Ok this is a question that has been bugging me for some time.
Suppose you got two ideal voltage sources, one 15V and one 10V say, and grounded both negative terminals. Then you connected both postives terminals to the same node.
What would the voltage be at the node? Since they are both ideal voltage sources wouldn’t they be “competing” to supply their respective voltages.
Also what would happen if you connected two ideal current sources in series on the same branch of a circuit? Wouldn’t the same problem arise.
I realise i’m talking about ideal sources so if you need to include non-idealalities to make the problem solveable then go ahead.
The voltage of the connection node will be somewhere between 15V and 10V; the exact value will depend on the internal resistances of the two batteries. There will be a current flow from the 15V battery to the 10V battery equal to 5 volts divided by the combined internal resistances of both batteries. This will generate a voltage drop across the effective internal resistances of both batteries, and the voltage at the connection node will be equal to 15V minus the current times the internal resistance of the 15V battery, which should also equal 10V plus the current times the internal resistance of the 10V battery.
The voltage at the connecting node will also change over time, as the 15V battery discharges and the 10V battery gets overcharged (and damaged, most likely.)
With theoretical zero-resistance batteries, and zero-resistance wire, the question becomes impossible to answer - you have to divide by zero to get the current.
The higher voltage source will flow into the lower voltage source, the current given by I=(V1-V2)/R where V1 and V2 are the voltages, and R is the resistance of the wire between. Since R should be very low, I should be large.
If you were to put a diode in front of each voltage source, so that current didn’t flow as described above, then the lower voltage source would not get used, and everything would flow from the higher voltage source. And the voltage seen by the circuit will be the voltage of the higher source.
AndrewL’s last paragraph answered the OP. I’ll also add my two cents.
You cannot connect an ideal 15 VDC source and ideal 10 VDC source together using ideal (0 Ohm) wires. You just can’t do it. The equations break down. It can’t be done in real life either, since there’s no such thing as an ideal voltage source[sup]1[/sup].
By the same token, you cannot short out an ideal voltage source with an ideal wire. And you cannot allow an ideal current source to remain “open circuited”; something must be connect across its terminals.
[sup]1[/sup] [sub]An ideal DC voltage source maintains a constant voltage, has no source impedance, and can produce any amount of current.[/sub]
The two voltage sources will act like a single source. If you connect a 15V and a 10V backwards in series, then they will combine to become a single 5V source.
Then you short out the 5V source with some 0-ohm wire. A current will rise rapidly (rise time depending on circuit inductance,) rising towards infinity, and a huge and growing amount of energy will move from the higher voltage source and into the lower. Also, some energy will be stored in the form of an increasing magnetic field surrounding the one-turn inductor which you have created, but this energy is small when compared to the V x I energy flow between the two voltage sources.
I’m with crafter man on this one. An ideal voltage source maintains its voltage no matter what (that’s why it’s called ideal). If you have them connected to the same nodes (i.e. connected together) then your nodes have two different voltages defined across them. This is like that whole irresistable force and immovable object thing. You can have one or the other in the system, but you can’t have both. You simply cannot have differing ideal voltage sources connected together with an ideal conductor.
The same thing happens with two different ideal current sources in series. All of the current flowing out of the first source goes into the second source, which means that they have to be equal, but then you’ve got that current defined as something else entirely. It just can’t happen.
The first case is a clear violation of Kirchoff’s voltage law. The second case is a clear violation of Kirchoff’s current law. In either case you can approach the situation, with the current/voltage increasing towards infinity, but once you actually get to the destination you’re dividing by zero somewhere.
bbeaty: You get points for creativity, but keep in mind we’re assuming first-order ideal circuit components here. This means the wires have an impedance of 0 + j0 Ohms. When you connect the 15 VDC ideal voltage source to the 10 VDC ideal voltage source using ideal wires, the current will go to infinity in 0 seconds. Not only that, but it is impossible to have two different voltages between two points in space simultaneously. A mathematician would say the arrangement represents an “undefined” state.
Exactly. The method available for analyzing the problem is Kirkhoff’s Laws. Add up all the voltages around the loop and divide by the impedance to get the current. So the equation is:
I = (15 - 10)/0
Which is improper because of the 0 in the denominator.
The OP describes a situation that is not analyzable by available methods and is impossible besides.