Dependent voltage source, find max. power to load in circuit & R(Th)

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img831/2617/homeworkprobsg213.jpg Find the load R_L that will result in maximum power delivered to the load of the circuit. Also determine P_max delivered

Homework Equations

V = IR, KVL, KCL,

voltage division, current division

Thevenin procedures,Pmax to a load resistor = (V Thevenin)²/(4 R_load)R(Th) = R(load) ??

The Attempt at a Solution

No idea what to do regarding dependent voltage source.

If I try to silence it with the 6V to find R(Th) I would get

[ (1/4) + (1/6) ]^-1 = 2.4Ω but that isn't right.And then I have no idea how I would find VTh afterwards either

(although I know it would typically be the voltage where a is with the 4 ohm load resistor cut out of the circuit) Any tips?

 

[gneill]

The 4Ω resistor is not the load resistor, it's part of the network "behind" the load. RL is the load resistor. The 4Ω resistor stays put. Vth would be the open circuit voltage with RL removed.

Try removing the resistor RL and writing a node equation for node a. Solve for Va.

 

[Color_of_Cyan]

I know the 4Ω wasn't part of the load (which is why I did 1/6 + 1/4 for parallel to try to find RTh, but to no avail)

If I take out the load, can I put 6 and 4 in series, even though there's a dependent voltage source?

With the load cut out and the above statement, all I could think of for it (and nodal equations) are:

Req (non thevenin) = 10Ω

Va = 2Vab + 6V

I = ( 6V + 2Vab )/10Ω

but I'm not sure

 

[gneill]

I know the 4Ω wasn't part of the load (which is why I did 1/6 + 1/4 for parallel to try to find RTh, but to no avail)

If I take out the load, can I put 6 and 4 in series, even though there's a dependent voltage source?

If they are in series, then they are in series. No argument. BUT. What are you trying to accomplish by putting them in series? You won't find the Thevenin resistance by considering the actual resistances alone; active sources can behave in ways that "mimic" resistances (even negative resistances!), so one needs to be a bit more clever when looking for the Thevenin resistance of a network with active sources (more below*).

With the load cut out and the above statement, all I could think of for it (and nodal equations) are:

Req (non thevenin) = 10Ω

Va = 2Vab + 6V

I = ( 6V + 2Vab )/10Ω

but I'm not sure

For a nodal equation I'd expect you to write a nodal equation for the node associated with terminal a (with RL removed), thus writing a KCL equation at node a that includes the potential at node a. Note that the potential at node a is identical to Vab so long as node b is the reference node.

With this nodal equation you should be able to determine the potential Vab. This will be the Thevenin voltage.

*note: The Thevenin resistance of a network can also be found by taking the ratio of the Thevenin voltage (open circuit voltage) to the short circuit current, also known as the Norton current. If you determine V_th and I_N, then R_th = V_th/_IN. The short circuit current should be easy to determine, since shorting the output pins Vab to zero.

 

[Color_of_Cyan]

Okay. Would i regard the 2Vab source though as the same kind of voltage source when doing nodal equations?

Finding VTh would still be a challenge for me for the same reason (because of that 2vab source)

And to find the Norton current as you say wouldn't the 4 ohm resistance be shorted too? It would still also leave the 2Vab current source too, wouldn't it?

And since Vab would be 0 then 2Vab would be 0 too, right?

So would I Norton just be 6v / 6 ohms ?

(1 amp?)

It seems too easy to be true. Still stuck on how to get VTh too though.

 

[gneill]

Okay. Would i regard the 2Vab source though as the same kind of voltage source when doing nodal equations?

Yes. It is a voltage source. Write it in as "2Vab". The node voltage happens to also be Vab.

Finding VTh would still be a challenge for me for the same reason (because of that 2vab source)

It's just another voltage term in the equation...

And to find the Norton current as you say wouldn't the 4 ohm resistance be shorted too? It would still also leave the 2Vab current source too, wouldn't it?

And since Vab would be 0 then 2Vab would be 0 too, right?

So would I Norton just be 6v / 6 ohms ?

(1 amp?)

It seems too easy to be true. Still stuck on how to get VTh too though.

Yes, the 4Ω resistor gets shorted, and yes, as a result Vab is zero. Nothing wrong with short circuit current being 1 amp, even if it is easy to find

For Vth, write the nodal equation and solve for Vab.

 

[The Electrician]

You could also try writing a mesh equation for the left mesh, which is the only mesh left if you remove Rload. Once you have the current in that mesh (call the current I1), then Vab is just 4*I1 = Vth.

