EG1216 · Physics 2 — Electricity & Magnetism
Theme 3 · Current & DC circuits

RC circuits

Put a resistor and a capacitor together and the charge builds — or drains — over time, set by the time constant \(\tau = RC\).

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §28.4.

Before you start

What you need first

  • Topic 12 — capacitance, \(Q = CV\).
  • Topic 17 — resistance and Ohm's law.
  • The idea of an exponential \(e^{-t/\tau}\).

What you'll be able to do

  • Find the time constant \(\tau = RC\).
  • Use the charge/discharge curves \(q = CV(1-e^{-t/\tau})\), \(q = Q_0 e^{-t/\tau}\).
  • Use the 63% / 37% and "5\(\tau\)" rules of thumb.
  • Find the final charge stored.

The setup

Charging a capacitor through a resistor

Connect a battery, a resistor \(R\), and a capacitor \(C\) in a loop. The capacitor doesn't fill instantly — the resistor slows the flow, so it charges up gradually: fast at first, then slower and slower, until it is full and the current stops.

R C
A battery charges \(C\) through \(R\) — gradually, not all at once.

The time constant

$$ \tau = RC $$
where:
SymbolMeaningSI unit
τtime constant — how fast it charges/dischargess
RresistanceΩ
CcapacitanceF
A bigger \(R\) or bigger \(C\) → bigger \(\tau\) → it charges more slowly. (Check: Ω × F = s.)
📐 Worked example 1

Time constant

Find the time constant for \(R = 1000\ \Omega\) (1 kΩ) and \(C = 2.0\ \mu\text{F}\).

1Use \(\tau = RC\):
$$ \tau = (1000)(2.0\times10^{-6}) $$
2Work it out:
$$ \tau = 2.0\times10^{-3}\ \text{s} = 2.0\ \text{ms} $$

How the charge changes over time

The charge follows an exponential. Charging up from empty, and discharging from a starting charge \(Q_0\):

$$ q = CV\big(1 - e^{-t/\tau}\big) \quad\text{(charging)} $$
$$ q = Q_0\,e^{-t/\tau} \quad\text{(discharging)} $$
After one \(\tau\): charging reaches 63% of full; discharging falls to 37%. After about 5\(\tau\) it is practically full (or empty).
t q full (Q = CV) τ 63%
Charge rises fast, then levels off toward \(Q = CV\); it is 63% full at \(t = \tau\).
📐 Worked example 2

Charging up

A \(C = 10\ \mu\text{F}\) capacitor is charged by a \(V = 9.0\) V battery through \(R = 50\ \text{k}\Omega\). Find (a) the time constant, (b) the final charge, (c) the charge after one time constant.

1Time constant \(\tau = RC\):
$$ \tau = (50\,000)(10\times10^{-6}) = 0.50\ \text{s} $$
2Final charge \(Q = CV\):
$$ Q = (10\times10^{-6})(9.0) = 90\ \mu\text{C} $$
3After one \(\tau\), it is 63% charged:
$$ q \approx 0.63 \times 90 \approx 57\ \mu\text{C} $$

✏️ Try it yourself

A \(C = 100\ \mu\text{F}\) capacitor charged to 12 V is discharged through \(R = 2.0\ \text{k}\Omega\).

(a) Find the time constant.
(b) Find the initial charge.
(c) Find the charge left after one time constant.

(a) \(\tau = RC = (2000)(100\times10^{-6}) = 0.20\ \text{s}.\) (b) \(Q_0 = CV = (100\times10^{-6})(12) = 1.2\times10^{-3}\ \text{C} = 1.2\ \text{mC}.\) (c) Discharging, after one \(\tau\) it is at 37%: \(q \approx 0.37 \times 1.2\ \text{mC} \approx 0.44\ \text{mC}.\)

Common mistakes

MistakeFix
Thinking the capacitor charges instantly.The resistor slows it; it takes about \(5\tau\) to fill.
Mixing up the 63% and 37% values.Charging rises to 63% in one \(\tau\); discharging falls to 37%.
Forgetting to convert µF / kΩ.Use farads and ohms so \(\tau\) comes out in seconds.
Thinking \(R\) changes the final charge.The final charge is \(Q = CV\); \(R\) only sets how long it takes.

Recap — the whole topic on one screen

IdeaWhat you own now
Time constant\(\tau = RC\) (in seconds).
Charging\(q = CV(1-e^{-t/\tau})\); 63% at one \(\tau\).
Discharging\(q = Q_0 e^{-t/\tau}\); 37% at one \(\tau\).
Final charge\(Q = CV\); \(R\) only sets the speed. Practically done by \(5\tau\).

Next topic · new theme (after the midterm)

Theme 4 — Magnetism

That completes electricity — and the midterm. Next we turn to magnetism: the force a magnetic field puts on moving charges and currents, and how currents make magnetic fields of their own.

→ Topic 23 · Magnetic force on a moving charge