Problem — charging a capacitor in an RC circuit

Problem

A $10\ \mu\text{F}$ capacitor is in series with a $10\ \text{k}\Omega$ resistor and a $12\ \text{V}$ battery (switch initially open). At instant $t = 0$ the switch is closed. Calculate the time constant of the circuit and the voltage across the capacitor after $0{,}15\ \text{s}$.

Solution

Time constant: $\tau = R\,C = 10\cdot10^{3}\ \Omega \cdot 10\cdot10^{-6}\ \text{F} = 0{,}10\ \text{s}$

Voltage across the capacitor. While charging, starting from uncharged, the capacitor follows $V_C(t) = \mathcal{E}\left(1 - e^{-t/\tau}\right)$ At $t = 0{,}15\ \text{s}$ the ratio is $t/\tau = 0{,}15/0{,}10 = 1{,}5$, and $e^{-1{,}5} \approx 0{,}223$: $V_C = 12\left(1 - 0{,}223\right) = 12 \cdot 0{,}777 \approx 9{,}3\ \text{V}$

After $1{,}5$ time constants the capacitor has reached about $78\%$ of the battery voltage.

$\ev{\tau = 0{,}10\ \text{s} \qquad V_C \approx 9{,}3\ \text{V}}$

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Topics: Circuiti elettrici Concepts: Circuito RC