Why a capacitor never fills up all at once: the RC time constant

How to use this simulation

1. Start with the defaults: a 12 V battery, a 10 kΩ resistor and a 100 µF capacitor. The switch is on “Charging”. The time constant shown in the readings is τ = 1 s.
2. Watch the top graph. The capacitor voltage climbs steeply and then flattens towards the dashed green line at 12 V. The orange dots on the wire move fast at the start and slow down as the current dies away.
3. Look at the marked point at 1τ on both graphs. The voltage has reached 63.2% of its final value, which is 7.59 V, while the current has dropped to 36.8% of its starting value of 1.20 mA.
4. Move the resistor slider up. The time constant grows, so everything happens more slowly, but the final voltage stays the same. Now move the capacitor slider up and see the same slowing, plus a larger charge in the readings.
5. Flip the switch to “Discharging” once the capacitor is nearly full. The battery is cut out, the plates push their charge back through the resistor, the current reverses and both curves fall away exponentially.
6. Change E in the middle of a run. The graphs restart from wherever the capacitor stands, and the voltage heads for the new target. Compare the energy reading with ½CV² on paper.

A capacitor stores charge, a resistor limits the flow

A capacitor is two metal plates with a gap between them. When a battery pushes charge on, one plate gathers positive charge and the other an equal negative charge. The more charge sits on the plates, the higher the voltage across them. The ratio of charge to voltage is a fixed number for each capacitor, called its capacitance C.

If the capacitor were wired straight to the battery it would fill almost instantly. A resistor in the path changes that. The resistor limits how much current can flow, so charge arrives gradually. The combination is called an RC circuit, and its speed is set by just two numbers, R and C.

RC circuits are everywhere: the delay before a phone screen dims, the flash on a camera, the smoothing of a power supply, the blinking of an indicator, and the timers in countless gadgets. This one picture explains all of them.

Why the charging slows down

Think about the voltage across the resistor at each moment. At the very start the capacitor is empty, so it has no voltage, and the whole battery voltage E falls across the resistor. Ohm’s law gives the starting current E ÷ R. With the defaults that is 12 ÷ 10 = 1.20 mA (volts divided by kilo-ohms gives milliamperes).

As charge builds up, the capacitor develops its own voltage V that opposes the battery. The voltage left over for the resistor is only E − V. A smaller voltage across the same resistor means a smaller current. So the more the capacitor has filled, the slower it fills.

That feedback, where the rate of filling is proportional to how much room is left, always produces an exponential curve. It is the same shape as a cup of tea cooling, a bucket with a hole draining, or a radioactive sample decaying.

The charging formulas

Solving the idea above exactly gives two short formulas. The first is the voltage across the capacitor, the second is the current in the circuit. Both are written with the number e ≈ 2.718 and the time t since the switch was closed.

• At t = 0 the exponential is 1, so V = 0 and I = E ÷ R. The capacitor looks like a plain wire at the first instant.
• As t becomes large, the exponential dies away, so V approaches E and I approaches zero. A fully charged capacitor blocks steady current, like a broken wire.
• The voltage never quite reaches E in finite time. In practice, after a few time constants the difference is too small to measure.

The time constant τ = RC

The combination RC appears inside the exponential. It has units of time, and we give it its own symbol: τ = RC, the time constant. It is the single number that tells you how quick or slow the circuit is.

Check the units with the simulation’s own values. R = 10 kΩ = 10,000 Ω and C = 100 µF = 0.0001 F. Then τ = 10,000 × 0.0001 = 1 s. Nothing else needs converting. In general, kilo-ohms multiplied by micro-farads gives milliseconds: 10 × 100 = 1,000 ms.

Put t = τ into the charging formula and the exponential becomes e⁻¹ = 0.368. So after one time constant the capacitor has covered 63.2% of the way to E, and the current has fallen to 36.8% of its starting value. This is the fact to remember. It holds for every R, every C and every E.

How many time constants until it is “full”?

Because the curve only approaches E, engineers use time constants as a measuring stick. The table gives the charge level reached and the voltage for the default circuit, where E = 12 V.

