# Finding the current through a circuit

1. Jan 31, 2012

### eurekameh

http://imageshack.us/photo/my-images/9/94290584.png/

I'm trying to find the current through this circuit.
I tried doing Kirchhoff's current law, but that would lead me to more unknown currents than equations.
Kirchhoff's voltage law also gives me more unknowns than equations.
Simplifying the circuit also doesn't seem to work. There are no parallel or series resistances.
How can I determine the current?

2. Jan 31, 2012

### Staff: Mentor

Here's your circuit drawn large enough for the human eye to perceive

Is it the current designated by the red arrow that you're trying to find?

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3. Jan 31, 2012

### eurekameh

Thanks. :D
Yes, it is.

4. Jan 31, 2012

### Staff: Mentor

Okay, the first thing you should ask yourself about that current is, "Where is the current going?", and second, "Is there a circuit (closed loop)?"

5. Jan 31, 2012

### eurekameh

The current is going through the right-most resistor.
And yes, this is a circuit, if that's what you're asking?

6. Jan 31, 2012

### Staff: Mentor

But to WHERE? Where can that current go?
Not quite. Is the rightmost resistor part of a circuit (a closed loop wherein current can flow from some source, through the circuit, and back again to the source)?

7. Jan 31, 2012

### eurekameh

The current can flow to other resistors that may not be shown.
Ah.
I'm guessing the current must be 9 A, since the right-most resistor is not part of the circuit and hence, all of the 9 A through the circuit (closed loop) must be leaving the circuit.

8. Jan 31, 2012

### Staff: Mentor

Resistors that are not shown don't exist unless the problem states explicitly that there is something beyond the end terminal, in which case they would also have to give you the numerical value of the current I.

No, all the current from the current source can ONLY flow back to the current source... There is no other closed path available. Current sources must reclaim exactly the same amount of current that they produce -- what flows out of the top of the current source must be exactly and identically balanced by what flows back into its bottom.

9. Jan 31, 2012

### eurekameh

If all the current flows back into its current source, doesn't that mean that I = 0 A?

10. Jan 31, 2012

### Staff: Mentor

Yes, yes it does

Open connections never carry current (at least, not any steady-state current).

11. Feb 1, 2012

### eurekameh

Does that mean that no current ever touches that wire? - Because my intuitive understanding of it is that it does.

12. Feb 1, 2012

### Staff: Mentor

For current to flow through a component there must be a potential difference across that component to drive the current, and there must be a path for the current to follow to return to the source of the potential/current. A "hanging branch" with no exit path has no exit for the current, and no source to provide a potential across it. So no current can flow.

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