An RC circuit has R = 1000\, and C = 5.010^-6 F. Find the time constant. [1 point]

College Board AP 2024 (style)-style practice: Section II (FRQ, derivation). A capacitor C, initially uncharged, is charged through a resistor R by a battery of EMF when a switch closes at t = 0. (a) Write the loop equation and the resulting differential equation for the charge q(t). (b) Solve it for q(t). (c) Find the current I(t) and state the time constant.

A 6-point FRQ deriving the charging solution from the differential equation. (a) Differential equation (2 points): loop rule, - IR - qC = 0, with I = dqdt: = Rdqdt + qC. (b) Solution (3 points): separating variables and integrating with q(0) = 0 gives q(t) = C(1 - e^-t/RC). (c) Current and time constant (1 point): I(t) = dqdt = Re^-t/RC; the time constant is = RC. Markers reward the differential equation, the exponential solution with the correct initial condition, and = RC.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How do charge, current and voltage evolve in time when a capacitor charges or discharges through a resistor?

Topic 11.8 Resistor-Capacitor (RC) Circuits: model the exponential charging and discharging of a capacitor through a resistor using the time constant.

A calculus-based answer to AP Physics C E&M Topic 11.8, covering the differential equation of an RC circuit, the exponential charge and discharge solutions, the time constant, and the initial and final behavior of the capacitor.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

In an RC circuit, a capacitor charges or discharges through a resistor, and the charge changes exponentially in time. Charging from a battery of EMF $\varepsilon$ , the loop rule $\varepsilon = IR + \dfrac{q}{C}$ with $I = \dfrac{dq}{dt}$ gives $q(t) = C\varepsilon\left(1 - e^{-t/RC}\right)$ and $I(t) = \dfrac{\varepsilon}{R}e^{-t/RC}$ . Discharging, $q(t) = q_0 e^{-t/RC}$ and $I(t) = \dfrac{q_0}{RC}e^{-t/RC}$ . The time constant $\tau = RC$ is the time to reach $1 - e^{-1} \approx 63\%$ of the final charge (charging) or to fall to $e^{-1} \approx 37\%$ (discharging). At $t = 0$ an uncharged capacitor acts like a wire (maximum current); after a long time it acts like an open switch (zero current, fully charged).

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What this topic is asking
The differential equation
Charging: the exponential approach
Discharging
The time constant and limiting behavior
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What this topic is asking

The College Board (Topic 11.8) wants you to model an RC circuit: a capacitor charging or discharging through a resistor. Because the current depends on the changing charge, the circuit obeys a first-order differential equation whose solution is an exponential governed by the time constant $\tau = RC$ . This is the calculus-based heart of the circuits unit.

The differential equation

For a capacitor charging through a resistor from a battery, Kirchhoff's loop rule gives

\varepsilon = IR + \frac{q}{C}, \qquad I = \frac{dq}{dt}

Substituting the current as the rate of change of charge yields a first-order linear differential equation:

\varepsilon = R\frac{dq}{dt} + \frac{q}{C}

This is the defining feature of an RC circuit: the current is not constant but tied to how fast the charge changes.

Charging: the exponential approach

Solving with the initial condition $q(0) = 0$ (uncharged capacitor) gives the charge and current:

q(t) = C\varepsilon\left(1 - e^{-t/RC}\right), \qquad I(t) = \frac{\varepsilon}{R}\,e^{-t/RC}

The charge rises from zero toward its final value $C\varepsilon$ , while the current starts at $\dfrac{\varepsilon}{R}$ and decays to zero. The voltage across the capacitor mirrors the charge: $V_C = \dfrac{q}{C} = \varepsilon\left(1 - e^{-t/RC}\right)$ .

Discharging

If a charged capacitor (initial charge $q_0$ ) discharges through a resistor with no battery, the loop equation $\dfrac{q}{C} + R\dfrac{dq}{dt} = 0$ solves to

q(t) = q_0\,e^{-t/RC}, \qquad I(t) = \frac{q_0}{RC}\,e^{-t/RC}

Both decay exponentially from their initial values toward zero.

The time constant and limiting behavior

Charging an RC circuit

A $2.0\,\mu\text{F}$ capacitor, initially uncharged, charges through a $5.0\times10^5\,\Omega$ resistor from a $10$ V battery. Find the time constant, the initial current, and the charge after one time constant.

step 1 Find the time constant

$\tau = RC = (5.0\times10^5)(2.0\times10^{-6}) = 1.0$ s.

step 2 Find the initial current

At $t = 0$ the capacitor is a wire, so $I_0 = \dfrac{\varepsilon}{R} = \dfrac{10}{5.0\times10^5} = 2.0\times10^{-5}$ A.

step 3 Find the final charge

Fully charged, $q_\infty = C\varepsilon = (2.0\times10^{-6})(10) = 2.0\times10^{-5}$ C.

step 4 Charge after one time constant

$q(\tau) = q_\infty\left(1 - e^{-1}\right) = (2.0\times10^{-5})(0.632) = 1.26\times10^{-5}$ C, that is $63\%$ of the full charge.

Try this

Q1. An RC circuit has $R = 1000\,\Omega$ and $C = 5.0\times10^{-6}$ F. Find the time constant. [1 point]

Cue. $\tau = RC = (1000)(5.0\times10^{-6}) = 5.0\times10^{-3}$ s.

Q2. State how a fully charged capacitor behaves in a DC circuit after a long time. [1 point]

Cue. Like an open switch: no current flows through its branch.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). In an RC charging circuit, a long time after the switch closes, the current through the resistor is (A) maximum (B)

\dfrac{\varepsilon}{R}

\dfrac{\varepsilon}{2R}

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the steady state of an RC circuit. The answer is (C).

As $t \to \infty$ the capacitor charges fully to the EMF, so no further charge flows: the current decays to zero. The maximum current $\varepsilon/R$ occurs at $t = 0$ , the instant the switch closes, when the capacitor is uncharged and acts like a wire. The trap is (B), which is the initial, not the final, current.

AP 2024 (style)6 marksSection II (FRQ, derivation). A capacitor

C

, initially uncharged, is charged through a resistor

R

by a battery of EMF

\varepsilon

when a switch closes at

t = 0

. (a) Write the loop equation and the resulting differential equation for the charge

q(t)

. (b) Solve it for

q(t)

. (c) Find the current

I(t)

and state the time constant.

Show worked answer →

A 6-point FRQ deriving the charging solution from the differential equation.

(a) Differential equation (2 points): loop rule, $\varepsilon - IR - \dfrac{q}{C} = 0$ , with $I = \dfrac{dq}{dt}$ : $\varepsilon = R\dfrac{dq}{dt} + \dfrac{q}{C}$ .
(b) Solution (3 points): separating variables and integrating with $q(0) = 0$ gives $q(t) = C\varepsilon\left(1 - e^{-t/RC}\right)$ .
(c) Current and time constant (1 point): $I(t) = \dfrac{dq}{dt} = \dfrac{\varepsilon}{R}e^{-t/RC}$ ; the time constant is $\tau = RC$ .

Markers reward the differential equation, the exponential solution with the correct initial condition, and $\tau = RC$ .

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)