Capacitors and RC Circuits: Charging, Discharging and the Time Constant

A camera flash charges for a second or two after you press the shutter, then dumps that stored energy in a fraction of a millisecond — bright enough to freeze motion, gone before your eye can register it as anything but a flash. Both halves of that story, the slow charge and the near-instant discharge, are governed by the same handful of equations: the physics of a resistor and a capacitor wired together, an RC circuit.

This page is three linked interactive calculators in one. Drag through an RC circuit's charge and discharge curve and read off the voltage, current and stored charge at any moment. Wire capacitors together in series and parallel and watch the combination rules run backwards from resistors. Then build a parallel-plate capacitor from scratch and see why real capacitors need a dielectric material to be useful at all.

What Is an RC Circuit and What Is the Time Constant?

An RC circuit is simply a resistor and a capacitor connected in a loop, usually with a battery and a switch to start the charging or discharging process. When a capacitor charges through a resistor, its voltage doesn't jump instantly to the supply voltage — the resistor limits how fast charge can flow onto the plates, so the voltage climbs smoothly, fast at first and progressively more slowly as it approaches the supply voltage.

The time constant, $\tau = RC$, measured in seconds when $R$ is in ohms and $C$ is in farads, sets the pace of that climb. It is the single number that answers "how fast?" for any RC circuit, and it shows up everywhere in the two governing equations:

$V_{charging}(t) = V_0\left(1 - e^{-t/\tau}\right) \qquad\qquad V_{discharging}(t) = V_0\, e^{-t/\tau}$

Why Is the Time Constant Useful? What Happens at t = τ?

Set $t = \tau$ in either formula and the exponential term becomes $e^{-1} \approx 0.368$. That gives two facts worth memorising outright:

• Charging: after one time constant, the capacitor has reached $1 - 0.368 = 63.2\%$ of the supply voltage.
• Discharging: after one time constant, $36.8\%$ of the original voltage still remains — equivalently, $63.2\%$ of it is gone.

This is the same shape of curve as radioactive decay, just with a different "half-life" analogue: instead of halving every fixed interval, an RC circuit crosses 63% of its journey every $\tau$. After five time constants ($5\tau$), the transient is $99.3\%$ complete — close enough to "done" that circuit designers almost universally treat $5\tau$ as "fully charged" or "fully discharged."

Interactive RC Circuit Simulator

Choose charging or discharging, set the resistor, capacitor and supply voltage, then drag the Elapsed Time slider — measured in units of $\tau$, so the same slider always means the same fraction of the way through the transient, no matter what R and C you've chosen. Watch the capacitor symbol in the schematic fill with colour exactly as fast as the number in the status readout climbs, and use the green marker on the curve below to read off the exact voltage at any instant.

How Do You Calculate Current During Charging and Discharging?

The current flowing through the resistor follows the same exponential shape as the voltage, in both directions:

$I(t) = \frac{V_0}{R}\,e^{-t/\tau}$

At the very instant charging begins ($t=0$), the capacitor looks like a dead short (zero voltage across it), so the current is at its absolute maximum, $I_0 = V_0/R$ — limited only by the resistor. As the capacitor's own voltage builds up and opposes the supply, the current dies away toward zero. Discharging tells the same story in reverse: current starts at $I_0$ (now flowing the opposite way, out of the capacitor) and decays toward zero as the capacitor empties.

Why Does the Charging Current Decrease Over Time?

Think of charging a capacitor like filling a balloon by mouth: at the start, the balloon is slack and easy to inflate quickly. As it fills, the balloon pushes back harder, and each additional breath adds air more slowly for the same effort. A capacitor's own voltage plays that "pushing back" role — as $V_C$ rises, it opposes more and more of the driving voltage $V_0$, leaving less and less "net push" ($V_0 - V_C$) to force current through the resistor.

How Do Capacitors Combine in Series and Parallel?

Wire two capacitors together and something backwards happens compared to resistors. Put resistors in series and their resistance adds; put capacitors in series and their combined capacitance drops below the smallest one in the chain. The rule flips again for parallel: parallel resistors give less resistance than the smallest one, while parallel capacitors give more capacitance than the largest.

$\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \qquad\qquad C_{parallel} = C_1 + C_2 + \cdots$

The reason traces back to what stays the same across each wiring. Capacitors in series all carry the same charge — whatever charge flows onto the first plate has nowhere else to go but through the whole chain, so every capacitor ends up storing an identical amount, and the supply voltage divides up across them according to each one's own $V=Q/C$. Capacitors in parallel all see the same voltage — they're wired directly across the same two points — so each stores charge in proportion to its own capacitance, and the charges simply add. It's the same structure as resistors with charge standing in for current — series shares the invariant quantity and divides voltage, parallel shares voltage and divides the invariant quantity — but the combination formula itself flips: resistors add directly in series and reciprocal-add in parallel, capacitors do the opposite. See our series and parallel circuits post for the resistor side of that comparison.

