Physics › Capacitance › The time constant and exponential decay
The time constant and exponential decay
One number, R times C, sets the clock for the whole process: 37% left after one time constant, half gone every 0.69 of one, and a pair of exponential equations that a log-linear plot flattens into a straight line you can measure.
Builds on Charging and discharging and Capacitors and energy stored.
IN THIS TOPIC
- Calculate the time constant RC and read it from graphs.
- Use the discharge and charging equations, and T½ = 0.69RC.
- Determine RC from a log-linear plot, as in required practical 9.
WHAT YOU PROBABLY THINK
After one time constant the capacitor is empty.
RC, the built-in clock
Multiply the resistance by the capacitance and, remarkably, the units of ohms times farads come out as seconds: volts per amp times coulombs per volt is coulombs per amp, which is time. The product RC is the time constant of the circuit, the single number that sets its pace. Big resistance throttles the current; big capacitance means more charge to move; either way the process stretches.
Two landmarks anchor every graph question. After one time constant, 37% of the charge remains, the fraction 1/e, which kills the lie above on the spot. And the charge halves in a fixed time,
the same halving again and again, exactly the constant-ratio behaviour you will meet once more in radioactive decay.
The equations
The discharge curve has an exact form:
and because V = Q/C and I = V/R, the pd and the current obey the same equation with their own starting values, V0 and I0 replacing Q0. Charging mirrors it:
climbing to 63% of the final value after one time constant, the complement of the 37% left behind in discharge. In both directions, five time constants is the practical rule of thumb for a process effectively complete, with under 1% of the change still to run.
Required practical 9: the straight-line test
The ninth required practical charges and discharges a capacitor while logging the decay, and its analysis is the exponential's fingerprint test. Taking natural logs of the discharge equation gives ln Q = ln Q0 − t/RC: a straight line of ln Q against t, with gradient −1/RC. Plot the logged data, and straightness confirms the decay is genuinely exponential while the gradient hands you the time constant, far more reliably than reading a single point off a curve. The same log-linear move will reappear, symbol for symbol, in the nuclear unit.
THE EXAM BIT
- RC in seconds needs base units in: ohms and farads, not kilo-ohms and microfarads. Convert first, multiply second.
- The landmark fractions are quotable: 37% left after RC on discharge, 63% reached after RC on charging, half gone every 0.69RC.
- One equation serves Q, V and I on discharge; swap in the matching starting value and say so, since the sentence itself carries a mark.
- From a graph, find RC either from T½ divided by 0.69 or from the time to fall to 37%: state which route you used.
- In the practical analysis, plot ln Q (or ln V) against t, quote the gradient as −1/RC, and use the straightness of the line as your evidence the decay is exponential.
CHECK YOURSELF
A 470 μF capacitor discharges through a 10 kΩ resistor. Find the time constant, the half-life, and the fraction of the charge remaining after 9.4 s.
Show a hint
Base units first; then notice what 9.4 s is in time constants.
Show the answer
RC = 10 000 × 470 × 10−6 = 4.7 s.
T½ = 0.69RC = 0.69 × 4.7 = 3.2 s.
9.4 s is two time constants, so the fraction left is e−2 = 0.135, about 13.5%: two rounds of keeping 37%.
RC is the clock: 37% left after one, half gone every 0.69.
Log the data and the exponential stands up straight.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.