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The loop rule with a capacitor, exponential charging and discharging with $\tau = RC$, initial and final states, and the energy lost while charging.
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By the end of this lesson you will be able to solve RC circuits for charges, currents and voltages as functions of time, and find times to reach given levels.
From the last lessons you can find the charge and energy of a capacitor and solve resistor circuits with Kirchhoff's rules. From mechanics you have solved $dv/dt = -v/\tau$, the drag equation. Putting a capacitor and a resistor together gives exactly that equation again, now for charge, and the exponentials you know return with a new meaning.
| Term | What it means |
|---|---|
| RC circuit | A circuit with a resistor and a capacitor, whose currents change with time. |
| Time constant | $\tau = RC$, the time for the charge to close 63 percent of the gap to its final value. |
| Charging | Filling a capacitor from a battery: $q = C\mathcal{E}(1 - e^{-t/\tau})$. |
| Discharging | Emptying a capacitor through a resistor: $q = Q_0e^{-t/\tau}$. |
| Transient | The changing behavior just after a switch is thrown, before a steady state. |
| Steady state | The long-run condition, when no current flows through a capacitor branch. |
| Half-life | $\tau\ln 2$, the time for the charge or current to halve. |
In a circuit with a battery, a resistor and a capacitor, the loop rule reads
$$\mathcal{E} - IR - \frac{q}{C} = 0, \qquad I = \frac{dq}{dt}.$$
The capacitor's voltage depends on its charge, which depends on the current, so this is a differential equation. Starting from an empty capacitor, its solution is
$$q = C\mathcal{E}\left(1 - e^{-t/RC}\right), \qquad I = \frac{\mathcal{E}}{R}e^{-t/RC}.$$
The time constant $\tau = RC$ sets how fast things happen. At the first instant, the empty capacitor has no voltage and the current is $\mathcal{E}/R$, as if the capacitor were a wire. As charge builds, the current falls; long afterward, the capacitor is full and blocks current. Discharging through $R$, the charge simply decays, $q = Q_0e^{-t/RC}$.
Another way: picture
Picture filling a balloon through a thin straw from a tank at steady pressure. At first air rushes in, since the balloon pushes back hardly at all. As it fills, its back-pressure grows and the flow slows, until the balloon's pressure matches the tank's and flow stops. The straw's resistance and the balloon's stretchiness set how long that takes: their product is the time constant.
Another way: steps
Rewrite the loop rule as $R\,dq/dt = \mathcal{E} - q/C = -(q - C\mathcal{E})/C$. Separate: $\dfrac{dq}{q - C\mathcal{E}} = -\dfrac{dt}{RC}$. Integrate from $q = 0$ at $t = 0$: $\ln\dfrac{C\mathcal{E} - q}{C\mathcal{E}} = -\dfrac{t}{RC}$. So the gap between the charge and its final value $C\mathcal{E}$ shrinks exponentially, and $q = C\mathcal{E}(1 - e^{-t/RC})$.
Differentiate for the current: $I = (\mathcal{E}/R)e^{-t/RC}$. The capacitor's voltage $q/C$ rises toward $\mathcal{E}$ while the resistor's voltage $IR$ falls toward zero; at every moment they add to $\mathcal{E}$. This is the same mathematics as a falling body with linear drag approaching terminal speed.
Disconnect the battery and connect the charged capacitor across the resistor. Now $R\,dq/dt + q/C = 0$, and $q = Q_0e^{-t/RC}$. The current, $I = -dq/dt = (Q_0/RC)e^{-t/RC} = (V_0/R)e^{-t/RC}$, starts at $V_0/R$ and dies away. After one time constant, 37 percent of the charge remains; after five, less than 1 percent.
To find when the voltage reaches a given level, take logarithms: $t = RC\ln(V_0/V)$. The time to halve is $RC\ln 2 \approx 0.69RC$, whatever the starting voltage, just as for radioactive decay.
Ohms times farads is seconds: $\Omega \cdot \text{F} = (\text{V/A})(\text{C/V}) = \text{C/A} = \text{s}$. A $1$ kΩ resistor with a $1$ μF capacitor gives $1$ ms; $1$ MΩ with $1$ μF gives a full second. Large $R$ limits the current, so filling takes longer; large $C$ needs more charge, so filling takes longer too.
