Posted by Home Guy on April 15, 2011, 2:47 pm
"trader4@optonline.net" wrote:
> Take a look at the dual voltage source circuit diagram that Jim
> Wilkins supplied a couple posts back. It's example #1.
>
> http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
I did, you dumbass #2.
Go and read my last post.
> It's a simple diagram of two ideal voltage sources with
> series resistors connected to a load.
And note what happens when the voltage sources have unequal voltages.
And note that we are not talking about batteries here in the case of a
municipal power grid and a PV system.
> And I'll bet if you do the equations with the voltage sources
> at the same value, you'll find that twice as much current
> flows from the voltage source with the 10 ohm resistor as
> the one with the 20 ohm resistor.
If that diagram shows reverse current flow because Battery 1 has a lower
voltage than Battery 2 (and current I1 is negative), then at what point
does current I1 become zero? What would the voltage of battery 1 have
to be for current I1 to be zero?
Posted by Jim Wilkins on April 15, 2011, 4:48 pm
> ...
> And note what happens when the voltage sources have unequal voltages.
> And note that we are not talking about batteries here in the case of a
> municipal power grid and a PV system.
> ...
Considering the level of this discussion when the solutions are simple
algebra, you aren't ready to handle the analytical geometry and
differential equations of AC circuit analysis.
http://www.allaboutcircuits.com/vol_2/index.html
http://www.circuit-magic.com/laws.htm
jsw
Posted by Jim Wilkins on April 15, 2011, 5:11 pm
> > ...
> > And note what happens when the voltage sources have unequal voltages.
> > And note that we are not talking about batteries here in the case of a
> > municipal power grid and a PV system.
> > ...
> Considering the level of this discussion when the solutions are simple
> algebra, you aren't ready to handle the analytical geometry and
> differential equations of AC circuit analysis.http://www.allaboutcircuits.com/vol_2/index.htmlhttp://www.circuit-magic.com/laws.htm
> jsw
This is a good example:
http://www.allaboutcircuits.com/vol_2/chpt_5/4.html
"Now that we've seen how series and parallel AC circuit analysis is
not fundamentally different than DC circuit analysis,..."
Electrical engineers use j instead of i for the square root of -1
because i has long been the standard for Current (intensity). In this
instance the imaginary number is an excellent tool to analyze real-
world physical phenomena.
http://www.microwaves101.com/encyclopedia/Smithchart.cfm
jsw
Posted by trader4@optonline.net on April 15, 2011, 6:12 pm
> "trad...@optonline.net" wrote:
> > Take a look at the dual voltage source circuit diagram that Jim
> > Wilkins supplied a couple posts back. It's example #1.
> >http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
> I did, you dumbass #2.
> Go and read my last post.
I have read your posts and like most other people here
have concluded you are wrong, so I don't see why
you're calling ME the dumbass.
> > It's a simple diagram of two ideal voltage sources with
> > series resistors connected to a load.
> And note what happens when the voltage sources have unequal voltages.
Uh, huh. I have noted that and if you have any questions about how
that works, I'll be happy to answer them. I'm still waiting for your
answer about the case with EQUAL voltages. That was the essence
of your argument, was it not? That a power source can't provide
power in parallel with another unless it raised the voltage? So,
leave everything else the same and just make the voltage sources
equal. Tell us what current flows through the load resistor and what
currents flow through each voltage source. This is EE course circuit
theory course 101, about the first week.
BTW, you have an engineering degree?
> And note that we are not talking about batteries here in the case of a
> municipal power grid and a PV system.
Yes and since you seem incapable of understanding a simple
circuit that represents 2 batteries connected in parallel to a
load, no need to add the additional complexities.
> > And I'll bet if you do the equations with the voltage sources
> > at the same value, you'll find that twice as much current
> > flows from the voltage source with the 10 ohm resistor as
> > the one with the 20 ohm resistor.
> If that diagram shows reverse current flow because Battery 1 has a lower
> voltage than Battery 2 (and current I1 is negative), then at what point
> does current I1 become zero? What would the voltage of battery 1 have
> to be for current I1 to be zero?
I see you do have a question.
Simple. With no current flow through battery 1, then the circuit
is reduced to an ideal voltage source connected to two resistors
in series. One of these represents the internal resistance of
battery 2 and is 20 ohms. The other is the load resistor of 40
ohms. So, we have 20 volts across 60 ohms, giving a
current of .333Amps. That .333amps produces a voltage of
13.34 across the load resistor. (.333 A X 40ohms.) That means
voltage source V1 would have to be at the same potential,
13.34 volts and when it is, no current flows through what
represents battery 1.
I've answered your question, now answer mine:
What are currents I1, I2, I3 when the voltage sources V1 and V2
are both 20 volts.
Posted by krw@att.bizzzzzzzzzzzz on April 15, 2011, 11:33 pm
>On Apr 15, 5:46 am, "k...@att.bizzzzzzzzzzzz"
>> >"k...@att.bizzzzzzzzzzzz" wrote:
>>
>> >> > We can forget about generators and distributions systems.
>> >> > Just take two 12V batteries and connect them in parallel
>>
>> >There's your problem right there.
>>
>> >It's usually not a good idea to connect batteries together in parallel,
>> >unless they are exactly of the same type, age, condition, etc. If you
>> >get a weak cell in one of the batteries it will turn into a load.
>>
>> >And what happens when they are not exactly at the same voltage before
>> >being connected together? How do you insure that you always get current
>> >flowing out of both of them?
>>
>> >> > But clearly he thinks if we put a second AC power source on
>> >> > a distributions system, it has to be at a higher voltage to
>> >> > "push" current out.
>>
>> >The IEEE paper I posted earlier today shows exactly that - that PV
>> >systems raise local grid voltage and the utility company must compensate
>> >by reducing primary supply voltage to down-regulate the secondary
>> >voltage coming from the distribution transformer.
>>
>> What a dumbass!
>>
>> >> > So, what happens with the two batteries? Under the
>> >> > laws of physics the rest of us use the voltage would
>> >> > remain at 12 volts and BOTH batteries would be supplying
>> >> > part of the 1 AMP flowing through the resistors.
>>
>> >Take a 12 V car battery and wire it up in parallel with 8 AAA batteries
>> >connected in series. Then connect a load and tell me how much current
>> >the AAA batteries will supply vs their what their potential current
>> >supply could be if they were connected to their own isolated load.
>>
>> They will share in proportion to their capacity. Electricity, water, nor shit
>> flow uphill. ...though you have been pumping enough of the latter here.
>>
>> >> Not on Homeguy's and Vaughn's planet. One of the batteries
>> >> will be charging the other.
>>
>> >If they are unequal in capacity, then yes that will eventually happen.
>>
>> Wrong!- Hide quoted text -
>>
>> - Show quoted text -
>He said unequal voltage. In which case he's right.
>Capacity is not voltage.
If he really means unequal voltage then he's proving my point. He's a dumbass
(second point proven).
> Wilkins supplied a couple posts back. It's example #1.
>
> http://www.electronics-tutorials.ws/dccircuits/dcp_4.html