superposition: looking at only voltage source, i = V/R = 5/3 looking at only current source, current division -> i = 4(2/5) = 8/5
Look carefully at the circuit. The voltage source is connected directly across the two resistors with (ideal) zero resistance conductors. The voltage across the two resistors cannot be anything other than 5 volts. When you have a current source in parallel with a voltage source, and with other loads also in parallel, the voltage source determines the voltage across everything it is in parallel with. The current source has no effect on the voltage across the loads. When you have a current source in series with a voltage source, and also in series with other loads, the current source determines the current through everything in the series circuit. The voltage source has no effect on the current through the various loads. Now, fully understanding this, what is the current i? And, if you get that right, you might ask yourself, what effect does the current source have in this problem?
i = v/r = 5/3 A No effect, the voltage at the node above the 3 ohm resistor is 5 volts... the voltage in parallel with the current source dominates the circuit... That was a good explanation and definitely cleared up some confusion thanks But to make sure though: When you have a current source in parallel with a voltage source, and with other loads also in parallel, the voltage source determines the voltage across everything it is in parallel with. The current source has no effect on the voltage across the loads. Can the current source have any effect on the current in the circuit? My guess according to this last example would be no... I now see that the voltage source dominates the voltage across everything it is in parallel with, but I don't quite yet see how that eliminates the current that would be produces by the current source... the voltage across the current source would be the same as the voltage source, but I guess that fact alone means the current from that current source isn't considered? When you have a current source in series with a voltage source, and also in series with other loads, the current source determines the current through everything in the series circuit. The voltage source has no effect on the current through the various loads. Can the current source have any effect on the voltage in the circuit? Now: http://img23.imageshack.us/img23/6836/234lbc.jpg answer: http://img23.imageshack.us/img23/9191/1234yqd.jpg if you'd like to throw a problem you make up my way to prove my understanding of what you just said be my guest
There's only one possible answer. Imagine the current out of the voltage source if the current source were absent. Now add the current source. Its current goes backwards, so to speak, into the voltage source, subtracting 4 amps from its output current. The current out of the voltage source without the current source present is 5/2+5/3 amps. With the current source present, the current out of the voltage source will be 5/2+5/3-4 amps, and the currents in the resistors will be unchanged.
The addition of the 6 Ω resistor in series with the current source doesn't change what happens to the rest of the circuit. The 4 amp current still only changes the current out of the 5 volt source. It has no effect on the fact that there is still exactly 5 volts across the resistors since they are in parallel with the 5 volt source. In the last part of your solution where you replaced the 5 volt source with a short, all of the 4 amps will go through the short, and none will go through the resistors. This is because a short has a resistance of zero ohms so that when you calculate the voltage across the resistors in that last part, it will be zero volts. Therefore, there will be no current through the resistors as a result of the 4 amp current source.
Ok honestly I've never had a formal introduction to any of this, I still have a hard time picturing what 'voltage' even is... I just think of it as a potential difference at two different points... I just picture current as charge flowing, but how voltage and current relate to each other in terms of sources and then when resistors come into play seems to be a nuisance of a problem that is hindering my understanding; so before I try to analyze your last reply can you answer these few questions to clear up the confusion I seem to be having: Does a voltage source by itself in a loop without any resistors or anything (just a wire connected to + & -) create a current? Does a current in the above scenario create a voltage? How does placing a resistor in the circuit affect it besides decreasing the current due to I = V/R, i.e. voltage etc
Ok this I'm not grasping totally.. Without just the current source Total current in the circuit -> i = 5/3 + 5/2 = 25/6 A With the current source Total current in the circuit -> i = 5/3 + 5/2 - 4 = 1/6 A So I don't see how the currents in the resistors would be unchanged once the current source is in
"A voltage source by itself...without any resistors or anything..." wouldn't be "...in a loop...". No, there has to be a load of some kind, such as a resistor, before there will be a current. In the real world, there will always be some leakage due to imperfect insulation, perhaps as small as picoamperes, but for your academic problems there is no current out of a voltage source without somewhere for it to go. [/QUOTE] Putting a resistor in series with a current source doesn't decrease the current. The current out of a current source is always just what the value of the current source is. The voltage just rises to what it has to be to make that much current pass through whatever is connected to it. If it is shorted, then the rated current passes through the short, with no voltage across the short, because a short is zero ohms, zero ohms * (however many amps) = zero volts. If a resistance is connected to a current source, then the voltage out of the current source adjusts itself to make just that much current pass through the resistor. If the value of the resistance is high, then the voltage across the resistor will be high; V = I*R A box with two terminals containing an ideal current source which didn't have anything connected to it would product a spark, an electric arc, which would pass between the terminals, carrying whatever the value of the current was. If the terminals of the box were many feet apart, the voltage would rise to millions of volts, if that's what it took to make that much current pass in the air between the terminals, because the effective resistance of the air would be many millions of ohms (once the arc begins, the effective resistance decreases, but it's still high). This property makes good current sources dangerous, but real world current sources usually have a maximum voltage they can produce.