A handy fact: the initial slope of the charging curve, extended as a straight line, would reach the final value in exactly one time constant. That gives a quick way to read $\tau$ off an oscilloscope trace, and a check on any calculated value.
Many problems ask only for the currents just after a switch closes and long afterward, and there is a shortcut. The charge on a capacitor cannot jump, since that would need infinite current, so just after a switch closes, each capacitor keeps its previous voltage: an empty one acts as a wire. Long afterward, no current flows into any capacitor, so each acts as a break.
With these two pictures, ordinary resistor analysis gives the initial and final currents and voltages everywhere, and the time constant, found from the resistance the capacitor sees, connects them with an exponential.
While charging, the battery moves charge $C\mathcal{E}$ across its full emf, doing work $C\mathcal{E}^2$. The capacitor ends with $\tfrac{1}{2}C\mathcal{E}^2$. The other half is dissipated in the resistor, which you can confirm by integrating $I^2R$ from zero to infinity: $\int_0^\infty (\mathcal{E}/R)^2e^{-2t/RC}R\,dt = \tfrac{1}{2}C\mathcal{E}^2$.
The resistance cancels. A small resistor passes a large current briefly; a large one a small current for longer; the heat is the same. Charging a capacitor from a fixed voltage always wastes half the energy, which is why efficient chargers ramp the voltage up gradually instead.
Checking an answer. The expression must give the right value at $t = 0$ and as $t \to \infty$. Its derivative must satisfy the loop rule. $RC$ must come out in seconds. And the energy supplied must equal the energy stored plus the heat.
Using $I = dq/dt$ assumes the current in the resistor is the rate charge builds on the capacitor's plate, which holds when they are in series with nothing else joining the wire between them. The loop rule itself holds at each instant because the circuit is small enough that changes spread through it almost instantly.
Saying a capacitor's voltage cannot jump follows from $I = C\,dV/dt$: a sudden jump in $V$ would need an infinite current, which a real resistor cannot supply. With no resistance at all, charge would slosh at high frequency instead, which the inductance lesson treats properly.
RC circuits set time everywhere in electronics. The delay on a car's intermittent windshield wipers, the blink rate of an LED, and the timing of the 555 timer chip, one of the best-selling integrated circuits ever, all come from a capacitor charging through a resistor to a threshold.
RC circuits also filter signals. A resistor in series with a capacitor to ground passes slow changes and smooths out fast ones; swapping them passes fast changes and blocks steady ones. Audio tone controls, the power supplies that turn wall current into steady voltage, and the input stages of sensors all rely on this.
A classic lab measures $\tau$ by charging a large capacitor, say $1000$ μF, through a $10$ kΩ resistor and timing the voltage with a meter. With $\tau = 10$ s, the changes are slow enough to follow by eye. Plotting $\ln V$ against $t$ during discharge gives a straight line of slope $-1/\tau$.
That straight line is the strongest test that the behavior is exponential. A curve instead of a line reveals leakage through the capacitor, or a meter whose own resistance is loading the circuit. Most digital meters have about $10$ MΩ of resistance, which in parallel with a $10$ kΩ resistor changes the time constant by only $0.1$ percent, but with megohm resistors the meter matters.
In a circuit with several resistors, the time constant is $R_{\text{th}}C$, where $R_{\text{th}}$ is the resistance measured between the capacitor's two terminals with the capacitor removed and every battery replaced by a plain wire. This works because, seen from the capacitor, the rest of any linear circuit behaves like a single battery in series with a single resistor, a result known as Thevenin's theorem.
For a capacitor in parallel with one resistor and fed through another, replacing the battery by a wire puts the two resistors in parallel with each other, so $R_{\text{th}} = R_1R_2/(R_1 + R_2)$, smaller than either. The capacitor therefore charges faster than it would through the feeding resistor alone, because the parallel resistor drains part of the current and the capacitor settles at a lower final voltage.
