A circuit that has 3 power sources connected in series, where

V1 = 0.2A at 20V

V2 = 0.05A at 5V

V3 = 0.2A at 20V



simulate this circuit – Schematic created using CircuitLab

V1 & V2 are in opposition to V3, problem here is that V2 has higher current. I could not predict the outcome correctly(probably catastrophic). However, considering that V3 is connected in series and opposing V1&V2, maybe it could cancel out with V1, leaving V2's current flow of 3A at 5V at point "A"?

  • \$\begingroup\$ You can't have a current without resistance. Where is the resistance? \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 9 '14 at 3:03
  • \$\begingroup\$ Added to the circuit. \$\endgroup\$ – Pupil Nov 9 '14 at 3:20

You seem to have some confusion regarding the voltage and current ratings regarding a voltage source. An ideal voltage source will have a constant voltage across it and will source or sink any amount of current to keep that voltage constant. In real life voltage sources cannot source infinite current, thus they have a current rating that represents the maximum current it can source before something bad happens. What will happen when you exceed the current capabilities of the voltage source depends on the device, it can blow a fuse, shut down, drop the output voltage or become damaged.

Regarding the circuit you have posted assuming point A is ground the voltage on the other side of V2 will be 5v, past V1 will be 25v and on the other side of V3 will be back down to 5v. Thus the voltage across R1 is 5V and that makes the current (via Ohm's law) 100ohm/5V = 50mA. If we go back and look at the current ratings of the voltage source, V1 and V3 are capable of 200mA and V2 can handle 50mA which means they can handle the 50mA current calculated above.

This is assuming that the current ratings on the voltage source you mention is for sourcing and sinking current, as in the case of V3 it is sinking 50mA of current, which may not actually work in real life depending on the specifications of the voltage source (many real life voltage sources don't like sinking current). If V3 is only capable of sourcing current and not of sinking current then the voltage source is being used outside of its specifications and all sorts of bad things may occur. It may have some sort of internal protection and thus not allow any current through, in which case there will be 25V across it. If it doesn't have any protection it may become damaged and act as a short and thus have 0V across it.

Also running V2 at the maximum current rating is ill advised as any small variation in the circuit can cause the current to exceed the rating of the voltage source.

  • \$\begingroup\$ But the current values are not equal? I mean, in series I(1) = I(2) = I(3), which in this case it's not. Let assume a case where V2 had 5A instead of that which is in the circuit? \$\endgroup\$ – Pupil Nov 9 '14 at 7:39
  • \$\begingroup\$ The focus is more with the idea of having multiple sources in series with different current ratings, what might be the result? Considering also, having a third source that is in opposition(V3). \$\endgroup\$ – Pupil Nov 9 '14 at 7:42
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    \$\begingroup\$ @Key As mentioned in the first paragraph the current values of the voltage sources are maximums, the voltages sources will supply up to that much current, it's perfectly fine doing less. Thus all three sources being in series will have 50mA going through them which is within the current ratings of them. Regarding V3 being opposite I updated the paragraph on that, basically voltages sources in real life don't always like sinking current, and unless it is rated for it it could cause damage to the device, but a ideal voltage source will handle it just fine. \$\endgroup\$ – Gorloth Nov 9 '14 at 21:21
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    \$\begingroup\$ It doesn't matter which voltage source has the lowest limit, when you calculate the current going through the voltage sources you need to make sure that it's within the limit of each source. Having two voltage sources that 'cancel out' can simplify the circuit analysis but it doesn't change the real circuit and it's limits. It's similar to when simplifying equations, x^2/x simplifies down to just x but there is still the restrictions that x cannot be zero from the original equation. \$\endgroup\$ – Gorloth Nov 9 '14 at 22:16
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    \$\begingroup\$ @Key A voltage source can (and should) have a current limit higher than what is required by the circuit. If you have multiple sources in series the effective current limit is the voltage source with the lowest rating. So in the example you gave with V1 rated at 100A and V2 rated at only 3A if they were in series then it could only handle up to 3A. \$\endgroup\$ – Gorloth Nov 9 '14 at 23:33

Your circuit violates the definition of a constant voltage source. V1 and V2 add up to 25 volts connected in parallel with V3, a 20 volt source. You can't simultaneously have 2 different voltage sources in parallel. For a real source with internal resistance, the circuit can make sense because the resistances can absorb the voltage difference. Also, note that the current rating of the source does not mean it will always source that amount of current. The current drain from any voltage source depends on its load. If you load V1 with a 1k resistor, it will source 20V/1k or 20 ma, not 1 amp. You would have to load it with a 20 ohm resistor for it to source 1A. This also applies to V2 and V3.

  • \$\begingroup\$ Adjusted that mistake, its in series now no parallel circuit here. That should allow V3 to cancel out with that 25V. \$\endgroup\$ – Pupil Nov 9 '14 at 3:25
  • \$\begingroup\$ I also adjusted the current based on the resistor in that circuit. I'm wondering as to what would happen where V3 in opposing V1&V2 also consider V2 having a lower current than V1. Also that this is a series circuit where Itot = I(V1)=I(V2)=I(V3). But as you can see V2 is not the same. \$\endgroup\$ – Pupil Nov 9 '14 at 3:31

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