# Ideal op amp circuit with variable voltage gain, how do I calculate the current?

I am currently calculating various quantities for the circuit shown below. However, I am stuck on how to calculate the current (Ix) and the input resistance (0 output current). The circuit is an ideal op amp circuit.

There should be a virtual short between the two opamp inputs so that in the active region, the are at nearly identical voltages. I am still unsure how to calculate the voltage at the non-inverting input.

• I wonder, did electronics.stackexchange.com exist 4 years ago? Pretty sure it did. Jul 27 at 10:25
• Ok, for new readers ... Jul 27 at 10:27

The key to doing these problems the "easy way" is to think how "cause and effect" works in the circuit.

The potentiometer is connected directly across the input voltage. An ideal op amp draws no input current, so the current in both parts of the potentiometer is the same. Therefore $V_x$ is just $\alpha V_i$.

Now look at R1. For an ideal op amp operating in the linear region, the two inputs are at the same voltage, so one end of R1 is at $V_x$. The other end is at $V_i$.

So you can calculate $I_x$ by Ohm's law.

Because the ideal op amp draws no input current, there is only one place for the current $I_x$ to go after it has gone through R1, and that is through R2. So you can use Ohm's Law again to find $V_o$.

The "hard way" to do the problem is set up all the circuit equations using KCL and KVL and then solve them. That's a good way for a computer program to simulate the circuit, but it's not what human engineers should be doing IMO - a bit of insight and understanding always beats a lot of grunt work!

With "little" equations ... for real and ideal opamp.