# Homework Help: Basic circuit law problem

1. Nov 16, 2016

### Reverend Shabazz

1. The problem statement, all variables and given/known data
This problem is Problem 2.24 from Fundamentals of Electric Circuits (5th edition).

The problem is as follows:

2. Relevant equations
KVL, KCL

3. The attempt at a solution
In the official solution, they assume that the total current in the lefthand loop is $I_o$, whereas the total current in the righthand loop is $αI_o$. So they say the current flowing through $R_4$ is $αI_o$. However, if that is true, then at the node that I circled in red wouldn't KCL be violated: $I_o + αI_o \neq 0$ (unless $α$=-1)?

#### Attached Files:

File size:
7.4 KB
Views:
72
2. Nov 16, 2016

### Staff: Mentor

Can you draw a complete circuit (closed path) for a current passing through that connection?

3. Nov 16, 2016

### Reverend Shabazz

Hmm, I see your point. I guess my only question would be: is it possible that current coming IN though that connection builds up at the current controlled current source (say, via a capacitor or some sort) and therefore wouldn't need a return path out of the righthand loop? Admittedly, I might be getting too much into the weeds for a chapter 2 problem..

4. Nov 16, 2016

### Staff: Mentor

No current will flow through that connection other than perhaps brief transients due to the movement of stray static electric charges from external encounters, so don't worry about charge build-up.

Charge could not build up anyways since any such build up on one side or the other would create a potential gradient forcing the charge back where it came from. The conductive path ensures that both sides of the circuit remain at the same potential.

5. Nov 16, 2016

### CWatters

That node in red has 4 wires connecting to it. Your equation only has two terms so two are missing.

6. Nov 16, 2016

### Reverend Shabazz

I would agree with that. Perhaps, then, my contention rises from the assumption that there is no potential gradient between one side vs the other. On the left side, the potential difference between the voltage source and the common ground (the red circled node) is $V_s$. On the right side, if we assume $α<0$, the current flows clockwise, and the potential difference between the current source and the ground is, by Ohm's law, $(αI_o)⋅(\frac {R_3⋅R_4} {R_3+R_4})$. So it would seem only when these two are equal will there be no charge flowing across the "bridge" from one loop to the other.

7. Nov 16, 2016

### Reverend Shabazz

That's true, technically. However, my contention with this problem is that it seems to assume that no current ever flows between the left hand loop and right hand loop and so they can be treated independently. For instance, maybe some of the current $I_o$ from the left side loop flows over the red circled bridge and into the right side loop and remains there. It would seem that gneill's point is well taken -- namely, that there won't be any current transfer if the two sides are at the same potential -- but I'm curious how we can know that.

8. Nov 16, 2016

### Staff: Mentor

No (sustained) current will never flow through a path unless there is also a return path. There must be a circuit -- literally a closed path. There is no circuit in this instance. The potential on either side of the connection is the same simply because the wire joins the two sides. Nothing that you hang onto either side can change that unless it provides another path between the two sides and connects at a location with a different potential on at least one of the sides.

The currents in each side of the indicated connection will be totally independent and self-contained.

You could remove the connection entirely and the circuit behavior would be unchanged, although it might cause some consternation for the technician who would have to build completely isolated "floating" power supplies for each side (the circles and diamonds on a circuit diagram conceal a lot of the practicalities of building real sources for circuits ).

9. Nov 16, 2016

### CWatters

It can't. gneil has given you the best reasons why not but...

Try applying KCL to the voltage source Vs. Io must flow in both terminals. eg out of the +ve and into the -ve terminal. That means Io must flow into and out of the node in the red circle. Similar applies on the right hand side.

So the correct KCL equation for the red node is..

+Io + (-Io) + αIo + (-αIo) = 0

10. Nov 17, 2016

### Staff: Mentor

That doesn't look right, not with $R_3=R_4$.

(Though it may be a reasonable approximation were $R_3≫R_4$.)

11. Nov 17, 2016

### Reverend Shabazz

Hah, I could imagine! I'm actually interested to learn more about this stuff. I work in a completely unrelated field, but I want to increase my knowledge of how circuits are developed, so I'm trying to learn these things on my own.

Ok, I agree with you in that the two sides are independent insofar as the potential is the same on both sides of that connecting wire. I guess when doing problems like this that is an assumption that one has to make -- namely that the circuit is already in an equilibrium state. So in this case, for instance, it seems to me that one could juice up one side of the connecting wire with electrons such that it is at a higher potential relative to the other side and, hence, create a flow across the connecting wire. But, agreeably, that isn't part of the problem, so there shouldn't be any flow across the connecting wire.