Topic 13.5: Circuits with Resistors and Inductors (LR Circuits)

What Topic 13.5 requires

Topic 13.5 has one learning objective and nine essential-knowledge statements, the most of any topic in Unit 13. It prints no boundary statement.

13.5.A, describe the physical and electrical properties of a circuit containing a combination of resistors and a single inductor.

• 13.5.A.1 states that a resistor will dissipate energy that was stored in an inductor as the current changes.
• 13.5.A.2 states that Kirchhoff's loop rule can be applied to a series LR circuit with a battery of emf $\mathcal{E}$, resulting in a differential equation that describes the current in the loop, with the derived equation $\mathcal{E} = IR + L\dfrac{dI}{dt}$.
• 13.5.A.3 states that the time constant is a significant feature of the behavior of an LR circuit.
• 13.5.A.3.i states that the time constant of a circuit is a measure of how quickly an LR circuit will reach a steady state and is described with the equation $\tau = \dfrac{L}{R_{\text{eq}}}$.
• 13.5.A.3.ii states that the time constant represents the time an LR circuit would take to reach a steady state if the system continued to change at the initial rate of change.
• 13.5.A.3.iii states that for an inductor that has zero initial current, the time constant represents the time required for the current in the inductor to reach approximately 63 percent of its final asymptotic value.
• 13.5.A.3.iv states that for an inductor with an initial current, the time constant represents the time required for the current in the inductor to reach approximately 37 percent of its initial value.
• 13.5.A.4 states that the electric properties of inductors change during the time interval in which the current in the inductor changes, but will exhibit steady state behavior after a long time interval.
• 13.5.A.4.i states that when a switch is initially closed or opened in a circuit containing an inductor, the induced emf will be equal in magnitude and opposite in direction to the applied potential difference across the branch containing the inductor.
• 13.5.A.4.ii states that the potential difference across an inductor, the current in the inductor, and the energy stored in the inductor are exponential with respect to time and have asymptotes that are determined by the initial conditions of the circuit.
• 13.5.A.4.iii states that after a time much greater than the time constant of the circuit, an inductor will behave as a conducting wire with zero resistance.

Notice the wording of the objective: a combination of resistors and a single inductor. Plural resistors, one inductor. That is why 13.5.A.3.i writes the time constant with $R_{\text{eq}}$ rather than $R$, and it is the single most common place to lose a mark in this topic.

The suggested skills are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

How far the CED takes the differential equation

Topic 13.5 prints no boundary statement, so nothing fences off the mathematics that way. What sets the level is the label on 13.5.A.2 and what the framework does and does not print elsewhere.

The framework's Required Equations page states that not all equations in the course framework appear on the equation sheet, that many are provided for reference and guidance or to demonstrate the final results of derivations expected of students on the exam, and that these are denoted as Derived Equations. Statement 13.5.A.2 carries that label. So:

The differential equation itself is the expected derivation. You are expected to apply Kirchhoff's loop rule to a series LR circuit with a battery and arrive at

It is not on the equation sheet. What is on the sheet is what you need to write it: $\mathcal{E} = -L\,dI/dt$ and $I = \Delta V/R$.

No solved exponential is printed anywhere in Unit 13. Look for a current-versus-time function in the framework and there is not one, for the rise or for the decay. What the CED prints instead is the behaviour. Statement 13.5.A.4.ii says the potential difference across an inductor, the current in the inductor, and the energy stored in the inductor "are exponential with respect to time and have asymptotes that are determined by the initial conditions of the circuit". Statements 13.5.A.3.iii and 13.5.A.3.iv then pin the two numbers you need from those exponentials, 63 percent and 37 percent, without ever writing an exponential function.

Deriving the current as a function of time is named as a skill. The unit's "Building the Science Practices" page says that describing how a solenoid resists a change in electrical current due to the required change in magnetic field within that solenoid is a valuable skill, "in addition to deriving a mathematical equation that shows the current in the solenoid as a function of time".

Put those three together and the honest reading is: set up the equation, know that the solution is exponential with the asymptotes the circuit dictates, be able to produce the function if asked, and expect most questions to be answerable from the two limits and the time constant instead.

One more calibration, from the CED's own sample free-response set rather than from the framework. Its Question 4, on an RC circuit, instructs students to "Derive, but do not solve, a differential equation" for a rate of change, and a later part asks them to justify a prediction "by referring to the differential equation you wrote". That is the shape of the expectation in this part of the course: the setup and the reasoning carry the points.

