Transient and Oscillatory Behavior with Inductors: AP Physics C E&M Notes

LR Circuits (Resistor–Inductor Transients)

What an LR circuit is (and why inductors are “time-dependent” components)

An LR circuit is typically a series combination of a resistor (resistance R) and an **inductor** (inductance L), connected to a source (often a DC battery of emf \mathcal{E}) and a switch. The defining feature is that the inductor resists changes in current: when you try to make the current change quickly, the inductor produces a voltage that opposes that change.

This matters because many electromagnetic systems are governed by induction. Inductors model coils, solenoids, motors, transformers (in certain limits), and even unintended “parasitic inductance” in wires. In circuits with inductance, you cannot generally treat the current as changing instantaneously after you flip a switch. Instead, the circuit exhibits a transient (a time-dependent adjustment) on a characteristic timescale.

At the physics level, the key idea is Faraday’s law: changing current changes magnetic flux through the coil, which induces an emf that opposes the change (Lenz’s law). In circuit form, that becomes the inductor voltage-current relation:

V_L = L\frac{dI}{dt}

Here V_L is the voltage across the inductor (with a sign depending on your chosen polarity), and I is the current through it.

A resistor, in contrast, obeys Ohm’s law at all times:

V_R = IR

So an LR circuit combines an element that relates voltage to the rate of change of current (the inductor) with an element that relates voltage to the current itself (the resistor). That combination naturally produces exponential behavior.

“Inductor as open/short” intuition (with a crucial caveat)

You’ll often hear rules like:

  • Right after switching (very early time), an inductor “acts like an open circuit.”
  • A long time after switching in DC, an inductor “acts like a wire” (short circuit).

What these really mean is:

  • Current through an inductor cannot jump instantly. So if the current was 0 before closing a switch, it is still 0 at t = 0^+. That can make it behave like an open circuit initially.
  • In steady-state DC after a long time, \frac{dI}{dt} = 0, so V_L = 0. With zero voltage drop across it, the inductor behaves like an ideal wire.

Common misconception: thinking the inductor “blocks current” forever. It only blocks changes in current; in DC steady state it allows constant current without an induced voltage.

Building the transient equation with Kirchhoff’s loop rule

Consider a series R–L circuit connected to a battery \mathcal{E}. Choose current I(t) around the loop. Apply Kirchhoff’s loop rule (sum of potential changes around a loop is zero). With a sign convention where drops across elements are positive in the direction of current, you get:

\mathcal{E} - IR - L\frac{dI}{dt} = 0

Rearrange into a standard first-order differential equation:

L\frac{dI}{dt} + RI = \mathcal{E}

This equation is the mathematical statement of the physical competition:

  • The battery tries to drive current.
  • The resistor dissipates energy and demands a voltage IR.
  • The inductor demands a voltage L\frac{dI}{dt} when current changes.

Time constant and current growth after closing the switch

Solve the differential equation for the common initial condition I(0)=0 (switch closes at t=0 with no prior current). The solution is:

I(t) = I_{\infty}\left(1 - e^{-t/\tau}\right)

where

I_{\infty} = \frac{\mathcal{E}}{R}

and the time constant is

\tau = \frac{L}{R}

Interpretation:

  • I_{\infty} is the final steady current after a long time, because in steady DC the inductor drop goes to zero.
  • \tau sets how quickly the transient happens. After one time constant, the current has reached 1 - e^{-1} \approx 0.632 (about 63%) of its final value.

This “63% rule” is the same exponential idea you may know from RC circuits, but here it’s current changing rather than capacitor voltage.

Voltages across the resistor and inductor during the transient

Once you know I(t), you can find element voltages.

Resistor voltage:

V_R(t) = I(t)R = \mathcal{E}\left(1 - e^{-t/\tau}\right)

Inductor voltage (using loop rule or derivative):

V_L(t) = \mathcal{E} - V_R(t) = \mathcal{E}e^{-t/\tau}

Key physical picture:

  • At t=0^+, I=0 so V_R=0 and the entire battery voltage appears across the inductor: V_L(0^+) = \mathcal{E}.
  • As current builds, the resistor drop increases and the inductor drop decays.
  • At long times, V_L \to 0 and V_R \to \mathcal{E}.

Common sign mistake: students sometimes write V_L = -L\frac{dI}{dt} without being consistent about polarity. The safest approach on exams is to pick a loop direction, assign element voltage drops consistently, write KVL, and then interpret signs from the resulting expressions.

