schematic diagram

I recently completed my final for my first year electrical Engineering Subject and I have the image of the question above.

I was really confused because it says when both switches are open, which leaves the inverting input of the op-amp (-) disconnected and ungrounded, which I now assume the op-amp has an internal ground at its negative terminal? (I might be wrong) so this would make the inverting terminal =0 and the non-inverting a fraction of 5V. Therefore I would figure that the op-amp keeps turning up the voltage until its limit. How on earth can you select the resistors to match a Vb of 0.5V as the question asks for?

Maybe I am missing something with how the internal value of the op-amp works, please help.

  • \$\begingroup\$ Haha, EGB120 at Queensland University of Technology I see. Hope your exam went well mate! \$\endgroup\$
    – DSWG
    Commented Nov 2, 2018 at 1:59

1 Answer 1


You are misinterpreting the schematic.

If both switches are open, the inverting terminal of the op amp will still be biased by the op amp output.

It will still be an inverting configuration amplifying voltage Va.

  • \$\begingroup\$ i see what you mean, thank you very much i completley missed where the input was indeed coming from. Too bad i cannot re-enter my exam, Thank you for your help though, much appreciated \$\endgroup\$ Commented Nov 2, 2018 at 1:49
  • \$\begingroup\$ @ReeceWoods technically the only input is at the non-inverting terminal as Va is floating. \$\endgroup\$ Commented Nov 2, 2018 at 1:53
  • \$\begingroup\$ so to set the vb to 0.5V the inverting input voltage would also be 0.5V? \$\endgroup\$ Commented Nov 2, 2018 at 2:05
  • \$\begingroup\$ That’s correct. \$\endgroup\$ Commented Nov 2, 2018 at 2:05
  • \$\begingroup\$ ahh what a relieft i actually did that as a sorta "this can't be right thing" in the exam, that's what i guessed, so maybe i will get some marks thank you again! \$\endgroup\$ Commented Nov 2, 2018 at 2:11

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