Lessons · circuits

Charging a capacitor: the current is the slope of the charge

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A capacitor $C = 1$ mF charges through a resistor $R = 1$ k $\Omega$ from a $V = 10$ V battery. The charge $Q(t)$ climbs toward $CV$ while the current $I(t)$ fades from $V/R$ to zero. Plotted as a stacked $Q$ – $t$ over $I$ – $t$ , the slope of the charge is the current at every instant — the same slope↔value pivot as position and velocity, one domain over. Drag $R$ and watch the time constant $\tau = RC$ stretch both curves.

Time constant

\tau = RC

1 s

Final charge

Q_\infty = CV

0.01 C

Initial current

I_0 = V/R

0.01 A

Charge after one

\tau

(

\approx 63\%

)

0.006321 C

Kirchhoff's voltage law is a differential equation

V = IR + \frac{Q}{C}, \quad I = \frac{dQ}{dt} \ \Longrightarrow\ R\,\frac{dQ}{dt} + \frac{Q}{C} = V

Around the loop, the battery EMF equals the resistor drop $IR$ plus the capacitor voltage $Q/C$ . Because the current is the rate of change of charge, $I = dQ/dt$ , this is a first-order differential equation for $Q(t)$ — not an algebra equation.

check $I = dQ/dt$ ; check $R\,dQ/dt + Q/C = V$ ; check $Q = \int_0^t I\,dt'$ ; $\tau = RC$ (63\% at $t=\tau$ ); steady state $Q\to CV$ , $I\to 0$

✓ The current is the slope of the charge: $I = \dfrac{dQ}{dt}$ — exactly as velocity is the slope of position. [structural]
✓ The charge solves Kirchhoff's voltage law $V = IR + Q/C$ , i.e. the RC equation $R\,\dfrac{dQ}{dt} + \dfrac{Q}{C} = V$ . [structural]
✓ The charge is the accumulated current — the area under $I$ : $Q(t) = \int_0^t I\,dt'$ . [structural]
✓ The time constant is $\tau = RC$ : at $t = \tau$ the capacitor is $1 - 1/e \approx 63\%$ charged, $Q(\tau) = CV(1 - e^{-1})$ . [structural]
✓ The steady state is full charge and no current: $Q \to CV$ and $I \to 0$ as $t \to \infty$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “The capacitor charges at a steady rate, so the current stays constant while it fills up.”

It does not. The current is largest at the start and decays exponentially as the capacitor fills: $I(t) = (V/R)\,e^{-t/RC}$ . As charge accumulates, the capacitor's voltage $Q/C$ rises and opposes the battery, leaving a smaller voltage across the resistor and so a smaller current. Because the current is the slope of the charge, $I = dQ/dt$ , a decaying current is exactly why the charging curve bends over and levels off instead of climbing in a straight line. Watch the panels: where $Q$ is steepest (at $t=0$ ) the $I$ curve is highest; where $Q$ flattens, $I$ approaches zero.

Modeling assumptions — author-asserted, disclosed not discharged

An ideal series RC circuit: a constant-EMF battery, an ohmic resistor, and an ideal capacitor, with no lead resistance or dielectric leakage.
The capacitor starts uncharged, $Q(0) = 0$ , so the initial current is the full $I_0 = V/R$ (the capacitor behaves momentarily like a wire).

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):