Current decay when the battery is removed (discharging an inductor)

If the inductor initially carries current I(0)=I_0 and then the battery is removed while the resistor and inductor remain in a closed loop, the loop rule becomes:

IR + L\frac{dI}{dt} = 0

Solution:

I(t) = I_0 e^{-t/\tau}

with the same time constant \tau = \frac{L}{R}.

This is the inductor “trying” to keep current flowing; it generates whatever voltage is necessary (within physical limits like insulation breakdown, arcing, etc.) to oppose the decrease in current.

Energy in an inductor (and where it goes in an LR transient)

An inductor stores energy in its magnetic field. The stored energy is:

U_L = \frac{1}{2}LI^2

During current growth, the battery supplies energy that is partly dissipated by the resistor and partly stored in the inductor’s magnetic field. During current decay, the magnetic field energy decreases and is dissipated as thermal energy in the resistor.

A useful power relation (conceptually) is that the inductor power is P_L = V_L I. When V_L and I have the same sign, energy is being stored; when opposite, energy is being released.

Worked example 1: Current growth and “how long until…?”

A series LR circuit has L = 0.50\ \text{H}, R = 5.0\ \Omega, and is connected to a battery \mathcal{E}=10\ \text{V} at t=0 with I(0)=0.

1) Find \tau and I_{\infty}.

\tau = \frac{L}{R} = \frac{0.50}{5.0} = 0.10\ \text{s}

I_{\infty} = \frac{\mathcal{E}}{R} = \frac{10}{5.0} = 2.0\ \text{A}

2) Find the current at t = 0.20\ \text{s}.

Use the growth solution:

I(t) = I_{\infty}\left(1 - e^{-t/\tau}\right)

I(0.20) = 2.0\left(1 - e^{-0.20/0.10}\right) = 2.0\left(1 - e^{-2}\right)

Numerically, e^{-2} \approx 0.135, so

I(0.20) \approx 2.0(1 - 0.135) = 1.73\ \text{A}

3) What is V_L at t = 0.20\ \text{s}?

V_L(t) = \mathcal{E}e^{-t/\tau}

V_L(0.20) = 10e^{-2} \approx 1.35\ \text{V}

Notice how the inductor voltage has dropped a lot by two time constants, even though it started equal to the full battery voltage.

Worked example 2: Inductor discharge and energy interpretation

An inductor L = 0.20\ \text{H} carries an initial current I_0 = 3.0\ \text{A}. It discharges through R = 10\ \Omega.

1) Find the time constant:

\tau = \frac{L}{R} = \frac{0.20}{10} = 0.020\ \text{s}

2) Find the current after 0.060\ \text{s}:

I(t) = I_0 e^{-t/\tau} = 3.0e^{-0.060/0.020} = 3.0e^{-3}

Since e^{-3} \approx 0.0498,

I \approx 0.149\ \text{A}

3) Compare stored magnetic energy initially and at that time.

U_L = \frac{1}{2}LI^2

Initial:

U_{L0} = \frac{1}{2}(0.20)(3.0)^2 = 0.90\ \text{J}

Later:

U_L = \frac{1}{2}(0.20)(0.149)^2 \approx 0.00222\ \text{J}

Almost all the original magnetic energy has been converted to thermal energy in the resistor by three time constants.

Exam Focus
  • Typical question patterns:
    • A switch is closed or opened in a series R–L circuit; you’re asked for I(t), V_L(t), V_R(t), or the time to reach a given fraction of the final current.
    • Conceptual prompts about initial vs long-time behavior (why current cannot change instantaneously, what “steady state” means for an inductor).
    • Energy questions: compute U_L at some time or reason about where energy goes during growth/decay.
  • Common mistakes:
    • Mixing up RL and RC time constants (using RC instead of L/R) or claiming the current approaches \mathcal{E}/L instead of \mathcal{E}/R.
    • Treating the inductor current as discontinuous at switching (writing I(0^+) = I_{\infty}). Remember: inductor current is continuous.
    • Sign errors in KVL for V_L. Avoid this by writing one consistent loop equation first, then solving.

LC Circuits and Oscillations

What an LC circuit is (and why it oscillates)

An LC circuit is an ideal circuit containing only a capacitor (capacitance C) and an **inductor** (inductance L). Unlike an LR circuit, there is no resistor to dissipate energy. Instead, energy sloshes back and forth between:

  • the capacitor’s electric field (stored as charge separation), and
  • the inductor’s magnetic field (stored due to current).

