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Introduction

This section will be dedicated to the difficult circuits that you will probably have to solve. Let's start by just investigating a complicated circuit. There are three sections to this: Step Response and Phasors
To begin with we must now consider the fact that circuits consist of varying, multiple elements. As such, we must figure out ways to integrate resistors with capacitors with inductors.

RC Circuits: Discharging: V = V₀e^{-t / RC}
Charging: V = V₀(1 - e^{-t / RC})
General: V = V_f - (V₀ - V_f)e^{-t / RC}
RL Circuits: Discharging: V = V₀e^{-Rt / L}
Charging: V = V₀(1 - e^{-Rt / L})
General: V = V_f - (V₀ - V_f)e^{-Rt / L}
RLC Circuits: This is a little more complicated and the derivations can be found in a Basic electrical circuits book or by adding up the the voltages using KVL.
ω² = 1 / (LC)
α = 1 / (2RC) for parallel circuits
α = R / (2L) for series circuits

Step Response

The step response system is described as the actions taken in a circuit when you have a pulse input voltage. That is, going from one voltage to another voltage instantly. First thing that must be realized is the actions taken by individual elements.

Resistors act the same at all times.
C = dv / dt
L = di / dt

These formulas state that across certain elements a step response cannot happen. If there is a capacitor and the voltage steps, dt = 0 and thus C→∞ which cannot happen. Same with inductance if current steps. Next we have to take into consideration the idea of how these interact with one another. By using the general formulas, and the idea of ∞, we can fill out the general formula. Let's first look at a simple RC circuit.

V^- = V⁰ = V⁺ since voltage cannot change instantly in a capacitor.
As such let's find the voltage across the capacitor at t^-. Since in these circumstances the switch has been closed for a long time, or ∞, and we know that a capacitor acts as a short at ∞, the voltage across the capacitor at t^- = R₂ / (R₁ + R₂). This can now go into the formula as V₀. Now lets look at t→∞ after the switch is open. This is V_f.
Since there is no power source in the new circuit with the switch open the capacitor will completly discharge and will have a voltage of zero. So the formula for the voltage across the capacitor of this circuit after the switch is opened is V = [R₂ / (R₁ + R₂)]e^{-t / (R₂C)}
Now let's investigate the same circuitbut replace the capacitor with an inductor.

We will examine this circuit in the same fasion finding a formula for voltage across the inductor with respect to time. But since current is the controlling factor with an inductor, we will be taking all our figures with the idea of current flow instead of voltage.
i(0) = v₁ / R₁ since an inductor as t→∞ acts as a short.
After the switch is opened and t→∞ all the power will be consumed so i = 0.
So
i(t) = (v₁ / R₁)e^{-t(R₂ / L)}

Now wasn't that simple. Now for the next level, an LRC circuit, which isn't that much harder. Let's first look at a circuit.
We will be writing the equation for v(t) across the capacitor. Again we will be merely evaluating the circuit as time goes to infinity before the the switch is opened, and then after the switch is opened. The only other idfference is that in an LRC circuit we have to evaluate in a different order.
Going by the formulas above we need ot determine α and ω.
α = 1 ⁄ (2R₂C)
ω = 1 / √(LC)
s_1,2 = -α±√(α² - ω²)
These two quantities are than evaluated to determine the formula with respect to time.
If α>ω is Overdamped and the general formula is v(t) = Ae^-s₁t + Be^-s₂t
If α<ω then it is Underdamped and the general formula is v(t) = Ae^αcos(βt) + Be^αsin(βt) where β = √(ω² - α²)
Finally..
If α=ω then it is critically damped and the general formula is Ate^-αt + Be^-αt
Phew. That was alot and we are not quite done yet. We got some general formulas and now we need to fill in the constants. To do this we need some initial conditions. We know that at t = 0 v = 0 for this circuit across the capacitor. Find the voltage as time goes to infinity after the switch is closed and you have two formulas and two unknowns that you can out into a matrix.

Phasors

This is my favorite way to solve formulas because you can easily do this for any circuit and it turns everything into resistors which are easy to solve because you can revert to the oh so easy Ohm's Law. In order to do this though we need to know a couple things that I will get into as I need to. First lets look how we can convert capacitors and inductors into Ohm's. We call this reactance. To do this we need complex numbers and because i is used in complex numbers but it also is current we use j instead.
X_R = R + j0 Ω
X_C = 0 - j(1⁄(ωC)) Ω
X_L = 0 + jωL Ω
Now we can treat them all as resistors and add them up depending on their series on parallel configurations. Of course adding and subtracting them is easy size you can just add and subtract the real and imaginary components. But if we need to multiply or divide things get a little more hairy. As such we need a way to determine the phasor relationship so we can multiply and divide phasors.
General phasor representation.
|X_x|∠(ArcTan[imaginary component / real component])
X_R = R∠0
X_C = 0∠-90
X_L = 0∠90
Now we can rock Ohm's Law. Basically that is it. Phasors make it so much easier because everything is a resistor. You can even do Delta-Wye conversions replaceing the resistors with their Reactance values.