The same idea explains why every current and voltage in a single-capacitor circuit changes with the same time constant: there is only one differential equation, for the capacitor's charge, and every other quantity follows from it by Ohm's law and Kirchhoff's rules at each instant. Circuits with two capacitors have two time constants, and their behavior is a sum of two exponentials.
The first implantable pacemakers, developed in Minnesota in the late 1950s by Earl Bakken's company Medtronic and surgeons at the University of Minnesota, used an RC timing circuit to fire electrical pulses at a steady rate. A capacitor discharging through a resistor to a threshold sets the interval between beats, around $0.8$ s for about $75$ beats per minute.
Modern pacemakers use digital clocks and sense the heart's own rhythm, pacing only when needed, but capacitors still deliver each pulse, charged from the battery and discharged through the heart muscle in about a millisecond. Hundreds of thousands of Americans have pacemakers implanted each year.
A camera flash stores energy in a capacitor of a few hundred microfarads charged to about $300$ V. The charging circuit, drawing on small batteries, takes a few seconds, which is why flashes need time to recharge between shots. The discharge through the flash tube's low resistance lasts about a millisecond.
The ratio of those times, seconds to charge and a millisecond to discharge, comes from the two different resistances the capacitor sees: the charging circuit's high resistance and the flash tube's low one. The same capacitor, with two time constants, turns a trickle of battery power into a burst bright enough to light a room.
In steady DC circuits, no current flows through a capacitor, and it is easy to conclude that capacitors always block current. But at the moment a switch closes, an empty capacitor has no voltage across it and passes current freely, like a wire. The current falls only as charge builds up.
A related error is to think a capacitor's voltage can change instantly when a switch is thrown. The voltage follows the charge, which changes only as current flows; it is the current that can jump, never the capacitor's voltage.
A $100$ μF capacitor charges through $20$ kΩ from a $9.0$ V battery. Find the time constant.
$\tau = 20 \times 10^3 \times 100 \times 10^{-6} = 2.0\ \text{s}$
$RC$.
Find the initial current.
$I_0 = \dfrac{9.0}{20 \times 10^3} = 0.45\ \text{mA}$
The empty capacitor acts as a wire.
Find the final charge.
$Q = 100 \times 10^{-6} \times 9.0 = 0.90\ \text{mC}$
Full battery voltage.
Find the voltage after $2.0$ s.
$V_C = 9.0(1 - e^{-1}) = 5.7\ \text{V}$
One time constant.
Find when it reaches $8.0$ V.
$t = 2.0\ln\dfrac{9.0}{9.0 - 8.0} = 2.0\ln 9 = 4.4\ \text{s}$
Solve $1 - e^{-t/\tau} = 8/9$.
A $50$ μF capacitor at $200$ V discharges through $40$ kΩ. Find the time constant.
$\tau = 40 \times 10^3 \times 50 \times 10^{-6} = 2.0\ \text{s}$
$RC$.
Find the initial current.
$I_0 = \dfrac{200}{40 \times 10^3} = 5.0\ \text{mA}$
$V_0/R$.
Find the voltage after $3.0$ s.
$V = 200e^{-1.5} = 44.6\ \text{V}$
Exponential decay.
Find when it reaches $10$ V.
$t = 2.0\ln 20 = 6.0\ \text{s}$
$t = \tau\ln(V_0/V)$.
Find the initial energy.
$U = \tfrac{1}{2} \times 50 \times 10^{-6} \times 200^2 = 1.0\ \text{J}$
All of it becomes heat in the resistor.
Find the energy left after one half-life.
$U = 1.0 \times \left(\tfrac{1}{2}\right)^2 = 0.25\ \text{J}$
Energy goes as $V^2$, so it falls twice as fast.
A $12$ V battery, a $4.0$ kΩ resistor and a switch are in series with a $6.0$ kΩ resistor, and a $10$ μF capacitor is across the $6.0$ kΩ resistor. The switch closes with the capacitor empty. Find the battery current just after.
$I = \dfrac{12}{4.0\ \text{k}\Omega} = 3.0\ \text{mA}$
The empty capacitor shorts out the $6.0$ kΩ resistor.
Find the current long after.