Discharging: the same curve, running downhill

Flip the switch to “Discharging” and the battery is out of the picture. The capacitor now acts as a small battery pushing its charge back through the resistor. If it starts at V₀, the voltage and current both die away exponentially with the same τ.

The current flows the opposite way compared with charging, so the current graph in the simulation goes below the zero line. The size still decays by the same factor. After one time constant the capacitor has lost 63.2% of its voltage and keeps 36.8% of it.

Energy stored, and where the rest goes

The energy stored in a charged capacitor is U = ½ C V². With the default circuit fully charged to 12 V, that is ½ × 100 µF × 12² = 7,200 µJ, or 7.20 mJ. The charge on a plate is Q = CV = 100 × 12 = 1,200 µC.

The battery, though, had to push the whole charge Q through a potential difference E, so it gave up Q × E = C E² = 14.40 mJ. Only half of that is stored. The other half, 7.20 mJ, was turned into heat in the resistor on the way.

This surprises most students, because it does not depend on R at all. A bigger resistor charges the capacitor more slowly, but it wastes exactly the same total energy. Try different values of R in the simulation and compare the final energy readout: it never changes.

Worked example 1: find the time for a given voltage

A 9 V supply charges a 470 µF capacitor through 47 kΩ. How long until the capacitor reaches 6 V?

Worked example 2: a discharge

A capacitor is charged to 12 V and then discharged through 10 kΩ with C = 100 µF, so τ = 1 s. Two questions.

What is the voltage after 2 s? Here t/τ = 2.00, so V = 12 × e^(−2.00) = 1.62 V.

How long until it falls to 1 V? From V = V₀ e^(−t/τ) we get t = τ ln(V₀/V) = 1 × ln(12) = 1 × 2.485 = 2.48 s.

The starting current is 12 ÷ 10 = 1.20 mA. After three time constants it is 1.20 × e⁻³ = 0.06 mA, only 5% of where it began.

Worked example 3: a flash capacitor

A camera-flash style capacitor of 1,000 µF is charged to 24 V. The charge on a plate is 1,000 µF × 24 V = 24.00 mC. The energy is ½ × 1,000 µF × 24² = 288 mJ, which is 0.288 J.

That is as much energy as lifting a 100 g apple through 29 cm. It sounds small, but it is released in a thousandth of a second or less, so the power is large. That is why a flash is bright and why a capacitor is the right tool for short bursts.

Rules of thumb for exams

These shortcuts save time in multiple-choice and in numericals.

• After 1τ: 63.2% charged. After 2τ: 86.5%. After 3τ: 95.0%. After 5τ: 99.3%.
• To reach 50% takes 0.693 τ = 0.69 s here. To reach 90% takes 2.303 τ = 2.30 s. To reach 99% takes 4.605 τ = 4.61 s.
• The initial current is E ÷ R no matter what C is. The final voltage is E no matter what R is.
• Doubling R or doubling C doubles τ. The final charge depends on C and E only.
• The slope of the voltage curve at t = 0 is E ÷ τ. If the capacitor kept charging at that starting rate it would reach E in exactly one τ. The tangent drawn at the start meets the final-value line at t = τ.
• The capacitor voltage cannot jump. The current through a resistor can.

Common mistakes

These four cost the most marks.

• Thinking the capacitor is full at t = τ. At one time constant it is only 63% charged.
• Mixing units. Use ohms with farads, or kilo-ohms with micro-farads (and read the answer in milliseconds). Never kilo-ohms with farads.
• Saying the battery’s energy all ends up in the capacitor. Exactly half goes into heat in the resistor.
• Forgetting the sign of the discharge current. It runs the opposite way to the charging current, even though it decays the same way.

Revision list

Say these out loud before the exam.

• τ = RC, measured in seconds when R is in ohms and C is in farads.
• Charging: V = E(1 − e^(−t/RC)) and I = (E/R) e^(−t/RC).
• Discharging: V = V₀ e^(−t/RC), current reversed.
• At t = τ: 63.2% charged, current down to 36.8%.
• Stored energy is ½CV². The battery gives CE²; half is lost as heat in R.
• The capacitor acts as a wire at t = 0 and as a gap when fully charged.

Frequently asked questions