Your answers for impedance would only be correct if the opamp was the only component.As the question said the Op-amp is Ideal so I thought this could be the approach.
This is homework. You should delete your post.so the input resistance is
the other 0 is a bit far, Sorry for the inconvenience, Sir. The voltage gain calculation is easy. And as Mr Bob said the input resistance will be considered the 1K Ohm. I've understood that now, But what about the output resistance?? will it be the reference voltage i.e. 1 M ohms?hi 24,
10^6 / 10^3 = ????
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It's unfortunate that you were able to read his post before it could be removed. That isn't the way homework help is supposed to work. We're supposed to give hints to make you remember what you were taught. Giving you answers won't help you on tests and most schools would consider getting answers from others to be cheating.as Mr Bob said the input resistance will be considered the 1K Ohm.
hi 24,will it be the reference voltage i.e. 1 M ohms?
Sir, I have calculated the input impedance.It's unfortunate that you were able to read his post before it could be removed. That isn't the way homework help is supposed to work. We're supposed to give hints to make you remember what you were taught. Giving you answers won't help you on tests and most schools would consider getting answers from others to be cheating.
Don't guess.But what about the output resistance?? will it be the reference voltage i.e. 1 M ohms?
The example op amp is ideal.But there is no data sheet that was supposed to contain Zout, Aol.
Your formula for Zout contains the feedback effect (reduction factor due to loop gain) - and this is totally correct.Sir, the 1 M Ohm and Op-amp are parallel to each other.
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But there is no data sheet that was supposed to contain Zout, Aol.
Like for this example:
View attachment 273254
No - consult your own result in post#11.I get it now, the Output resistance of the circuit equals the output resistance of the OP-AMP i.e., zero ohms.
hi @studyhard24I get it now, the Output resistance of the circuit equals the output resistance of the OP-AMP i.e., zero ohms.
Yes, thats correct as far as the original question (idealized opamp model) is concerned.hi @studyhard24
In the original question the OPA is stated as Ideal, so your answer to this question, saying the output resistance is Zero ohms, is correct.
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