Op-Amp transient modeling

This was already mentioned by Russel, but I hope to present it in a different way.

The main problem here, it seems to me, is that your book (or whatever source of information you're using) missed one important point: The voltage between inverting and non-inverting inputs of an ideal operational amplifier should always be zero with this and similar setups. If we include that assumption and take a look at the circuit, we can get a logical answer.

The output of an op-amp is modelled as an ideal controlled voltage source. The input impedance is infinite and no current flows into the op-amp. So far so good. Next, we know that the voltage between the inputs is zero, so we know that the voltage with respect to ground and the inverting input is same as the one on the non-inverting input. That voltage comes from the ideal controlled voltage source at the output. Next, let's take a look at the current issue. Since we have infinite input impedance, no current flows into the operational amplifier, so from where does the output current come? Well from the ideal controlled voltage source at the output.

As I said, the voltage source is ideal, so it can source infinite current, it's controlled so you have your gain, the current is set by resistor and there's no contradiction there at all. In reality, the current will come from the power supply pins and be limited by construction of the operational amplifier, but this is a mathematical model. So let's take a look at a pretty pictures now:

The first image may seem a bit drastic:

I've crossed out the op-amp on purpose here. It seems to me that trees are obstructing your view of the forest here. If we remove the op-amp symbol and take a look at how we're supposed to model it instead (note the \$ 100 \mbox{ }G \Omega \$ resistor):

We can clearly see that the current is coming from the one terminal voltage source which is the output of the op-amp.

Next, I'll show a bit more complex version of the same circuit and explain how it degenerates into what you've shown:

Let's see what we can see here:

We've got the input voltage \$U_i\$, the output voltage \$U_o\$ and the resistors \$R_1\$ and \$R_2\$.

Now we know from our model that the voltage between the inputs is zero, so we can write following safely: \$U_i-R_1I=0\$, since the resistor \$R_1\$ has a short circuit to inverting input. From that we get the current: \$I=\frac{U_i}{R_1}\$. The current can only come from the op-amp output in this case, so we know that it is the current going through the resistor \$R_2\$ too. From that we get the equation for the output voltage of the op-amp: \$ U_o-R_2I-U_i=0\$ and after that: \$U_o=R_2 I + U_i= R_2 \frac{U_i}{R_1} + U_i=U_i(\frac{R_2}{R_1}+1)\$. From this, we have \$ \frac{U_o}{U_i}=1+ \frac{R_2}{R_1}\$. In the circuit you showed, equivalent elements would be \$R_2=0\$ and \$R_1=\infty\$. As you can see, the output current isn't a problem with this setup and again, there's no contradiction here.

With the few assumptions I've shown and few equations, you can do basic op-amp circuits without any problems. I recommend that you read from freely downloadable books Amplifiers and Bits: An Introduction to Selecting Amplifiers for Data Converters pages 6 and 7 and from Op Amps for Everyone Design Guide chapter 3 (or at least take a good look at the pictures there). Both books (well, a book and an application report) are by Texas Instruments (a major op-amp manufacturer) and should come up on most popular search engines as the first response.

Since this is a homework problem, I'm not just going to give you the answer. However, consider simplifying the circuit first. You have a current source and a voltage source. Do you remember some techniques for merging them to a single equivalent source with a single resistance? Look up "Thevenin" and "Norton" if not.

Also consider the inductor in steady state. Inductors and capacitors can both be greatly simplified for the purpose of steady state analisys. What is the simplification of a inductor in steady state? Using that, what is the inductor voltage and current in steady state in both cases?

You need to show some work on your own before we can continue and eventually get to the solution.

Added:

It seems you have made some correct observations, but I'm not following all you are saying. You are right in that the inductor is a short circuit in steady state.

There are really two parts to this problem. Find the inductor current before the switch closes, then apply that as the initial condition for the circuit with the switch closed. The circuit with the switch open is irrelevant other than to find the inductor current.

For the switch open steady state case, the voltage accross the inductor and R2 is 0. Note that R1 is irrelevant as it is in series with a current source. Since the inductor voltage is 0, the two sources are essentially decoupled and the current contribution from each can be simply added. The current source tells you its current outright, so that's 10 A thru the inductor. The 30 V source has 2 Ω load on it, so that's another 15 A. The inductor current is therefore 25 A. That's the only relevant piece of information from before the switch closes.

For calculating the response, I would first calculate the steady state inductor current with the switch closed. Then, transform the two voltage sources and various resistors into a single Norton or Thevenin source. Actually you only need the resistance of that source. You know the inductor will exponentially decay from its initial value to the steady state value with a time constant of L/R.

So go ahead and perform these simplifications, tell us the final inductor current and the effective impedance it sees. The answer is then just a simple exponential.