RC charging time constant theory
Understand rc charging time constant: the equations, assumptions, supporting concepts and measurement method.
01Charge storage and voltage
A capacitor stores equal and opposite charge on its conductors. For a linear capacitor, stored charge is proportional to its terminal voltage. Initially discharged means Vc(0) = 0. With finite series resistance the capacitor voltage changes continuously; current initially has its largest value and then decreases.
Can the capacitor voltage jump instantly in this finite-current model?
No. An instantaneous voltage step would require an impulse of charging current.
02Derive the charging equation
Kirchhoff’s loop rule gives E = IR + Vc. Substitute I = C dVc/dt and solve the first-order differential equation with Vc(0) = 0. The voltage difference E − Vc decays exponentially even though Vc itself grows. These equations require a constant source, constant R and C, and negligible leakage and instrument loading.
Which quantity decays exponentially during charging?
The remaining difference E − Vc, and also the charging current.
03What one time constant means
Define τ = RC. After one τ the voltage has reached 1 − 1/e, about 63.2%, of E and current has dropped to 1/e of its initial value. After three τ the voltage is about 95.0% of E. It approaches E asymptotically; no finite ideal time makes it exactly equal to E. For 10,000 Ω and 100 × 10⁻⁶ F, τ is 1.000 s.
Is τ the time to reach 100% of the final voltage?
No. It is the time to reach about 63.2% from an initially discharged state.
04Make the logarithm dimensionless
Rearrange E − Vc = E exp(−t/τ). Taking a natural logarithm of E/(E − Vc) gives a positive transformed variable y = t/τ. This ratio is dimensionless. Do not use log base ten without its conversion factor and do not take the logarithm of an unnormalized voltage value. Valid charging readings satisfy 0 ≤ Vc < E.
What happens to ln[E/(E − Vc)] at Vc = E?
It is undefined because the denominator is zero; such a reading cannot be transformed.
05Fit the correct sign and gradient
Plot time t in seconds on the horizontal axis and y = ln[E/(E − Vc)] vertically. Fit y = mt + b. The gradient is positive, has units s⁻¹, and τ = 1/m. If instead plotting ln[(E − Vc)/E], the slope would be negative and τ = −1/m. Retain an intercept: a timing offset or incomplete initial discharge can shift it.
For the positive logarithm used here, what does a gradient 2 s⁻¹ imply?
τ = 0.5 s.
06Sampling, transformation and uncertainty
With τ near one second, manual stopwatch-and-meter readings may be too slow; use synchronized data acquisition or an oscilloscope with adequate sampling. Rounding voltage to 0.001 V becomes increasingly influential as Vc approaches E, because dy/dVc = 1/(E − Vc). Source uncertainty is also amplified near the asymptote. The transformed observations therefore need not have equal uncertainty; an unweighted fit is a teaching simplification. Fit-only u(τ) = u(m)/m² excludes resistor tolerance, capacitor tolerance, loading and clock calibration.
Why avoid treating near-final transformed readings as equally precise?
A small voltage error is magnified when E − Vc is small.
The generator calculates Vc = 5[1 − exp(−t/1.000)] V and rounds to 0.001 V. Analysis uses x = t and y = ln[5/(5 − Vc)], estimating τ = 1/gradient. The reference 1.000 s is the product of the fixed model R and C, not a measurement of actual components.