Exercise 1: Charging a capacitor: initial, final and intermediate values

A $220$ µF capacitor, initially discharged, is connected at time $t=0$ to an ideal $12.0$ V source through a $4.70$ kΩ resistor. The voltage across it is $v_{C}(t)=\mathcal{E}\left(1-e^{-t/\tau}\right)$ and the current is $i(t)=\dfrac{\mathcal{E}}{R}e^{-t/\tau}$ , with $\tau=RC$ .

a) Calculate the time constant.
b) Calculate the current right after the switch is closed, in mA.
c) Calculate the final charge of the capacitor, in mC.
d) Calculate the voltage across the capacitor at $t=2.00$ s.
e) How long does it take for the voltage to reach $99$ % of its final value?

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Answers

a) $\tau=1.03$ s
b) $i_{0}=2.55$ mA
c) $Q=2.64$ mC
d) $v_{C}\approx 10.3$ V
e) $t=\tau\ln 100\approx 4.76$ s

a)

$\tau=RC=4.70\times 10^{3}\times 220\times 10^{-6}=1.03$ s. Convert the kilohms and the microfarads BEFORE multiplying; one ohm times one farad gives one second.

b)

At $t=0$ the capacitor is discharged, $v_{C}=0$ : the whole voltage is across the resistor. $i_{0}=\dfrac{\mathcal{E}}{R}=\dfrac{12.0}{4700}=2.55\times 10^{-3}$ A $=2.55$ mA. Right after the switch closes, a discharged capacitor behaves like a WIRE.

c)

In steady state the current vanishes and $v_{C}=\mathcal{E}$ : $Q=C\mathcal{E}=220\times 10^{-6}\times 12.0=2.64\times 10^{-3}$ C $=2.64$ mC. A charged capacitor behaves like an OPEN circuit.

d)

$v_{C}=12.0\times\left(1-e^{-2.00/1.034}\right)=12.0\times(1-0.145)=10.3$ V.

e)

$1-e^{-t/\tau}=0.99$ , so $e^{-t/\tau}=0.01$ and $t=\tau\ln 100=4.61\tau=4.76$ s. The practical rule “charged after $5\tau$ ” comes from here: at $5\tau$ , less than $1$ % remains to be gained.

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Exercise 2: Discharging: setting up and solving the differential equation

A $47.0$ µF capacitor charged at $9.00$ V is connected at $t=0$ to a $100$ kΩ resistor. Let $q$ be the charge on the positive plate and $i$ the current leaving that plate through the resistor, so that $i=-\dfrac{dq}{dt}$ .

a) What differential equation does the loop rule give for this circuit?
b) Separating the variables gives $q(t)=Q_{0}e^{-t/RC}$ . Calculate the time constant.
c) Calculate the current at $t=0$ , in µA.
d) Calculate the voltage across the capacitor at $t=10.0$ s.
e) How long does it take for the voltage to fall to $1.00$ V?

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Answers

a) $R\dfrac{dq}{dt}+\dfrac{q}{C}=0$
b) $\tau=4.70$ s
c) $i(0)=90.0$ µA
d) $v_{C}\approx 1.07$ V
e) $t=\tau\ln 9\approx 10.3$ s

a)

Going around the loop, the capacitor voltage $\dfrac{q}{C}$ equals the voltage drop across the resistor $Ri$ : $\dfrac{q}{C}=Ri=-R\dfrac{dq}{dt}$ , so $R\dfrac{dq}{dt}+\dfrac{q}{C}=0$ . The minus sign expresses that the charge DECREASES while the current flows.

b)

$\dfrac{dq}{q}=-\dfrac{dt}{RC}$ , then $\ln q=-\dfrac{t}{RC}+\text{const}$ , hence $q=Q_{0}e^{-t/RC}$ . $\tau=RC=100\times 10^{3}\times 47.0\times 10^{-6}=4.70$ s.

c)

$i(0)=\dfrac{V_{0}}{R}=\dfrac{9.00}{1.00\times 10^{5}}=9.00\times 10^{-5}$ A $=90.0$ µA. The same result comes from differentiating: $i=-\dfrac{dq}{dt}=\dfrac{Q_{0}}{RC}e^{-t/RC}$ .

d)

$v_{C}=9.00\,e^{-10.0/4.70}=9.00\times 0.119=1.07$ V.

e)

$e^{-t/\tau}=\dfrac{1.00}{9.00}$ , so $t=\tau\ln 9=4.70\times 2.197=10.3$ s. Consistent with d): at $10.0$ s the voltage is barely above $1$ V.

