In short

Connect a battery of EMF \varepsilon, a resistor R, and an initially uncharged capacitor C in a single loop and close the switch. Charge accumulates on the capacitor as

\boxed{\;q(t) \;=\; C\varepsilon\left(1 - e^{-t/RC}\right)\;}

and the current in the loop decays as

\boxed{\;i(t) \;=\; \frac{\varepsilon}{R}\,e^{-t/RC}\;}.

If you instead start with a capacitor already charged to Q_{0} and short it through R (no battery), the charge bleeds away:

\boxed{\;q(t) \;=\; Q_{0}\, e^{-t/RC}\;}, \qquad i(t) \;=\; \frac{Q_{0}}{RC}\,e^{-t/RC}.

The single number that controls how fast either process happens is the time constant

\boxed{\;\tau \;=\; RC\;}.

Its unit is seconds (ohm \times farad = second). After t = \tau, the charging capacitor has reached 1 - e^{-1} \approx 63\% of its final charge; after t = 5\tau, it is at 99.3\% — close enough to "fully charged" for any engineering purpose.

During charging, the battery delivers total energy Q\varepsilon = CV^{2}. Exactly half, \tfrac{1}{2}CV^{2}, ends up stored in the capacitor as field energy. The other half is dissipated as heat in the resistor — and this split is independent of R. A 1 \Omega resistor wastes the same total energy as a 1 k\Omega resistor for the same capacitor and voltage; only the charging time differs.

Press the shutter button on the Canon point-and-shoot that comes out at every Indian wedding. You hear a faint whine. Wait two seconds. Now you can press again. The whine, the wait, the red "ready" LED: that is a capacitor charging through a resistor. Inside the camera, a tiny 3-volt battery feeds a step-up transformer that drives a 100 \muF electrolytic up to about 300 volts. The charging path has a resistance of about 4 k\Omega, set by the transformer's output impedance. The time constant is

\tau \;=\; RC \;=\; (4000\ \Omega)(100 \times 10^{-6}\ \text{F}) \;=\; 0.4\ \text{s}.

After two seconds — five time constants — the capacitor is at 1 - e^{-5} = 99.3\% of 300 V, and the ready light flicks on. The physics that decides whether the flash is ready is exactly the physics of this chapter.

The same equation runs the defibrillator in AIIMS's cardiac ward. A 32 \muF capacitor bank is charged from a rectifier at around 1000 ohms of source resistance to about 1600 V. Charging time: \tau = 1000 \times 32 \times 10^{-6} = 32 ms per time constant, so the "charged" indicator comes on roughly 160 ms later. When the doctor presses "shock", a relay disconnects the charger and connects the capacitor bank across the patient's chest, whose torso looks like about 50 \Omega. Discharge time constant: \tau_{\text{discharge}} = 50 \times 32 \times 10^{-6} = 1.6 ms. Same capacitor, different R, two very different \tau — slow charge, fast discharge. Understand \tau = RC and you understand why a defibrillator looks like it does.

A third example, closer to home: the intermittent-wipe mode on a Maruti Alto's wiper stalk. Twist the stalk and you get a wipe every few seconds instead of continuously. The delay between wipes is set by an RC timer on the wiper relay board — twisting the stalk varies a potentiometer that changes R, which changes \tau, which changes the delay. You already owned this circuit; you just did not know its equation.

The circuit and its equation

Take the charging case first. A battery of EMF \varepsilon, a resistor R, an initially uncharged capacitor C, and a switch, all in series.

RC charging circuitA battery of EMF epsilon in series with a switch, a resistor R, and an initially uncharged capacitor C. When the switch closes, current flows and the capacitor charges.εSRC+q−qi
A battery of EMF $\varepsilon$ drives current through a resistor $R$ into a capacitor $C$. At any moment the capacitor holds charge $q$ on its upper plate and $-q$ on its lower plate, and the loop current is $i$. The switch S is closed at $t=0$.