$I = \dfrac{12}{10\ \text{k}\Omega} = 1.2\ \text{mA}$
The full capacitor passes no current.
Find the final capacitor voltage.
$V_C = 1.2 \times 6.0 = 7.2\ \text{V}$
Across the $6.0$ kΩ resistor.
Find the resistance the capacitor sees.
$R_{\text{th}} = \dfrac{4.0 \times 6.0}{10} = 2.4\ \text{k}\Omega$
With the battery replaced by a wire, the resistors are in parallel.
Find the time constant.
$\tau = 2.4 \times 10^3 \times 10 \times 10^{-6} = 24\ \text{ms}$
$R_{\text{th}}C$.
Write the capacitor voltage.
$V_C = 7.2\left(1 - e^{-t/24\ \text{ms}}\right)\ \text{V}$
From zero to its final value.
Find the battery current at $24$ ms.
$I = 1.2 + (3.0 - 1.2)e^{-1} = 1.86\ \text{mA}$
Every current moves from initial to final with the same $\tau$.
Find the time constant.
$\tau = 5.0 \times 10^3 \times 20 \times 10^{-6} = 0.10\ \text{s}$
$RC$.
Find the initial current.
$I_0 = \dfrac{50}{5000} = 10\ \text{mA}$
$V_0/R$.
Decay it by one time constant.
A $5$ kΩ resistor and a $4$ μF capacitor are connected in series with a battery. What is the circuit's time constant?
Complete the worked solution: an uncharged $7$ μF capacitor is charged fully through a resistor by a $5$ V battery. Find the final charge in μC, the work done by the battery in μJ, and the energy stored in the capacitor in μJ.
Multiply the capacitance by the emf.
$Q = C\mathcal{E} =$ q
At the end no current flows, so $V_C = \mathcal{E}$.
Find the battery's work.
$W = Q\mathcal{E} =$ w
Every bit of charge crosses the full emf.
Find the stored energy.
$U = \tfrac{1}{2}C\mathcal{E}^2 =$ u
The capacitor's voltage rose from zero.
Account for the difference.
$W - U = \tfrac{1}{2}C\mathcal{E}^2 \text{ as heat}$
Lost in the resistor, whatever its size.
Match each RC quantity to its expression.
| $RC$ | $C\mathcal{E}(1 - e^{-t/\tau})$ | $(V_0/R)e^{-t/\tau}$ | $\mathcal{E}/R$ | |
|---|---|---|---|---|
| time constant | ||||
| charge while charging | ||||
| current while discharging | ||||
| first-instant charging current |
An uncharged $4$ μF capacitor is connected through a $2$ kΩ resistor to a $6$ V battery. Fill in the current at the moment of connection in mA, the charge on the capacitor long afterward in μC, and the time constant in ms.
| value | |
|---|---|
| initial current (mA) | |
| final charge (μC) | |
| time constant (ms) |
An uncharged $4$ μF capacitor is connected at $t = 0$ through a $5$ kΩ resistor to a $6$ V battery. Write the voltage across the capacitor, in V, as a formula in $t$ (ms).
Answer:
A capacitor of $2$ μF, charged to $100$ V, discharges through a $8$ kΩ resistor. How long does it take for its voltage to fall to $10$ V, in ms?
Answer: ms to fall to 10 V
A biomedical engineer in Minneapolis models a simple pacemaker timing circuit: a $0.5$ μF capacitor discharges through a $1.5$ MΩ resistor, and a pulse fires when its voltage has fallen by a factor of $3$, after which it recharges instantly. At how many beats per minute does it pace the heart?
Answer: beats per minute
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An uncharged $2$ μF capacitor is connected at $t = 0$ through a $1$ kΩ resistor to a $7$ V battery. Write the voltage across the capacitor, in V, as a formula in $t$ (ms).
Answer:
You can analyze RC circuits. Explain to someone why an empty capacitor passes current at first but a full one does not.
21. Your turn: a $20$ μF capacitor at $50$ V discharges through $5.0$ kΩ. What is the current after $0.10$ s?, step 3
$I = 10e^{-1} = 3.7\ \text{mA}$
37 percent remains.