Exercise 3: Two resistors: which time constant does the capacitor see?

In the circuit below, a $10.0$ V source feeds, through $R_{1}=6.00$ kΩ, the resistor $R_{2}=3.00$ kΩ in parallel with an initially discharged capacitor $C=100$ µF. The switch K is closed at $t=0$ .

a) What voltage does the capacitor reach in steady state?
b) Seen from the capacitor, the rest of the circuit is equivalent to a source in series with a resistance $R_{eq}$ , obtained by turning the source off. Calculate $R_{eq}$ .
c) Calculate the time constant of the charging.
d) Once the capacitor is charged, K is opened. Calculate the new time constant.
e) Calculate the capacitor voltage $0.500$ s after K is opened.

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Answers

a) $v_{C}=3.33$ V
b) $R_{eq}=2.00$ kΩ
c) $\tau=0.200$ s
d) $\tau'=0.300$ s
e) $v_{C}\approx 0.630$ V

a)

In steady state the capacitor no longer lets any current through: $R_{1}$ and $R_{2}$ form a voltage divider carrying $\dfrac{10.0}{9.00\times 10^{3}}=1.11$ mA. $v_{C}=R_{2}I=\dfrac{R_{2}}{R_{1}+R_{2}}\mathcal{E}=3.33$ V, not $10.0$ V.

b)

With the source turned off (replaced by a wire), the capacitor sees $R_{1}$ and $R_{2}$ in PARALLEL: $R_{eq}=\dfrac{6.00\times 3.00}{6.00+3.00}=2.00$ kΩ.

c)

$\tau=R_{eq}C=2.00\times 10^{3}\times 100\times 10^{-6}=0.200$ s. The trap is to take $R_{1}C=0.600$ s: part of the supplied current escapes through $R_{2}$ , so the charging is faster and levels off lower.

d)

With K open, $R_{1}$ is out of the circuit: the capacitor discharges through $R_{2}$ alone, $\tau'=R_{2}C=0.300$ s. Charging and discharging do not have the same time constant.

e)

$v_{C}=3.33\,e^{-0.500/0.300}=3.33\times 0.189=0.630$ V.

Part b): the source turned off becomes a wire. Seen from the terminals of C,

R_{1}

and

R_{2}

connect the same two nodes A and B: they are in parallel,

R_{eq}=2.00

kΩ.

Part d): K open,

R_{1}

leads nowhere. The capacitor discharges through

R_{2}

alone, along the arrowed loop:

\tau'=R_{2}C=0.300

s.

Exercise 4: Kirchhoff at the initial instant and in steady state

In the circuit below, the $18.0$ V source is connected at $t=0$ , the capacitor being discharged. The right-hand branch contains $R_{3}$ in series with the capacitor; it is in parallel with $R_{2}$ .

a) Right after the connection, calculate the current supplied by the source, in mA.
b) Right after the connection, calculate the current in the capacitor branch, in mA.
c) In steady state, calculate the current supplied by the source, in mA.
d) In steady state, calculate the voltage across the capacitor.
e) Calculate the energy finally stored in the capacitor, in mJ.

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Answers

a) $I_{0}=7.71$ mA
b) $2.57$ mA
c) $6.00$ mA
d) $v_{C}=12.0$ V
e) $U=0.72$ mJ

a)

Discharged capacitor: it behaves like a wire. $R_{2}$ and $R_{3}$ are in parallel, $\dfrac{2\times 4}{2+4}=1.33$ kΩ, in series with $R_{1}$ : $R_{total}=2.33$ kΩ. $I_{0}=\dfrac{18.0}{2333}=7.71\times 10^{-3}$ A $=7.71$ mA.

b)

Voltage across the parallel block: $7.714\times 1.333=10.29$ V. Capacitor branch: $\dfrac{10.29}{4.00}=2.57$ mA (and $5.14$ mA in $R_{2}$ ; the junction rule does give $7.71$ mA).

c)

Charged capacitor: its branch no longer conducts. What remains is $R_{1}$ and $R_{2}$ in series: $I=\dfrac{18.0}{3.00\times 10^{3}}=6.00$ mA.

d)

No more current in $R_{3}$ , so no voltage drop across it: the capacitor has the voltage of $R_{2}$ , $v_{C}=2.00\times 10^{3}\times 6.00\times 10^{-3}=12.0$ V.

e)

$U=\dfrac{1}{2}Cv_{C}^{2}=\dfrac{1}{2}\times 10\times 10^{-6}\times 12.0^{2}=7.2\times 10^{-4}$ J $=0.72$ mJ.