At the instant you close the switch, there is no charge on the capacitor, and therefore no voltage across it. The whole EMF sits across the resistor, so the current is as large as it can ever be: i(0) = \varepsilon/R. Charge begins to accumulate. As q grows, the voltage across the capacitor, q/C, grows with it, and the voltage left over for the resistor — \varepsilon - q/C — shrinks. So the current, i = (\varepsilon - q/C)/R, also shrinks. Eventually the capacitor voltage reaches \varepsilon, no voltage is left for the resistor, and the current is zero. The capacitor is "full".

That description is the physics. Now the algebra.

Kirchhoff's loop law

At time t, let the charge on the capacitor be q(t) and the loop current be i(t). Walk once around the loop starting from the negative terminal of the battery, in the direction of the current:

For the sum of voltage changes around a closed loop to be zero (Kirchhoff's voltage law),

\varepsilon - iR - \frac{q}{C} \;=\; 0.

Why: Kirchhoff's voltage law is conservation of energy for a unit charge making one lap around the loop. The charge cannot gain or lose net energy by returning to where it started.

The current is the rate at which charge piles onto the capacitor's positive plate, i = dq/dt. Substitute:

\varepsilon \;-\; R\,\frac{dq}{dt} \;-\; \frac{q}{C} \;=\; 0.

Rearrange to isolate the derivative:

\boxed{\;R\,\frac{dq}{dt} \;+\; \frac{q}{C} \;=\; \varepsilon.\;}

Why: this is a first-order linear ordinary differential equation. Its one unknown is q(t); the one piece of extra information you need to pin down a unique solution is the initial condition, q(0)=0 (uncharged at the moment of switching).

Solving the differential equation

Move the q/C term to the right-hand side and absorb the \varepsilon into it by factoring:

R\,\frac{dq}{dt} \;=\; \varepsilon - \frac{q}{C} \;=\; \frac{1}{C}\bigl(C\varepsilon - q\bigr).

Why: the quantity C\varepsilon is the final charge the capacitor will carry once the current stops — we may as well give it a name. Call it Q_{\infty} = C\varepsilon, the equilibrium charge.

So

R\,\frac{dq}{dt} \;=\; \frac{Q_{\infty} - q}{C}, \qquad \text{or}\qquad \frac{dq}{Q_{\infty} - q} \;=\; \frac{dt}{RC}.

Why: separating variables puts everything involving q on the left and everything involving t on the right. The equation is now ready to integrate term by term.

Integrate both sides, from t=0 (when q=0) to a generic t (when q = q):

\int_{0}^{q} \frac{dq'}{Q_{\infty} - q'} \;=\; \int_{0}^{t} \frac{dt'}{RC}.

The left-hand integral is -\ln(Q_{\infty} - q') evaluated between limits; the right-hand is t/(RC):

-\ln(Q_{\infty} - q) + \ln(Q_{\infty}) \;=\; \frac{t}{RC}.

Why: \int du/(a - u) = -\ln(a - u) (because d/du\,[-\ln(a-u)] = 1/(a-u), the negative signs cancel). Plug in the upper and lower limits — at t=0, q=0, so the integral contributes \ln Q_{\infty}.

Combine the logs using \ln A - \ln B = \ln(A/B):

\ln\!\left(\frac{Q_{\infty}}{Q_{\infty} - q}\right) \;=\; \frac{t}{RC}.

Exponentiate:

\frac{Q_{\infty}}{Q_{\infty} - q} \;=\; e^{t/RC}, \qquad\Longrightarrow\qquad Q_{\infty} - q \;=\; Q_{\infty}\,e^{-t/RC}.

Solve for q:

\boxed{\;q(t) \;=\; Q_{\infty}\left(1 - e^{-t/RC}\right) \;=\; C\varepsilon\left(1 - e^{-t/RC}\right).\;}

Why: this is the charging formula. At t=0, e^{0}=1, so q(0) = C\varepsilon(1-1) = 0 — consistent with an initially uncharged capacitor. At t \to \infty, e^{-t/RC} \to 0, so q \to C\varepsilon = Q_{\infty} — the capacitor settles at its final charge.

