An LR circuit has L = 0.10 H and R = 50\,. Find the time constant. [1 point]

College Board AP 2024 (style)-style practice: Section II (FRQ, derivation). An inductor L and resistor R are connected in series to a battery of EMF when a switch closes at t = 0. (a) Write the loop equation and the differential equation for the current. (b) Solve for I(t). (c) State the time constant and the final current.

A 6-point FRQ deriving the LR rise. (a) Differential equation (2 points): loop rule, - IR - LdIdt = 0, that is = IR + LdIdt. (b) Solution (3 points): solving with I(0) = 0 gives I(t) = R(1 - e^-Rt/L). (c) Time constant and final current (1 point): = LR; as t→∞, I → R. Markers reward the differential equation, the exponential solution with the right initial condition, and = L/R.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How do current and voltage evolve in time in a circuit with a resistor and an inductor?

Topic 13.5 Circuits with Resistors and Inductors (LR Circuits): model the exponential growth and decay of current in an LR circuit using the time constant.

A calculus-based answer to AP Physics C E&M Topic 13.5, covering the differential equation of an LR circuit, the exponential rise and decay of current, the time constant L/R, and the initial and final behavior of the inductor.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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

Have a quick question? Jump to the Q&A page

Quick answer

In an LR circuit, an inductor resists changes in current, so when a battery is connected the current rises exponentially rather than jumping. The loop rule $\varepsilon = IR + L\dfrac{dI}{dt}$ solves to $I(t) = \dfrac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)$ , rising from zero toward the final value $\dfrac{\varepsilon}{R}$ . When the source is removed, the current decays: $I(t) = I_0\,e^{-Rt/L}$ . The time constant is $\tau = \dfrac{L}{R}$ , the time to reach $63\%$ of the final current (rising) or fall to $37\%$ (decaying). At $t = 0$ the inductor acts like an open switch (current cannot jump, so it starts at zero); after a long time it acts like a plain wire (steady current, no back-EMF). This is the exact dual of the RC circuit, swapping the roles of current and charge.

Jump to a section

What this topic is asking
The differential equation
Current rise
Current decay
The time constant and limiting behavior
Try this

What this topic is asking

The College Board (Topic 13.5) wants you to model an LR circuit: a resistor and inductor in series with a battery. The inductor's back-EMF makes the current rise (or fall) exponentially, governed by a first-order differential equation and a time constant $\tau = \dfrac{L}{R}$ . This mirrors the RC circuit, with current playing the role that charge played there.

The differential equation

For an inductor and resistor in series with a battery, Kirchhoff's loop rule, including the inductor's back-EMF $L\dfrac{dI}{dt}$ , gives

\varepsilon = IR + L\frac{dI}{dt}

This is a first-order linear differential equation for the current. The inductor term ties the current's behavior to its rate of change, exactly as the capacitor tied the RC circuit to the rate of change of charge.

Current rise

Solving with the initial condition $I(0) = 0$ (the inductor forbids a sudden jump) gives

I(t) = \frac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)

The current climbs from zero, slowed by the back-EMF, toward its final value $\dfrac{\varepsilon}{R}$ (set by the resistor alone, since a steady current produces no back-EMF). The voltage across the inductor is $V_L = L\dfrac{dI}{dt} = \varepsilon\,e^{-Rt/L}$ , starting at $\varepsilon$ and decaying to zero.

Current decay

If the battery is shorted out (the inductor now drives the circuit), the loop equation becomes $IR + L\dfrac{dI}{dt} = 0$ , with solution

I(t) = I_0\,e^{-Rt/L}

The current decays exponentially from its initial value $I_0$ toward zero, the energy $\tfrac{1}{2}LI^2$ dissipating in the resistor.

The time constant and limiting behavior

Current rise in an LR circuit

A $0.40$ H inductor and a $20\,\Omega$ resistor are connected in series to a $10$ V battery at $t = 0$ . Find the time constant, the final current, and the current after one time constant.

step 1 Find the time constant

$\tau = \dfrac{L}{R} = \dfrac{0.40}{20} = 0.020$ s.

step 2 Find the final current

After a long time the inductor is a plain wire: $I_\infty = \dfrac{\varepsilon}{R} = \dfrac{10}{20} = 0.50$ A.

step 3 Find the current after one time constant

$I(\tau) = I_\infty\left(1 - e^{-1}\right) = (0.50)(0.632) = 0.316$ A, that is $63\%$ of the final current.

step 4 Check the initial current

At $t = 0$ , $I = I_\infty(1 - e^0) = 0$ , confirming the inductor blocks an instantaneous jump.

Try this

Q1. An LR circuit has $L = 0.10$ H and $R = 50\,\Omega$ . Find the time constant. [1 point]

Cue. $\tau = \dfrac{L}{R} = \dfrac{0.10}{50} = 2.0\times10^{-3}$ s.

Q2. State how an inductor behaves a long time after the switch closes in a DC circuit. [1 point]

Cue. Like a plain wire: a steady current flows with no back-EMF.

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 2023 (style)1 marksSection I (multiple choice). In an LR circuit, immediately after the switch connects the battery, the current through the inductor is (A) maximum,

\varepsilon/R

(B) zero (C)

\varepsilon/2R

(D) infinite. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the initial state of an LR circuit. The answer is (B).

The inductor opposes any sudden change in current, so the current cannot jump from zero: at $t = 0$ it is still zero (the inductor acts like an open switch). The current then rises toward its final value $\varepsilon/R$ . The trap is (A), which is the final, not the initial, current.

AP 2024 (style)6 marksSection II (FRQ, derivation). An inductor

L

and resistor

R

are connected in series to a battery of EMF

\varepsilon

when a switch closes at

t = 0

. (a) Write the loop equation and the differential equation for the current. (b) Solve for

I(t)

. (c) State the time constant and the final current.

Show worked answer →

A 6-point FRQ deriving the LR rise.

(a) Differential equation (2 points): loop rule, $\varepsilon - IR - L\dfrac{dI}{dt} = 0$ , that is $\varepsilon = IR + L\dfrac{dI}{dt}$ .
(b) Solution (3 points): solving with $I(0) = 0$ gives $I(t) = \dfrac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)$ .
(c) Time constant and final current (1 point): $\tau = \dfrac{L}{R}$ ; as $t\to\infty$ , $I \to \dfrac{\varepsilon}{R}$ .

Markers reward the differential equation, the exponential solution with the right initial condition, and $\tau = L/R$ .

Related dot points

Sources & how we know this

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