Right after the connection, the discharged capacitor behaves like a wire:

I_{0}=7.71

mA splits into

I_{2}=5.14

mA in

R_{2}

and

I_{3}=2.57

mA in the capacitor branch.

In steady state, the capacitor branch no longer conducts: the same

I=6.00

mA flows through

R_{1}

and

R_{2}

, and

v_{C}=12.0

V, the voltage of

R_{2}

, since

R_{3}

carries no drop.

Exercise 5: Where the capacitor's energy goes: the integral of Joule heating

A $2200$ µF capacitor charged at $25.0$ V discharges through a $50.0$ Ω resistor. The current is $i(t)=\dfrac{V_{0}}{R}e^{-t/\tau}$ with $\tau=RC$ .

a) Calculate the energy initially stored.
b) Calculate the power dissipated in the resistor at $t=0$ .
c) Show that the energy dissipated up to the end, $\displaystyle\int_{0}^{\infty}Ri^{2}\,dt$ , equals $\dfrac{1}{2}CV_{0}^{2}$ .
d) Calculate the time constant.
e) How long does it take for $90$ % of the energy to be dissipated, in ms?

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Answers

a) $U_{0}\approx 0.688$ J
b) $P_{0}=12.5$ W
c) $\int Ri^{2}dt=\frac{1}{2}CV_{0}^{2}$
d) $\tau=0.110$ s
e) $t=\dfrac{\tau\ln 10}{2}\approx 127$ ms

a)

$U_{0}=\dfrac{1}{2}CV_{0}^{2}=\dfrac{1}{2}\times 2200\times 10^{-6}\times 25.0^{2}=0.688$ J.

b)

$P_{0}=\dfrac{V_{0}^{2}}{R}=\dfrac{625}{50.0}=12.5$ W.

c)

$\displaystyle\int_{0}^{\infty}R\dfrac{V_{0}^{2}}{R^{2}}e^{-2t/\tau}dt=\dfrac{V_{0}^{2}}{R}\left[-\dfrac{\tau}{2}e^{-2t/\tau}\right]_{0}^{\infty}=\dfrac{V_{0}^{2}}{R}\cdot\dfrac{RC}{2}=\dfrac{1}{2}CV_{0}^{2}$ . All the energy of the capacitor ends up as heat, whatever the value of $R$ : the resistance changes only the DURATION of the discharge.

d)

$\tau=50.0\times 2200\times 10^{-6}=0.110$ s.

e)

The remaining energy is $\dfrac{1}{2}Cv_{C}^{2}\propto e^{-2t/\tau}$ : it decays TWICE as fast as the voltage. $e^{-2t/\tau}=0.10$ gives $t=\dfrac{\tau\ln 10}{2}=0.127$ s $=127$ ms, whereas the voltage has only fallen to $32$ % of its initial value.

Exercise 6: Measuring a capacitance by the half-life, and the voltmeter trap

In the laboratory, an unknown capacitor is charged, then left to discharge through a resistor $R=1.0$ MΩ. A digital multimeter connected across it shows that the voltage is cut in half in $t_{1/2}=6.9$ s. The multimeter's manual gives an internal resistance $R_{V}=10$ MΩ in voltmeter mode.

a) Show that $t_{1/2}=\tau\ln 2$ .
b) Ignoring the voltmeter, calculate the capacitance, in µF.
c) The voltmeter is connected in parallel with $R$ . Calculate the resistance actually seen by the capacitor, in MΩ.
d) Calculate the corrected capacitance, in µF.
e) What relative error was made by ignoring the voltmeter, in %?

