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5 Charging & discharging (RC)

Time constant τ = R C Time to reach 63% of final charge (charging) or fall to 37% (discharging). Discharging Q=Q_0e^{-t/RC}\qquad I=I_0e^{-t/RC} Exponential decay of charge, current and voltage. Charging Q=Q_0\left(1-e^{-t/RC}\right) Approaches the final charge asymptotically.

Reading exponential graphs

  • After 1 time constant τ: charged to ~63%, or discharged to ~37% of start.
  • After ~5τ the capacitor is essentially fully charged/discharged.
  • Larger R or C ⟹ slower charging (bigger τ).
Worked example Time constant A 100 µF capacitor discharges through a 47 kΩ resistor. Find τ and the charge after 10 s if Q₀ = 5.0 mC. τ = RC = 47×10³ × 100×10⁻⁶ = 4.7 s. Q = Q₀ e^(−t/τ) = 5.0×10⁻³ × e^(−10/4.7). Q = 5.0×10⁻³ × e^(−2.13) = 5.0×10⁻³ × 0.119 = 5.9×10⁻⁴ C. ✓ τ = 4.7 s; Q ≈ 0.59 mC after 10 s.

How to sketch a charging or discharging curve

  • Mark the STARTING value: I₀ = V/R (discharging) or the final asymptote value (charging).
  • Mark the value at ONE time constant: ≈37% of the start (discharging) or ≈63% of the final value (charging).
  • The curve is a smooth EXPONENTIAL, not a straight line — steepest at the start, flattening as it approaches its limit.
  • By about 5τ the curve is indistinguishable from its final value — draw it levelling off, never quite touching the axis (discharging) or the supply line (charging).
✕

A charging or discharging curve never touches the axis

The exponential only APPROACHES zero (discharging) or the supply voltage (charging) — it never actually reaches it, mathematically or on your sketch. Drawing the curve flat against the axis, or as a straight diagonal line, are the two most common ways to lose the mark for "correct shape." Worked example Comparing series and parallel combinations Three identical 60 µF capacitors are each charged individually to 5.0 V. Compare the initial current and the time constant when they are (a) connected in series and discharged through a 5.0 Ω resistor, versus (b) connected in parallel and discharged through the same resistor. Series: C_T = 60/3 = 20 µF. The three 5.0 V charges add (aiding), giving a combined 15 V driving the discharge. I₀ = 15/5.0 = 3.0 A. τ = RC_T = 5.0 × 20×10⁻⁶ = 100 µs. Parallel: C_T = 3 × 60 = 180 µF. The voltage stays at 5.0 V (the same as one capacitor alone). I₀ = 5.0/5.0 = 1.0 A. τ = RC_T = 5.0 × 180×10⁻⁶ = 900 µs. ✓ Series: I₀ = 3.0 A, τ = 100 µs (discharges fast). Parallel: I₀ = 1.0 A, τ = 900 µs (discharges 9× slower). Combining n identical capacitors in series divides C by n; in parallel it multiplies C by n — and τ moves the same way.

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