Deriving the loop equation

Statement 13.5.A.2 names the method: Kirchhoff's loop rule applied to a series LR circuit with a battery. Do it carefully once and the sign convention stops being mysterious.

Take a battery of emf $\mathcal{E}$ in series with a resistor $R$ and an inductor $L$. Define the positive current direction as the one the battery drives, and traverse the loop in that direction.

1. Across the battery, from the negative to the positive terminal, the potential rises by $\mathcal{E}$.
2. Across the resistor, travelling with the current, the potential drops by $IR$. That is Ohm's law in the form $I = \Delta V / R$ that the sheet prints.
3. Across the inductor, the self-induced emf is $-L\,dI/dt$ by statement 13.4.A.3, and it opposes the change. Travelling with the current while the current is rising, the potential drops by $L\,dI/dt$.
4. Sum to zero around the loop: $\mathcal{E} - IR - L\dfrac{dI}{dt} = 0$, which rearranges to the printed derived equation.

Read the equation rather than solving it and it gives up almost everything.

At $t = 0$, with the inductor starting from zero current, the $IR$ term vanishes and the whole battery emf sits across the inductor: $L\,dI/dt = \mathcal{E}$, so the initial slope is $dI/dt = \mathcal{E}/L$. That is statement 13.5.A.4.i in symbols, and it is where the initial rate of change comes from.

Long after, the current stops changing, so $L\,dI/dt = 0$ and $\mathcal{E} = IR$, giving the final current $I_\infty = \mathcal{E}/R$. That is 13.5.A.4.iii: the inductor has become a conducting wire with zero resistance.

In between, every joule per coulomb the battery supplies is split between the resistor and the inductor, and the split shifts continuously from all-inductor to all-resistor.

If you do want the solution, separating variables gives it in a few lines. Write the equation as $L\,dI/dt = \mathcal{E} - IR$, separate to $\dfrac{dI}{\mathcal{E} - IR} = \dfrac{dt}{L}$, integrate both sides from $I = 0$ at $t = 0$, and exponentiate:

That function is not printed in the framework or on the sheet. Its two asymptotes, $I = 0$ at the start and $I = \mathcal{E}/R$ at the end, are exactly the "asymptotes determined by the initial conditions of the circuit" that 13.5.A.4.ii refers to.

The time constant, and the four things the CED says about it

The CED spends four separate statements on this one quantity, 13.5.A.3 and its three sub-statements, which is more attention than it gives any other single idea in the unit. Each one says something different, and they are each worth being able to state.

13.5.A.3.i: it measures how quickly the circuit reaches a steady state. Larger inductance means more electrical inertia and a slower approach. Larger resistance means a smaller final current to reach and a faster approach. The subscript on $R_{\text{eq}}$ is not decoration.

13.5.A.3.ii: it is the time the circuit would take if it kept changing at its initial rate. This is the tangent-line reading, and it is the definition that explains the formula. Starting from zero current, the initial slope is $\mathcal{E}/L$ and the target is $\mathcal{E}/R$, so a straight line at that slope would arrive after

The emf cancels, which is why the time constant does not depend on the battery. Draw the tangent to the current curve at $t = 0$ and extend it until it meets the final value: the horizontal coordinate where they cross is one time constant. That is a standard skill 1.C sketching task.

13.5.A.3.iii: for zero initial current, it is the time to reach about 63 percent of the final value. The exact figure is $1 - e^{-1} = 0.6321$, so 63 percent is the CED's rounding of it.

13.5.A.3.iv: for an inductor with an initial current, it is the time to fall to about 37 percent of the initial value. The exact figure is $e^{-1} = 0.3679$. Note that the two statements describe different situations, a rise and a decay, and the two percentages add to 100 because they are the same exponential read from opposite ends.

A table of the multiples worth remembering, computed rather than recalled:

And the piece of vocabulary that goes with it: an exponential never actually arrives. "Steady state" means close enough that nothing measurable is still changing, which is why 13.5.A.4.iii says "after a time much greater than the time constant" rather than naming a moment.

Now the equivalent resistance. The objective says a combination of resistors and a single inductor, so the resistance in $\tau$ is the one the inductor actually sees. The reliable procedure is to look out from the inductor's two terminals into the rest of the circuit, with any ideal battery treated as a zero-resistance connection, and reduce what you see using the series and parallel rules that the sheet prints. In a plain series loop that is just $R$. In the second worked example, with a resistor in series and another in parallel, it is the parallel combination of the two, and it equals neither of the printed values.

