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

[RoKr93]

Homework Statement

Here's the problem:

Find the equivalent conductance G_eq and then the equivalent resistance R_eq "seen" by the current source I_s in the circuit in terms of the literals R₁, R₂, and g_m.

Figure:

Homework Equations

Ohm's Law: V = IR

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

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

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

The Attempt at a Solution

I wasn't really sure how to go about this one. I know that R_eq = V_s/I_s, 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.

 

[gneill]

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

 

[RoKr93]

So correct me if I'm wrong, but if I do KCL on the top node, I get I_s - V_s/R₁ + g_mV_x = 0. I'm not sure how that helps...if I solve for I_s and substitute, that just throws a V_x into my equation for R_eq.

 

[gneill]

So correct me if I'm wrong, but if I do KCL on the top node, I get I_s - V_s/R₁ + g_mV_x = 0. I'm not sure how that helps...if I solve for I_s and substitute, that just throws a V_x into my equation for R_eq.

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 (V_x)?

 

[RoKr93]

Okay...I tried going with I_s - V_s/R₁ - V_x/R₂ + g_mV_x = 0 and doing KVL on the left loop to get V_x = V_s - I_sR₁, then plugging all that into the R_eq 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...

 

[gneill]

Okay...I tried going with I_s - V_s/R₁ - V_x/R₂ + g_mV_x = 0 and doing KVL on the left loop to get V_x = V_s - I_sR₁, then plugging all that into the R_eq 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...

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).

 

[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.

 

[gneill]

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.

You're welcome