# Output Resistance for this circuit?

Discussion in 'Homework Help' started by jegues, Feb 26, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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See figures attached for problem statement as well as my attempt.

I don't see how I can find the output resistance for this circuit.

I made an attempt to solve for $\frac{V_{o}}{i_{L}}$, but it's still in terms of one unknown, RL.

How do I obtain the output resistance? Also, does anyone see any problems with my answer for part a)?

Thanks again!

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I don't think they want the output resistance at the Vo node.

They want the resistance seen by RL. Calculate the voltage across RL and then divide that by IL. That will be the equivalent of RL in parallel with the output resistance from which you can calculate the output resistance.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Is the answer simply RL then?

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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That's not what I get. Show your calculations.

5. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Here's my best attempt at figuring it out. Please respond ASAP, I have a test on this today and I need to know how to do this.

Can you show me your solution?

Thanks again!

EDIT: Does Rout simply approach infinity?

• ###### Rout.JPG
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Last edited: Mar 4, 2011
6. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Bump, still looking to clear this up ASAP

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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In the left hand column of your image, you've made an algebra error. Part way down you have:

iL - iL/10 - (iL/10)*(RL/R) = 0

which in the next line becomes:

iL*(1 - 1/10 - RL/R) = 0 which is wrong; it should be:

iL*(1 - 1/10 - (1/10)RL/R) = 0

You then draw the conclusion that this implies that iL = 0. This is not so, because if the quantity in parentheses is zero the equation will also be satisfied. In other words, if:

iL*(1 - 1/10 - (1/10)RL/R) = 0

this could imply that iL = 0.
But it also could be because (1 - 1/10 - (1/10)RL/R) = 0, which is the case here. This expression allows you to calculate R so that iL = 10*iI

However, this is not needed to calculate the output resistance which is driving RL. Let us call that output resistance Ro. We know that the resistance seen between Vy and Vo is just RL || Ro; that is, RL in parallel with Ro, or (RL*Ro)/(RL+Ro).

You have VL/iL = (Vy-Vo)/((Vy-Vo)/RL) = RL; but VL/iL is equal to RL || Ro.

Thus, 1/RL = 1/RL + 1/Ro. From this we get Ro = ∞