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Answers

a) $t_{1/2}=\tau\ln 2$
b) $C\approx 9.95$ µF
c) $R_{p}=0.909$ MΩ
d) $C\approx 10.95$ µF
e) About $9$ %

a)

$v(t_{1/2})=\dfrac{V_{0}}{2}$ gives $e^{-t_{1/2}/\tau}=\dfrac{1}{2}$ , so $t_{1/2}=\tau\ln 2=0.693\,\tau$ . The half-life does not depend on the starting voltage: that is what makes it a good measurement.

b)

$\tau=\dfrac{6.9}{0.693}=9.95$ s and $C=\dfrac{\tau}{R}=\dfrac{9.95}{1.0\times 10^{6}}=9.95\times 10^{-6}$ F $=9.95$ µF.

c)

The capacitor discharges through $R$ AND through the voltmeter: $R_{p}=\dfrac{1.0\times 10}{1.0+10}=0.909$ MΩ.

d)

$C=\dfrac{9.95}{0.909\times 10^{6}}=1.095\times 10^{-5}$ F $=10.95$ µF.

e)

$\dfrac{10.95-9.95}{10.95}=9.1$ %. A measuring instrument is part of the circuit: a $10$ MΩ voltmeter is “perfect” next to kilohms, not next to a megohm. Choose $R\ll R_{V}$ , or correct for it.

Part c): the voltmeter is modelled by its internal resistance

R_{V}=10

MΩ, in parallel with

R

. The capacitor discharges through both at once:

R_{p}=0.909

MΩ.

Exercise 7: The pacemaker: a relaxation oscillator

In a simplified pacemaker, a $6.0$ V battery charges a $10$ µF capacitor through a resistor $R$ . As soon as the capacitor voltage reaches $4.0$ V, a circuit discharges it almost instantly into the heart, and the charging starts over from zero.

a) Show that the duration of a cycle is $T=RC\ln\dfrac{\mathcal{E}}{\mathcal{E}-V_{th}}$ and calculate $\ln\dfrac{\mathcal{E}}{\mathcal{E}-V_{th}}$ .
b) We want $72$ beats per minute. Calculate the required period.
c) Calculate the required resistance, in kΩ.
d) Near the end of its life, the battery supplies only $5.5$ V, the threshold remaining at $4.0$ V. Calculate the new period.
e) What heart rate is then imposed, in beats per minute?

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Answers

a) $\ln 3\approx 1.099$
b) $T=0.833$ s
c) $R\approx 75.9$ kΩ
d) $T'\approx 0.985$ s
e) About $61$ beats per minute

a)

$v_{C}(T)=\mathcal{E}\left(1-e^{-T/RC}\right)=V_{th}$ gives $e^{-T/RC}=\dfrac{\mathcal{E}-V_{th}}{\mathcal{E}}$ , so $T=RC\ln\dfrac{\mathcal{E}}{\mathcal{E}-V_{th}}$ . Here $\ln\dfrac{6.0}{2.0}=\ln 3=1.099$ .

b)

$T=\dfrac{60}{72}=0.833$ s.

c)

$R=\dfrac{T}{C\ln 3}=\dfrac{0.833}{10\times 10^{-6}\times 1.099}=7.59\times 10^{4}$ Ω $=75.9$ kΩ.

d)

$T'=RC\ln\dfrac{5.5}{1.5}=0.759\times 1.299=0.985$ s. The threshold now represents a larger fraction of the battery voltage: it takes longer to reach.

e)

$\dfrac{60}{0.985}=61$ beats per minute. The drift in rate is an indicator of battery wear, watched at check-ups; real devices regulate the voltage to avoid it.

Exercise 8: Five statements to correct

Each of the following statements is false, or false by a decisive detail. Choose the correct statement or give the value requested. Two marks each.

a) “During charging, electrons cross the capacitor from one plate to the other.”
b) “Right after the switch is closed, the voltage across a discharged capacitor already equals $\mathcal{E}$ .”
c) “Doubling the resistance doubles the energy finally stored in the capacitor.”
d) “After $2\tau$ , a capacitor is $99$ % charged.” After how many time constants is it really?
e) “With a source voltage twice as large, the capacitor takes twice as long to charge.”

