# Finding dependent source for Norton circuit

Discussion in 'Homework Help' started by Loonie, Nov 23, 2011.

1. ### Loonie Thread Starter New Member

Aug 30, 2011
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Hi, I would appreciate some directions on how to proceed to solve this Norton equivalent circuit.

It is Q1 (b).

I have found the Rth using test source method of 10V.

However, I am now stuck to how to find the dependent current source Ix...
Can I do source conversion for the 80V 20omhs part? Wouldn't it "eat" up the short circuit? If so, then should I use nodal analysis to find Ix? If using nodal analysis to find Ix, assuming that V1 is on the left, then what is V2, because it faces the 80V source as well as the 2A current source (which becomes a 120V source after conversion).

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2. ### samski New Member

Nov 23, 2011
20
1
hey, you can kill all the right hand side of the circuit i think, the Ix is just sunk straight into the variable current source so it has no effect on a or b, so all that remains is the 2A in parallel with the 60ohm, the 80V and the 20ohm. answer i think is 40V & 80ohm?

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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Focus for the moment on the node where the two 60Ω resistors, the 2A source and the controlled source meet. You'll notice that whatever current Ix flows in the left hand 60Ω it is exactly matched by the controlled current source. Hence the controlled source forces the condition [node voltage] whereby the current value into the node from the left-hand 60Ω is exactly balanced by the current flowing out of the controlled source branch.

From the point of view of Kirchoff's current law we must then conclude that the nett current from all other branches (excluding the left-hand 60Ω and the controlled source) connecting to that node must be zero.

We also have a 2A source driving into the node. That 2A must be balanced by 2A flowing out of the node in the remaining branches. There is only one other branch - the 60Ω down to the common rail. So that 60Ω is carrying the 2A. That implies the node voltage is 120V. If you know that node voltage you can then easily determine the open circuit voltage across terminals a-b. This will give the Thevenin equivalent volatge at terminals a-b. The Thevenin resistance can be readily determined looking back into terminals a-b with all voltage sources shorted and all current sources opened.

One can readily convert from Thevenin to Norton equivalent.

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4. ### Loonie Thread Starter New Member

Aug 30, 2011
17
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Wow yes, answer scheme states that Isc is 0.5A

Can you check my diagram again? To make sure I correctly understood your answer.

I did consider the Ix being eliminated, but that that leaves me with dealing with this part. (Fig 3) Where 2A + 4A = 6A?? But I guess that you cannot make the "short circuit" because the 80V and 20ohms elements share the same line?
I didn't see that the 80V and 120V conversion will cut out each other and hence giving 40V.

However, I notice that the resistances 60+20 = 80ohms. Does it mean that using test source conversion was unnecessary? Or only apply to this particular case because it cancels itself out.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I should have been more careful when describing how to find the Thevenin / Norton resistance. The unusual result of the disconnect between the two sides of the circuit at the main node just means the effective resistance looking in terminals a-b is indeed (20+60)Ω. The consequences of having a dependent source can be tricky sometimes.

As an alternative, you can find the a-b short circuit current (I_Norton) by applying the same logic I mentioned previously.

The 2A source is still driving the aforementioned node but the incoming 2A is now split between the 60Ω branch to ground and the branch including a-b shorted. The condition for this to work is

V_node=20*Iab+80=(2-Iab)*60

Or

20*Iab+80=120-60*Iab

OR

80*Iab=120-80=40

OR

Iab[short-circuit]=I_Norton=40/80=0.5A

The required Norton resistance would be given by

R_Norton=V_Thevenin/I_Norton=40/0.5=80Ω

6. ### Loonie Thread Starter New Member

Aug 30, 2011
17
0
Ok, looks like there is lots of ways to solve circuits!
Thanks for your help