# Currents in circuits

I know how to find the currents as functions of time when I have various circuits involving inductors, resistors and capacitors. Like for instance, if I have the following circuit:

Then I have 4 currents to analyze. Using kirchoff's laws, I can obtain 4 equations, two involving time derivatives thanks to the inductors. Since there are 4 equations and 4 functions, I get a system of differential equations, which I may or may not be able to solve using various methods.

However, I've now come across many different problems where you're supposed to only find the currents "immediately after closing S" and "a long time after closing S". It feels like this approach that I'm using is rather complex, and that there is a better way.

What I've tried to do is to substitute something that I know is true of the circuit immediately after closing S and a while after closing S. However, I really cannot find anything that I think is obvious. After a long time, it usually helps to assume that $\frac{di}{dt} \longrightarrow 0$ for most currents. Even though this assumption is based on the circuit reaching some sort of equilibrium state asymptotically, I don't think you can substitute anything with a stronger argument unless you actually solve the differential equations.

However, I really cannot figure out what to substitute in order to know the currents at time $t = 0$. It's hard because it feels like you need to know what $\frac{di}{dt}$ is at $t = 0$, but I don't see how you can find that out. Note that I've tried to find an explanation in the provided answers, but they don't show how the problem is solved. Actually, I've tried looking for this everywhere, and I really cannot find any resource where they explain how you should think.

Another example of a problem where we're supposed to do something similar is:

In this case, after a long time, you can probably assume that the current into the capacitors is $0$, meaning that the voltage over the resistors in series with them is $0$. However, what happens immediately after closing the circuit?

Basically, it feels like there's no easy way of thinking that works for most circuits. Therefore, to summarize my questions:

1. In the first picture, how do you go about finding the currents at $t = 0$?
2. In the second picture, how can you find the currents at $t = 0$?
3. Most importantly though: Is there any sort of general way of solving these problems? How should you think? Are there any basic rules you can check when you stumble into a problem like this, in order to find the "initial" and "long time" currents?

Switching on and thinking in the $\mathrm{d}/\mathrm{d}t$ properties of $C$ and $L$ means that the resistance of an inductance is basically infinity and that of a capacitance is zero. For long times the exact opposite is the case. In both cases you just have to solve a simple circuit of resistors. In the first case you open the circuit diagram at every coil and substitute every capacitor by just a cable. In the second case the other way around.
• Oh, I understand! For the capacitor, the initial charge is $0$, which I assume is the reason why it's like a cable at $t = 0$ (because $V = Q/C$). However, I don't understand why an inductance "acts" like a infinite resistance-resistor. I mean $di/dt$ can't be infinite, because that wouldn't be consistent with kirchoff's 2nd law. Which assumption is it then based off? – Max May 22 '17 at 11:25
• Now that I think of it, is it really obvious that the capacitor is like a cable at $t = 0$ just because the charge is $0$? I mean technically, since the cables are assumed to have zero resistance, couldn't it be possible that the charge is rearranged "infinitely fast", meaning that the initial charge wouldn't be $0$? Like, if you connect a capacitor directly to a battery. – Max May 22 '17 at 11:28
• @Max For the capacitor you have $i = C \mathrm{d}u/\mathrm{d}t$. If you agree with this formula for the ideal capacitor, it is clear that if you switch on the voltage (infinite slope) you get an infinite current. This is independent of the charge already present on the capacitor. It does not need to be zero. For "switching on", hence, the capacitor is like a short circuit. – mikuszefski May 22 '17 at 12:52
• @Max And again it is the opposite for the inductance. With $u = L \mathrm{d}i/\mathrm{d}t$ you see that a jump in current would result in an infinite voltage. So the current in an inductor is not discontinuous. As it is zero in the beginning it will stay like this for the first instance of time. – mikuszefski May 22 '17 at 12:59