The topic of this problem is Operational Amplifier Circuits and the problem is to determine Vo and Io in the circuits shown below. In our circuit we have one independent voltage source which is tied to the non-inverting input of our op-amp. And we have some resistors as part of our resistor network at the output of the op-amp. We have an output current I sub 0 which is measured through the 10 kilo ohm resistor. And we have Vo which is measured across the 10 kilo ohm resistor from the top to the bottom where the bottom is our ground node. So, we're trying to find Vo and Io in this problem. We need to use our properties of an ideal op-amp that are important for the newer circuit analysis. First of all, we draw our symbol for the op-amp. It's got its inverting input at the top and non-inverting input at the bottom. We have currents associated with both of these inputs We have voltages associated with each one of these inputs, and we know for the ideal op-amp that the currents into the op-amp are equal to zero. And we know the voltages, whether it's the inverting or non inverting terminal are equal to one another. So V+ = V-. So we're going to use those properties to solve these op-amp circuits. So if we use those we know that, first of all our voltage at the non inverting input of our op-amp is going to be 12V. We also know that the voltage at the inverting input of our op-amp is also 12V through our properties of op-amp. And if it's 12V at the inverting input, then it's 12V at this node between the 2k and the 12k resistor, so that's also at 12V. So if that's at 12V, then we can find this current through the 2 kilo ohm resistor because we also know the voltage down here, it's ground. So it's 12V on the top of the resistor and 0V at the bottom. So the current through here, which we might call I2k. It's going to be equal to (12V- 0V) / 2 kilo ohms. So we end up with 6 mA for the current through the 2 kilo ohm resistor. We also know that the current that's flowing down through the 2 kilo ohm resistor is the same current which is flowing through the 12 kilo ohm resistor. And the reason we know that is because there is no current flowing back into the op-amp. The other property of our op-amp that we need to use to solve this problem is that the currents into the op-amp are equal to 0. So there's no current through this particular connection between the op-amp and the 12 and 2k resistors. So I2k is also flowing through this 12 kilo ohm resistor. So if we're looking for Vo which is the voltage from the top of the circuit of the output to ground, then we can find it by summing of the voltage across the 12k resistor and the voltage across the 2k resistor together. So I2k is 6 mA, so we can easily find the voltage across the 2 kilo ohm resistor and we can also find the voltage across the 12k. So, Vo = V12k + V2k = 6 mA (14K) total, the 12K and the 2K. So if we multiply this out we end up with a Vo which is 84V. Put 12V into our op-amp circuit, we're getting 84V out of it across our 10k load resistance. If we know what Vo is then we can find Io as well. Because we know that Vo / 10k, through Ohms Law is going to be equal to Io. So Io comes out to be 8.4 mA.
