# Ideal Op Amp and its Input Voltage

In an ideal operational amplifier, I understand that the inputs draw no current (due to infinite impedance). So, in the example below, the value of Vp would equal the value of the EMF (because there is no voltage drop across the resistance), which is 1V.

Unless I'm mistaken, the definition of voltage at a particular point is the amount of electric-potential energy per coulomb of charge. I'm wondering how exactly the electric-potential energy from the EMF gets "transferred" (for lack of a better of term) across the resistor and to another section of the wire if there is no movement of electrons?

Also, since there is no current, then what's the purpose of the 8.9kΩ resistor if there is no voltage drop?

Image Source: http://raise.spd.louisville.edu/EE220/L13.html

• The 8.9k resistor is to minimize the impact of bias currents, which are a property of real op amps, and don't enter into ideal op amp discussion Mar 9, 2016 at 1:11
• From the image source: "There is a reason for the presence of the 8.9 kΩ resistor. It is the parallel combination of the resistors connected to the negative input. Real op-amp circuits have tiny "bias currents" that flow into the positive and negative inputs. By balancing the resistances seen by these two terminals, the effects of the bias currents tend to cancel out." Mar 9, 2016 at 1:37
• Ah thank you! @tangrs I tried scanning through the webpage to ensure that I'm not being redundant, but managed to miss that piece of information. Mar 9, 2016 at 1:47

If it helps, also remember there is a small parasitic capacitance from the node labelled $v_p$ to any other nearby conductive objects (which we usually model as just being connected to the ground node). So when you first connect the 1 V battery, current does flow through the resistor, but only until those parasitic capacitors are charged up to the point that the potential is equal and current stops flowing through the resistor.