The topic of this problem is Operational Amplifiers, and
we're going to work with a circuit that is an inverting amplifier.
The problem is to find V sub out in terms of the input voltages.
In our case, we have one input voltage, it's V sub n,
that's attached to a branch with Rs of 1,
and ultimately, to the inverting input of our op-amp.
We know that, for op-amps, we have a number of properties,
which help us solve our circuits.
So, if we look at, just a general op-amp configuration,
and this is the circuit symbol for it, we have currents associated
with the inverting input and with the non-inverting input.
We also have voltages associated with those inverting and
non-inverting inputs as well.
And we know that some of the properties of ideal op amps
are that the current into the op amp are equal to 0.
They're equal to each other and they're equal 0.
We also know that another important property for us when we're solving for
these op amps is that the voltage at the inverting input
of the op am is equal to the voltage at the non inverting input of the op amp.
So V minus is equal to V plus.
So we're going to use these two properties when we solve circuits
with op amp in them for linear circuit applications.
So let's look and see how we can solve the problem that we have in this example,
using our the two properties of the op amp.
If we use those properties, one thing that we know is that the voltage at this point
is equal to the voltage at ground, which is zero volts.
So we're at 0 volts at this non-inverting input.
So if it's 0 volts at the non-inverting input, using our second
characteristic of op amps, then we're also at 0 volts at the inverting inputs.
So that's 0 volts as well at that point.
So once we have that,
we can use tools that we've already learned in the past to solve this problem.
Again, we're looking for V out in terms of the input voltages.
In our case the one input voltage which is tied to our inverting input.
So if we use Kirchhoff's current law at that node, we'll call it node one.
This is Kirchhoff's current law at node one,
then we can sum the currents into that node.
We have a current through R1, we have a current through R sub F and
we have the zero current associated with the non-inverting or the inverting input.
So let's first look at the current through R sub 1.
It's going to be the N minus 0 over R1.
