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oliverlines1234567
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Ideal current source is modeled as current source with infinite resistance in parallel. This way, no current flows through the parallel resistance and all the current flows through the load. You are asked to find the Thevenin's resistance between A and B. Hence, you need to replace the current source by it's internal impedance, which is infinity in this case. Hence, while calculating RAB, you just open the current source.oliverlines1234567 said:Ah, so an ideal current source has infinite internal resistance, and hence its just normal series and parallel resistors that are left? But I don't understand, how does current go through something with an infinite resistance?
It doesn't. That's why the circuit you posted is so easily solved. Just ignore the current source.oliverlines1234567 said:Ah, so an ideal current source has infinite internal resistance, and hence its just normal series and parallel resistors that are left? But I don't understand, how does current go through something with an infinite resistance?
Well, an ideal voltage source in parallel with an ideal current source is an invalid combination in network analysis. Indeed the current source is designed to provide a constant current. If any load is connected to a current source and all the current passes through the load, then there is no "leakage" current and hence, internal (parallel) resistance of an ideal CS is infinite.oliverlines1234567 said:Am I correct if I say, no extra current (say if I connected a battery across A and B) can go through the current source because the current source is designed so that it maintains a constant current.
Regarding a voltage source in parallel with a current source, actually it is quite valid, if I may disagree. If the constant voltage source (CVS) is 1.0 volts, and the constant current source (CCS) is 1.0 amps, they can be connected together in parallel. One will transfer power to the other depending on polarity. If the CVS has the plus terminal on top, CCS has arrow upward, then the CCS charges the CVS, with 1.0 watts transferring from CCS to CVS. Reverse the polarity of just 1 or the other, not both, and the CVS transfers 1.0 watt of power into the CCS. Otherwise I agree with the rest of your post, thanks.cnh1995 said:Well, an ideal voltage source in parallel with an ideal current source is an invalid combination in network analysis. Indeed the current source is designed to provide a constant current. If any load is connected to a current source and all the current passes through the load, then there is no "leakage" current and hence, internal (parallel) resistance of an ideal CS is infinite.
This is a diagram showing source transformation with 'practical' sources.
View attachment 94890
Practical current source with parallel resistance is equivalent to practical voltage source with a series(intetnal) resistance. Note that both the resistances are same. In CS, it appears in parallel(current division takes place) while is VS, it is in series(voltage division takes place). If you use load resistance as 2Ω, you can see, in VS diagram, voltage across load is 5V. It is same as in the CS diagram, where current through the load is 2.5A.
Right! I was just about to edit! Actually, two ideal voltage sources in parallel or two ideal current sources in series are the invalid cases. Thanks for the correction!cabraham said:Regarding a voltage source in parallel with a current source, actually it is quite valid, if I may disagree. If the constant voltage source (CVS) is 1.0 volts, and the constant current source (CCS) is 1.0 amps, they can be connected together in parallel. One will transfer power to the other depending on polarity. If the CVS has the plus terminal on top, CCS has arrow upward, then the CCS charges the CVS, with 1.0 watts transferring from CCS to CVS. Reverse the polarity of just 1 or the other, not both, and the CVS transfers 1.0 watt of power into the CCS. Otherwise I agree with the rest of your post, thanks.
Claude
Okay, yeah that makes sense. So was what I said last time about the excess current can't enter the current source (because the current needs to stay constant), hence it has effectively infinite resistance a good/correct way of thinking about it? Because that's the conclusion I have kind of come up with from what you've said. Because then that would explain why the total resistance of the current source as seen by the other components is infinite, when actually only the internal parallel resistance of the current source is infinite and the part where the current actually travels through has zero resistance.cnh1995 said:Well, an ideal voltage source in parallel with an ideal current source is an invalid combination in network analysis. Indeed the current source is designed to provide a constant current. If any load is connected to a current source and all the current passes through the load, then there is no "leakage" current and hence, internal (parallel) resistance of an ideal CS is infinite.
This is a diagram showing source transformation with 'practical' sources.
View attachment 94890
Practical current source with parallel resistance is equivalent to practical voltage source with a series(intetnal) resistance. Note that both the resistances are same. In CS, it appears in parallel(current division takes place) while is VS, it is in series(voltage division takes place). If you use load resistance as 2Ω, you can see, in VS diagram, voltage across load is 5V. It is same as in the CS diagram, where current through the load is 2.5A.
Edit: First line corrected by Cabraham. It is valid actually.
Sounds right. If the current source is not ideal, the excess current will flow through it's internal resistance. Hence, the voltage source connected across it will "see" the internal resistance instead of infinite resistance.oliverlines1234567 said:So was what I said last time about the excess current can't enter the current source (because the current needs to stay constant), hence it has effectively infinite resistance a good/correct way of thinking about it?
Great. Thanks a lot :)cnh1995 said:Sounds right. If the current source is not ideal, the excess current will flow through it's internal resistance. Hence, the voltage source connected across it will "see" the internal resistance instead of infinite resistance.
Actually, this would be a battery charger. A real world practical example of such a network.cnh1995 said:Well, an ideal voltage source in parallel with an ideal current source is an invalid combination in network analysis.
cnh1995 said:Right! I was just about to edit! Actually, two ideal voltage sources in parallel or two ideal current sources in series are the invalid cases. Thanks for the correction!
That's what I wrote.Averagesupernova said:If they were in series I would say that is invalid.
I really need to see my optometrist. LOLcnh1995 said:That's what I wrote.
To calculate the equivalent resistance between two points, you need to use Ohm's law, which states that resistance is equal to voltage divided by current. You will also need to determine the resistances of all components in the circuit between the two points and use the appropriate formula to find the total resistance.
The main factors that affect the equivalent resistance between two points are the number and type of components in the circuit, the arrangement of those components, and the material and dimensions of the conductors used. Temperature and the presence of any resistors in the circuit can also impact the equivalent resistance.
When resistors are added in series, their resistances simply add up to give the total equivalent resistance. For resistors in parallel, the total resistance is calculated using the formula 1/Rtotal = 1/R1 + 1/R2 + 1/R3... etc. As more resistors are added in parallel, the total resistance decreases.
Yes, the equivalent resistance between two points can be lower than the individual resistances if the resistors are connected in parallel. In this case, the total resistance decreases as more resistors are added in parallel.
Equivalent resistance refers to the overall resistance in a circuit between two points, which takes into account the effects of all components in the circuit. Total resistance, on the other hand, only considers the resistances of the individual components between the two points and does not take into account other factors such as temperature or the arrangement of the components.