Why Am I Misunderstanding the Output Resistance in This Circuit Analysis?

[gl0ck]

Output resistance problem

Homework Statement

Hi,
I am not getting the answers right, and I think I misunderstand the logic behind this circuit.

http://imageshack.us/photo/my-images/577/screenshotfvm.png/

Homework Equations

I know that the equation of Vin is Vin=(VsRin)/(Rs+Rin)
and the equation of Vout is Vout=(GVinRl)/(Rout+Rl)

The Attempt at a Solution

I have found Vin right just from the equation to be equal to 5V.
But when I had to calculate the Vout I found to be 25V instead of 50V(given answer)
Also on the other question where I have to find the output resistance, I think it will be 200ohms, but it is given to be 100Ohms. The logic I found here is the 2 final resistors to be calculated as Parallel and then I get the answer of 100Ohms. Also the final question where is said that if we remove the Rl the what should be the output resistance. And I think it will be 100 because of the Rout is 100Ohms but Again when the previous part is not correct with my logic I expect here to misunderstand something.

Thanks very much!

 

[NascentOxygen]

I have found Vin right just from the equation to be equal to 5V.
But when I had to calculate the Vout I found to be 25V ✔ [/size][/color]

Also on the other question where I have to find the output resistance, I think it will be 200ohms, but it is given to be 100Ohms. The logic I found here is the 2 final resistors to be calculated as Parallel and then I get the answer of 100Ohms.

A parallel pair of 100Ω resistors does not make 100Ω.

The poor wording of the question is reprehensible.http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon8.gif I think in (d) they should be asking you to determine the output impedance with R_L in situ. Output impedance is going to have a value less than R_L, anyway, so 200Ω can't possibly be right. (Side note: the only way to make output impedance greater than the load is through the use of positive feedback from the output, and as you can see, there is no feedback at all from the output.) This has the hallmarks of a trick question: what is the resistance connected across the output terminals of the amplifier. Well, with the amplifier being labelled as what's inside the box, anyone can see there's a 100Ω resistance connected across that box's terminals, so the logical answer must be 100Ω. But is that worth 3 marks? Unlikely, so I infer that what they asked is not what they intended to ask. For 3 marks they probably want you to determine the impedance measured across R_L with R_L in situ. And the answer won't be 100Ω.

Also the final question where is said that if we remove the Rl the what should be the output resistance. And I think it will be 100 because of the Rout is 100Ohms ✔[/size][/color]

 

[gl0ck]

Thanks,
sorry for the mistake the output resistance is given to be 50 Ohms.
which gives parallel pair of 2 100ohm resistors but I am not sure this is the right logic.