The current

The current is the derivative of the charge:

i(t) \;=\; \frac{dq}{dt} \;=\; C\varepsilon \cdot \frac{d}{dt}\!\left(1 - e^{-t/RC}\right) \;=\; C\varepsilon \cdot \frac{1}{RC}\,e^{-t/RC}.

The C cancels, leaving

\boxed{\;i(t) \;=\; \frac{\varepsilon}{R}\,e^{-t/RC}.\;}

Why: at t=0, i(0) = \varepsilon/R — the same current you would get if the capacitor were replaced by a plain wire. That makes sense: an uncharged capacitor has zero voltage across it, so at the instant of closing, it behaves like a short. As t \to \infty, i \to 0 — a fully charged capacitor blocks DC completely.

Charge grows, current decays, and both are governed by the same exponential, e^{-t/RC}. They are two sides of the same coin.

The time constant \tau = RC

The quantity RC that appears in every exponent has units of seconds. Check:

[R][C] \;=\; \Omega \cdot \text{F} \;=\; \frac{\text{V}}{\text{A}}\cdot\frac{\text{C}}{\text{V}} \;=\; \frac{\text{C}}{\text{A}} \;=\; \frac{\text{C}}{\text{C/s}} \;=\; \text{s}.

Why: a farad is a coulomb per volt; an ohm is a volt per ampere; an ampere is a coulomb per second. The volts cancel, and the ratio of a coulomb to a coulomb-per-second is a second.

Give RC the symbol \tau. It is the time constant of the circuit — the single number that says "how fast" or "how slow".

Interpreting \tau

At t = \tau, the exponent is -1, and e^{-1} \approx 0.368. For the charging capacitor,

q(\tau) \;=\; C\varepsilon\,(1 - e^{-1}) \;\approx\; 0.632 \, C\varepsilon.

In one time constant the capacitor has climbed to about 63% of its final charge. In two time constants it is at 1 - e^{-2} \approx 86.5\%. In five time constants it is at 1 - e^{-5} \approx 99.3\% — close enough that in engineering you call it "fully charged".

t e^{-t/\tau} q/Q_{\infty} (charging) i/(\varepsilon/R)
0 1.000 0% 100%
\tau 0.368 63.2% 36.8%
2\tau 0.135 86.5% 13.5%
3\tau 0.050 95.0% 5.0%
5\tau 0.007 99.3% 0.7%
10\tau 4.5\times10^{-5} 99.995% 0.005%

Watching q(t) grow

The charging curve is an exponential approach to Q_{\infty}. Watch the animation: the red dot traces the charge q(t), the red dashed line marks the asymptote Q_{\infty}, and the tick at \tau shows where the 63% line crosses.

Animated charge q(t) on a charging capacitorThe charge on the capacitor rises from zero, reaching 63 percent of its asymptote at one time constant, 95 percent at three, and 99 percent at five. A red dot traces the position of the charge at the current time.time t (in units of τ)q(t) / Q∞Q∞ (fully charged)0.6301.00τ2τ3τ4τ5τ
The charge $q(t)$ climbs along $1 - e^{-t/\tau}$, reaching 63% of the asymptote at $t = \tau$, 86.5% at $2\tau$, 95% at $3\tau$, and 99.3% at $5\tau$. The red dot is the live value; ghost markers appear at each integer multiple of $\tau$. Click replay to watch again.

Explore what happens when R or C changes

The only way the time constant \tau = RC can change is if R or C changes. Drag the slider below to scale \tau from 0.5 to 3.0 and watch the charging curve stretch or compress. The shape is always the same exponential; only the horizontal scale shifts.

Interactive: charging curve as tau variesA charging capacitor curve that reshapes as the user drags a slider for the time constant tau. A live readout shows the percentage charged at the fixed observation time of one second.time t (s)q(t) / Q∞01.000.631235t = 1 s(5τ is "fully charged")drag the red dot along t-axis to change τ
Drag the red dot along the $t$ axis to scale the time constant $\tau$. The curve stretches or compresses but never changes shape. The readout shows what percentage of $Q_{\infty}$ the capacitor has reached at the fixed observation time $t = 1$ s. When $\tau$ is small (fast charging), you are already near 100% at $t=1$ s. When $\tau$ is large (slow charging), you are still on the rising part of the curve.

