# Circuit Theory Question (KCL/KVL/Dependent Source)

1. Jun 16, 2013

### RoKr93

1. The problem statement, all variables and given/known data
Here's the problem:

Find the equivalent conductance Geq and then the equivalent resistance Req "seen" by the current source Is in the circuit in terms of the literals R1, R2, and gm.

Figure:

2. Relevant equations

Ohm's Law: V = IR

KVL: V1 + V2 + ... Vn = 0 for closed loops

KCL: I1 + I2 + ... In = 0 going in an out of a node

Voltage Division: V1 = Vsource * (R1/(R1+R2))

3. The attempt at a solution

I wasn't really sure how to go about this one. I know that Req = Vs/Is, so I set about trying to find a way to get that in terms of the given values, but to no avail- I don't know whether to treat this as a series circuit or parallel based on the independent source (since if it's parallel one of the paths is just the dependent current source with no resistors).

I tried using voltage division to solve for Vx, but I don't really know what good that will do me, or if I even can do that in this situation.

I'd really appreciate some direction here. Thanks.

2. Jun 16, 2013

### Staff: Mentor

Consider writing KCL (nodal analysis) for the top node. The node potential will be Vs...

3. Jun 16, 2013

### RoKr93

So correct me if I'm wrong, but if I do KCL on the top node, I get Is - Vs/R1 + gmVx = 0. I'm not sure how that helps...if I solve for Is and substitute, that just throws a Vx into my equation for Req.

4. Jun 16, 2013

### Staff: Mentor

Your equation is not correct; You have to take into account R2 for the current in the middle branch. That current in the middle branch flows through R1 and R2, so what's an expression for the potential across R2 (Vx)?

Last edited: Jun 16, 2013
5. Jun 16, 2013

### RoKr93

Okay...I tried going with Is - Vs/R1 - Vx/R2 + gmVx = 0 and doing KVL on the left loop to get Vx = Vs - IsR1, then plugging all that into the Req formula. I was left with only the proper variables, which is good, but my answer (after putting in given values for the numbers) was way off, so clearly I did something wrong. I'm still not certain about my handling of KCL for the middle branch; am I missing something again there? I don't think I want to combine the two resistors because I have a defined voltage drop across one of them...

6. Jun 16, 2013

### Staff: Mentor

The current in the middle branch passes through two resistors. They are in series. You can't avoid that. Use the total resistance of the branch to write its current.

Once you've written an expression for the current in the branch you can use that expression to determine an expression to replace Vx (use Ohm's law).

7. Jun 16, 2013

### RoKr93

I got it! Thank you very much for your help- I definitely would have kept on futilely separating R1 and R2 without it, heh. I don't know why I got it into my head that I couldn't add them up...I'll definitely have to keep that in mind for future problems.

Thanks again.

8. Jun 16, 2013

### Staff: Mentor

You're welcome