Help with intro circuit analysis problem Thevenin and max power prob

[nchin]

Problem #1:

What i did so far was I source transformed the left side with the I_o and 4Ω into a voltage of 4I_o. Then i did mesh analysis, i_o and i₂.

the eq'ns i got are
4v_x - 2I₂ - 2I_o = 0
10i₂ - v_x - 2i_o = 0

IM not sure if this is right but does i₂ = v_x/4?? because of I = V/R?

I worked everything out and I got confused.
7v_x/4 = i_o
V_x(10/4 - 14/4) = v_x

i need help please! (see attached pic)

 

[gneill]

Problem #1:

What i did so far was I source transformed the left side with the I_o and 4Ω into a voltage of 4I_o. Then i did mesh analysis, i_o and i₂.

the eq'ns i got are
4v_x - 2I₂ - 2I_o = 0
10i₂ - v_x - 2i_o = 0

IM not sure if this is right but does i₂ = v_x/4?? because of I = V/R?

I worked everything out and I got confused.
7v_x/4 = i_o
V_x(10/4 - 14/4) = v_x

i need help please! (see attached pic)

Please follow the posting template.

Your loop equations do not appear to be correct; why is there an I_o term in both? Can you show your work detailing how you arrived at them?

 

[nchin]

Please follow the posting template.

Your loop equations do not appear to be correct; why is there an I_o term in both? Can you show your work detailing how you arrived at them?

i used two mesh loops. instead of i₁, i used i_o. so the two loops are i_o and i₂.

so after that the mesh equations would be,

For the i_o loop:
-4I_o + 4V_x + 2(I_o-I₂) = 0

For the i₂ loop:
-V_x + 2(I₂-I_o) + 2I₂ + 6I₂ = 0

after simplifying I get:
4Vx - 2I2 - 2Io = 0
10i2 - Vx - 2io = 0

and on the i₂ loop there is a voltage drop of V_x across the 4Ω resistor so i₂ = V_x/4 ??

Am i correct?

 

[gneill]

i used two mesh loops. instead of i₁, i used i_o. so the two loops are i_o and i₂.

so after that the mesh equations would be,

For the i_o loop:
-4I_o + 4V_x + 2(I_o-I₂) = 0

For the i₂ loop:
-V_x + 2(I₂-I_o) + 2I₂ + 6I₂ = 0

after simplifying I get:
4Vx - 2I2 - 2Io = 0
10i2 - Vx - 2io = 0

and on the i₂ loop there is a voltage drop of V_x across the 4Ω resistor so i₂ = V_x/4 ??

Am i correct?

I_o is a fixed source, and after the Thevenin equivalent conversion, it becomes a fixed voltage source of magnitude 4I_o. So it is NOT a mesh current. You should choose another variable name for the mesh current in the first loop.

Yes, Vx is related to the current in the second loop. Be sure to take note of the direction of the potential drop caused by that mesh current when you write the expression for Vx; Vx's potential is 'measured' in a particular way as designated by the "+ -" designation.

 

[nchin]

I_o is a fixed source, and after the Thevenin equivalent conversion, it becomes a fixed voltage source of magnitude 4I_o. So it is NOT a mesh current. You should choose another variable name for the mesh current in the first loop.

Yes, Vx is related to the current in the second loop. Be sure to take note of the direction of the potential drop caused by that mesh current when you write the expression for Vx; Vx's potential is 'measured' in a particular way as designated by the "+ -" designation.

if i use another variable name then i would have 4 unknowns and and only 2 equations, i_o, i₁, i₂ and Vx? How would i solve that?

 

[gneill]

if i use another variable name then i would have 4 unknowns and and only 2 equations, i_o, i₁, i₂ and Vx? How would i solve that?

In this context i_o is a given value (granted, it's not numerical value, but it is a constant in this context). i₁ and i₂ are the unknown mesh currents, and Vx is strictly dependent on i₂. So only the two mesh currents are unknowns.

I'm not sure where you're going with this particular analysis. Sure, you can determine the Thevenin voltage (as a function of i_o) by multiplying the mesh current i₂ by the 6Ω resistor in the second loop, but you won't have the Thevenin resistance. How were you planning to find that?

