Find Maximum Power to Resistor R

[Gwozdzilla]

Homework Statement

Homework Equations

P_max = ((V_Th)²)/(4R_Th)

R_parallel = (R₁R₂)/(R₁ + R₂)

KCL ΣI = 0
KVL ΣV = 0

The Attempt at a Solution

Okay, so to find R_Th, I remove all independent sources and set the "terminals" at R to have a 1A current running through them. Since there is a 30kΩ resistor in parallel with the 1A current that I added, I assumed that the voltage across the 1A that I added was 30kV and that R_Th = 30kV/1A = 30kΩ.

Then I returned to the initial circuit to find V_Th. I applied nodal analysis such that v₁ = v₀ and v₂ = V_Th.

Applying KCL @ v₁:
(v₁ - 100)/10 + v₁/40 + (v₁ - v₂)/22 = 0

Applying KCL @ v₂:
(v₁ - v₂)/22 - .003v₁ + v₂/30 = 0

After simplifying and solving the system, I got v₁ = 70.1V and v₂ = 42.7V. Then I plugged v₂ into the equation for p_max and I calculated .0152W. However, the solution in the back of the textbook says that the power should be infinity. What have I missed?

 

[gneill]

The Attempt at a Solution

Okay, so to find R_Th, I remove all independent sources and set the "terminals" at R to have a 1A current running through them. Since there is a 30kΩ resistor in parallel with the 1A current that I added, I assumed that the voltage across the 1A that I added was 30kV and that R_Th = 30kV/1A = 30kΩ.

Why did you ignore the rest of the circuit? The dependent current source may well influence the Thevenin resistance (hint: It does!).

Stick on the 1A external current source and determine the voltage at the output with the rest of the circuit intact (including the 100 V source). Find your Rth from there by the ratio of I/V.

Then I returned to the initial circuit to find V_Th. I applied nodal analysis such that v₁ = v₀ and v₂ = V_Th.

Applying KCL @ v₁:
(v₁ - 100)/10 + v₁/40 + (v₁ - v₂)/22 = 0

Applying KCL @ v₂:
(v₁ - v₂)/22 - .003v₁ + v₂/30 = 0

Your first node equation looks fine, but the second seems to be mixing incoming and outgoing current terms in the sum. Check it.

After simplifying and solving the system, I got v₁ = 70.1V and v₂ = 42.7V. Then I plugged v₂ into the equation for p_max and I calculated .0152W. However, the solution in the back of the textbook says that the power should be infinity. What have I missed?

You have a couple of glitches to iron out in your derivations, but I must admit that I don't understand their claim to infinite power. Now, the Thevenin resistance may turn out to be a bit "different", but in operation with a load resistor attached the voltage and current should both be finite according to my reckoning.

Let's see how you get on after addressing the items I pointed out above.

EDIT: I just realized why they claim infinite power for the maximum delivered to the load. Let's see if you can spot the reason

 

[Gwozdzilla]

I ignored most of the circuit while calculating R_Th because that had worked a few times in the past when there was a resistor in parallel with the 1A of added current, but I guess that was just a coincidence.

I've never calculated an R_Th before while leaving an independent source in tact. Is the 100V supposed to stay in tact while calculating R_Th because of the dependent voltage source? How do I know when the independent sources should or shouldn't be removed when finding a Thevenin equivalent?

Attempting to calculate R_Th without ignoring most of the circuit:
I added the 1A of current, and then since the whole circuit is there, I just adjusted my nodal analysis from earlier, remembering that v₂ will instead equal the voltage that has a relationship with R_Th (v₂ ≠ v_Th ... v₂/1A = R_Th).

I believe I have found the error at v₂...
KCL @ v₂:
(v₂ - v₁)/22 - .003v₁ + v₂/30 - 1 = 0

KCL @ v₁:
(v₁ - 100)/10 + v₁/40 + (v₁ - v₂)/22 = 0

Simplifying and solving the system, I got that v₁ = 67.9V and v₂ = 34.5V, which would mean that R_Th = v₂/1 = 34.5kΩ...

When I remove the -1 from the equation for KCL @ v₂ and resolve the matrix, I get similar values for v₁ and v₂, which I believe would equal V_Th...I still don't see how the power would equal infinity.

 

[gneill]

I ignored most of the circuit while calculating R_Th because that had worked a few times in the past when there was a resistor in parallel with the 1A of added current, but I guess that was just a coincidence.

I've never calculated an R_Th before while leaving an independent source in tact. Is the 100V supposed to stay in tact while calculating R_Th because of the dependent voltage source? How do I know when the independent sources should or shouldn't be removed when finding a Thevenin equivalent?

Okay, that was not the best advice on my part. You should suppress the independent sources when you drive the circuit with an external stimulation. Sorry about that.

Attempting to calculate R_Th without ignoring most of the circuit:
I added the 1A of current, and then since the whole circuit is there, I just adjusted my nodal analysis from earlier, remembering that v₂ will instead equal the voltage that has a relationship with R_Th (v₂ ≠ v_Th ... v₂/1A = R_Th).

I believe I have found the error at v₂...
KCL @ v₂:
(v₂ - v₁)/22 - .003v₁ + v₂/30 - 1 = 0

KCL @ v₁:
(v₁ - 100)/10 + v₁/40 + (v₁ - v₂)/22 = 0

Simplifying and solving the system, I got that v₁ = 67.9V and v₂ = 34.5V, which would mean that R_Th = v₂/1 = 34.5kΩ...