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Answers

a) No charge crosses the insulator
b) $v_{C}(0)=0$
c) $U=\frac{1}{2}C\mathcal{E}^{2}$ does not depend on $R$
d) $\ln 100\approx 4.6\,\tau$
e) $\tau=RC$ , independent of $\mathcal{E}$

a)

No charge crosses the insulator. Electrons leave one plate and, at the same moment, as many electrons arrive on the other through the rest of the circuit: the current is the same everywhere in the loop, but nothing passes through the dielectric.

b)

A capacitor's voltage cannot jump: $v_{C}=\dfrac{q}{C}$ and the charge needs time to build up. Right after the switch closes, $v_{C}=0$ and the whole source voltage is across the resistor, hence the maximum current $\dfrac{\mathcal{E}}{R}$ .

c)

The final energy $\dfrac{1}{2}C\mathcal{E}^{2}$ does not depend on $R$ . Doubling $R$ only doubles the time constant: the charging is twice as SLOW, not larger.

d)

$1-e^{-n}=0.99$ gives $n=\ln 100=4.6$ . After $2\tau$ , the capacitor is only $1-e^{-2}=86.5$ % charged.

e)

$\tau=RC$ does not depend on the source: the final voltage doubles, but the fraction reached at each instant is the same. The charge takes the same time to reach $63$ % of its final value, whatever $\mathcal{E}$ is.

Statement a): during charging, electrons, purple arrows, leave the top plate toward the + terminal; as many arrive on the bottom plate from the − terminal. The current

i

is the same everywhere, but no electron crosses the insulator.

Exercise 9: Problem: the intermittent windshield wiper

The control of an intermittent windshield wiper charges a $100$ µF capacitor from zero, at $12.0$ V, through an adjustable resistance $R$ made of a fixed $10.0$ kΩ resistor in series with a $0$ to $500$ kΩ potentiometer. A comparator triggers a sweep when $v_{C}$ reaches $\dfrac{2}{3}$ of the supply voltage; the capacitor is then emptied and the cycle starts over.

a) Show that the time between two sweeps is $T=RC\ln 3$ .
b) Calculate the minimum time, in s.
c) Calculate the maximum time, in s.
d) The driver wants a sweep every $15$ s. What value must the potentiometer be set to, in kΩ?
e) Why does a threshold set as a fraction of the supply voltage make the time insensitive to variations in the battery?

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Answers

a) $T=RC\ln 3$
b) $1.10$ s
c) $56.0$ s
d) $126.5$ kΩ on the potentiometer
e) $\mathcal{E}$ cancels out

a)

$\mathcal{E}\left(1-e^{-T/RC}\right)=\dfrac{2}{3}\mathcal{E}$ gives $e^{-T/RC}=\dfrac{1}{3}$ and $T=RC\ln 3$ .

b)

$R_{min}=10.0$ kΩ: $T=10.0\times 10^{3}\times 100\times 10^{-6}\times\ln 3=1.10$ s.

c)

$R_{max}=510$ kΩ: $T=51.0\times 1.099=56.0$ s.

d)

$R=\dfrac{15}{100\times 10^{-6}\times 1.099}=1.365\times 10^{5}$ Ω, that is $136.5$ kΩ in all, so $126.5$ kΩ on the potentiometer. Forgetting the fixed resistor gives a $10$ kΩ error.

e)

If the battery goes from $12.0$ to $13.5$ V, the threshold goes from $8.0$ to $9.0$ V: the FRACTION to reach stays $\dfrac{2}{3}$ , and $\mathcal{E}$ cancels out of the equation. This is the principle of the 555 integrated circuit, unlike the pacemaker of the previous exercise, whose threshold was fixed.

Exercise 10: Problem: the defibrillator

An external defibrillator charges a $100$ µF capacitor at $2.00$ kV, then discharges it through the patient's chest, modelled as a $50.0$ Ω resistor. A battery-powered converter charges the capacitor at an average power of $50.0$ W.

a) Calculate the stored energy, in J.
b) Calculate the charging time.
c) Calculate the current at the start of the discharge.
d) Calculate the time constant of the discharge, in ms.
e) What fraction of the energy is delivered during the first $5.00$ milliseconds, in %?