 

[nchin]

In this context i_o is a given value (granted, it's not numerical value, but it is a constant in this context). i₁ and i₂ are the unknown mesh currents, and Vx is strictly dependent on i₂. So only the two mesh currents are unknowns.

so would this be correct now using the mesh variables as i₁ & i₂?

For the i₁ loop:
-4Io + 4Vx + 2(I₁-I₂) = 0

for i₂:
-Vx + 2(I₂-I₁) + 2I₂ + 6I₂ = 0

so how would we find a numerical value of i₁ and i₂ if i_o and V_x are not a numerical value?

I'm not sure where you're going with this particular analysis. Sure, you can determine the Thevenin voltage (as a function of i_o) by multiplying the mesh current i₂ by the 6Ω resistor in the second loop, but you won't have the Thevenin resistance. How were you planning to find that?

I assume i simplify the resistors? like for example, the 2Ω parallel with 6Ω + 4Ω...? some thing like that?

 

[gneill]

so would this be correct now using the mesh variables as i₁ & i₂?

For the i₁ loop:
-4Io + 4Vx + 2(I₁-I₂) = 0

for i₂:
-Vx + 2(I₂-I₁) + 2I₂ + 6I₂ = 0

They look okay. Note that your Vx really is a voltage drop across a real resistor in loop two.
It can be treated as just another resistor in loop two. And you can substitute the appropriate voltage drop for Vx in loop one. Thus the "variable" Vx disappears.

so how would we find a numerical value of i₁ and i₂ if i_o and V_x are not a numerical value?
Vx disappears as described above; it's just a voltage drop measured across a resistor, and that voltage drop depends upon the mesh current i₂ flowing through a fixed value resistor.

You DON'T find a numerical value for the currents. You find expressions for them in terms of the value of I_o. The loop currents depend upon the output of the current source I_o. Remember, I_o is something you're going to be solving for later for a given power dissipated by RL.

I assume i simplify the resistors? like for example, the 2Ω parallel with 6Ω + 4Ω...? some thing like that?

That works when you have only independent sources in the circuit. Here there's the dependent source 4Vx. You need to find another method to find the Thevenin resistance of the circuit. What methods have you been taught?

 

[nchin]

They look okay. Note that your Vx really is a voltage drop across a real resistor in loop two.
It can be treated as just another resistor in loop two. And you can substitute the appropriate voltage drop for Vx in loop one. Thus the "variable" Vx disappears.

I see. so i can just substitute everywhere with Vx with I₂/4?

That works when you have only independent sources in the circuit. Here there's the dependent source 4Vx. You need to find another method to find the Thevenin resistance of the circuit. What methods have you been taught?

Hmm looking over my notes I see a way to solve the R_th is Dividing the open circuit voltage with the short circuit current?

R_TH = V_OC/I_SC?

 

[gneill]

I see. so i can just substitute everywhere with Vx with I₂/4?

Yes, sort of, if you meant I₂*4, but BE SURE TO PAY ATTENTION TO THE SIGNS. Vx is NOT +4I₂. Note the direction of i₂ and the resulting polarity of the potential drop across the 4Ω resistor. Note also the indicated polarity for measuring Vx. Are they the same?

Hmm looking over my notes I see a way to solve the R_th is Dividing the open circuit voltage with the short circuit current?

R_TH = V_OC/I_SC?

Yes, that's one way. It's quite a bit of work though. If I may make a suggestion, why not leave RL in the circuit, adding another (but very simple) mesh. If you solve for the voltage across RL by finding the current in its mesh, you'll have an expression from which you can easily pick out both the Thevenin voltage and the Thevenin resistance. You'll also have an expression for the current through RL which you can use to determine the power dissipated... Three birds with one stone, to paraphrase an old saying.

