i = v/r = 5/3 ANow, fully understanding this, what is the current i?
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...And, if you get that right, you might ask yourself, what effect does the current source have in this problem?
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.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?
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.
Ok this I'm not grasping totally..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.
"A voltage source by itself...without any resistors or anything..." wouldn't be "...in a loop...".Does a voltage source by itself in a loop without any resistors or anything (just a wire connected to + & -) create a current?
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.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
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.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
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.2)
I understand what you explained about current sources... voltages do not coexist with currents unless there is some resistance.
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.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?
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.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.
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