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Answers

a) $U=200$ J
b) $t=4.00$ s
c) $I_{0}=40.0$ A
d) $\tau=5.00$ ms
e) $1-e^{-2}\approx 86.5$ %

a)

$U=\dfrac{1}{2}CV^{2}=\dfrac{1}{2}\times 100\times 10^{-6}\times(2.00\times 10^{3})^{2}=200$ J. The voltage must be squared in VOLTS: $2.00$ kV $=2000$ V.

b)

$t=\dfrac{U}{P}=\dfrac{200}{50.0}=4.00$ s. This is why the device asks for a few seconds before allowing the shock.

c)

$I_{0}=\dfrac{V}{R}=\dfrac{2000}{50.0}=40.0$ A.

d)

$\tau=RC=50.0\times 100\times 10^{-6}=5.00\times 10^{-3}$ s $=5.00$ ms.

e)

The remaining energy decays as $e^{-2t/\tau}$ : at $t=\tau$ , $e^{-2}=13.5$ % is left, so $86.5$ % has been delivered. A charge built up over four seconds is released in a few milliseconds: that is the very function of the capacitor, a reservoir that fills slowly and empties fast.

Exercise 11: Two capacitors, in series then in parallel, in an RC circuit

Two discharged capacitors $C_{1}=30.0$ µF and $C_{2}=60.0$ µF are charged at $12.0$ V through a $5.00$ kΩ resistor, first connected in series, then connected in parallel.

a) Calculate the equivalent capacitance of the series arrangement, in µF.
b) Calculate the time constant of the series arrangement.
c) Calculate the time constant of the parallel arrangement.
d) Series arrangement, steady state: calculate the charge on each capacitor, in µC.
e) Series arrangement: calculate the voltage across $C_{1}$ , then across $C_{2}$ .

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Answers

a) $C_{s}=20.0$ µF
b) $\tau_{s}=0.100$ s
c) $\tau_{p}=0.450$ s
d) $Q=240$ µC on each
e) $V_{1}=8.00$ V, $V_{2}=4.00$ V

a)

In series, the reciprocals add: $\dfrac{1}{C_{s}}=\dfrac{1}{30}+\dfrac{1}{60}=\dfrac{3}{60}$ , so $C_{s}=20.0$ µF, smaller than either one. This is the REVERSE of the rule for resistors.

b)

$\tau_{s}=RC_{s}=5.00\times 10^{3}\times 20.0\times 10^{-6}=0.100$ s.

c)

In parallel, the capacitances add: $C_{p}=90.0$ µF and $\tau_{p}=0.450$ s, four and a half times longer.

d)

In series, both capacitors carry the SAME charge, the one that has flowed through the resistor: $Q=C_{s}\mathcal{E}=20.0\times 12.0=240$ µC each. The inner plate of $C_{1}$ and that of $C_{2}$ , connected together and isolated from the rest, carry opposite charges whose sum stays zero.

e)

$V_{1}=\dfrac{240}{30.0}=8.00$ V and $V_{2}=\dfrac{240}{60.0}=4.00$ V. The SMALLER capacitor takes the larger voltage, and the sum does give back $12.0$ V.

Exercise 12: The debounce filter of a push-button

When a push-button is released, its contacts bounce for about $5$ ms. So that the microcontroller does not count several presses, a capacitor is placed in parallel with the button: with the button closed, the capacitor is short-circuited and $v_{C}=0$ ; with the button released, it charges at $3.30$ V through $R=10.0$ kΩ. The logic input reads “released” only when $v_{C}$ exceeds $1.50$ V.

a) With $C=100$ nF, calculate the time constant, in ms.
b) Show that the threshold is reached after $t=\tau\ln\dfrac{3.30}{3.30-1.50}$ and calculate $t$ , in ms.
c) Is this filter enough to mask $5$ ms bounces?
d) What minimum time constant is needed so that the threshold is reached only after $5.0$ ms, in ms?
e) What minimum capacitance is then needed, in nF?

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Answers

a) $\tau=1.00$ ms
b) $t\approx 0.606$ ms
c) No, the threshold falls in the middle of the bounces
d) $\tau\ge 8.25$ ms
e) $C\ge 825$ nF

a)

$\tau=RC=10.0\times 10^{3}\times 100\times 10^{-9}=1.00\times 10^{-3}$ s $=1.00$ ms. Watch the prefix: one nanofarad is $10^{-9}$ F.

b)

$3.30\left(1-e^{-t/\tau}\right)=1.50$ gives $e^{-t/\tau}=\dfrac{1.80}{3.30}$ , so $t=\tau\ln\dfrac{3.30}{1.80}=1.00\times 0.606=0.606$ ms.

c)

No: the input crosses the threshold after $0.6$ ms, right in the middle of the bounces. Each bounce that closes the contact again empties the capacitor and restarts the charging, and several threshold crossings remain possible.

d)

We need $\tau\times 0.606\ge 5.0$ ms, so $\tau\ge\dfrac{5.0}{0.606}=8.25$ ms.

e)

$C\ge\dfrac{8.25\times 10^{-3}}{10.0\times 10^{3}}=8.25\times 10^{-7}$ F $=825$ nF. We would take the next standard value up, $1$ µF. Too large a capacitance, however, would make the button sluggish: the user would feel the delay beyond about a hundred milliseconds.