Watch also that opening a switch usually changes $R_{\text{eq}}$, so the decay after a switch opens generally has a different time constant from the rise before it. The third worked example has a decay eight times faster than the rise that preceded it, in the same circuit.

The two instants that need no calculus

Most LR questions on the multiple-choice section are about one of two moments, and neither requires solving anything.

The instant a switch is thrown. The current through an inductor cannot jump, because a jump would need an infinite $dI/dt$ and so an infinite emf. So the inductor's current is whatever it was an instant earlier, and statement 13.5.A.4.i gives the consequence: the induced emf is equal in magnitude and opposite in direction to the applied potential difference across the branch containing the inductor.

• If the inductor was carrying no current, it carries none at that instant either. Treat that branch as an open circuit for the purpose of finding the currents elsewhere, then find the inductor's potential difference from the rest of the circuit.
• If the inductor was carrying some current, it carries exactly that current at the first instant after the switch changes. Treat it as a current source of that value.

A long time later. With the current steady, $dI/dt = 0$ and the inductor's potential difference is zero. Statement 13.5.A.4.iii: it behaves as a conducting wire with zero resistance. So it short-circuits anything in parallel with it, and all of that branch's current goes through the inductor instead.

Side by side with a capacitor, which is the comparison questions exploit, the two are exact opposites:

Both time constants are printed on the sheet, in that form, with the equivalent subscripts. Note the structural difference between them, because it is a good check: resistance is in the denominator for an inductor and in the numerator for a capacitor. More resistance slows a capacitor down and speeds an inductor up. If you find yourself writing $\tau = LR$, that check catches it.

A good habit for any switching question: draw the circuit twice, once for $t = 0$ with the inductor replaced by the appropriate model, and once for $t \to \infty$ with it replaced by a plain wire. Skill 1.C is listed for this topic and the two redrawn circuits are the fastest route to a correct sketch of what happens in between.

Sketching the four graphs

Skill 1.C is listed first for this topic, and the CED's fifth sample activity for Unit 13 is built entirely on it: it puts students in pairs, has Student A construct quantitative graphs of current against time and voltage against time for the inductor, resistor 1 and resistor 2, showing the behaviour before and after a switch opens or closes, and then has Student B construct the circuit with the switch, both resistors and the inductor from those graphs alone.

So be able to draw all four curves for a simple series LR circuit closing onto a battery at $t = 0$.

Four features to get right, because they are the ones a scoring guideline can identify:

1. The initial value and the asymptote. Statement 13.5.A.4.ii says the asymptotes are determined by the initial conditions of the circuit, so label both ends.
2. The initial slope. For the current it is $\mathcal{E}/L$, and the tangent at $t = 0$ meets the asymptote at $t = \tau$. Marking that construction on the graph is often worth a point on its own.
3. The concavity. A rising exponential approaching an asymptote is concave down everywhere. A decaying one is concave up.
4. The two voltages add to the battery emf at every instant. $IR + L\,dI/dt = \mathcal{E}$ is the loop equation, so if you draw both curves on the same axes their sum must be a horizontal line at $\mathcal{E}$. That is a free self-check and it catches a mis-drawn pair immediately.

The energy curve is the one that is usually drawn wrong. Because $U_L$ goes as $I^2$, it starts off flatter than the current does and then catches up, so it approaches its asymptote more slowly than the current approaches $\mathcal{E}/R$. In fact the energy is at half its final value when the current is at $1/\sqrt{2}$, about 71 percent, of its final value, which happens later than $t = \tau$.

If the graph is a skill 1.B quantitative graph rather than a 1.C sketch, label the axes with units and put a scale on them. The distinction between the two skills is exactly that.

Where the energy goes

Statement 13.5.A.1 is the topic's first line for a reason: a resistor will dissipate energy that was stored in an inductor as the current changes. Two audits are worth being able to run.

During the rise. The battery supplies energy at a rate $\mathcal{E}I$. Some of it heats the resistor at $I^2R$ and the rest goes into the magnetic field at $LI\,dI/dt$. Multiply the loop equation through by $I$ and that is exactly what you get:

Every term is a power. The left is supplied, the first on the right is dissipated, and the second is $dU_L/dt$, the rate at which the field is being built.