Where you have said "Total current in the circuit", it's not clear what you mean by that. You have to be more precise in your description of where the "total current" is. There are several loops in this circuit, and just saying "total current in the circuit" insufficiently specifies what you're talking about. The total current in the voltage source changes when you add the current source, but not the current in the resistors. It's as though the 4 amp current only goes around in the loop consisting of the voltage source and the current source. Imagine that you remove the resistors. Now you have a single loop, consisting of the voltage source and the current source. The 4 amps will flow around in the loop without the resistors. Adding the resistors will take some more current from the voltage source, but the 4 amps from the current source will still just be part of the current in the voltage source. It won't pass into the resistors. (The point of view I describe in the previous paragraph isn't the only way to imagine that currents would partition themselves, but it gives the correct result.) (Also, we all should remember that it's charge that flows, not current, but it's common usage to speak of current flow. )
Ok your latest explanation definitely cleared up some confusion But just to make sure we're on the right page: 1) http://img27.imageshack.us/img27/3711/5576u.jpg In an ideal case, i.e. my academic purposes, there is no current in that 'loop'/circuit, right? 2) I understand what you explained about current sources... voltages do not coexist with currents unless there is some resistance.
Unfortunately, that isn't right. Since the resistance of a short is zero ohms, the current is V/R = V/zero = ∞. So the current would be infinite (in the academic context). In the real world, the current could be very high, often high enough to cause damage; that's why we have fuses and circuit breakers. You've got this exactly right. That's just what 2-terminal impedances such as resistors, capacitors and inductors do; they establish a relationship between the voltage across their terminals and the current through them.
1) Maybe I should think of a way to generalize it for the time being (the next few days) just so I am sure to get the right answer in reasoning that: The reason why all of the charge from the current source flows only into the voltage source and not into the resistors is because there is no resistance through the voltage source so I can think of it as a short and all of the charge will flow through it? In other words I don't know the reason why all the charge from the current source goes only into the voltage source, so that's my (probably incorrect) reasoning 2) Ok so I was misusing Ohm's I = V/R relationship for a current source because anything in series has the same current.. so putting a resistor in series does not reduce the current... HOWEVER, for a given voltage, the current varies with varying resistance, so I = V/R applies in this case.
In a sense, that's what happens. It's because the additional current through the voltage source doesn't change its voltage. This property is usually described by saying that the internal resistance of an ideal voltage source is zero ohms. It's sort of like a short that has a voltage across it (that's a little weird). The main thing is that the voltage of a voltage source doesn't change no matter what current flows through it. Real world voltage sources have an internal resistance that is greater than zero, but is still typically very small. Have you ever seen what happens when you allow a low resistance to connect the terminals of a car battery? You get a lot of sparks and melted metal. The current into a short is enough to melt wires. On the other hand, current sources have an internal impedance of ∞. Your only misuse of I = V/R was to apply it to the current source itself. It doesn't apply to ideal voltage and current sources directly. It really only applies to 2-terminal impedances such as resistors. Putting a resistor in series with a current source doesn't change the current, but the voltage across the resistor (V = IR) is added to whatever voltage is across the current source. And, remember, the voltage across the current source is determined by the total effective resistance it is looking into.
Ya I learned about internal resistance in Physics To clarify on what you talked about: Putting a resistor in series with a current source doesn't change the current, but the voltage across the resistor (V = IR) is added to whatever voltage is across the current source. And, remember, the voltage across the current source is determined by the total effective resistance it is looking into. http://img14.imageshack.us/img14/9161/5555vys.jpg The voltage between points a and b in that circuit is V = IReff = 6*9 = 54V The part I highlighted in bold I don't quite understand... the voltage across the 2 ohm resistor for example is V = IR = 6*2 = 12V, this voltage is added to whatever is across the current source? 12V + 54V? I don't think that's what you meant or I'm reading it wrong
What I mean is this: if you have a black box containing a current source, it will have some voltage across the current source due to the circuit it's connected to. If you put a resistor in the black box with the current source, the voltage across the resistor will be added to the voltage across the current source (which may change due to the addition of the resistor), and that final voltage will be what appears across the terminals of the black box. I'm not sure it's particularly helpful here. It's just that if a lone resistor is in series with a current source, the voltage across that resistor is really easy to calculate, and the voltage across the combination of that one resistor and the source, is just the sum of the voltage across the current source and the voltage across the resistor. I was just trying to explain that in an earlier example where you added a 6 Ω resistor in series with a current source, the only effect was to put another voltage drop (the drop across the 6 Ω resistor) in series with the current source. It really had no effect on the currents in the rest of the circuit.
Amazing how such a simple looking circuit can give me such a hard time Linked here for convenience: http://img154.imageshack.us/img154/5195/57507543.jpg 1) Would I be able to apply the node method to the single node at the top? Node method always works to my understanding, but I'm sure I would get the equation screwed up because the voltage source confuses me. e/2 + e/3 - 4 = 0? 2) So using superposition and all we talked about the answer is indeed i = 5/3 A?
For this problem the equation e/2 + e/3 - 4 = 0 wouldn't get you anything even if it were correct, which it isn't, because e is already known; it's 5 volts. Substitute e = 5 in the equation, and you can see that it's incorrect; it doesn't take into account the current in the 5 volt source. The sum of the resistor currents isn't 4 amps, its 5/2+5/3. The node method has problems when one of the nodes is connected to a voltage source, and likewise the loop method has problems when a loop contains a current source. You have to use the concept of supernodes and superloops. For this problem, just realize that since everything is in parallel, the voltage source determines the voltage across everything, and the current source contributes nothing to the current in the resistors. Since you then have only one source, it isn't even superposition. Superposition assumes at least two sources that have an effect; the current is without effect in this circuit (unless you care about the current in the voltage source; this was discussed earlier in the thread). So, yes, i = 5/3 amp