Circuit network with multiple ports

[magnifik]

Find V₂/V₁ for the network below:

What I did was try to solve by reducing to thevenin equivalents for V₁ and V₂ separately
For V₁:
I got V_th = V₁
Z_th1 = 41/15
V₁ = I₁*(41/15)

For V₂:
I got V_th = V₂
...but I am having trouble trying to find the Z_eq2:

I know once I find that V₂ = I₂*Z_eq2
and i also need help with relating I₁ to I₂..

thanks in advance

 

[gneill]

I think you'll find it better to just find the Thevenin equivalent for the circuit looking into the V2 port. That is, find the open circuit voltage there due to V1 and the Thevenin resistance looking into that port. That will reduce the circuit to a single voltage source (which will be V1 scaled by some factor) and a single series resistance. Then you can use these to determine the output voltage V2 across the load.

You can perform a series of Thevenin transformations, working your way across the circuit until you've "consumed" all of the circuit between V1 and the load.

An alternative is to use mesh analysis to solve for the current in the loop containing the load.

 

[magnifik]

I think you'll find it better to just find the Thevenin equivalent for the circuit looking into the V2 port. That is, find the open circuit voltage there due to V1 and the Thevenin resistance looking into that port. That will reduce the circuit to a single voltage source (which will be V1 scaled by some factor) and a single series resistance. Then you can use these to determine the output voltage V2 across the load.

You can perform a series of Thevenin transformations, working your way across the circuit until you've "consumed" all of the circuit between V1 and the load.

An alternative is to use mesh analysis to solve for the current in the loop containing the load.

I am attempting to do the problem in the first way you described by looking into the V2 port. I know if I do that I'll get something like V2 = (some resistance)*V1...but my problem is how to get that resistance. I'm not sure what's in series/parallel with each other

i know V2/V1 should *supposedly* be 1/41 but i am not getting that answer

when i try to solve for equivalent resistance looking into V2 i get 15/11

 

[gneill]

You can proceed stepwise across the circuit from V1, creating a series of Thevenin equivalents as you go. As you go you'll take up the resistances in bite-size portions that are easy to digest. Here's one such suggested set of 'slices', the first one broken out for inspection. Hint: Be careful what you do with that first 1 Ohm resistor in parallel with the voltage source when you are calculating Rth₁.

 

[magnifik]

You can proceed stepwise across the circuit from V1, creating a series of Thevenin equivalents as you go. As you go you'll take up the resistances in bite-size portions that are easy to digest. Here's one such suggested set of 'slices', the first one broken out for inspection. Hint: Be careful what you do with that first 1 Ohm resistor in parallel with the voltage source when you are calculating Rth₁.

in the last step, i get an equivalent resistance of 11/4 in parallel with the 1ohm resistor.

 

[gneill]

Can you show your work step by step? Let's begin with the very first "slice". What do you get for the Thevenin voltage and resistance?

 

[magnifik]

Starting from the left...

For my first slice, I get 3 (1 in series with 2)
Combining this with slice 2, i get 11/4 ((3 || 1) + 2)
Combining this with the last slice, I am getting 41/15 ((11/4 || 1) + 2)

 

[gneill]

Starting from the left...

For my first slice, I get 3 (1 in series with 2)
Combining this with slice 2, i get 11/4 ((3 || 1) + 2)
Combining this with the last slice, I am getting 41/15 ((11/4 || 1) + 2)

For the first slice, the 1Ohm resistor is in parallel with the voltage source. When you short out the voltage source to find the Thevenin resistance, that resistor is also shorted out (remember I said to be careful with that one!). So for the first slice the Thevenin resistance is just 2 Ohms, and the Thevenin voltage is V1.

 

[magnifik]

ok. now i got 10/3 as the equivalent resistance

 

[gneill]

ok. now i got 10/3 as the equivalent resistance

Is that your Thevenin resistance for the whole network (up to but not including the load resistor)? If so, that's not what I calculate it to be. I think you'll have to show your work.

 

[magnifik]

1) i have the 2 ohm resistor in series with 1 || 2 = 8/3
2) i have the 8/3 in series in 1 || 2 = 10/3

 

[gneill]

1) i have the 2 ohm resistor in series with 1 || 2 = 8/3
2) i have the 8/3 in series in 1 || 2 = 10/3

I think you're mixing up what's in series and what's in parallel.

Presuming that you're calculating the Thevenin resistance for the second "slice", then the circuit you're looking at should look something like this:

When you short out Vth₁, then Rth₁ will be in parallel with the 1Ω resistor, and then the 2Ω resistor will be in series with them. A similar situation will hold for the next "slice".

 

[magnifik]

series/parallel is my enemy.

so..
1) (2||1) + 2 = 8/3
2) (8/3 || 1) + 2 = 30/11

 

[gneill]

series/parallel is my enemy.

so..
1) (2||1) + 2 = 8/3
2) (8/3 || 1) + 2 = 30/11

That looks better!

 

[magnifik]

ok..so after that i get
V2 = 1/(1+(30/11))V1
which gives V2/V1 = 11/41.. this is what i had originally :\

 

[gneill]

What did you get for your final Thevenin voltage? (it's not V1!). You should calculate a new Thevenin voltage for every "slice" as you move across the circuit. So you have a new Thevenin resistance value and a new Thevenin voltage at the completion of each "slice".

 

[magnifik]

What did you get for your final Thevenin voltage? (it's not V1!). You should calculate a new Thevenin voltage for every "slice" as you move across the circuit. So you have a new Thevenin resistance value and a new Thevenin voltage at the completion of each "slice".

*le sigh* i did not do that

 

[gneill]

"Alas" or "at last"? The result looks okay to me

 

[magnifik]

is this correct for the ratio of the current (I2/I1)
V2 = -I2
I1 = 56V2
I2/I1 = -V2/(56/V2) = -1/56 ??

 

[gneill]

is this correct for the ratio of the current (I2/I1)
V2 = -I2
I1 = 56V2
I2/I1 = -V2/(56/V2) = -1/56 ??

Yes, that looks okay for the current ratio.