 

[Color_of_Cyan]

So would one equation be

Vab = 6V - 2Vab

by any chance?I don't have to worry about current right now for this either, right?

 

[The Electrician]

You do have to worry about current.

Write a mesh equation starting at the bottom left of the left hand mesh, and going clockwise.

Start out like this:

-6 volts - 6Ω*I1 amps - 2*Vab volts...= 0

Substitute Vab = 4*I1

and finish it up.

 

[gneill]

So would one equation be

Vab = 6V - 2Vab

by any chance?


I don't have to worry about current right now for this either, right?

That is neither a nodal equation nor a mesh equation for the circuit. The current is important, as it causes potential drops across the resistances.

Pretend that the dependent voltage source is just another voltage source, but use "2Vab" for its value when writing the node equation (or mesh equation if you wish to go that route).

I have been assuming that you are familiar with writing node equations (or mesh equations). If you aren't, it would be helpful to know what circuit analysis techniques you are familiar with.

 

[Color_of_Cyan]

I don't think I've really done either before with 1 loop before like this (or just not enough with actual sources like this either).

To be honest, I'm not even sure how the potential will stack up with the dependent source either (since there's a resistor "behind" the 2Vab and another "in front")

I'll try though:

6V/6Ω = (6V + 2Va)/4Ω ?(I know at least the current will be the same in the resistors because they're in series)

 

[gneill]

No, that's not right.

In nodal analysis you assume that each node has some potential as represented by a variable. In this case there's one independent node and we give it the variable "Va". We also assume some common reference node, and here we choose b to be the zero reference.

Writing a node equation consists of doing a KCL sum of currents at the node. Usually one chooses to write the sum assuming that all currents are entering or all currents are leaving the node, so the sum will be zero. The currents for individual branches are determined by way of a KVL path walk of the branch given the two potentials at either end of the branch; assume current I is flowing in the branch in the assumed direction and sum the potential changes along the path. Solve for I and that's the term for that branch in the nodal equation.

In this case the node potential for the node of interest is taken to be Va, and at the other end of the branches it's 0 (the reference node). So, for example, if we want the term for the current leaving node though the 4Ω resistor branch we'd write the KVL: Va - I*4Ω = 0, so that I = (Va - 0)/4Ω, or simply Va/4Ω, where I is the current for that branch. Terms for each branch connected to the node should be written, the sum of terms being zero.

Can you write the term for the current leaving the node via the 6Ω resistor branch? The potential starts at Va, drops by 2Va, then by I*6Ω, then by 6V where it finally reaches the 0V potential of the reference node.

 

[Color_of_Cyan]

Sorry for getting back to this so late.

I honestly don't think I've seen anything like that before / yet or had no idea you could do it this way (for nodal). It gets confusing at times, and you make it look so easy.

It's helpful though, so thanks for showing me. So would the other equation then be

Va - 2Va - I*6Ω - 6V = 0 ? (I just tried setting this equation equal to I = Va / 4Ω though and then solving for Va, plugging in everything but the answer was still wrong).

 

[gneill]

Sorry for getting back to this so late.

I honestly don't think I've seen anything like that before / yet or had no idea you could do it this way (for nodal). It gets confusing at times, and you make it look so easy.

It's helpful though, so thanks for showing me.

I'm surprised that your course would introduce dependent sources before thoroughly covering nodal and mesh analysis techniques.

So would the other equation then be

Va - 2Va - I*6Ω - 6V = 0 ?

Yes. Solve it for I also to yield the current term for that branch.

Note that these I's are (usually) different for different branches -- the I's that we're using here are just "temporary variables" used as an aid to help you find the current terms for each branch. After some practice you will be able to write down the branch current terms by inspection without having to use these "I" placeholders to construct the expression via KVL walk.

(I just tried setting this equation equal to I = Va / 4Ω though and then solving for Va, plugging in everything but the answer was still wrong).

Yes, that's because I is not an actual variable to be used here; The two I's represent the branch current expressions (Maybe you want to give them different names, I1 and I2, just to make it clear that they represent different current terms). I'm sorry, I should have clarified this before: it's the expressions for the individual branch currents that you want to find, since they form the terms of the node equation you wish to construct. Introducing I as a temporary variable in the KVL walk for a branch is meant to help you solve for the branch current expressions in a (hopefully) familiar way.

Sum the two current terms and set it to zero (KCL for the node). Then solve for Va.

