What would happen in this circuit? inductor/reactor series

[Electric-Light]

Another thing that plagues me is that when a 1.5kVA capacitive load is placed at the end of an extension cord, there's definitely I^2R loss going on in the extension cord. The voltage measured at input minus 2*voltage drop on neutral does not equal the load side voltage.

The load side voltage is higher than expected.

What gives.... ?

 

[steve66]

I haven't done a breadboard in a long time. If you have a sine wave signal gen. you can use small currents and scale up the result.

There's a hifi set in my basement that gets ratty sound in one channel after a half hour and I still haven't assembled the function gen. chip so I can input a sine into this channel and trace it through.
And there's a car radio/cassette player I have that has waited since 2003 to be installed.


And tonight I will maybe go out with long underwear into 0C weather onto the driveway and finish that head gasket job.
:thumbsup:

You know Pspice is available for free on the Web?

Something you said earlier made me think you had set up a circuit and measured the current through the 1 ohm resistor. But looking back, I see you meant you calculated the current. (2 amps, same result I get )

I'm with you - it would be fun to set this up and verify. But it will have to wait until the end of a long list of things to do.

P.S. I think you can forget about the radio/cassette player. Even the CD's seem to be close to obsolete.

Now that all the music and videos are stored on the "cloud", I loose sleep at night worrying about how the recording industry is going to make money when they can't change the media format every few years.

 

[Electric-Light]

Now that all the music and videos are stored on the "cloud", I loose sleep at night worrying about how the recording industry is going to make money when they can't change the media format every few years.

The big battle for them is digital rights management. These days, they're getting better about locking down the song you buy to one specific device.

 

[Haji]

I do not know what the typical percent power loss in the circuit described by the OP in the single phase as well as three phase multiwire circuit with no added resistance in the neutral/third phase..

 

[gar]

121128-0630 EST

Haji:

Suppose the branch circuit is 100 ft long from the main panel to the point of interest, and the wire is #12 copper for each of the three wires.

Each 100 ft length has a resistance of 0.159 ohms at 20 deg C. Each hot wire has a power loss of 1²*0.159 = 0.159 W, and the neutral is 2²*0.159 = 0.636 W.

.

 

[Haji]

gar:
Thanks for your reply.
What will then be power factor on the primary side of single / three phase hv transformer feeding the said loads?

 

[gar]

121128-0751 EST

Haji:

The input power factor to the transformer primary will be very low because the amount of dissipated power is small compared to the input VA.

If the input primary was 120 V, then the input current is approximately 2 A, or 240 VA. Assuming no losses in the transformer, then the dissipated power is 2*0.156 + 4*0.156 = 6*0.156 = 0.936 W. Approximate power factor at transformer primary is PF = 0.936/240 = 0.004 .

.

 

[Electric-Light]

gar, any idea about what I asked in post #21?

http://forums.mikeholt.com/showthread.php?t=149256&p=1442804#post1442804

 

[steve66]

Another thing that plagues me is that when a 1.5kVA capacitive load is placed at the end of an extension cord, there's definitely I^2R loss going on in the extension cord. The voltage measured at input minus 2*voltage drop on neutral does not equal the load side voltage.

The load side voltage is higher than expected.

What gives.... ?

The voltages are not in phase. In fact, the voltage drop across the extension cord will be almost 90 degrees out of phase with the voltage source and load voltage.

Say you have a 120V source, and 10V dropped across the extension cord. You should still have approx.:

sqrt (120^2 - 10^2) = 119.6 volts at the load.

 

[gar]

121128-0935 EDT

I do not have an answer for your post #21. The likely cause is series resonance, but also might be treated as parallel resonance. Other measurements would be needed, and careful experimental procedures.

When measuring small voltage drops along a conductor in an AC circuit you must be concerned with the measurement wires (test leads) forming a loop in which an error voltage is magnetically induced. This is true of transients in DC circuits as well.

.

 

[G._S._Ohm]

Its a little suprising, but yes, it would actually short out as soon as you remove the neutral wire.

The neutal wire keeps the inductor and capacitor from trading energy back and forth each cycle (which is what normally happens in a resonant circuit.)

Try this:
with Va =Vb = 120, and Xla = j120 and Xcb= -j120 and Rn = the neutral resistance, the loop current
Ia = (2Rn-j120)/120
and
Ib = (120^2 + [2Rn^2]-[Rnj120])/{(Rn-j120)120}

It works for Rn = 0 and I believe it also works for Rn = infinite.
To do the middle values I'll need to write polar/rect and rect/polar routines for Excel.

 

[Electric-Light]

test test.. did this thread get corrupted? I see 4 pages, but I can't see page 4

so there's some technical issue. i coudn't access page 4, which has post #31. then I added this test post and now I can see page 4... what the heck?