some current will flow due to the capacitance between the wires (or components on the circuit). This current will cause power loss in the wires due to their finite resistance.

11. ### justrob Thread Starter New Member

Apr 25, 2009
3
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Mik3: Many thanks for your response.

Are there any formulae for calculating the additional AC losses in the wire over and above the resistance losses? Am I correct in saying that there are no additional losses in the equivalent DC circuit?

Thanks again!

12. ### mik3 Senior Member

Feb 4, 2008
4,846
65
Losses are caused only by resistance. When you talk about reactive 'losses' is the efficiency (power factor) of the system that is affected.

13. ### AlexR Well-Known Member

Jan 16, 2008
735
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As mik3 says the losses are resistive or to be exact I$^{2}$R losses but the reactive characteristics of the transmission line cause an increase in current which in turn greatly increases the power losses.

Take a look at http://en.wikipedia.org/wiki/Electric_power_transmission for a good run down on both AC and DC power transmission losses.

14. ### b.shahvir Active Member

Jan 6, 2009
444
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Let me focus on the self-inductance of the transmission line which is lumped in series with the load for analysis. Does this series inductance contribute to power loss? If so, in what way....as the series inductance is a reactive (wattless) component and power loss is due to active (wattful) components of a transmission line! plz elaborate on this concept.
Thanx

15. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
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No. It does contribute to a reduced power factor.

16. ### b.shahvir Active Member

Jan 6, 2009
444
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Agreed! but doesn't a reduced p.f. partially reduce the active (wattful) power transfer thru the line and hence indirectly contribute to power loss?

17. ### b.shahvir Active Member

Jan 6, 2009
444
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The following are my thoughts on the topic;

In my opinion, power loss in transmission line means when the receiving end voltage Vr is less than the sending end voltage Vs. This happens as Vs is dropped across line components resulting in reduced voltage Vr. Reduction in Vr accounts for power loss.

The power loss is due to series line resistance, skin effect, series line loop inductance, shunt leakage capacitances (in long lines) shunt inductances (load side), power loss due to radiation and induction, etc.

Presently my topic of interest is power loss due to series line inductance which causes reduced Vr and hence in my opinion is equally responsible for power loss akin to series line resistance and skin effect.

Suitable guidance for the same will be greatly appreciated. Thanx.

18. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Power loss is unusable power dissipated as heat. Reduced power available to the load as a result of poor power factor is not a power loss.

19. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,053
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Reactive currents by definition aren't lossy, but they cause a phase shift that reduces the power factor. Unlike "true" losses, reactive currents can be compensated by reciprocal reactances in the circuit.

However, in many cases, reactances can cause excessive currents to flow in some parts of a circuit, in which case the resistance (I^2 R) losses are increased. The loss is still caused by resistive heating, but it is directly affected by reactance.

Hope this helps.
eric

20. ### b.shahvir Active Member

Jan 6, 2009
444
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Thanx, Actually, this is what i'm trying to convey, that series inductance reduces the p.f. indirectly causing the KVar (reactive component) to increase but decreasing the transfered KW (active component) in the process! Reduction in transfered KW is, in my opinion, also an indirect power loss due to series inductance drop.

Also, in any case, the size of switchgear and power equipment will have to be increased to compensate for voltage drop at Vr....whatever may be the reason for the loss of voltage or power at the receiving end.