3-22You can compute the values for view (C) and prove that point A in that circuit is also at virtualground.The whole point is that the inverting input to the operational amplifier shown in figure 3-13 is atvirtual ground since it is at 0 volts (for all practical purposes). Because the inverting input is at 0 volts,there will be no current (for all practical purposes) flowing into the operational amplifier from theconnection point of R1 and R2.Given these conditions, the characteristics of this circuit are determined almost entirely by the valuesof R1 and R2. Figure 3-15 should help show how the values of R1 and R2 determine the circuitcharacteristics.Figure 3-15.—Current flow in the operational circuit.NOTE: It should be stressed at this point that for purpose of explanation the operational amplifier isa theoretically perfect amplifier. In actual practice we are dealing with less than perfect. In the practicaloperational amplifier there will be a slight input current with a resultant power loss. This small signal canbe measured at the theoretical point of virtual ground. This does not indicate faulty operation.The input signal causes current to flow through R1. (Only the positive half cycle of the input signalis shown and will be discussed.) Since the voltage at the inverting input of the operational amplifier is at 0volts, the input current (I_{in}) is computed by:The output signal (which is opposite in phase to the input signal) causes a feedback current (I_{fdbk}) toflow through R2. The left-hand side of R2 is at 0 volts (point A) and the right-hand side is at E_{out}.Therefore, the feedback current is computed by:(The minus sign indicates that E_{out} is 180 degrees out of phase with E_{in} and should not be confusedwith output polarity.)