Small \tau means either a small resistor (lots of current can flow) or a small capacitor (little charge needed to fill it). Large \tau means the opposite. This is the whole qualitative story: \tau sets the clock of the circuit.

Discharging — the time-reversed problem

Now take the opposite setup. No battery. The capacitor already carries charge Q_{0}. A switch connects it across the resistor.

RC discharge circuitA capacitor with initial charge Q0 is connected through a switch and a resistor R in a loop. When the switch closes, the charge bleeds off through R.C+Q₀−Q₀SRi
A charged capacitor connected across a resistor. When the switch closes, charge flows from the positive plate through $R$ back to the negative plate. There is no battery, so the loop law becomes $q/C + iR = 0$.

Kirchhoff's loop law around this loop (going from the positive plate, through the resistor, back to the negative plate):

\frac{q}{C} - iR \;=\; 0 \qquad\text{but now}\qquad i \;=\; -\frac{dq}{dt}

Why: because the positive plate is losing charge during discharge, the rate of change of q is negative — and the current is still positive if we measure it in the direction of actual flow. So i = -dq/dt keeps the conventions consistent.

Substitute:

\frac{q}{C} + R\,\frac{dq}{dt} \;=\; 0 \qquad\Longrightarrow\qquad \frac{dq}{dt} \;=\; -\frac{q}{RC}.

Why: the equation "dq/dt is proportional to -q" is the definition of exponential decay. Whatever q is, it is shrinking at a rate proportional to itself.

Separate and integrate:

\int_{Q_{0}}^{q} \frac{dq'}{q'} \;=\; -\int_{0}^{t} \frac{dt'}{RC}.
\ln q - \ln Q_{0} \;=\; -\frac{t}{RC} \qquad\Longrightarrow\qquad \ln\!\left(\frac{q}{Q_{0}}\right) = -\frac{t}{RC}.

Exponentiate:

\boxed{\;q(t) \;=\; Q_{0}\,e^{-t/RC}.\;}

Why: this is the discharge formula. At t=0, q = Q_{0} (the initial charge). As t \to \infty, q \to 0 — the capacitor bleeds completely dry. At t = \tau, q = Q_{0}/e \approx 0.368\,Q_{0} — the charge has fallen to 37% of its initial value in one time constant.

The discharge current is

i(t) \;=\; -\frac{dq}{dt} \;=\; -Q_{0}\,\frac{d}{dt}\,e^{-t/RC} \;=\; \frac{Q_{0}}{RC}\,e^{-t/RC}.

At t=0, the current is i_{0} = Q_{0}/(RC) = V_{0}/R, where V_{0} = Q_{0}/C is the initial voltage. This matches the physical picture: at the instant of closing, the capacitor voltage drives a current V_{0}/R through the resistor. Thereafter the voltage drops exponentially, and so does the current — same time constant, same curve, just falling instead of rising.

Energy bookkeeping — the universal 50–50 split

Here is a fact that looks surprising until you derive it: during the charging of a capacitor through a resistor, exactly half of the energy the battery delivers ends up stored in the capacitor, and exactly half is dissipated as heat in the resistor. The 50-50 split does not depend on R, or on C, or on \varepsilon. It is a pure consequence of q(t) = C\varepsilon(1 - e^{-t/RC}).

Total energy delivered by the battery

During the charging, the battery pushes a total charge Q = C\varepsilon through itself (that is how much ends up on the capacitor's positive plate). Each coulomb crosses the EMF, gaining energy \varepsilon. So the total work the battery does is

W_{\text{battery}} \;=\; Q\varepsilon \;=\; C\varepsilon^{2}.

Why: the battery is a constant-voltage source. It does work \varepsilon\,dq on each increment of charge pushed. Integrating over the full charge gives \varepsilon\,Q.

Energy stored in the capacitor

From the Energy Stored in a Capacitor chapter, the final energy in the capacitor is

U_{\text{cap}} \;=\; \tfrac{1}{2}C\varepsilon^{2}.

