Originally posted by Rakarskiy
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"How is the force of current in a closed circuit formed?"
Obviously the potential difference.
I took time and read your article thoroughly.
In your Wise Eye article EMF, Current, Voltage, Resistance, you say "Ohm's law doesn't say anything about voltage drop, only about voltage or applied voltage." This statement of yours is false. Ohm's Law always applies, including the potential difference across a circuit element, such as "voltage drop."
There are also some other issues with your article on which I'll elaborate.
With your circuit example you say "Provided that the bulbs have a resistance of 15 ohms and the first voltmeter will show 4.5V, which corresponds to the battery voltage, the other two voltmeters will show 2.25V. The current will have a value of 0.15A." That would be 15 ohms for each of the 2 bulbs. So each bulb has a 2.25V drop and 0.15A per Ohm's Law. And now the load resistance for the source (4.5V) is 2R or 30 ohms, so Ohm's Law yields 0.15A, same as the calculation for voltage drop.
You seem to imply an issue here. What is it? Then you abruptly jump to a 220VAC supply and then attempt to insert 0.02 ohm 'r' for a generator resistance. Such a low r value is a correct order of magnitude for a low voltage (4.5 or so) battery but not for a 220V generator, as you run into.
Then you proceed to confuse the (R+r) terminology. This leads to several problems like your "current according to the calculation was 0.425 A, which is (4.545A / 0.425=) 10.6 times less than the required value for the operation of our load (TEN 1 kW)." And "the EMF of the generator phase is equal to 462.7 Volts". Both are erroneous.
Also you miss the fact that the DC voltage resulting from rectified filtered AC is not the same value as the RMS. It is the peak, or 0.707 factor.
Another issue is that you add resistance from two circuits (220VAC vs. ~325VDC) for some calculations, (R + r).
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