Switch between series and parallel above and watch the equivalent capacitance move in the opposite direction from what the same numbers would do as resistors.

What Determines a Capacitor's Capacitance?

For the simplest capacitor geometry — two flat parallel plates facing each other — capacitance depends only on geometry and what fills the gap:

$C = \frac{\varepsilon_0 \varepsilon_r A}{d}$

where $A$ is the overlapping plate area, $d$ is the plate separation, $\varepsilon_0 = 8.85\times10^{-12}$ F/m is the permittivity of free space, and $\varepsilon_r$ is the relative permittivity (dielectric constant) of the insulating material between the plates. Bigger plates or a thinner gap both raise capacitance; so does a better dielectric — which is why real capacitors are packed with a thin insulating film rather than left as an air gap. See how the electric field between charged plates is what a capacitor stores its energy in.

Push the sliders toward realistic lab-scale numbers — a plate the size of a large stamp, a millimetre of air gap — and the result lands in the picofarad range, thousands of times smaller than the microfarad capacitors used above. That gap is why practical capacitors don't use flat air-gapped plates at all: manufacturers roll or stack many layers of thin foil separated by a dielectric film only micrometres thick, packing a huge effective area and a tiny separation into a component the size of a fingertip. The energy stored, $U = \frac{1}{2}CV^2$, grows with the square of voltage, which is why the same small capacitor charged to a higher voltage can release a disproportionately bigger burst of energy — exactly the trick a camera flash relies on.

Worked Examples for Physics Exams

Example 1: Voltage at one time constant

A 100 μF capacitor charges through a 1 kΩ resistor from a 9 V supply. Find the time constant and the voltage after one time constant has passed.

$\tau = RC = 1000 \times (100\times10^{-6}) = 0.1$ s. At $t = \tau$ ($m=1$): $V_C = 9(1 - e^{-1}) = 9 \times 0.6321 \approx 5.69$ V — the universal 63.2% figure. Verify in the simulator: these are the default settings; set the Elapsed Time slider to 1.00 × τ and the readout shows V ≈ 5.69 V.

Example 2: Current at the same instant

Using the same circuit as Example 1, find the current flowing at $t = \tau$.

$I = (V_0/R)\,e^{-1} = (9/1000) \times 0.3679 \approx 0.00331$ A $= 3.31$ mA — this can also be found from Ohm's law applied to the resistor at that instant (see our Ohm's law calculator for the underlying $V=IR$ relationship): the voltage across the resistor is $V_0 - V_C = 9 - 5.69 = 3.31$ V, so $I = 3.31\text{V}/1000\,\Omega = 3.31$ mA — matching.

Example 3: Finding an unknown resistor

A 220 μF capacitor is charged from a 12 V supply, and the circuit's time constant is measured to be 2.2 s. Find the resistance.

Rearranging $\tau = RC$ for $R$: $R = \tau/C = 2.2 / (220\times10^{-6}) = 10{,}000\ \Omega = 10\ \text{k}\Omega$.

Example 4: Two capacitors in series

A 4 μF and a 12 μF capacitor are connected in series across a 9 V supply. Find the equivalent capacitance, the charge on each capacitor, and the voltage across each.

$1/C_{eq} = 1/4 + 1/12 = 3/12 + 1/12 = 4/12$, so $C_{eq} = 3$ μF — smaller than either individual capacitor, as series combinations always are. Both capacitors carry the same charge, $Q = C_{eq}V = 3\times9=27$ μC. Their voltages: $V_1 = Q/C_1 = 27/4 = 6.75$ V, $V_2 = Q/C_2 = 27/12 = 2.25$ V — summing to $9$ V, matching the supply. Verify in the simulator: these are the default settings.

Example 5: Energy stored in a parallel-plate capacitor

A parallel-plate capacitor has an area of 100 cm² and a plate separation of 1 mm, with air between the plates ($\varepsilon_r=1$). Find its capacitance and the energy stored when charged to 12 V.

$C = \varepsilon_0\varepsilon_r A/d = (8.85\times10^{-12})(1)(0.01)/(0.001) = 8.85\times10^{-11}$ F $\approx 88.5$ pF. Energy: $U = \frac12 CV^2 = 0.5\times(8.85\times10^{-11})\times12^2 \approx 6.37\times10^{-9}$ J $= 6.37$ nJ. Verify in the simulator: these are the default settings.