Problem — discharge of a capacitor

Problem

During the discharge of a capacitor (initial value $V_0 = 10\ \text{V}$) the voltage $V_C(t)$ was measured at the following times:

$t$ (s)	0	2	4	6	10
$V_C$ (V)	10,0	6,07	3,68	2,23	0,82

(a) Verify that the trend is exponential by calculating the ratios $V_C(t+2)/V_C(t)$. (b) Determine the time constant $\tau = RC$.

Solution

(a) Verifying the exponential trend — in a discharge $V_C(t) = V_0\,e^{-t/\tau}$, the ratio between two values separated by $\Delta t = 2\ \text{s}$ is constant, because $\frac{V_C(t+2)}{V_C(t)} = \frac{V_0\,e^{-(t+2)/\tau}}{V_0\,e^{-t/\tau}} = e^{-2/\tau}$ independent of $t$. Let us calculate from the data: $\frac{6{,}07}{10{,}0} = 0{,}607, \quad \frac{3{,}68}{6{,}07} = 0{,}606, \quad \frac{2{,}23}{3{,}68} = 0{,}606$

The ratio is constant ($\approx 0{,}607$): the trend is exponential. ✓

(b) Time constant — from the constant ratio $e^{-2/\tau} = 0{,}607$, take the logarithm: $-\frac{2}{\tau} = \ln(0{,}607) \approx -0{,}50 \quad\Rightarrow\quad \tau = \frac{2}{0{,}50} = 4\ \text{s}$

Check at the distant point $t = 10\ \text{s}$: $V_0\,e^{-10/4} = 10\cdot e^{-2{,}5} \approx 10\cdot 0{,}082 = 0{,}82\ \text{V}$, in agreement with the data. ✓

$\ev{\text{exponential trend}; \quad \tau = RC \approx 4\ \text{s}}$

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Topics: Circuiti elettrici Concepts: Circuito RC Skills: Lettura dei grafici