 

[nchin]

Yes, sort of, if you meant I₂*4, but BE SURE TO PAY ATTENTION TO THE SIGNS. Vx is NOT +4I₂. Note the direction of i₂ and the resulting polarity of the potential drop across the 4Ω resistor. Note also the indicated polarity for measuring Vx. Are they the same?

i see. so then it would be, for the i2 loop:

-(-4I₂) + 2(I₂ - I₁) + 2I₂ + 6I₂ = 0 ?

since the mesh loop goes through the neg side first

 

[gneill]

i see. so then it would be, for the i2 loop:

-(-4I₂) + 2(I₂ - I₁) + 2I₂ + 6I₂ = 0 ?

since the mesh loop goes through the neg side first

Yes. Of course, the 4Ω resistor is just a 4Ω resistor in its loop. Vx is just a measurement taken there; Vx has no effect at all on that loop, since it is only a measurement.

 

[nchin]

Yes, that's one way. It's quite a bit of work though. If I may make a suggestion, why not leave RL in the circuit, adding another (but very simple) mesh. If you solve for the voltage across RL by finding the current in its mesh, you'll have an expression from which you can easily pick out both the Thevenin voltage and the Thevenin resistance. You'll also have an expression for the current through RL which you can use to determine the power dissipated... Three birds with one stone, to paraphrase an old saying.

i really don't know what to do with these equations. So i leave R_L in there so now i have three mesh eq'ns.

After simplifying the mesh eq'ns i have:

I₁:
-4Io - 18I₂ + 2I₁ = 0

I₂:
14I₂ - 2I₁ = 0

I₃:
6I₃ - 6I₂ + R_LI₃ = 0

what can i do now?

 

[gneill]

i really don't know what to do with these equations. So i leave R_L in there so now i have three mesh eq'ns.

After simplifying the mesh eq'ns i have:

I₁:
-4Io - 18I₂ + 2I₁ = 0

I₂:
14I₂ - 2I₁ = 0

I₃:
6I₃ - 6I₂ + R_LI₃ = 0

what can i do now?

Check your equation for mesh 1; in particular the coefficient for the i₁ term. In the second equation there should be a term for [/SUB]I₃. The third mesh equation looks fine.

Now you want to solve for an expression for i₃. Hint: if you set up the equations in matrix form and use Cramer's Rule, you can solve for i₃ without having to deal with the other two...

 

[nchin]

Check your equation for mesh 1; in particular the coefficient for the i₁ term. In the second equation there should be a term for [/SUB]I₃. The third mesh equation looks fine.

im not sure what's wrong with mesh 1?

i checked it and i got:
-4Io + 4Vx + 2(I₁-I₂) = 0
-4Io + 4(-4I₂) + 2(I₁-I₂) = 0
-4Io - 16I₂ + 2(I₁-I₂) = 0
-4Io - 18I₂ + 2I₁ = 0
??

my mistake i did forget the I₃.
i now have
14I₂ - 2I₁ - 6I₃ = 0

 

[gneill]

im not sure what's wrong with mesh 1?

i checked it and i got:
-4Io + 4Vx + 2(I₁-I₂) = 0
-4Io + 4(-4I₂) + 2(I₁-I₂) = 0
-4Io - 16I₂ + 2(I₁-I₂) = 0
-4Io - 18I₂ + 2I₁ = 0
??

What happened to the 4Ω resistor in loop 1? Here's the circuit we're dealing with:

my mistake i did forget the I₃.

No worries, these things happen.

 

[nchin]

What happened to the 4Ω resistor in loop 1? Here's the circuit we're dealing with:

Ah you're right! I forgot about that as well.

so for mesh 1 i have
-4Io + 6I₁ - 18I₂ = 0

 

[nchin]

Now you want to solve for an expression for i₃. Hint: if you set up the equations in matrix form and use Cramer's Rule, you can solve for i₃ without having to deal with the other two...

So i can solve i₃ just from using mesh 3 equation alone?

6I₃ - 6I₂ + RLI₃ = 0

i understand how to do 2x2 or 3x3 cramer rule but how would i do 1x1 with cramer?

 

[gneill]

So i can solve i₃ just from using mesh 3 equation alone?

6I₃ - 6I₂ + RLI₃ = 0

i understand how to do 2x2 or 3x3 cramer rule but how would i do 1x1 with cramer?

