The prediction we will test
- With 10 kΩ and 100 µF, inspect voltage at 1 s. Compare 63.2% with the 90% target.
- Select “Go to the 90% point” and read the required time.
- Return to 1 s and double R to 20 kΩ while keeping C fixed. Observe the fraction reached.
- Reset and double C to 200 µF while keeping R fixed. Compare voltage at that same one-second instant.
Solution: the remaining error must fall to one tenth
Vc/Vs = 1 − e^(−t/τ)
0.9 = 1 − e^(−t/τ)
t90 = −τ ln(0.1) ≈ 2.302585 τ
0.9 = 1 − e^(−t/τ)
t90 = −τ ln(0.1) ≈ 2.302585 τ
Initially, τ = 10,000 Ω × 100 µF = 1 s. At 1 s the voltage is 3.1606 V; at 2.3026 s it is about 4.5 V. Doubling R or C makes τ = 2 s: after one second the voltage reaches 39.347%, and 90% takes about 4.6052 s.
On a t/τ scale, these circuits have the same shape. On a seconds scale, one takes twice as long. That is why both time representations are useful.
The initial condition is part of the result
This activity models charging from zero with a constant 5 V supply only. It does not represent an already charged capacitor or discharging. Percentages refer to that supply, not a universal maximum component charge.
With R = 1 kΩ and C = 1 µF, must reaching 90% take a whole number of seconds?
No. τ = 0.001 s and t90 ≈ 0.0023026 s, about 2.303 ms. The numeric field accepts fractions of a second.