RC-RL Circuits

In general, for an RC circuit, $V_{c} (t) = k e^{- t / τ}$ , where $k$ and $τ$ are constants
For a circuit with no independent sources:
- $τ = R_{T H} C$ , if you consider the capacitor the load
  - $τ$ is the time constant in seconds
- $k = V_{c} (0)$
We say that after $5 τ$ , it's pretty much stable

Simple RC Loop

Consider a charged up capacitor and a resistor in a loop
$V_{c} (t) = V_{0} e^{- t / R C}$
In this case, $τ = R C$ , and we can find $k$ by setting $t = 0$ where $V_{c} (0) = V_{0}$
$i_{c} (t) = C \frac{d V_{c} (t)}{d t} = - (\frac{V_{0}}{R}) e^{- t / R C}$
The capacitor initially acts like a voltage source, then turns into an open circuit

All in series
Consider the current $i (t) = \frac{V_{s} - V_{c} (t)}{R} = C \frac{d V_{c} (t)}{d t}$
Rearranging gives $\frac{d V_{c} (t)}{d t} + \frac{1}{R C} V_{c} (t) - \frac{1}{R C} V_{s}$
By the inability to solve differential equations, we get $V_{c} (t) = k_{1} + k_{2} e^{- t / τ}$
- If we bring time to $\infty$ , then $V_{c} (\infty) = k_{1}$
- Then at $t = 0$ , $V_{c} (0) = V_{c} (\infty) + k_{2}$ , so $k_{2} = V_{0} - V_{c} (\infty)$
- So $V_{c} (t) = V_{c} (\infty) + (V_{0} - V_{c} (\infty)) e^{- t / R C}$
- $V_{c} (t) = V_{0} e^{- t / τ} + V_{c} (\infty) (1 - e^{- t / τ})$ , so you exponentially go from $V_{0}$ to $V_{c} (\infty)$

Take a voltage source, resistor, and inductor in series, for example

When trying to solve ODEs for these circuits in AC, you can assume the solution is of the form $i (t) = I \sin (ω t + θ)$
Actually that's stupid, instead do $i (t) = ℑ {I e^{j (ω t + θ)}}$ , and $v_{s} (t) = ℑ {V_{p p} e^{j ω t + ϕ}}$