Worked exercise — charging RC circuit

Problem

A capacitor $C = 200\;\mu\text{F}$, initially uncharged, is charged through a resistor $R = 10\;\text{k}\Omega$ by a battery $V_0 = 12\;\text{V}$. Calculate: (a) the time constant $\tau$; (b) the voltage across the capacitor at times $t = \tau$, $2\tau$, $5\tau$; (c) the energy stored at steady state.

Battery $V_0$, resistor $R$ and capacitor $C$ in series in a single loop.

Solution

(a) Time constant. $\tau = RC = 10^4 \cdot 2\cdot 10^{-4} = 2\;\text{s}$

(b) Voltage across the capacitor. While charging it follows the law $V_C(t) = V_0\,(1 - e^{-t/\tau})$: $V_C(\tau) = 12\,(1 - e^{-1}) \approx 7{,}59\;\text{V}$ $V_C(2\tau) = 12\,(1 - e^{-2}) \approx 10{,}4\;\text{V} \qquad V_C(5\tau) = 12\,(1 - e^{-5}) \approx 11{,}9\;\text{V}$ After five time constants the capacitor is practically fully charged (deviation from $V_0$ below 1%).

(c) Energy at steady state. With full charge $V_C \to V_0$: $U_\infty = \tfrac{1}{2}\,C\,V_0^2 = \tfrac{1}{2}\cdot 2\cdot 10^{-4} \cdot 144 = 1{,}44\cdot 10^{-2}\;\text{J} = 14{,}4\;\text{mJ}$ Note: during charging an equal amount of energy is dissipated in $R$ (a general property of RC circuits).

$\ev{\tau = 2\;\text{s},\quad V_C(\tau)\approx 7{,}59\;\text{V},\ V_C(2\tau)\approx 10{,}4\;\text{V},\ V_C(5\tau)\approx 11{,}9\;\text{V},\quad U_\infty = 14{,}4\;\text{mJ}}$

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Topics: Electric circuits Concepts: RC circuit