Exercise 13: DRAM memory: a leaky capacitor

A dynamic random-access memory cell stores one bit as a charge on a $30$ fF capacitor ( $1$ fF $=10^{-15}$ F), charged at $1.0$ V for a “1”. Leakage through the transistor is equivalent to a resistance of $1.0\times 10^{14}$ Ω. The cell is read as a “1” as long as its voltage exceeds $0.50$ V. $e=1.60\times 10^{-19}$ C.

a) Calculate the number of electrons stored in the cell.
b) Calculate the time constant of the leakage.
c) After how long would the cell be read as a “0” without refreshing?
d) The memory is refreshed every $64$ ms. What fraction of the voltage remains just before the refresh, in %?
e) What safety margin, in number of refresh periods, does this represent before the bit is lost?

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Answers

a) $N\approx 1.9\times 10^{5}$ electrons
b) $\tau=3.0$ s
c) $t=\tau\ln 2\approx 2.1$ s
d) $97.9$ %
e) About $32$ periods

a)

$Q=CV=30\times 10^{-15}\times 1.0=3.0\times 10^{-14}$ C, that is $N=\dfrac{3.0\times 10^{-14}}{1.60\times 10^{-19}}=1.9\times 10^{5}$ electrons. One bit is fewer than two hundred thousand electrons.

b)

$\tau=RC=1.0\times 10^{14}\times 30\times 10^{-15}=3.0$ s.

c)

$e^{-t/\tau}=0.50$ , so $t=\tau\ln 2=2.1$ s.

d)

$e^{-0.064/3.0}=0.979$ : $97.9$ % of the voltage remains, well above the threshold.

e)

$\dfrac{2.08}{0.064}=32$ periods. Manufacturers keep this margin because the leakage resistance drops sharply when the chip heats up: at high temperature, $\tau$ can be divided by ten, and the margin shrinks accordingly.

Exercise 14: Timing with a capacitor: the linear regime at the start of charging

To measure a very short duration, a switch is closed at the start of the event and opened at the end: in between, a capacitor charges at $\mathcal{E}=10.0$ V through a resistor, with $\tau=RC=1.00$ s. Its voltage, which then no longer changes, is read afterwards. Measurement: $v_{C}=0.470$ V.

a) Calculate the exact duration of the event, in ms.
b) At the start of charging, $e^{-x}\approx 1-x$ for small $x$ . Show that $v_{C}\approx\mathcal{E}\dfrac{t}{\tau}$ and calculate the duration estimated this way, in ms.
c) Calculate the relative error of the approximation, in %.
d) The next term of the expansion gives $v_{C}\approx\mathcal{E}\left(\dfrac{t}{\tau}-\dfrac{t^{2}}{2\tau^{2}}\right)$ . The relative error of the linear approximation is therefore about $\dfrac{t}{2\tau}$ . Evaluate this term, in %.
e) To keep the error of the linear approximation under $1$ %, what maximum fraction of $\mathcal{E}$ can the measured voltage reach, in %?

Show the solution

Answers

a) $t\approx 48.1$ ms
b) $t\approx 47.0$ ms
c) $2.4$ %
d) $\dfrac{t}{2\tau}\approx 2.4$ %
e) $v_{C}\le 2$ % of $\mathcal{E}$

a)

$v_{C}=\mathcal{E}\left(1-e^{-t/\tau}\right)$ gives $t=-\tau\ln\left(1-\dfrac{v_{C}}{\mathcal{E}}\right)=-1.00\times\ln(0.953)=4.81\times 10^{-2}$ s $=48.1$ ms.

b)

