Output Resistance for this circuit?

Thread Starter

jegues

Joined Sep 13, 2010
733
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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The Electrician

Joined Oct 9, 2007
2,971
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.
 

Thread Starter

jegues

Joined Sep 13, 2010
733
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.
Is the answer simply RL then?
 

Thread Starter

jegues

Joined Sep 13, 2010
733
That's not what I get. Show your calculations.
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?
 

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Last edited:

The Electrician

Joined Oct 9, 2007
2,971
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 = ∞
 
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