Capacitance and transients

• Calculate an RC time constant, estimate transient voltages, and diagnose a capacitor polarity, leakage, or connection fault from a measured curve.

• circuits-kirchhoff
• rates-integration

A capacitor stores separated charge and therefore energy in an electric field. Its defining relation is Q = CV: at a given voltage V, a larger capacitance C stores more charge Q. Current describes how quickly charge changes, so i = C dV/dt. A capacitor's voltage cannot jump instantaneously in the ideal model unless infinite current is available. This makes capacitors useful for smoothing noise, holding a reference briefly, and shaping timing.

In a resistor-capacitor circuit driven by a DC step from 0 to V_s, capacitor voltage is V_C(t) = V_s(1 − e^(−t/RC)). The time constant τ = RC has units Ω·F = s. After one τ, voltage reaches about 63% of its final value; after five τ, it is above 99% in the ideal model. During discharge from V_0 through R, V_C(t) = V_0e^(−t/RC). There is no sharp “finished” instant.

Suppose a sensor input chatters from short glitches. A 10 kΩ resistor with 10 µF gives τ = 0.10 s, which may suppress very brief events but also delays genuine changes. That latency is a build tradeoff, not free filtering. Real capacitors have tolerance, leakage, equivalent series resistance, and voltage limits. Polarized electrolytics must be oriented correctly.

A common misconception is that capacitance blocks all DC immediately. During charging, current flows; after settling in the ideal DC case, it approaches zero. Another failure is reading 100 µF as 100 F or assuming the nominal time constant is an exact measured delay. Use the curve, define a threshold, and account for component tolerance and source/load resistance. The associated RC simulation is a bounded ideal first-order model: it does not include every parasitic or instrument-loading effect.

Problem

A 47 kΩ resistor charges a 10 µF capacitor from a 5 V step. Find τ and ideal capacitor voltage after one τ and two τ.

Steps

1. Convert values: 47 kΩ = 47,000 Ω and 10 µF = 10×10^-6 F.
2. τ = RC = 47,000 × 10×10^-6 s = 0.47 s.
3. At t = τ, V_C = 5(1 − e^-1) ≈ 3.16 V.
4. At t = 2τ, V_C = 5(1 − e^-2) ≈ 4.32 V.

Result

The time constant is 0.47 s; ideal voltage is about 3.16 V at 0.47 s and 4.32 V at 0.94 s.

Sanity check

Both voltages lie between 0 and 5 V, increase with time, and approach rather than exceed the supply.

If R doubles while C and the source stay unchanged, what happens to the ideal RC charging curve?

A. The time constant doubles and each voltage fraction takes twice as long B. The final voltage doubles C. The time constant halves D. The capacitor voltage jumps instantly

Your answer: ______________________________

Reasoning:

Goal

Compare a measured low-voltage RC charge curve with the ideal time-constant prediction.

Equipment

• 3×AA battery holder
• 47 kΩ resistor
• 10 µF electrolytic capacitor rated above 5 V
• multimeter
• breadboard
• timer
• jumper wires

Procedure

1. With power absent, confirm capacitor polarity and build the series resistor-capacitor circuit; calculate τ from measured or nominal values.
2. Discharge the capacitor through the resistor, then apply the battery and record actual capacitor voltage at 0, 0.5τ, 1τ, 2τ, and 5τ.
3. Compare readings with ideal expected values and note meter timing and component tolerance.
4. Remove power, discharge safely, then deliberately move the resistor so the capacitor has no charging path; repower and diagnose the flat or erratic trace as an open-path fault.
5. Restore the path and repeat one checkpoint.

What to observe

• Expected ideal values are about 39%, 63%, 86%, and 99% of final voltage at 0.5τ, 1τ, 2τ, and 5τ; actual readings must be kept distinct.
• Expected fault clue: an open charging path prevents the smooth rise and may leave a residual voltage determined by leakage and the meter.

Transfer challenge

Choose R or C to target a 50 ms time constant for suppressing a brief button glitch, then state the response delay that choice introduces.

Cautions

• Use only the low-voltage battery holder and orient the electrolytic capacitor correctly.
• Remove power and discharge through the resistor before rewiring; never short a charged capacitor directly.

Learner: ____________________ Date: ____________________

Evidence artifact or file: ____________________________________________

Notes: