Thevenin and Norton Equivalent

[xlu2]

Homework Statement

Find Vt and In

Homework Equations

KCL

The Attempt at a Solution

I found Rt to be 5.94 ohms (1/(1/7+1/9)+2)
Then I am stuck.
I know to find Vt (which is equal to Vopen circuit), I have to do a KCL in an open circuit, so my KCL equation is
2+Voc/3=Voc/4
So Voc=-12V=Vt?

If the Voc is correct, then In is just Voc/Rt.

Would you anyone help me with finding Voc?

Many thanks in advance!

 

[milesyoung]

You found the Norton resistance correctly and you know the Norton current, so that means you've found the Norton equivalent circuit. Then you have all you need to convert it to its Thévenin equivalent:
http://en.wikipedia.org/wiki/Norton's_theorem#Conversion_to_a_Th.C3.A9venin_equivalent

 

[xlu2]

You found the Norton resistance correctly and you know the Norton current, so that means you've found the Norton equivalent circuit. Then you have all you need to convert it to its Thévenin equivalent:
http://en.wikipedia.org/wiki/Norton's_theorem#Conversion_to_a_Th.C3.A9venin_equivalent

I did not realize I found Iin. Would you please tell me which number was it?

 

[gneill]

Homework Statement

Find Vt and In

View attachment 59103

Homework Equations

KCL

The Attempt at a Solution

I found Rt to be 5.94 ohms (1/(1/7+1/9)+2)
Then I am stuck.
I know to find Vt (which is equal to Vopen circuit), I have to do a KCL in an open circuit, so my KCL equation is
2+Voc/3=Voc/4
So Voc=-12V=Vt?

If the Voc is correct, then In is just Voc/Rt.

Would you anyone help me with finding Voc?

Many thanks in advance!

Your KCL to find Voc doesn't look right. If you want to use nodal analysis and make a node where three resistors meet, then you'll need another node, too, at the top of the 4Ω resistor.

There are several different ways to find Voc. For example, you could use the current divider rule to find the current through the path with the 9Ω resistor (and hence the potential across it). Or perhaps first convert the 2A source and 4Ω resistor into their Thevenin equivalent and then use the voltage divider rule to find the potential at the top of the 9Ω resistor...

 

[milesyoung]

I did not realize I found Iin. Would you please tell me which number was it?

Sorry that was a complete brainfart on my part. In my mind there was a connection between the upper nodes of the 3 and 2 ohm resistors.

 

[xlu2]

Your KCL to find Voc doesn't look right. If you want to use nodal analysis and make a node where three resistors meet, then you'll need another node, too, at the top of the 4Ω resistor.

There are several different ways to find Voc. For example, you could use the current divider rule to find the current through the path with the 9Ω resistor (and hence the potential across it). Or perhaps first convert the 2A source and 4Ω resistor into their Thevenin equivalent and then use the voltage divider rule to find the potential at the top of the 9Ω resistor...

KCL for where the three resistor meet: Voc/2=Voc/9+Voc/3?
KCL for 4 ohms node: 2+Voc/3=Voc/4?

Is the current through 9 ohms 2*9/(1/(1/(4+3)+1/9))? (4 and 3 are in series which are then in parallel with 9)

How do you convert 2A and 4 ohms resitor into their Thevenin equivalent?

 

[gneill]

KCL for where the three resistor meet: Voc/2=Voc/9+Voc/3?
KCL for 4 ohms node: 2+Voc/3=Voc/4?

No, those aren't right. In your equations you've neglected the potentials that lie at the remote ends of the branches. You are summing currents, and the current through a given branch depends upon the potential across that branch.

That means, if a branch (even a single resistor) lies between two nodes, the potential across that branch is equal to the difference in potential of the two nodes. So if there are two nodes A and B with potentials VA and VB, and they have a resistor R between them, then the current through R going from node A to node B is (VA - VB)/R. Note that the potentials of both nodes where R connect are referenced.

Is the current through 9 ohms 2*9/(1/(1/(4+3)+1/9))? (4 and 3 are in series which are then in parallel with 9)

No, that doesn't look right. The 4Ω resistor is in parallel with the (3+9)Ω branch.

How do you convert 2A and 4 ohms resitor into their Thevenin equivalent?

The same way you would find the Thevenin equivalent for any circuit. Consider the current source and resistor in isolation. What's the open-circuit voltage? What's the Thevenin resistance?

 

[xlu2]

No, those aren't right. In your equations you've neglected the potentials that lie at the remote ends of the branches. You are summing currents, and the current through a given branch depends upon the potential across that branch.

That means, if a branch (even a single resistor) lies between two nodes, the potential across that branch is equal to the difference in potential of the two nodes. So if there are two nodes A and B with potentials VA and VB, and they have a resistor R between them, then the current through R going from node A to node B is (VA - VB)/R. Note that the potentials of both nodes where R connect are referenced.

No, that doesn't look right. The 4Ω resistor is in parallel with the (3+9)Ω branch.

The same way you would find the Thevenin equivalent for any circuit. Consider the current source and resistor in isolation. What's the open-circuit voltage? What's the Thevenin resistance?

So (Voc-V1)/2+Voc/9+(Voc-V2)/3+Voc/4=2? But how would I find V1 and V2 if I label V1 to be the node joining the 3 resistors and V2 joining 3 and 4?

For Thevenin equivalent, Voc=2*4=8V and R=4 ohms?

 

[gneill]

So (Voc-V1)/2+Voc/9+(Voc-V2)/3+Voc/4=2? But how would I find V1 and V2 if I label V1 to be the node joining the 3 resistors and V2 joining 3 and 4?

No, there's still something wrong with your equations.

Let's redraw the circuit to remove the "Y" artistry, label the nodes, and establish the ground reference:

Note that Voc means Open Circuit voltage. That means NO CURRENT can flow through the 2Ω resistor, and it can be ignored. Voc will be the same potential as V2. Now can you write the two node equations?

You might note that with the 2Ω resistor being open circuited that the second node (V2) is not really required to "solve" the circuit; it's added in the middle of the branch ( combined (3+9)Ω branch) in order to facilitate finding Voc. Otherwise you could solve for the one node voltage V1 and then use it to apply the voltage divider rule to the branch.

For Thevenin equivalent, Voc=2*4=8V and R=4 ohms?

Yes, that's right. If you do that you should be able to find the potential across the 9Ω resistor (which is the same as V2 in the diagram above) using the voltage divider rule.