# AC current direction

I'm designing a system to inject an AC current into an AC bus. I'm trying to get my head around something. To get current injected INTO the bus, would
the current have to be in phase or 180degrees out of phase with the voltage? At first thought it would be to have it in phase but if you think about this: Take an AC voltage and apply to a resistor. The current is in phase with the voltage and current flows from voltage to resistor so it would make sense that to inject current 180degrees out of phase, the current will go into the bus. Is this correct?
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Can you explain a little further on what you mean by "inject an AC current into an AC bus"? What are you trying to accomplish?
The more current and voltage are "out of phase" the worse your power factor (PF) becomes.
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What i mean by injecting an AC current into an AC bus is this. If you connect a load to the bus, that load takes AC current from the grid. I'm trying to put more AC current onto the bus.

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Warren Thai wrote:

What it sound like you are trying to do is to parallel a generator with an existing bus. Try researching that - there are loads of stuff on the internet about it.
--
Sue

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Way back at the dawn of the computer age there was a FORTRAN IV based program called ECAP. This "modeled' an electronic circuit by breaking down all electronic components into inductive and capacitive reactance's plus resistance. It also provided for current and voltage sources.
If you use this or one of its more modern cousins you can plug in different values and have it calculate the results.
If you try a "lab" experiment with real AC power you are likely to hurt someone real bad.
I'd recommend your attention to http://free-ed.net/sweethaven/ModElec/acee/default.asp for a start in AC theory.
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Tim Perry wrote:

If you are talking about a 'SPICE' program, there are plenty of free programs out there.
http://www.linear.com/software/ is a popular program.
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
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And it probably doesn't use punch cards either <g>
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Tim Perry wrote:

How many days would it take to load a program that big? :)
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
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On Fri, 12 Oct 2007 06:21:14 -0400, "Michael A. Terrell"

A typical "txt" card (the format for loading programs) had about 60 bytes of object code on it so say 60,000 bytes a minute. (2540 or 2501 card reader)
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snipped-for-privacy@aol.com says...

Also figure that memory was rarely larger than that and processors weren't exactly swift either.
--
Keith

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krw wrote:

Don't forget the room full of low density 9 track tape drives.
The product would be obsolete before you could finish the simulation. ;-)
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
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Looking at it that way, things have actually slowed down! Products are now obsolete before manufacturing even sees them. ;-)
--
Keith

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----------------- Yes, a load will draw a current from the bus- and this current will depend on the load and the bus voltage. You don't "inject" a current. If the load changes then its current will change. Whatever current that loads draw from the bus is determined by the loads -all the sources do is If you are wanting to connect another source to the bus then, as suggested by Palindrome, look at parallel operation of generators and load sharing between generators. A bit more circuit theory might also help before you try to "design" something.
In connecting the incoming generator it is necessary to match the <voltage> magnitude and phase to that of the bus. If done correctly, there is no current. If the voltage magnitude of the incoming generator is increased, the main effect will be a shifting of reactive from the other sources on the bus (not affecting the load except for secondary effects) not always what you want. If you try to speed up the machine you will advance the phase of the voltage (by a small amount, please) and some of the total power will be provided by this machine- again without changing the load current or power. Either way will result in a current from the incoming machine. It is better to think of complex power flows than currents.
--

Don Kelly snipped-for-privacy@shawcross.ca