Why: this is \tfrac{1}{2}CV^{2} with V = \varepsilon. The factor of \tfrac{1}{2} comes from the charge piling on against a voltage that rises linearly from 0 to \varepsilon — the average voltage during charging is \varepsilon/2.

Energy dissipated in the resistor

This is the interesting bit — let us compute it directly, not just by subtraction. The power dissipated in the resistor at time t is i^{2}(t)R. Using i(t) = (\varepsilon/R)e^{-t/RC},

P_{R}(t) \;=\; i^{2}(t)\,R \;=\; \left(\frac{\varepsilon}{R}\right)^{2} e^{-2t/RC}\cdot R \;=\; \frac{\varepsilon^{2}}{R}\,e^{-2t/RC}.

The total energy dissipated is the integral of power over all time:

W_{R} \;=\; \int_{0}^{\infty} P_{R}(t)\,dt \;=\; \frac{\varepsilon^{2}}{R}\int_{0}^{\infty} e^{-2t/RC}\,dt.

Why: the integral of e^{-2t/RC} from 0 to \infty is [-\tfrac{RC}{2}\,e^{-2t/RC}]_{0}^{\infty} = \tfrac{RC}{2}. The exponential dies off completely for large t, so the upper limit contributes 0, and the lower limit contributes -RC/2; the negative sign of the antiderivative flips it to +RC/2.

W_{R} \;=\; \frac{\varepsilon^{2}}{R} \cdot \frac{RC}{2} \;=\; \tfrac{1}{2}C\varepsilon^{2}.

And there it is:

W_{\text{battery}} \;=\; C\varepsilon^{2}, \qquad U_{\text{cap}} \;=\; \tfrac{1}{2}C\varepsilon^{2}, \qquad W_{R} \;=\; \tfrac{1}{2}C\varepsilon^{2}.

The resistor dissipates as much heat as the capacitor stores. Notice that R has cancelled out of W_{R}: a small resistor runs hot for a short time, a big resistor warm for a long time, but the total heat is the same.

Why this is surprising: you would think a bigger resistor "wastes" more energy. It does not. It just takes longer. For the same final state (capacitor charged to \varepsilon), the total waste is fixed. The only way to charge a capacitor without wasting half the energy is to charge it slowly through a source whose voltage ramps along with the capacitor — which is what a switching power supply does when it uses an inductor instead of a plain resistor.

Worked examples

Example 1: Camera flash capacitor

A point-and-shoot camera has a 100 μF capacitor that must be charged to 300 V by a step-up circuit whose Thevenin source voltage is 300 V and whose source resistance is 4.0 kΩ. The switch closes at t=0; the LED "ready" light comes on when the capacitor has reached 95% of its final charge. How long does the reader wait between shots?

Camera flash charging circuitA 300 V source through a 4 kilohm resistor charges a 100 microfarad capacitor. The ready LED lights when the capacitor reaches 95 percent of the supply voltage.300 V4 kΩ100 μFready LED
The charging loop inside a camera. The LED lights up when the capacitor voltage crosses the threshold set by a comparator.

Step 1. Identify the knowns.

R = 4.0 \times 10^{3}\ \Omega, C = 100 \times 10^{-6}\ \text{F}, \varepsilon = 300 V, target q/Q_{\infty} = 0.95.

Why: the EMF value never enters the time calculation — the fractional approach to full charge is set entirely by R, C, and the threshold percentage.

Step 2. Compute the time constant.

\tau \;=\; RC \;=\; (4.0\times10^{3})(100\times10^{-6}) \;=\; 0.40\ \text{s}.

Step 3. Solve for t at which q/Q_{\infty} = 0.95.

The charging law says

\frac{q(t)}{Q_{\infty}} \;=\; 1 - e^{-t/\tau} \;=\; 0.95 \qquad\Longrightarrow\qquad e^{-t/\tau} \;=\; 0.05.

Take natural logs:

-\frac{t}{\tau} \;=\; \ln(0.05) \;=\; -2.996 \qquad\Longrightarrow\qquad t \;=\; 2.996\,\tau.

