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CED objectives

Resistor-Capacitor (RC) Circuits

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletRC Lab · a clock in units of R1 C, a second resistor across the capacitor, and the energy books between battery, capacitor and heat

A capacitor's voltage cannot jump, because $V_C = Q/C$ and charge takes time to arrive. That continuity turns an RC circuit into two solvable resistor networks: at the switching instant an uncharged capacitor is a wire, and after a long time it is an open circuit. The loop rule then gives $R\,dq/dt = \varepsilon - q/C$, whose solutions are $q = C\varepsilon(1 - e^{-t/\tau})$ for charging and $q = Q_0e^{-t/\tau}$ for discharging, with $\tau = R_{\text{eq}}C$ built from the resistance seen looking out of the capacitor's terminals with ideal batteries replaced by wires.

Five errors dominate. Letting $V_C$ snap to its final value the instant a switch closes, which deletes the entire transient. Swapping the two snapshots, so $t = 0$ is analysed with the branch open and steady state with it conducting. Building $\tau$ from whichever resistor is drawn beside the capacitor rather than the equivalent resistance it charges through. Reading the exponential as if it were linear, calling the capacitor full at one time constant or treating $\tau$ as a half-life. And crediting the capacitor with all the energy the battery supplied, when exactly half of $C\varepsilon^2$ is dissipated in the resistance no matter how small that resistance is.

the two snapshots are two DIFFERENT resistor circuits t = 0: uncharged C is a WIRE 12 V 2 Ω 4 Ω 4 Ω is shorted: I = 12/2 = 6 A t → infinity: C is an OPEN 12 V 2 Ω 4 Ω series pair: I = 12/6 = 2 A
Same hardware, two replacements, two different resistor problems. Swapping which replacement belongs to which instant makes every current in both snapshots wrong at once.
Q(final) straight-line reading: full at one tau t = tau 63% 2 tau 86% 3 tau 95% each tau closes 63% of WHAT REMAINS. tau is not a half-life: the half point is 0.693 tau.
The tangent at the origin does reach the final value at one time constant, which is exactly why the linear reading feels right. The curve leaves that tangent immediately and arrives at 63 percent instead.

The work

3 ways in · any order
Lesson
Resistor-Capacitor (RC) Circuits

Fixes the continuity of capacitor voltage and the two snapshots it produces, derives the charging and discharging exponentials from the loop rule, builds the time constant from the resistance seen at the capacitor's terminals, and settles the fifty-fifty energy split.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: snapping the capacitor to its final voltage at the switching instant, swapping the wire and open-circuit snapshots, building the time constant from the wrong resistance, calling the capacitor full at one time constant, and crediting it with all the energy the battery supplied. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions