Charging a capacitor

Problem

Reading a graph: charging a capacitor. The voltage $V_C(t)$ across a capacitor being charged through a resistor is measured:

$t$ (s)	$0$	$1$	$3$	$10$
$V_C$ (V)	$0$	$6{,}3$	$9{,}5$	$10{,}0$

Knowing that $V_C(t) = V_0(1 - e^{-t/\tau})$, determine $V_0$ (battery voltage) and $\tau$ (time constant $RC$). Hint: $1 - e^{-1} \approx 0{,}63$.

Solution

Battery voltage $V_0$. As $t\to\infty$ the exponential vanishes and $V_C \to V_0$: the capacitor charges up to the generator’s voltage. From the table, at $t = 10$ s the voltage has settled at $10{,}0$ V (it no longer increases), so $\ev{V_0 \approx 10\;\text{V}}.$

Time constant $\tau$. The time constant is, by definition, the instant at which the capacitor reaches a fraction $1 - e^{-1} \approx 0{,}63$ of the final value: $V_C(\tau) = V_0(1 - e^{-1}) \approx 0{,}63\,V_0 = 0{,}63\cdot 10 = 6{,}3\;\text{V}.$ The table shows exactly $V_C = 6{,}3$ V at $t = 1$ s. Hence $\ev{\tau = RC \approx 1\;\text{s}}.$

Check. Let’s verify the point at $t = 3$ s $= 3\tau$: $V_C = 10\,(1 - e^{-3}) = 10\,(1 - 0{,}050) = 9{,}5$ V, in perfect agreement with the table. After $\sim 3$ time constants the capacitor is at $95\%$ of full charge; after $\sim 5\tau$ it is essentially fully charged. This is the universal behaviour of $RC$ charging.

Links

Topics: Campo elettrico e potenziale Concepts: Capacità e condensatore · Densità di energia del campo Skills: Conservazione dell’energia Methods: Metodo del gradiente Objects: Condensatore piano