Why: you want 1 - e^{-t/\tau} = 0.95, so e^{-t/\tau} = 0.05. The natural log of 0.05 is -\ln 20 \approx -2.996. About three time constants.

Step 4. Multiply through.

t \;=\; 2.996 \times 0.40\ \text{s} \;\approx\; 1.2\ \text{s}.

Result. The reader must wait about 1.2 seconds between shots for the 100 μF, 4 kΩ flash circuit to reach 95% charge. At 5τ = 2.0 s the flash is at 99.3%, indistinguishable from full.

What this shows. The wait time between flashes is engineered — the camera designer picked R and C together so that \tau comes out to roughly 0.4 s, giving the user a photograph-ready flash in about 1-2 s. Halve R and you halve the wait, but the charging current doubles and the little step-up transformer might not survive. Everything in an RC circuit is a balance of \tau versus current.

Example 2: Defibrillator discharge through the torso

A defibrillator bank of 32 μF is charged to 1600 V. When the paddles are placed on the patient's chest (effective resistance 50 Ω), the bank is connected across the torso and discharges through it. (a) Find the initial current. (b) Find the time at which the voltage has dropped to 10% of its initial value. (c) How much energy is delivered to the patient?

Animated decay of voltage on a defibrillator capacitorThe fraction V(t)/V0 on a 32 microfarad capacitor discharging through a 50 ohm torso starts at 1.0 at t=0 and decays to about 0.01 at 5 time constants. A red dot traces the decaying voltage.time (ms)V/V₀101.63.28
The voltage on the defibrillator capacitor as a function of time in milliseconds. The red dot marks the current value; in one time constant $\tau = 1.6$ ms it has fallen to 37%; in 5$\tau$ = 8 ms, essentially to zero.

Step 1. Compute the time constant.

\tau \;=\; RC \;=\; 50\ \Omega \times 32 \times 10^{-6}\ \text{F} \;=\; 1.6\times10^{-3}\ \text{s} \;=\; 1.6\ \text{ms}.

Step 2. Initial current.

At t=0, the capacitor voltage is the full 1600 V, and it drives the current through 50 Ω:

i_{0} \;=\; \frac{V_{0}}{R} \;=\; \frac{1600}{50} \;=\; 32\ \text{A}.

Why: a capacitor, in this instant, looks exactly like a battery of EMF V_{0}. Current is V/R.

Step 3. Time to reach 10% of initial voltage.

V(t)/V_{0} \;=\; e^{-t/\tau} \;=\; 0.10 \quad\Longrightarrow\quad t \;=\; \tau\ln(10) \;=\; 1.6\,\text{ms} \times 2.303 \;\approx\; 3.7\ \text{ms}.

Step 4. Energy delivered.

All of the capacitor's stored energy goes into the patient (the only resistor in the circuit):

U \;=\; \tfrac{1}{2}CV_{0}^{2} \;=\; \tfrac{1}{2}(32\times10^{-6})(1600)^{2} \;=\; \tfrac{1}{2}\times 32 \times 10^{-6} \times 2.56\times 10^{6} \;=\; 41\ \text{J}.

Commercial defibrillators are typically set for 200 J in adult shock mode; to reach that, the manufacturer uses a bigger capacitor or higher voltage. The (32\ \mu\text{F}, 1600\ \text{V}) pair above gives one of the smaller 40-J training units.

Why: U = \tfrac{1}{2}CV^{2} counts every joule stored in the field between the plates; once connected across the torso, no other element stores energy, so all 41 J heats the tissue (and, crucially, stimulates the heart).

Result. (a) i_{0} = 32 A. (b) t \approx 3.7 ms. (c) U = 41 J.

What this shows. A few milliamps for a long time can pass through a human without harm; 32 amps for a few milliseconds, on the other hand, is enough to interrupt the uncoordinated electrical activity of ventricular fibrillation and let the heart re-synchronise. It is the peak current that matters, and the peak current is V_{0}/R — no \tau anywhere. But the total charge delivered is Q = CV_{0} = 0.051 C, and the energy is \tfrac{1}{2}CV_{0}^{2}, both of which care about the capacitor.

