Timeline for Current sources and voltage sources connection
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 20, 2022 at 17:07 | comment | added | Eugene Sh. | And no, bad and good are not "subjective" here. A good voltage source will provide a constant voltage within the rated tolerance when connected the full range of rated loads. That is it's internal resistance must be few orders of magnitude lower than any of the possible loads. The dual of this holds for a current source. If we are operating outside of the rated conditions, we can't call it "bad" because the conditions are "bad". | |
| Jun 20, 2022 at 16:57 | comment | added | Eugene Sh. | But you have already got your answer. You can connect "bad" current sources in series, as you can connect "bad" voltage sources in parallel. But if these are not that "bad" (that is close to ideal) you will do more harm than good by that. | |
| Jun 20, 2022 at 10:49 | comment | added | terahertz | The point is, a good current source is a bad voltage source, and vice versa. But, "good" and "bad" are subjective, and my original question still remains the same. Given that any voltage source can be transformed into a current source, and vice versa, how do we connect them, in series or in parallel? | |
| Jun 20, 2022 at 10:40 | comment | added | terahertz | Sh I get the drift of your argument. But there is a flaw in it. First consider a 10 A (real) current source with a 1000 ohm parallel (shunt) resistance. If we connect a 10 ohm load at the output terminals, approximately 9.9 amperes will flow through it, with 0.1 amperes going through the source's shunt resistance. However, if we connect a 1000 ohm load, the current will divide equally with 5 amperes in the shunt resistance and 5 amperes in the load. | |
| Jun 18, 2022 at 16:39 | comment | added | Eugene Sh. | I told you exactly how they are distinguishable. The real current source will have a very high internal resistance, so it will affect the output current very little. The real voltage source will have internal resistance that is very low in order to affect the voltage on the load very little. Try experimenting with a battery and few different resistors, and try convincing me that it is a current source. | |
| Jun 18, 2022 at 4:21 | comment | added | terahertz | Also, if you connect two different loads to the same (real) current source, the currents will be different in the two cases. | |
| Jun 18, 2022 at 4:18 | comment | added | terahertz | Sh I don't think the sources are distinguishable. The whole point of source transformation is that sources can be transformed (for ease of analysis) without the load impedance "knowing" about it. Furthermore, voltage and current sources are just mathematical devices that help us in our analysis. Remember, the whole of circuit theory is an approximation to field theory. | |
| Jun 17, 2022 at 14:59 | comment | added | Eugene Sh. | To clarify further, it is indeed possible to consider voltage source as a current source and vice versa, according to Thevenin/Norton transformations. But if you look at these, you will see that in order a voltage source to be closer to ideal one, it's internal resistance should be low, and for the current source the opposite is true. This is the basic difference between them. | |
| Jun 17, 2022 at 13:30 | comment | added | Eugene Sh. | So if you want to connect them in certain topology - you know their type and connect accordingly. It wouldn't make sense to connect current sources in series as it wouldn't make sense to connect voltage sources in parallel (in case of sufficiently different voltages), because in both cases it will cause "useless" currents fed by one source into the other and a waste of energy, or even damaging the devices. | |
| Jun 17, 2022 at 13:30 | comment | added | Eugene Sh. | @terahertz You must know what these sources are. They are distinguishable. If you connect two different loads (one at a time) to a current source in the black box, you will see the same current \$I\$ flowing through each (up to internal resistance correction, unless it is built to compensate it), and the voltage on the load will be different according to the Ohm's law (\$V=I\cdot R\$). In case there is a voltage source in the box, you will see the same voltage \$V\$ on the different loads, but different currents (again, according to Ohm's law). | |
| Jun 17, 2022 at 9:56 | comment | added | terahertz | Sh Thanks a lot for your detailed response. The question remains, if we have two energy sources (real, not ideal) encased inside black boxes with only their terminals accessible to us, should we connect them in series or in parallel? | |
| Jun 16, 2022 at 14:35 | history | answered | Eugene Sh. | CC BY-SA 4.0 |