A fact worth knowing: over the whole rise, the battery supplies exactly twice the energy that ends up stored in the inductor, with the other half dissipated in the resistor. That is the same 50 percent result as charging a capacitor through a resistor, and it does not depend on $R$ or $L$ at all.

During the decay. Remove the battery and leave the inductor in a loop with resistance. There is no source now, so the resistor's heat has to come from the field. Integrating the dissipated power over all time gives back exactly the stored energy:

That identity is 13.4.A.2.ii, which says the transfer of energy generated in an inductor to other forms obeys conservation laws, made into a calculation. And it is the fastest way to answer "how much energy is dissipated in the resistor after the switch opens": you do not integrate, you just compute $\frac{1}{2}LI_0^2$.

One thing to be careful about when the switch opens. If the inductor's current is suddenly forced through a much larger resistance, statement 13.5.A.4.i still holds and the induced emf matches the potential difference across that branch. With a current $I_0$ now flowing through a resistance $R'$, the potential difference across that branch is $I_0 R'$, and if $R'$ is large that can be far larger than the battery's emf ever was. The energy is unchanged, at $\frac{1}{2}LI_0^2$; it is just delivered faster and at a higher voltage. This is the spark across an opening switch, and it is why real inductive circuits include a path for the current to decay through.

LR against RC, and the AP Physics 2 sibling

There is no AP Physics 2 topic on LR circuits, because that course contains no inductors at all: the words inductance and inductor appear nowhere in the AP Physics 2 course and exam description. The structurally closest algebra-based page is AP Physics 2 Topic 11.8, Resistor-Capacitor (RC) Circuits.

That page is for students in AP Physics 2, and it is about a different circuit element. This page is for students in AP Physics C: Electricity and Magnetism. If you are in Physics 2, RC circuits are as far as transient circuit behaviour goes for you, and reaching for an inductor is reaching outside your course. If you are in Physics C, reading the RC page first is genuinely useful, because the two circuits share their entire mathematical structure and swapping the roles is most of the work.

Both time constants are on the C: E&M sheet. The row that catches people is the fourth: adding resistance makes an RC circuit slower and an LR circuit faster, because the resistance multiplies in one case and divides in the other.

AP Physics C also has its own RC circuit topic, in Unit 11, and the two are examined in the same register. The CED's sample Question 4 is an RC question that asks for a differential equation to be derived but not solved, which is a fair guide to what an LR free-response question would ask for.

One last comparison, because it is the difference students actually feel. In an RC circuit the current is largest at the start and decays; in an LR circuit the current starts at zero and grows. The two look like opposites on a graph, and they are, because a capacitor and an inductor resist opposite things.

How Topic 13.5 is tested

The question patterns:

1. Find the current and the potential differences at $t = 0$ and at $t \to \infty$ (skills 2.C, 3.C). This is the highest-frequency pattern and it needs no calculus.
2. Compute the time constant, taking care over $R_{\text{eq}}$ (skill 2.C).
3. Derive the loop equation from Kirchhoff's loop rule (skill 2.A), and reason from it without necessarily solving it.
4. Sketch current or voltage against time for a switch closing or opening, with the asymptotes labelled and the initial tangent marked (skill 1.C).
5. Compute stored or dissipated energy using $\frac{1}{2}LI^2$ rather than an integral.
6. Justify a claim about the circuit's behaviour from the equation or from a graph (skill 3.C), which is the only skill listed here that is not listed for Topic 13.4.

Two of the CED's five optional sample activities for Unit 13 are attached to this topic, and both are laboratory work. The first has students construct their own solenoid, or use a provided one, and measure its inductance using an LR circuit by measuring the time constant and the resistance to get $L$, then repeat with iron or steel in the core to get the increased inductance. The second is the graph-and-switch pairing described above.

That first activity is a complete skill 3.A and 1.B question in miniature, and it is worth thinking through as an experimental design. You cannot measure inductance directly, so you measure something you can: build a series LR circuit of known resistance, record the current against time after closing a switch, read off the time at which the current reaches 63 percent of its final value, and that time is $\tau$. Then $L = \tau R_{\text{eq}}$. The 63 percent figure is essential knowledge 13.5.A.3.iii being used as a measurement technique rather than as a fact to recall.

The CED's exam-preparation note for Unit 13 backs this up directly: it says laboratory investigations about the behavior of circuits when a solenoid is in the circuit are valuable exercises that concretely demonstrate principles and ideas that are often abstract.