Example 3: Wiper delay timer on a Maruti

The intermittent-wipe relay on a car uses a 470 μF capacitor discharging through a variable resistor to set the delay between wipes. When q(t) falls to half its initial value, a comparator fires and the wiper sweeps once. The user can twist the stalk to set R anywhere from 1 kΩ (fastest wipe) to 30 kΩ (slowest wipe). What is the range of delays?

Intermittent wiper timing circuitA 470 microfarad capacitor is charged and then discharged through a variable resistance R that the user selects with the wiper stalk. When the capacitor voltage falls to half of its charged value, a comparator fires the wiper motor for one sweep.+R (1 to 30 kΩ)470 μFcomparator fires at V/2
The timing side of the wiper relay board. The capacitor is topped up between sweeps and bleeds through $R$ until the comparator detects that its voltage has fallen to half.

Step 1. Set up the condition.

The capacitor discharges by q(t) = Q_{0}\,e^{-t/RC}. The comparator fires when q = Q_{0}/2, so

e^{-t/RC} \;=\; \tfrac{1}{2} \quad\Longrightarrow\quad t \;=\; RC\,\ln 2 \;\approx\; 0.693\,RC.

Why: "half-life" of an exponential is always \ln 2 \approx 0.693 time constants. The same number that governs nuclear decay governs this wiper timer.

Step 2. Fastest setting (R = 1 kΩ).

t_{\text{fast}} \;=\; 0.693 \times 1000 \times 470\times10^{-6} \;=\; 0.33\ \text{s}.

Step 3. Slowest setting (R = 30 kΩ).

t_{\text{slow}} \;=\; 0.693 \times 30{,}000 \times 470\times10^{-6} \;=\; 9.8\ \text{s}.

Result. The wiper delay ranges from about 0.33 s (fastest) to 9.8 s (slowest).

What this shows. Two orders of magnitude of delay from a single twist of a knob, because \tau scales linearly with R. The \ln 2 \approx 0.693 factor — same number you remember from radioactivity — sets the half-life. Every RC timer in every electronic device uses exactly this relationship.

Common confusions

If you came here to handle RC circuits in JEE and circuit analysis, you have what you need. What follows is (a) the phasor / complex-impedance picture for sinusoidal AC, (b) the RC integrator/differentiator — the key signal-processing identity that drops out of the same equation, and (c) a quick look at multi-loop RC problems and Thevenin reduction.

The complex impedance of a capacitor

In an AC circuit driven at angular frequency \omega, the instantaneous voltage v(t) = V_{m}\cos(\omega t) across a capacitor is related to the instantaneous current by i = C\,dv/dt, so

i(t) \;=\; -C V_{m}\omega \sin(\omega t) \;=\; C V_{m}\omega\cos(\omega t + \pi/2).

Why: the capacitor current leads the voltage by \pi/2 radians (90°). Differentiating a cosine produces a negative sine, which is cosine shifted by +\pi/2.

Define the capacitive reactance X_{C} = 1/(\omega C) and the complex impedance Z_{C} = -j/(\omega C) = 1/(j\omega C). A resistor-capacitor loop driven at angular frequency \omega has total impedance

Z \;=\; R + \frac{1}{j\omega C} \;=\; R - \frac{j}{\omega C}.

The magnitude is |Z| = \sqrt{R^{2} + 1/(\omega C)^{2}}, and the phase angle between voltage and current is \tan\phi = -1/(\omega R C) = -1/(\omega\tau). When \omega\tau \gg 1 (high frequency), the capacitor dominates; when \omega\tau \ll 1 (low frequency), the resistor dominates.

The transient-response world and the AC world are the same physics in different clothes: in both cases the relevant scale is \tau = RC. At \omega = 1/\tau, the RC network hits its corner frequency — and the phase flips smoothly from 0° to −90° across a factor-of-ten range around it.

RC integrator and differentiator

Consider the voltage across the capacitor in the charging loop. From q = Cv_{C} and Kirchhoff, v_{R} = iR and v_{R} + v_{C} = v_{\text{in}}, so

v_{C} + RC\,\frac{dv_{C}}{dt} \;=\; v_{\text{in}}(t).

