I asked this question in another forum ..
Here is one of the replies ...
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The output voltage is:
Vo = Vi / ( 1- 1/A)
where A is the open loop gain of the op amp. so A>>1
you can disregard 1/A then you have.
Vo=Vi
That is a voltage follower. In theory the same as the standard one with negative feedback.
But in real life the output will smack against one rail or the other and stay there. Any difference between the two inputs is amplified and increases the difference rather than reducing it.
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So what he is saying ..
If the Op-amp is ideal .. ( I+ = I = 0 ) and ( V+ = V- ) .. Then , Vin = Vout in this circuit ..
and if it is not ideal ,, the op-amp will be saturated and Vout will be equal to the value of the external power supply ..
What do you think of this???