One time constant is not one third of the job: exponentials lose a fixed fraction, not a fixed amount

01The mistake

A capacitor discharges through a resistor with time constant $\tau$. After one $\tau$ it holds 37% of its initial charge. Ask how much is left after $3\tau$. A common answer is zero or near zero, from subtracting 63% three times. The actual value is about 5%, and the quantity never reaches zero at any finite time.

The charging version produces the same error in the other direction. Students report the capacitor as fully charged after one $\tau$, since that is when the formula's characteristic time has elapsed. It is at 63% of final charge then, and roughly 95% at $3\tau$.

A third version misreads the half-life relationship. Students treat $\tau$ as the time to fall to half, which is a different constant: $t_{1/2} = \tau \ln 2 \approx 0.69\tau$. The two are close enough that numerical answers look nearly right, which is exactly what keeps the confusion alive.

The tell is a student who adds or subtracts percentages across intervals instead of multiplying them. “63% gone in the first $\tau$, so 126% gone in two” is arithmetic that produces an impossible number, and a student who writes it has a linear model regardless of what formula they quote.

02Why it makes sense to the student

Linear extrapolation is the default for every rate students have met. Constant velocity, constant acceleration, constant current — all of them permit a rate times a time. RC decay is often the first process in the course where the rate itself depends on the quantity, which is what makes it exponential.

Reading an exponential graph rewards the linear instinct at first. Over a short interval the curve looks like a line, and students estimate from a tangent without realizing it. The error only becomes large over intervals of a time constant or more, which is where every problem lives.

The name “time constant” implies completion. It sounds like the time for the process to happen, rather than the time for it to progress by a factor of $1/e$ — which is not a round number and has no everyday meaning. Students substitute the interpretation the name suggests.

And $e$ is unmotivated at this point in the course. A student who has not seen that it comes from solving $dq/dt = -q/RC$ has no reason to expect repeated multiplication rather than repeated subtraction. The base of the exponential is where the fixed-fraction behavior comes from, so a student treating $e$ as an arbitrary number has lost the mechanism.

03The correction

Teach the fixed-fraction property as the defining behavior, before any numbers. Every interval of one $\tau$ multiplies the remaining charge by $1/e \approx 0.37$. Multiplication is the operation, so the fraction is what repeats. That single sentence prevents both the linear extrapolation and the reach-zero conclusion.

Build the table and leave it up: $37\%$, $14\%$, $5\%$, $1.8\%$ at one through four time constants, each 37% of the one before. Students who have seen the pattern as repeated multiplication stop subtracting, and the fact that the numbers never hit zero is visible in the column rather than asserted.

Derive it from the differential equation, since this is a calculus-based course and the mechanism is short. $dq/dt = -q/RC$ says the rate of loss is proportional to what is left, so a nearly empty capacitor discharges slowly. That is the physical reason the process never finishes, and it is more convincing than the formula.

Separate $\tau$ from the half-life explicitly and give both numbers. $\tau$ is the time to fall to $1/e$, about 37%; the half-life is $\tau\ln 2$, about $0.69\tau$. Students who have the two constants written side by side stop using one for the other.

Then connect it to practice. Engineers call a capacitor settled after about $5\tau$, at which point 0.7% remains, and that is a convention rather than a completion. Giving students the real rule of thumb replaces the invented one they were using.

04A sample question

05What each wrong answer reveals

• A Fixed amount subtracted instead of fixed fraction multiplied. The dominant wrong answer, and the justification shows the arithmetic: three times 63% exceeds 100%, which is already impossible. Ask what 63% of the remaining charge is during the second interval. Reframing the loss as a fraction of what is left turns the subtraction into a multiplication.
• B Correct. $e^{-3} \approx 0.050$. Each time constant multiplies by about 0.37, and $0.37^3 \approx 0.05$.
• C Division where multiplication belongs. This student knows the result cannot be zero and is not reasoning linearly in the same way as A, which is progress. Dividing the remaining fraction by the number of intervals is still an arithmetic stand-in for repeated multiplication. Having them compute $0.37 \times 0.37 \times 0.37$ directly is the fastest repair.
• D The time constant treated as the whole decay. This student has $1/e$ attached to the process rather than to one interval of it, so time does not appear in their model at all. Ask what their answer would be at $t = 10\tau$; if it is still 37%, the exponent is not being used. This is the most basic of the four and needs the exponent read as a count of intervals.

A and C both substitute an arithmetic operation for repeated multiplication and differ only in which one, so both are fixed by computing the product explicitly. D has no time-dependence at all, which is a different and more basic gap. Only A's answer is impossible on its face, which makes it the easiest to open the conversation with.

06Try it in Mistake Master