Op-Amp Circuit

It's a simple voltage divider. Remember that the input impedance of the op-amp is (ideally) infinite. Therefore, no current flows into the + input. As a result, the current starts from the node that is at potential V_A and flows straight to ground across the 100 ohm and 600 ohm resistors. In other words, these two resistors are in series, which means that the current is the same across both of them. From Ohm's law, you know that this current drawn is the voltage divided by the total resistance:

I = V_A / R = V_A / (100 Ω + 600 Ω) ​

Now, the voltage at the + input is just equal to the voltage drop across the 600 ohm resistor (this is clear from the circuit). By Ohm's law, this voltage drop is equal to the current flowing across it times its resistance:

V₊ = I(600 Ω)​

Now just substitute the original expression for I:

V₊ = [ V_A / (100 Ω + 600 Ω) ] * 600 Ω ​

V₊ = (V_A)(600 Ω)/ (700 Ω) ​

This is just a re-iteration of how a voltage divider works. The voltage is divided amongst each resistor in the series circuit in proportion to its ratio to the total resistance. Hence, 1/7 of the voltage appears across the 100 ohm resistor, and 6/7 of the voltage appears across the 600 ohm resistor.

 

a simple solution is by applying nodal analysis at V+ and V- and then u'll get the solution by applying V+ = V-