This matters because LC oscillations are the core of resonance phenomena: radio tuning circuits, oscillators, and many electromagnetic wave concepts rely on the idea that inductive and capacitive storage can exchange energy periodically.

A mechanical analogy that helps many students: an LC circuit behaves like a frictionless mass-spring system. The capacitor is like the spring (potential energy), the inductor is like the mass (kinetic energy), and charge/current oscillate like position/velocity.

The two storage relations: capacitor and inductor

For a capacitor:

V_C = \frac{q}{C}

where q(t) is the capacitor charge (with sign).

Energy stored in a capacitor:

U_C = \frac{q^2}{2C}

For an inductor:

V_L = L\frac{dI}{dt}

Energy stored in an inductor:

U_L = \frac{1}{2}LI^2

The current and charge are connected because current is the time rate of change of charge on the capacitor:

I = \frac{dq}{dt}

A frequent conceptual error is mixing up the symbol conventions: in many AP problems, q is defined as the charge on one plate (with sign), so when the plates discharge, q can decrease through zero and become negative as the capacitor “reverses polarity.” That sign change is not a mistake; it’s the essence of oscillation.

Deriving the LC differential equation (KVL leads to simple harmonic motion)

Consider an ideal series LC loop with no battery after an initial charge is placed on the capacitor. Apply Kirchhoff’s loop rule around the loop:

V_L + V_C = 0

Substitute element relations:

L\frac{dI}{dt} + \frac{q}{C} = 0

Using I = \frac{dq}{dt}, you get a second-order equation in q:

L\frac{d^2q}{dt^2} + \frac{1}{C}q = 0

Rewrite it in the standard simple harmonic oscillator form:

\frac{d^2q}{dt^2} + \omega^2 q = 0

where the angular frequency \omega is:

\omega = \frac{1}{\sqrt{LC}}

So the charge on the capacitor oscillates sinusoidally, just like the position of a mass on a spring.

Solutions for charge, current, and voltage (and what the phase means)

A general solution for charge is:

q(t) = Q\cos(\omega t + \phi)

where:

  • Q is the maximum charge magnitude,
  • \phi is a phase constant set by initial conditions.

Current is the derivative of charge:

I(t) = \frac{dq}{dt} = -\omega Q\sin(\omega t + \phi)

So current is shifted by a quarter cycle relative to charge (a phase difference of \pi/2). Physically:

  • When the capacitor’s charge magnitude is maximum, the current is zero (the system is momentarily “stopped” before reversing).
  • When the capacitor is momentarily uncharged (passing through q=0), the current magnitude is maximum (all energy is in the inductor).

The period and frequency are:

T = 2\pi\sqrt{LC}

f = \frac{1}{2\pi\sqrt{LC}}

Voltage across the capacitor is:

V_C(t) = \frac{q(t)}{C}

Voltage across the inductor is:

V_L(t) = -V_C(t)

in the ideal loop (their sum is zero at all times).

Common misconception: thinking the capacitor “discharges once and then stops.” In an ideal LC circuit there is nothing to remove energy, so the capacitor overshoots, recharges with opposite polarity, and oscillates indefinitely.

Energy exchange and conservation in an ideal LC circuit

In an ideal LC circuit (no resistance, no radiation losses), total energy is constant:

U = U_C + U_L = \frac{q^2}{2C} + \frac{1}{2}LI^2

You can see the exchange explicitly by using the sinusoidal forms:

  • When q = \pm Q, then I = 0, so U = \frac{Q^2}{2C} (all electric energy).
  • When q = 0, then |I| is maximum. From energy conservation:

\frac{1}{2}LI_{\max}^2 = \frac{Q^2}{2C}

So:

I_{\max} = \frac{Q}{\sqrt{LC}}

This relation is especially useful on AP-style problems: if you know the initial charge (or initial capacitor voltage), you can quickly find the maximum current without solving for phase constants.

Initial conditions: how to set up the specific oscillation

Typical setups:

1) Initially charged capacitor, zero current (common): at t=0, q(0)=Q and I(0)=0.

That matches:

q(t) = Q\cos(\omega t)

I(t) = -\omega Q\sin(\omega t)

2) Initially maximum current, uncharged capacitor: at t=0, q(0)=0 and I(0)=I_{\max}.

That corresponds to a sine form for charge, such as:

q(t) = Q\sin(\omega t)

The main skill is translating “what’s true at t=0” into a choice of cosine vs sine (or a phase shift \phi).

