# Homework Help: Thevenin equivalent

1. Sep 8, 2014

### orangeincup

1. The problem statement, all variables and given/known data
Find the Thevenin equivalent circuit with respect to terminals a, b by finding open-circuit voltage and the short-circuit current.

2. Relevant equations
i1+i2+i3..=0
v1+v2+v3..=0
i=v/R

3. The attempt at a solution
I was trying to shove this on my own for practice, but when I looked at the solution I don't understand something. When they're calculating Vth, why did they completely ignore v1 and one side of the equation?

Also, when they calculate...
(v2-9)/8 + v2/10, why is it v2/70 in the first equation, and v2/10 in the second? I know one is open and one is short circuited, but why is the 60 being ignored?

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2. Sep 8, 2014

### phinds

When you do a Th equivalent to get the Rth, what do you with voltages sources and what do you do with current sources? Did you do that in this circuit? Have you looked hard at the redrawn circuit?

3. Sep 8, 2014

### orangeincup

I was told you pretend the voltage sources aren't part of the circuit to find Rth. For Vth I believe you have it set as an "open circuit" but I'm not sure how that changes the drawing. Do I take out the 60ohm resistor when it's open?

4. Sep 8, 2014

### Staff: Mentor

The current source between nodes v1 and v2 fixes the current in that branch regardless of what v1 and v2 happen to be. This is made apparent when they write the node equation for v2 and v1 does not appear in it, rendering an equation in one unknown only (v2). So a node equation for v1 is not required for finding Vth!

Shorting the output shorts out the 60Ω resistor. (What's the equivalent resistance of a 60Ω resistor in parallel with a 0Ω resistor?)

5. Sep 8, 2014

### Staff: Mentor

Yes that is one method. There is also a quite different-looking method, so don't confuse the two.

Is that a short-circuit I see across the 60Ω resistor?

Seeing that you need practice, why don't you solve this two different ways, and see whether you get the same answer?

Perhaps you should explain here, in words, what is involved in each of those separate methods, so you have it clear in your mind?

6. Sep 9, 2014

### orangeincup

Is one method finding the open source voltage to find V, then calculating the short circuit current to find i, and solving Rth from that?

Is the other method shorting out the voltage, and opening the current sources, to calculate Rth based off adding resistances?

7. Sep 9, 2014

### Staff: Mentor

That sounds right.

8. Sep 9, 2014

### orangeincup

I calculated it the first way correctly.

When I tried to calculate it the second way, I had trouble knowing which resistors are in parallel or series.

I redrew the circuit below.

Is 5+25, 10+60, both zero since they are parallel with a 0 resistor? So would Rth be 20?

9. Sep 9, 2014

### Staff: Mentor

No circuit is attached.

Rth is not 20Ω.

10. Sep 9, 2014

### Staff: Mentor

The 5+25 pair is shorted by the suppressed voltage source, but the 10Ω and 60Ω resistors are not. They are not in series, either! Look more carefully at the circuit when the supplies are suppressed. When you redraw the circuit you should remove any components made redundant (say by having been shorted by a suppressed voltage source), and feel free to rearrange the layout of the other components on the paper to make the connections clear.

Hmm. Looks like 20Ω to me...

11. Sep 9, 2014

### Staff: Mentor

Let me rephrase that ....

Rth is not the 20Ω resistor you see in the circuit.

But it is true that Rth evaluates to 20Ω, the same value as using the first method.

12. Sep 9, 2014

### orangeincup

Is (20Ω+10Ω)||60Ω=Rth correct?

Is the 5 and 25 resistors cancelled out because they no longer have a voltage?

#### Attached Files:

• ###### redrth.png
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13. Sep 9, 2014

### Staff: Mentor

Correct. They are shorted by the suppressed voltage source.

In your figure, if it's meant to show the analysis of Rth, you shouldn't include the short across the output terminals; That's for determining the short circuit current when any power supplies are still active.

14. Sep 9, 2014

### phinds

Well, not quite ... ONE END is but you can't just leave it at that. It matters a great deal that the other ends have been left open by the removal of the current source, thus removing them from the circuit. This is what I was talking about in post #2

orangeincup, "cancelled out" is not a good way to look at it, I think. Had they been completely short-circuited, then perhaps it would be a reasonable description but "cancelled out" doesn't seem to imply that you realize that they have been removed from the circuit by having one end disconnected.