PX157 - B14c - circuits - RC circuit

charging a capacitor:

at $t = 0 :$
- capacitor is not charged
- the circuit is closed:

\begin{aligned} I_{0} = \frac{V_{a b}}{R} & = \frac{ϵ}{R} \\ Q_{0} & = 0 \end{aligned}

at $t = t^{'} :$
- from kirchhoff's loop rule:

\begin{aligned} ϵ & = R I + \frac{Q}{C}, & I & = \frac{d Q}{d t} \\ ϵ & = R \frac{d Q}{d t} + \frac{Q}{C} \end{aligned}

- for $t \to \infty :$ the capacitor to be fully charged:

ϵ = V_{b c} = \frac{Q_{f i n a l}}{C} ⟹ Q_{f i n a l} = ϵ C

- solving a first order ODE : try $Q (t) = \tilde{Q} (t) e^{- t / R C} :$
$$\tilde Q(t) = \int_{0}^{t} \frac{\epsilon}{R} e^{- \frac{t'}{RC}},dt' = \epsilon C (e^\frac{t}{RC}-1)$$

⟹ Q (t) = ϵ C (1 - e^{- \frac{t}{R C}})

I (t) = \frac{d Q}{d t} = \frac{ϵ}{R} e^{- \frac{t}{R C}}

typical rate of charging is given by the time constant, or the relaxation time :

τ = R C

discharging a capacitor

at $t = 0 :$
- the battery is disconnected
- $I_{0} = 0$ , $Q_{0} = ϵ C$
- loop rule:

\begin{aligned} R I + \frac{Q}{C} & = 0 \\ R \frac{d Q}{d t} + \frac{Q}{C} & = 0 \end{aligned}

- solving the C - first order ODEs :

\begin{aligned} Q (t) & = Q_{0} e^{- \frac{t}{R C}} \\ = ϵ C e^{- \frac{t}{R C}} \end{aligned}

- current:

I (t) = \frac{d Q}{d t} = - \frac{ϵ C}{R C} e^{- \frac{t}{R C}} = - \frac{ϵ}{R} e^{- \frac{t}{R C}}