Real-world note: why real LC circuits don’t oscillate forever

Even if a problem says “ideal LC,” it’s good to understand reality: real circuits always have some resistance (wires aren’t perfectly conducting, components have losses), and accelerating charges can radiate electromagnetic waves. Those effects turn an LC circuit into a damped oscillator (an RLC circuit). In AP Physics C E\&M, you may be asked qualitatively why oscillations decay, even if you are not required to solve the full damped equation in this section.

Worked example 1: Oscillation frequency and period

An LC circuit has L = 2.0\ \text{mH} and C = 8.0\ \mu\text{F}. Find \omega, f, and T.

First convert units:

  • L = 2.0\times 10^{-3}\ \text{H}
  • C = 8.0\times 10^{-6}\ \text{F}

Compute angular frequency:

\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(2.0\times 10^{-3})(8.0\times 10^{-6})}}

LC = 1.6\times 10^{-8}

\sqrt{LC} = 1.26\times 10^{-4}

\omega \approx 7.9\times 10^{3}\ \text{rad/s}

Then:

f = \frac{\omega}{2\pi} \approx \frac{7.9\times 10^{3}}{2\pi} \approx 1.26\times 10^{3}\ \text{Hz}

T = \frac{1}{f} \approx 7.9\times 10^{-4}\ \text{s}

Worked example 2: Maximum current from initial capacitor voltage

A capacitor C = 4.0\ \mu\text{F} is initially charged to V_0 = 12\ \text{V}, then connected in series with an inductor L = 1.0\ \text{mH} to form an ideal LC circuit.

1) Find the initial charge magnitude Q.

Q = CV_0 = (4.0\times 10^{-6})(12) = 4.8\times 10^{-5}\ \text{C}

2) Find the total energy (initially all in the capacitor).

U = \frac{Q^2}{2C} = \frac{(4.8\times 10^{-5})^2}{2(4.0\times 10^{-6})}

U = \frac{2.30\times 10^{-9}}{8.0\times 10^{-6}} = 2.88\times 10^{-4}\ \text{J}

3) Find the maximum current.

At maximum current, all energy is magnetic:

\frac{1}{2}LI_{\max}^2 = U

I_{\max} = \sqrt{\frac{2U}{L}} = \sqrt{\frac{2(2.88\times 10^{-4})}{1.0\times 10^{-3}}}

I_{\max} = \sqrt{0.576} = 0.759\ \text{A}

You could also use the shortcut relation I_{\max} = \frac{Q}{\sqrt{LC}}, which comes from energy conservation.

Worked example 3: Charge and current at a specific time

In an ideal LC circuit, L = 0.40\ \text{H} and C = 10\ \mu\text{F}. At t=0 the capacitor has maximum charge Q = 2.0\times 10^{-4}\ \text{C} and the current is zero.

1) Write q(t).

First find \omega:

\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(0.40)(10\times 10^{-6})}}

LC = 4.0\times 10^{-6}

\sqrt{LC} = 2.0\times 10^{-3}

\omega = 500\ \text{rad/s}

With q(0)=Q and I(0)=0, use cosine:

q(t) = Q\cos(\omega t) = (2.0\times 10^{-4})\cos(500t)

2) Find I(t).

I(t) = \frac{dq}{dt} = -\omega Q\sin(\omega t) = -(500)(2.0\times 10^{-4})\sin(500t)

I(t) = -0.100\sin(500t)\ \text{A}

3) Evaluate at t = 3.14\ \text{ms}.

500t = 500(3.14\times 10^{-3}) = 1.57

So \cos(1.57) \approx 0 and \sin(1.57) \approx 1.

q \approx 0

I \approx -0.100\ \text{A}

The negative sign means the current direction is opposite to the direction you defined as positive, which is normal as the circuit oscillates.

Exam Focus
  • Typical question patterns:
    • Given L and C, find \omega, f, or T, often in the context of a capacitor initially charged and then connected to an inductor.
    • Energy-transfer questions: find I_{\max} from an initial voltage/charge, or determine when energy is all in the inductor vs all in the capacitor.
    • Write or interpret q(t) and I(t) using initial conditions (maximum charge at t=0 vs maximum current at t=0).
  • Common mistakes:
    • Confusing angular frequency \omega with ordinary frequency f, or forgetting the 2\pi when converting between them.
    • Using the wrong energy formula for the capacitor (mixing \frac{1}{2}CV^2 with \frac{q^2}{2C} without relating q and V first).
    • Assuming current and charge peak at the same time. They are out of phase by a quarter period: when q is maximum, I=0.