RC ChargingJEE Main

Capacitor charge through a resistor — interactive Physics simulation for IIT-JEE.

Concept

Charging a capacitor through a resistor is not instant. Charge grows — and current decays — exponentially, governed by a single time constant $\tau = RC$ . After one $\tau$ the capacitor reaches about 63% of its final charge.

Key formula

q(t) = Q_0\left(1 - e^{-t/RC}\right), \quad i(t) = \frac{\varepsilon}{R} e^{-t/RC}, \quad \tau = RC

Derivation

Kirchhoff's voltage law gives $\varepsilon = iR + q/C$ with $i = dq/dt$ . Solving this first-order equation yields $q(t) = C\varepsilon\,(1 - e^{-t/RC})$ while charging, and $q(t) = Q_0 e^{-t/RC}$ while discharging.

At $t = \tau$ : $1 - e^{-1} \approx 0.632$ , so the capacitor is 63.2% charged; after $5\tau$ it is essentially full (>99%).

Scenarios to explore

RC Charging — Exponential charge & the time constant τ = RC.

Real-world applications

Timing circuits (555 timer), camera flash charging.
Smoothing/filtering in power supplies.
Debouncing switches and RC delay lines.

JEE exam tips

Remember the milestones: 63% at $1\tau$ , 86% at $2\tau$ , ~99% at $5\tau$ .
During discharge, the half-life is $t_{1/2} = \tau\ln 2 \approx 0.693\,RC$ .

Common mistakes

Treating charging as linear instead of exponential.
Forgetting current is maximum at $t=0$ (capacitor acts as a wire) and zero when fully charged.
Unit slips: μF must become farads, so $\tau = RC$ comes out in seconds.

Exam traps to avoid

A capacitor blocks steady (DC) current once charged, but passes the initial surge.
Energy dissipated in $R$ while charging equals the energy stored ( $\tfrac12 CV^2$ ) — half the battery's work is lost as heat.