No, set up the 3x3 impedance matrix and the voltage vector as usual, then solve for i₃.

 

[nchin]

No, set up the 3x3 impedance matrix and the voltage vector as usual, then solve for i₃.

Will I get a numerical value for I3?

 

[nchin]

No, set up the 3x3 impedance matrix and the voltage vector as usual, then solve for i₃.

Im confused becasue I don't have the usual variables for cramers.

my equations are

6I1 - 18I2 - 4Io = 0
-2I1 + 14I2 - 6I3 = 0
-6I2 + R_LI₃ + 6I3 = 0

im confused because i have an Io, and R_{L variable}

 

[gneill]

Will I get a numerical value for I3?

No, you'll get an expression for I3 in terms of R_L and I_o. That expression will allow you to pick out the Thevenin voltage and resistance. With the Thevenin resistance you known the value of R_L that will maximize the power delivered to R_L.

 

[nchin]

No, you'll get an expression for I3 in terms of R_L and I_o. That expression will allow you to pick out the Thevenin voltage and resistance. With the Thevenin resistance you known the value of R_L that will maximize the power delivered to R_L.

I am not sure how to continue because I've never done an impedance matrix before. I've only done regular cramers rule with only variables of i1, i2, i3. I've never encountered this situation before. what do i do?

 

[gneill]

I am not sure how to continue because I've never done an impedance matrix before

The nomenclature isn't important. Just set up your equations in matrix form. You have three equations with three current variables.

 

[nchin]

The nomenclature isn't important. Just set up your equations in matrix form. You have three equations with three current variables.

6I1 - 18I2 - 4Io = 0
-2I1 + 14I2 - 6I3 = 0
-6I2 + RLI3 + 6I3 = 0

For values of i1 we have 6, -2, 0
i2: -18, 14, -6
i3: 0, -6, 6

I1 I2 I3
6 -18 0 4Io
-2 14 -6 0
0 -6 6 0

??

 

[gneill]

6I1 - 18I2 - 4Io = 0
-2I1 + 14I2 - 6I3 = 0
-6I2 + RLI3 + 6I3 = 0 <---- RL is in the third equation

For values of i1 we have 6, -2, 0
i2: -18, 14, -6
i3: 0, -6, 6

I1 I2 I3
6 -18 0 4Io
-2 14 -6 0
0 -6 6 0

??

What happened to RL in the third equation? The I3 coefficient is (6 + RL), right?

You're solving for I3 in terms of Io and RL.

 

[nchin]

What happened to RL in the third equation? The I3 coefficient is (6 + RL), right?

You're solving for I3 in terms of Io and RL.

is this right, so the final answer is suppose to be variables and not numerical values?

 

[gneill]

is this right, so the final answer is suppose to be variables and not numerical values?

It looks like the entries -2 and -18 have there places swapped. Surely the -18 should correspond to the coefficient of i2 in the first loop.

Yes, the result for i3 should be an expression involving variables Io and RL.

 

[nchin]

It looks like the entries -2 and -18 have there places swapped. Surely the -18 should correspond to the coefficient of i2 in the first loop.

Yes, the result for i3 should be an expression involving variables Io and RL.

is the final answer for the entire problem suppose to be an expression? not just the result of i3

 

[gneill]

The problem asks for the value of resistance for RL that maximizes the power dissipated by RL, and it asks for the value of Io that results in 54W being dissipated by that RL. So, two numerical values. They will be found AFTER you've got the expression for i3.

 

[nchin]

No, you'll get an expression for I3 in terms of R_L and I_o. That expression will allow you to pick out the Thevenin voltage and resistance. With the Thevenin resistance you known the value of R_L that will maximize the power delivered to R_L.

When we solve for i3 we will have two unknown variables in it right? Rl and Io

 

[gneill]

When we solve for i3 we will have two unknown variables in it right? Rl and Io

Yes. But we have the advantage of knowing something about RL and its relationship to Rth for maximum power transfer, regardless of the value of Io. As I stated previously, we can pick out Rth and Vth from the result of finding the expression for i3*RL.