So, the first thing we're going to use is this first property of the circuit that
there's no current flowing into the ideal op amp.
So if there's no current flowing into this non inverting input of the op amp,
that means that I out is a current which is confined to this loop that we have
at the bottom of our circuit.
So if we want to find I out, we can use Kirchhoff's Voltage Law to do that and
we can use Kirchhoff's Voltage Law around that mesh.
And in this case I out is measured in a counter clock wise direction.
So we're going to sum our voltages up counter clock wise around this circuit.
So if we start at the bottom right hand corner and start around the circuit
we first encounter the negative polarity of the minus 5 volts source.
So it's minus 5 volts.
And then we have the volts that drop across 20 kilohm resistor which
is 20k(I out) plus the voltage drop across 10k as we continue around this loop.
And the voltage around the 10k is going to be 10k(I out),
and then we return back to the starting point.
And so we know through cross voltage law the the sum of those
voltages is equal to 0.
So if we use that equation and we solve for I out,
then we end up with an I out which is equal one-sixth milliamps.
