What is the time constant of an RC circuit?

It is the product RC, measured in seconds when R is in ohms and C in farads. After one time constant a charging capacitor has reached 63.2% of the supply voltage, and a discharging one has fallen to 36.8% of where it started.

Why is the curve exponential?

The current is set by the voltage still left across the resistor, I = (E − V)/R, and that current is the rate at which charge builds up. So the rate of change is proportional to the gap that remains, and any quantity that shrinks in proportion to itself decays exponentially.

How do I find RC from a graph?

Plot ln V (for discharge) or ln I against t. The points lie on a straight line with gradient −1/RC, so RC = −1/gradient. You can also read the time for V to fall to 37% of its starting value, or the half-life and divide it by 0.693.

Where does half the energy go when a capacitor charges?

The battery moves charge Q through voltage E, so it supplies QE = CE². The capacitor ends up storing ½CE². The other half is turned into heat in the resistor, whatever the value of R.

How long does a capacitor take to charge to 90%?

t = −RC ln(1 − 0.9) = 2.30RC. With R = 10 kΩ and C = 500 μF, RC = 5.0 s, so it takes 11.5 s. Pick Time to a % in the Solve for panel to get it, with the working, for any values.

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Capacitor charging and discharging

A capacitor fills through a resistor quickly at first, then ever more slowly. One number, the time constant RC, sets the whole curve.

Capacitor voltage and current against time

Readouts

What's happening

When the switch joins the battery, charge flows onto the capacitor plates and the voltage across them rises. The bigger that voltage, the less is left across the resistor, so the current shrinks and the charging slows down: the rate of change is proportional to how far there is still to go, which is exactly what makes an exponential. After one time constant RC the capacitor has covered 63.2% of the way to the battery voltage; after 5RC it is over 99% charged. Flip the switch to B and the capacitor drives current back round through the resistor on its own, falling to 36.8% of its starting voltage after RC. The half-life, the time to cover half the remaining gap, is RC ln 2, about 0.693RC, whatever the starting voltage. The charge stored is Q = CV and the energy is ½CV², which comes from the battery: the battery supplies QE in total, and the other half is turned into heat in the resistor.

V = E(1 − e^−t/RC)V = V₀ e^−t/RCI = I₀ e^−t/RCQ = CVW = ½CV²t½ = RC ln 2

A-Level Physics (AQA, OCR A and B, Edexcel, WJEC): capacitance, charging and discharging through a resistor, the time constant, exponential decay and ln graphs, energy stored in a capacitor.

Work through the numbers with Ohm's Law and Circuits and Physics Formulas.

Challenge

Predict first: with R = 10 kΩ and C = 500 μF, how long does the capacitor take to reach half the battery voltage? Type your answer in the box, then press Start and watch for the moment V passes 3 V.

FAQ

What is the time constant of an RC circuit?: It is the product RC, measured in seconds when R is in ohms and C in farads. After one time constant a charging capacitor has reached 63.2% of the supply voltage, and a discharging one has fallen to 36.8% of where it started.
Why is the curve exponential?: The current is set by the voltage still left across the resistor, I = (E − V)/R, and that current is the rate at which charge builds up. So the rate of change is proportional to the gap that remains, and any quantity that shrinks in proportion to itself decays exponentially.
How do I find RC from a graph?: Plot ln V (for discharge) or ln I against t. The points lie on a straight line with gradient −1/RC, so RC = −1/gradient. You can also read the time for V to fall to 37% of its starting value, or the half-life and divide it by 0.693.
Where does half the energy go when a capacitor charges?: The battery moves charge Q through voltage E, so it supplies QE = CE². The capacitor ends up storing ½CE². The other half is turned into heat in the resistor, whatever the value of R.
How long does a capacitor take to charge to 90%?: t = −RC ln(1 − 0.9) = 2.30RC. With R = 10 kΩ and C = 500 μF, RC = 5.0 s, so it takes 11.5 s. Pick Time to a % in the Solve for panel to get it, with the working, for any values.