 

[Color_of_Cyan]

So, you pretty much always have to do more different / careful circuit techniques (ie nodal, mesh) when you deal with a dependent source then, right? I thought the currents would just directly be the same since all the elements are in series with the load cut out. I couldn't see that Va = 4Ω*(current through 4Ω) either at the beginning either because of the dependent voltage source, yet it has to still do with finding Va in the first place. So I guess that's something to keep in mind next time too.

Seems like I have the correct answer too now though, thanks (although signs are different).

Rth = 12 ohms, Va = -12V

and Pmax = 3W.

 

[gneill]

So, you pretty much always have to do more different / careful circuit techniques (ie nodal, mesh) when you deal with a dependent source then, right? I thought the currents would just directly be the same since all the elements are in series with the load cut out.

You bring whatever machinery is required to bear; sometimes Ohm's law will suffice. There's no hard and fast rule about what MUST be used. Use what you're familiar with that gets the job done.

I suggested nodal analysis because it would directly give you Va, which is Vth, by forming and solving a single equation. You could also have done KVL around the loop to find the current, then calculated Vth as the voltage drop caused by that current passing through the 4Ω resistor.

I couldn't see that Va = 4Ω*(current through 4Ω) either at the beginning either because of the dependent voltage source, yet it has to still do with finding Va in the first place. So I guess that's something to keep in mind next time too.

Seems like I have the correct answer too now though, thanks (although signs are different).

Rth = 12 ohms, Va = -12V

and Pmax = 3W.

Looks like something must've happened to a sign when you wrote/solved the node equation; you should get +12V as a result.

 

[Color_of_Cyan]

Here was the work
Va - I1*4Ω = 0

I1 = Va / 4ΩVa - 2Va - I2*6Ω - 6V = 0

I2*6Ω = Va - 2Va - 6V

I2 = (Va/6Ω) - (Va/3Ω) - 1AI1 + I2 = 0

Va/4Ω + Va/6Ω - Va/3 - 1A = 0

Va(1/4 + 1/6 + 1/3) = 1A

Va(1/4 - 1/3) = 1A

Va(3/12 - 4/12) = 1A

Va(-1/12) = 1A

Va = V Th = -12V

I Nort= 1ARTh = V Th / I Nort

Would RTh actually be negative too? (Or did I do something wrong again?)

 

[gneill]

Here was the work



Va - I1*4Ω = 0

I1 = Va / 4Ω


Va - 2Va - I2*6Ω - 6V = 0

I2*6Ω = Va - 2Va - 6V

I2 = (Va/6Ω) - (Va/3Ω) - 1A


I1 + I2 = 0

Va/4Ω + Va/6Ω - Va/3 - 1A = 0

Va(1/4 + 1/6 + 1/3) = 1A <--- How did -Va/3 become +Va/3 ?

Va(1/4 - 1/3) = 1A

Va(3/12 - 4/12) = 1A

Va(-1/12) = 1A

Va = V Th = -12V

I Nort= 1A


RTh = V Th / I Nort

Would RTh actually be negative too? (Or did I do something wrong again?)

Rth should be positive in this problem.

 

[Color_of_Cyan]

That was a typo, heh.

I fixed it in the next line though (subtracted 2/6 from 1/6). The answer for Vth isn't given though, just RTh and the power (and both were positive as you say)

Maybe he just meant absolute value or something? Even though you can also have negative resistance?

Not sure what else to do there, but again, thanks.

 

[gneill]

Here was the work



Va - I1*4Ω = 0

I1 = Va / 4Ω


Va - 2Va - I2*6Ω - 6V = 0

I2*6Ω = Va - 2Va - 6V

I2 = (Va/6Ω) - (Va/3Ω) - 1A


I1 + I2 = 0

Va/4Ω + Va/6Ω - Va/3 - 1A = 0

Va(1/4 + 1/6 + 1/3) = 1A <--- How did -Va/3 become +Va/3 ?

Va(1/4 - 1/3) = 1A <-- 1/6 - 1/3 ≠ -1/3

Va(3/12 - 4/12) = 1A

Va(-1/12) = 1A

Va = V Th = -12V

I Nort= 1A


RTh = V Th / I Nort

Would RTh actually be negative too? (Or did I do something wrong again?)

Oops. Missed the math error on the next line.

 

[Color_of_Cyan]

Okay I did miss that. Weird, lucky it still -1/12 the other time then

(1/4 - 1/6) = 2/24 = 1/12

Va(1/12) = 1A

Va = 12V now

my bad,

Everything's right now, thanks again.