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Why is i20 = (Va-Vb)/20k, not Va/20k?

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    \$\begingroup\$ Given what you see there, what voltage should appear at the (-) input node of the opamp? (This isn't a hard question.) \$\endgroup\$
    – jonk
    May 5 at 0:03
  • \$\begingroup\$ Ask yourself what's the voltage drop across the 40 kOhm resistor? What do you know about the voltage between the + and - terminals of an op-amp with high gain and negative feedback? \$\endgroup\$
    – John D
    May 5 at 0:04
  • \$\begingroup\$ It might be misleading to see this i100 in the opposite direction of i20 because this is the inverting input and you assume no significant input current. So the output goes down with up in i20 so here i20= - i100 or if shown in the same direction are actually the same currents. But Vin+ is your null input reference so feedback forces Vin to follow Vin+=Vin- which are not 0V. \$\endgroup\$ May 5 at 0:25
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    \$\begingroup\$ @samlikesham Because the opamp OUTPUT is hooked backwards towards the (-) node and can PULL current. Nothing like that exists at the (+) input node. So it cannot happen. But it sure can happen at the (-) node. And the opamp will wiggle its OUTPUT in order to make sure that the (-) node is the same as the (+) node. Yes? \$\endgroup\$
    – jonk
    May 5 at 1:24
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    \$\begingroup\$ @samlikesham Glad that helped. :) It's not so complicated once you grasp a few things and keep them in mind. \$\endgroup\$
    – jonk
    May 5 at 1:37

1 Answer 1


When an OpAmp is in equilibrium (ie: not in saturation) and in a negative feedback configuration, the + and - are practically the same. If we neglect the bias current for + and - it is fair to assume that vb = v+, is also = to v-.
So if the R20K sees va at one end and vb at the other end it is fair to mention that V(R20k) = va-vb,
since i20=V(R20k)/20K then the following is true: i20 = (va-vb)/20K.

It cannot be va/20k because v- is not zero, rather it is equal to vb
(See first postulate in phrase 1).
Of course, it would be true only if vb = zero

  • 1
    \$\begingroup\$ Note that your first statement is only true if the op amp is in a circuit with negative feedback and the output has not saturated...it is not true of op amps in general. \$\endgroup\$ May 5 at 9:36
  • 1
    \$\begingroup\$ Point well taken. I am editing my answer. \$\endgroup\$ May 5 at 13:15

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