Andrew Mason said:
The only way this could occur is if voltage and power output decreased as load increased. That would not be very practical for a system that is supposed to provide more power if load increases.
For example, suppose you have a generator producing 100 watts of power at 100 volts and 1 amps current supplying a 100 w. lightbulb. If you added another 100 watt lightbulb in parallel, thereby doubling the load, the voltage would have to drop to 50 volts in order to keep the current a constant 1 amp (1/2 amp for each light bulb). This means the power supplied to each lightbulb would decrease to 1/4 of its previous value.
What does this statement mean?
I have to agree with Integral that you need to learn about Ohm's law before we can have a meaningful discussion here.
AM
Dude, I hate to tell you, but you are the one in need of studying. I am in the final stage of the Ph.D. in EE, & I am in my 34th year as a pro EE. You're lecturing me on what's what, so how much education do you have. BSEE, MSEE, Ph.D.?
With a CCS operating mode, the lamps are wired in series. The switch which turns the lamp on & off is placed across the lamp (in parallel). With both switches shorted, both lamps are off, since the constant current goes through the 2 switches. Opening switch A results in constant current through lamp A. Lamp B remains off since the constant current is diverted through switch B. If the lamp is a 100 ohm value, and the CCS is 1.0 amp, we get 1.0 amp always, & the voltage is 100 volts. Total power is 100 watts.
If we wish to turn on lamp B, we open switch B. Now there are 2 lamps in series, with a 1.0 amp constant current. The voltage is now 200 volts, & the power is now 200 watts. Constant current can be made to work very well, but there is a problem with it.
In the miles of transmission lines from the power plant to the home, there are insulator losses, & conductor losses. The insulator loss is V
2G, & the conductor loss is I
2R. It so happens that insulator loss is way smaller than that for the conductor. If we generated at transmitted at full current all the time, & variable voltage depending on load, loss is higher than the CVS method.
It's less lossy to generate & transmit at full voltage all the time with current depending on load. This is why we do so. In addition we get another benefit with CVS operation. Constant voltage is a result of constant turbine speed. A side benefit is constant frequency. Since the CVS bus is also CF (constant frequency), synchronous motors can be used for applications where fixed speed is needed. Clocks are a great example.
In addition, it pays to use transformers to raise V & lower I reducing I
2R loss further. In doing so we increase V
2G loss, but that loss is so tiny in comparison, it's worth it. All of these issues have been thoroughly examined for over a century. Many people do not have a need to know this, but if you have never dealt with power, sources, energy conversion, etc., how can you tell an old pro that they got it wrong?
As far as my needing to learn Ohm's law goes, all Ohm states is that the relation between V & I is R. The V/I ratio equals R. That doesn't explain the pros & cons of CVS vs. CCS operation. What I explained above requires more than Ohm's law gives us. I don't mind people asking for clarification, but I advise all reading these posts to be careful before telling someone they are wrong.
A person w/ limited technical education should think twice before rebuking someone. Learning this stuff takes many years of very intense study. Very few know this topic as well as they would like to believe. Nothing personal.
Claude