$v_{C}\approx\mathcal{E}\left(1-\left(1-\dfrac{t}{\tau}\right)\right)=\mathcal{E}\dfrac{t}{\tau}$ : the capacitor charges at almost constant current $\dfrac{\mathcal{E}}{R}$ . Estimated duration: $t\approx\tau\dfrac{v_{C}}{\mathcal{E}}=47.0$ ms.

c)

$\dfrac{48.14-47.0}{48.14}=2.4$ %. The approximation underestimates the duration, because the current is already dropping a little during the charging.

d)

$\dfrac{t}{2\tau}=\dfrac{0.0481}{2}=0.024$ , that is $2.4$ %: the first neglected term accounts for the whole error.

e)

We need $\dfrac{t}{2\tau}\le 0.01$ , that is $\dfrac{t}{\tau}\le 0.02$ and therefore $\dfrac{v_{C}}{\mathcal{E}}\approx\dfrac{t}{\tau}\le 2$ %. The stopwatch is made linear by choosing $\tau$ at least fifty times larger than the durations to be measured.

Exercise 15: Problem: who receives the energy when two resistors share the discharge?

A $470$ µF capacitor charged at $20.0$ V discharges through two resistors $R_{1}=100$ Ω and $R_{2}=300$ Ω, connected first in series, then in parallel.

a) Calculate the stored energy, in mJ.
b) Series arrangement: calculate the time constant, in ms.
c) Series arrangement: calculate the energy dissipated in $R_{1}$ , in mJ.
d) Parallel arrangement: calculate the time constant, in ms.
e) Parallel arrangement: calculate the energy dissipated in $R_{1}$ , in mJ.

Show the solution

Answers

a) $U=94.0$ mJ
b) $\tau=188$ ms
c) $23.5$ mJ
d) $\tau\approx 35.3$ ms
e) $70.5$ mJ

a)

$U=\dfrac{1}{2}\times 470\times 10^{-6}\times 20.0^{2}=9.40\times 10^{-2}$ J $=94.0$ mJ.

b)

$\tau=(R_{1}+R_{2})C=400\times 470\times 10^{-6}=0.188$ s $=188$ ms.

c)

In series, the same current flows through both resistors: at each instant $P_{1}=R_{1}i^{2}$ and $P_{2}=R_{2}i^{2}$ , in the ratio $\dfrac{R_{1}}{R_{2}}$ . Over the whole discharge, $R_{1}$ receives $\dfrac{R_{1}}{R_{1}+R_{2}}U=\dfrac{100}{400}\times 94.0=23.5$ mJ. The LARGER resistor heats up the most.

d)

$R_{p}=\dfrac{100\times 300}{400}=75.0$ Ω, so $\tau=75.0\times 470\times 10^{-6}=3.53\times 10^{-2}$ s $=35.3$ ms.

e)

In parallel, it is the same VOLTAGE: $P_{1}=\dfrac{v^{2}}{R_{1}}$ and $P_{2}=\dfrac{v^{2}}{R_{2}}$ , in the ratio $\dfrac{R_{2}}{R_{1}}=3$ . $R_{1}$ receives $\dfrac{3}{4}U=70.5$ mJ: now it is the SMALLER resistor that heats up the most. In both arrangements, all $94.0$ mJ end up as heat.

Corrected exercises: RC circuits in direct current (203-SN2-RE)

Course recap

Part A: the basics (/50)

Exercise 1: Charging a capacitor: initial, final and intermediate values

Exercise 2: Discharging: setting up and solving the differential equation

Exercise 3: Two resistors: which time constant does the capacitor see?

Exercise 4: Kirchhoff at the initial instant and in steady state

Exercise 5: Where the capacitor's energy goes: the integral of Joule heating

Part B: problems and reasoning (/50)

Exercise 6: Measuring a capacitance by the half-life, and the voltmeter trap

Exercise 7: The pacemaker: a relaxation oscillator

Exercise 8: Five statements to correct

Exercise 9: Problem: the intermittent windshield wiper

Exercise 10: Problem: the defibrillator

Part C: the classics (/50)

Exercise 11: Two capacitors, in series then in parallel, in an RC circuit

Exercise 12: The debounce filter of a push-button

Exercise 13: DRAM memory: a leaky capacitor

Exercise 14: Timing with a capacitor: the linear regime at the start of charging

Exercise 15: Problem: who receives the energy when two resistors share the discharge?

See also

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