What do we need the op amp for in this current-to-voltage converter?

I'm still very new to electrical engineering and at the moment trying to get my head around the concept of the ideal op amp.

In Fundamentals of Electric Circuits by Alexander and Sadiku the following diagram is given as a "current-to-voltage converter".

simulate this circuit – Schematic created using CircuitLab

The reader is asked to prove $$v=-IR,$$ which I think I did. But what do we need the op amp for? Wouldn't the following diagram be exactly the same?

simulate this circuit

• No it is not the same ! What is the voltage across the current source in the 2nd circuit ? Now what is it in the 1st circuit, you already know that an opamp will make the voltage between it's inputs ...? – Bimpelrekkie Jul 1 '15 at 9:27
• @IC_designer_Rimpelbekkie: OK, I see what you're getting at. In the first circuit the voltage across the current source is zero. In the second circuit it need not be. So from the second diagram we cannot conclude anything about v. Correct? – Stefan Jul 1 '15 at 9:42
• 100% correct ! :-) – Bimpelrekkie Jul 1 '15 at 9:44
• The most important property of the opamp based circuit is that it constitutes a voltage SOURCE - which means: It provides the output voltage with an extremly low output resistance (Mikro...Milliohm range). Hence, any load does not change the voltage. – LvW Jul 1 '15 at 10:12