If \tau = RC is much larger than the timescale over which v_{\text{in}} varies, the v_{C} term is small compared to the derivative term, and v_{R} \approx v_{\text{in}}. Then i \approx v_{\text{in}}/R, and

v_{C}(t) \;=\; \frac{1}{C}\int i\,dt \;\approx\; \frac{1}{RC}\int v_{\text{in}}(t)\,dt.

The output across the capacitor is (approximately) the integral of the input. This is the RC integrator, and it is the simplest way to smooth a signal or to generate a triangular wave from a square wave.

Conversely, if \tau is much smaller than the input timescale, then v_{C} \approx v_{\text{in}} and the voltage across the resistor becomes

v_{R}(t) \;=\; iR \;=\; RC\,\frac{dv_{C}}{dt} \;\approx\; RC\,\frac{dv_{\text{in}}}{dt}.

The output across the resistor is (approximately) the derivative of the input — an RC differentiator. Take a square wave in; spikes come out at every edge.

Every oscilloscope has a front-end with both these circuits buried inside. The physics is the same differential equation; only the choice of which node to observe changes.

Thevenin reduction for RC circuits

Suppose a capacitor is buried in a network of many resistors and sources, with a switch that changes the network at t=0. For the purpose of predicting q(t), you can always reduce the rest of the network to a single Thevenin equivalent.

Procedure.

  1. Remove the capacitor. Look at the two nodes where it was connected.
  2. Compute the open-circuit voltage across those nodes, V_{\text{Th}}. This is the asymptotic (steady-state) voltage the capacitor will charge to.
  3. Compute the resistance seen between those nodes with all independent sources turned off (voltages shorted, currents opened), R_{\text{Th}}.
  4. The circuit is then equivalent to a single EMF V_{\text{Th}} in series with R_{\text{Th}} and C. Apply the charging/discharging formulas with \tau = R_{\text{Th}}\,C.

This reduction is exact. No matter how tangled the network, the capacitor's response is a single exponential with a single time constant. The reason is that a first-order linear differential equation has only one eigenvalue, and Thevenin's theorem guarantees you can always write the equation in first-order form.

Why the 50-50 split is universal

Return to the charging computation. The battery delivers C\varepsilon^{2}; the capacitor stores \tfrac{1}{2}C\varepsilon^{2}; the resistor dissipates \tfrac{1}{2}C\varepsilon^{2}. Where did the asymmetry come from?

The answer is subtle: the battery delivers energy \varepsilon per coulomb to every coulomb it pushes. But the capacitor stores energy v(t)\,dq per coulomb, where v starts at 0 and rises linearly to \varepsilon. So the capacitor stores, on average, \varepsilon/2 per coulomb. The other \varepsilon/2 per coulomb must go somewhere — and the only dissipative element is the resistor.

So the 50-50 split is really a statement about the shape of the voltage-versus-charge curve during charging. A straight line from 0 to \varepsilon has a triangle under it (area \tfrac{1}{2}Q\varepsilon) and a triangle above it (also area \tfrac{1}{2}Q\varepsilon), both adding to the full rectangle Q\varepsilon of battery work. The below-the-line triangle is stored; the above-the-line triangle is dissipated.

To beat 50-50, you must make the source voltage track the capacitor voltage — that is, charge in many small steps, each from a slightly higher source voltage. In the limit of infinitely many steps, the dissipation goes to zero. This is the principle of adiabatic switching and is the reason modern DC-DC converters use inductors and switches rather than plain resistors.

The RC circuit and the atomic nucleus

A final beautiful coincidence. The equation dq/dt = -q/\tau — exponential decay with a single decay constant — is the same equation that governs radioactive decay. In both cases, the rate at which "amount" decreases is proportional to the amount currently present. The half-life of carbon-14 is 5730 years; the half-life of a 1 MΩ, 1 μF RC circuit is \ln 2 \times RC = 0.693 seconds. Different physics, same mathematics, same \ln 2 factor. The exponential is the universal answer to "how does something that decays at a rate proportional to itself decay?"

Where this leads next