Irodov 3.170: Ninety Percent Needs 2.3 Time Constants

Problem Irodov 3.170

A capacitor of capacitance $C = 400$ pF is connected through a resistance $R = 650$ $\Omega$ to a source of constant voltage $V_0$ . How long after the connection does the voltage across the capacitor reach $V = 0.90\,V_0$ ?

The source V_0 and the capacitor C with the resistance R in series, before their terminals are joined. — The source $V_0$ and the capacitor $C$ with the resistance $R$ in series, before their terminals are joined.

Given

$C = 400$ pF
$R = 650~\Omega$
source voltage $V_0$
$V = 0.90\,V_0$
capacitor uncharged at the connection (assumed)

Find

the time

t

at which the capacitor voltage reaches

V

Solution

Step 1The charging equation

Once the terminals are joined, the source, $R$ and $C$ form one loop. Let $q$ be the capacitor’s charge, zero at the moment of connection $t = 0$ (assumed), and $I = dq/dt$ the charging current. Going round the loop, the source voltage is shared between the resistance and the capacitor:

V_0 = R\,\frac{dq}{dt} + \frac{q}{C} \tag{1}

Step 2Solving it

Separating the variables in (1) and integrating from the uncharged start:

\begin{aligned} \int_0^q \frac{dq'}{CV_0 - q'} &= \int_0^t \frac{dt'}{RC}\\[2pt] -\ln\frac{CV_0 - q}{CV_0} &= \frac{t}{RC} \end{aligned}

The capacitor voltage $q/C$ therefore rises as

V(t) = V_0\left(1 - e^{-t/RC}\right) \tag{2}

The capacitor voltage after the connection: it starts from zero and approaches V_0 (dotted level) with the time constant RC. — The capacitor voltage after the connection: it starts from zero and approaches $V_0$ (dotted level) with the time constant $RC$ .

Step 3The time to reach the given voltage

Setting $V(t) = V$ in (2):

e^{-t/RC} = 1 - \frac{V}{V_0} \tag{3}

The time constant is

RC = 400\times 10^{-12}~\text{F}\times 650~\Omega = 2.6\times 10^{-7}~\text{s}

Taking the logarithm of (3) gives the time; with $1 - V/V_0 = 0.10$ , numerically:

t = 2.6\times 10^{-7}~\text{s}\times\ln 10

t = -RC\ln\left(1 - \frac{V}{V_0}\right) \approx 0.60~\mu\text{s}

(4)

Check

As $V \to 0$ the logarithm tends to $-V/V_0$ and $t \to RC\,V/V_0$ : at first the capacitor charges at the full initial current $V_0/R$ , so it gains the voltage $V$ in the time $CV/(V_0/R)$ , the same. As $V \to V_0$ the time grows without bound: the exponential never quite arrives.

Remark

The answer does not depend on $V_0$ , only on the fraction reached. Each further factor of ten closer to $V_0$ costs another $RC\ln 10 \approx 2.3\,RC$ , which is why engineers count about five time constants (99.3 percent) as fully charged.