When I remove the -1 from the equation for KCL @ v₂ and resolve the matrix, I get similar values for v₁ and v₂, which I believe would equal V_Th...I still don't see how the power would equal infinity.

You need to be a bit careful when you write the resistance values in k Ohms in your equations. You are effectively dividing through the denominators by 1000, or equivalently, multiplying the whole equation by 1000. This will affect the sizes of current supplies too! So instead of a 1 amp source stimulating your circuit it's 1000 mA. And the dependent source needs to be multiplied by 1000 as well.

I think you'll find different results when you do that.

 

[Gwozdzilla]

Okay, so if I remove the independent voltage source and send a 1mA current through the terminals such that v₂/1mA = R_Th, would these equations be correct?

KCL @ v₁:
(v₂ - v₁)/22 - 3v₁ + v₂/30 - 1 = 0

KCL @ v₂:
v₁/10 + v₁/40 + (v₁ - v₂)/22 = 0

 

[gneill]

Yes they should work.

 

[Gwozdzilla]

Okay, so when I simplified and solved the above equations, I got v₁ = 4.33E-4 and v₂ = .0016, which means .0016V/1mA = 1.6Ω, I think. Then I adjusted my nodal analysis equations by removing the 1mA and returning the 100V, and these are the equations that I got:

KCL @ v₂:
(v₂ - v₁)/22 - 3v₁ + v₂/30 = 0

KCL @ v₁:
(v₁ - 100)/10 + v₁/40 + (v₁ - v₂)/22 = 0

Simplifying and solving these, I got v₁ = 125.7V and v₂ = 251.4V.

When I square 251.4 and divide by 4R_Th, I'm going to have a very large power, but it won't be infinite. What am I still missing?

 

[gneill]

I think you need to show step by step how you solved the previous two node equations and found Rth. Something's amiss.

 

[Gwozdzilla]

For R_Th:

KCL @ v₁:
(v₂ - v₁)/22 - 3v₁ + v₂/30 - 1 = 0

KCL @ v₂:
v₁/10 + v₁/40 + (v₁ - v₂)/22 = 0

For the first equation, I used 660 as a common denominator...
(30/30)((v₂ - v₁)/22) - ((30*22)/(30*22))3v₁ + (22/22)(v₂/30) - 1((30*22)/(30*22)) = 0

30v₂ - 30v₁ - 1980v₁ + 22v₂ = 660
-2010v₁ + 52v₂ = 660 ← Final Equation 1

For the second equation, I used 880 as my common denominator...
((4*22)/(4*22))(v₁/10) + (22/22)(v₁/40) + (40/40)((v₁ - v₂)/22) = 0

88v₁ + 22v₁ + 40₁ - 40v₂ = 0

150v₁ - 40v₂ = 0 ← Final Equation 2

Then I put these final two equations into my calculator as matrices, this time getting v₁ = -.364V and v₂ = -1.363V.

To find V_Th:
First, I adjusted Final Equation 1 by removing the term caused by the 1mA current:
-2010v₁ + 52v₂ = 0

Then I adjusted Final Equation 2 by adding a term for the 100V source:
150v₁ - 40v₂ = 8800

Then I put this into my calculator as matricies to get v₁ = -6.3V and v₂ = V_Th = -243.6V

 

[gneill]

So you're pushing 1 mA into the circuit across a voltage of v2 = -1.363 V. What does that make Rth?

Your Vth looks good.

 

[Gwozdzilla]

Doesn't that make R_Th = -1.363kΩ?

 

[gneill]

Doesn't that make R_Th = -1.363kΩ?

Yes it does!

 

[Gwozdzilla]

Okay so...

P_max = V_Th²/(4R_Th)

P = (-243.6)²/(4*(-1363)) = -10.8W

Are you allowed to have negative power?

 

[gneill]

Negative power implies that the circuit is absorbing power from the load, which can happen. But there's more to it here.

Draw the circuit as a load resistor connected to your Thevenin equivalent model. You have a voltage divider consisting of the Thevenin resistance and the load resistor. The load resistor is a standard one with positive resistance. Can you find a value that might lead to "interesting" behavior?

 

[Gwozdzilla]

If R_L = -R_Th, then when I apply KVL to the circuit you told me to draw...

V_Th + R_Thi + R_Li = 0
243.6 - 1363i + 1363i = 0
243.6 = 0
Which is wrong, which might have something to do with the power being infinity?

 

[gneill]

If R_L = -R_Th, then when I apply KVL to the circuit you told me to draw...

V_Th + R_Thi + R_Li = 0
243.6 - 1363i + 1363i = 0
243.6 = 0
Which is wrong, which might have something to do with the power being infinity?

Indeed. Leave the load resistance as a variable R and write an expression for the power. Take a limit as R approaches Rth.

 

[Gwozdzilla]

I think I'm confused or maybe using the wrong formula... regular power P = V²/R

P = V_Th²/R_L
lim R_L → R_Th V_Th²/R_L = constant, at least in this case where V_Th and R_Th are constants...

Or should I be using this formula:

P = ((V_Th)/(R_Th + R_L))²R_L

In which case, I can take the limit as R_L → -R_Th and get infinity.

 

[gneill]

The latter is closer. The voltage should be squared.
The first one ignores the Thevenin resistance which is in series with the load.