Fig. 2
Solution:
We begin by writing the KCL equations at both the + and – terminals of the op-amp.
For the negative terminal,

Therefore,
15 v- = vout
For the positive terminal,

This yields two equations in three unknowns, vout, v+ and v-. The third equation is the relationship between v+ and v- for the ideal OPAMP,
v+ = v-
Solving these equations, we find
vout = 10 vin = 5 sinωt V
Since 2 kΩ resistor forms the load of the op-amp, then the current iout is given by

For the different amplifier shown in fig. 3, verify that

Fig. 3
Solution:
Since the differential input voltage of OPAMP is negligible, therefore,
v1= vx
and v2 = vy
The input impedance of OPAMP is very large and, therefore, the input current of OPAMP is negligible.
Thus
And
From equation (E-1)
or
From equation (E-2)
or 
The OPAMP3 is working as differential amplifier, therefore,
GOTO >> 1 || 2 || 3 || Home