Mastering Thevenin/Norton Problem Analysis: Tips and Tricks | Homework Help

[RoKr93]

Homework Statement

Homework Equations

V = IR
I₁ + I₂ ... + I_n = 0 for a node (KCL)
V₁ + V₂ ... + V_n = 0 for a loop (KVL)

The Attempt at a Solution

I've really been struggling with this problem and Thevenin/Norton problems in general. I just can't seem to perform the proper circuit analysis. I know that I can set I_A = 0 and V_AB = VV_oc in order to find Thevenin values, but I just keep getting stuck. I tried to do KCL on the top node, which got me -kv_x - (v_x/50Ω) + ((v_s - v_x)/200Ω), but that appears to be getting me nowhere. I've been trying to find a relation between v_x and v_AB so I can substitute in, but to no avail. I'd really appreciate a nudge in the right direction here as well as general advice, as I'm pretty lost.

 

[gneill]

Are you familiar with nodal analysis? If so, use it to find the potential at the top of the controlled current source when the output is open-circuited. That'll be your Thevenin voltage. Show your work so we can see how you're doing.

 

[RoKr93]

-v_x/50Ω + v_AB/200Ω -kv_x = 0

v_AB/200Ω = v_x/50Ω + kv_x

v_AB = 4*v_x + 4*kv_x

additionally,

v_AB - v_s + v_x = 0

v_x = v_s - v_AB

so

v_AB = 4*(v_s - v_AB) + 4*k*(v_s - v_AB)

v_AB = 4*v_s - 4*v_AB + 4*0.025*v_s - 4*0.025*v_AB

5.1*v_AB = 4.1*v_s

v_AB = 16.078... V

I don't think that's right. Things aren't adding up when I plug it back in. Where'd I go wrong?

 

[gneill]

Okay, for starters don't treat V_x as "known" value for your node equation, except where it is referenced on controlled sources. V_x is a variable that you'll find an expression for based upon fixed circuit parameters. Write your node equation for the 50 Ω branch as though V_x wasn't labeled.

What's the expression for the current in the 50 Ω branch supposing that the node voltage is V_AB? Given that current flowing through the 50Ω resistor, what's an expression to replace V_x?

 

[RoKr93]

Isn't that what I did? Can't v_x be written as -(v_AB - v_s) or v_s - v_AB if we're treating it as the voltage difference between the nodal voltage and v_s?

 

[gneill]

Isn't that what I did? Can't v_x be written as -(v_AB - v_s) or v_s - v_AB if we're treating it as the voltage difference between the nodal voltage and v_s?

Okay, you're right, you can do it that way. My own preference is to write the node equations ignoring the "sampled" values except for their use in the controlled sources, then later replace those instances with expressions derived from the circuit.

Looking at your node equation manipulations:

-v_x/50Ω + v_AB/200Ω -kv_x = 0

v_AB/200Ω = v_x/50Ω + kv_x

vAB = 4*v_x + 4*kv_x​

check the coefficient for the last term on the right hand side.

 

[RoKr93]

Well that was stupid of me. Figures that I thought to check my circuit before I thought to check my actual math, lol. V_AB = V_oc = 18 V.

So now I need either R_th or I_sc. If I do the same nodal analysis on the top node but this time include a -I_A, does that make my V_AB = 0? It would be a short circuit, correct?

 

[gneill]

Well that was stupid of me. Figures that I thought to check my circuit before I thought to check my actual math, lol. V_AB = V_oc = 18 V.

So now I need either R_th or I_sc. If I do the same nodal analysis on the top node but this time include a -I_A, does that make my V_AB = 0? It would be a short circuit, correct?

Your open-circuit voltage looks fine.

Yes, you can repeat the analysis with the output shorted to determine the Norton current. Just find the node voltage and divide by the 40Ω "load" to find the current in that branch.

Note that for part (b) you're going to need a version of the solution for the Thevenin voltage that leaves k as a variable...

 

[RoKr93]

Got it. Thanks so much for your help- it's much appreciated.