4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A battery, a switch and a resistor R₁ are connected in series with a parallel combination of an ideal inductor and a resistor R₂. The switch has been closed for a long time and is then opened. Which statement correctly describes what happens to the energy stored in the inductor after the switch is opened?
Answer and reasoning
AIt stays in the inductor's magnetic field, since the battery is no longer connected. A student who treats the inductor like a charged capacitor, which keeps its energy when it is isolated, picks this. An inductor stores energy only while it carries a current; its current continues through R₂ and decreases, and R₂ dissipates the energy.
BIt decreases as the current in the inductor and R₂ decreases, and R₂ dissipates it.Correct Just before the switch opens, the inductor acts as a wire and carries a steady current, storing (1/2)LI². Its current cannot stop instantly, so after the switch opens it continues around the loop formed by the inductor and R₂ and decreases exponentially. As the current decreases, the stored energy (1/2)LI² decreases, and R₂ dissipates it as thermal energy. R₁ is in the branch that the open switch disconnects, so it carries no current.
CIt is turned into thermal energy in the inductor, which opposes the change in current. A student who thinks the inductor's opposition to the change works like a resistor's opposition to the current picks this. An ideal inductor has no resistance and dissipates nothing: it returns its stored energy to the circuit, and R₂ dissipates it.
DIt is gone at the instant the switch opens, because every current stops at that instant. A student who thinks the battery must drive every current picks this. The inductor's current cannot change instantaneously; it continues through R₂ and decreases over a time set by τ = L/R₂, so the energy is transferred to R₂ gradually.
A series circuit contains an ideal battery of emf ε, a resistor of resistance R, an ideal inductor of inductance L and a switch. The switch is closed at t = 0. At a later instant, the current in the circuit is I₁. What is the rate of change of the current, dI/dt, at that instant?
Answer and reasoning
Aε/L A student who thinks the current rises at a constant rate, its initial rate, picks this. ε/L is dI/dt only at t = 0, when I = 0; once there is a current, part of ε is across the resistor, less is left across the inductor, and dI/dt is smaller.
B−(ε − I₁R)/L A student who treats the inductor's emf as acting with the battery while the current grows writes ε + L dI/dt − I₁R = 0 and picks this. The result is negative, a decreasing current, which contradicts the growing current; the induced emf opposes the increase.
C(ε − I₁R)/LCorrect Kirchhoff's loop rule around the circuit, in the direction of the current: ε − I₁R − L dI/dt = 0, because the inductor's emf opposes the increase in current. So dI/dt = (ε − I₁R)/L: largest, ε/L, at t = 0, and zero once I₁ reaches ε/R.
D−ε/L A student who reads the ε in ε = −L dI/dt as the battery's emf picks this. That equation gives the inductor's own induced emf; the battery's emf enters through the loop rule, ε = I₁R + L dI/dt.
Working Loop rule, going around in the direction of the current: ε − I₁R − L dI/dt = 0 (the inductor's emf, −L dI/dt, opposes the increase). Solving: dI/dt = (ε − I₁R)/L. Checks: I₁ = 0 gives ε/L; I₁ = ε/R gives 0; units V/H = A/s.
A series circuit contains a battery, a resistor, an ideal inductor and a switch. What does the time constant τ of the circuit describe?
Answer and reasoning
AHow far the inductor lowers the steady current below the value set by the resistor A student who thinks an inductor opposes the current itself, even when it is steady, picks this. In the steady state dI/dt = 0 and an ideal inductor behaves as a wire, so it does not lower the steady current, ε/R.
BHow rapidly the current approaches its final value after the switch is closedCorrect τ = L/Req sets the time scale of the exponential change: after one τ the current has made about 63 percent of its change, and after a time much greater than τ it is effectively steady. The smaller τ is, the sooner the circuit approaches its steady state.
CHow long the inductor's magnetic field lasts once its current stops changing A student who thinks an inductor has a magnetic field only while its current changes picks this. A steady current keeps a steady field in the inductor for as long as it flows; τ describes how the current changes, not how long the field lasts.
DHow long the current takes to reach its final value and then stop changing A student who thinks τ is the time to reach the steady state picks this. After one τ the current has made only about 63 percent of its change; it approaches its final value more and more slowly and is effectively steady only after a time much greater than τ.
A series circuit contains an ideal battery of emf ε, a resistor, an ideal inductor and an open switch. The switch is closed at t = 0. Which statement correctly describes the emf induced in the inductor just after the switch is closed?
Answer and reasoning
AIts magnitude is ε, and it is directed opposite to the battery's emf around the loop.Correct Just after closing, the current is still zero, so the potential difference across the resistor, IR, is zero, and the loop rule puts the whole of ε across the inductor. The induced emf opposes the increase in current, so it is directed opposite to the battery's emf: equal in magnitude, opposite in direction.
BIts magnitude is ε, and it acts in the same direction as the battery's emf around the loop. A student who thinks the induced emf helps the battery drive the growing current picks this. By Lenz's law the induced emf opposes the change, here an increase in current, so it acts against the battery's emf.
CIt is zero, because at that instant there is not yet any current in the inductor. A student who thinks an inductor's emf depends on its current picks this. The emf depends on dI/dt, which is largest, ε/L, at the instant the switch closes, even though the current itself is zero.
DIts magnitude is less than ε, because the resistor takes part of the battery's potential difference. A student who thinks the resistor always takes its share of the battery's potential difference picks this. With zero current, the resistor's potential difference, IR, is zero, so the inductor's emf has the full magnitude ε.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
13.5.A.1 Energy dissipated by a resistor in an LR circuit Fix
Energy dissipated by a resistor in an LR circuit
As the current in an inductor decreases, the energy stored in its magnetic field, UL = (1/2)LI², decreases, and the resistors the current passes through dissipate that energy as thermal energy at the rate I²R. When a current I₀ decays to zero through resistors, all of the stored energy (1/2)LI₀² is dissipated in them. Unit: joule (J).
Students often think An inductor disconnected from its battery keeps its stored energy, as a charged capacitor keeps its charge and energy when it is disconnected. In fact Not while its current has a path through a resistor. The energy is stored in the magnetic field of the inductor's current, UL = (1/2)LI². When the battery is disconnected and the current continues around a loop containing a resistor, the current decreases, and the resistor dissipates the stored energy. Unlike a charged capacitor, an inductor cannot hold its energy without a current.
Students often think An inductor converts its stored energy into thermal energy itself, because it opposes the change in current in the way a resistor opposes the current. In fact No. An ideal inductor has no resistance, so it dissipates no energy. As its current decreases, it gives up the energy stored in its magnetic field, and the resistors in the loop dissipate that energy as thermal energy.
13.5.A.2 Loop rule for a series LR circuit Fix
Loop rule for a series LR circuit
Going around a series loop in the direction of the current, the potential rises by ε across the battery and drops by IR across the resistor and by L dI/dt across the inductor, so ε − IR − L dI/dt = 0, or ε = IR + L dI/dt: a first-order differential equation for the current I(t).
Induced emf of an inductor, εL
The emf induced in an inductor by its own changing current, εL = −L dI/dt. Its magnitude depends on the rate of change of the current, not on the current, and it is directed so as to oppose the change in current. Unit: volt (V).
Students often think While the current increases, the inductor's induced emf acts in the same direction as the battery's emf, adding to it and helping to drive the current. In fact Against it. The induced emf, εL = −L dI/dt, opposes the change in current. While the current increases, the inductor's emf is directed opposite to the battery's emf around the loop, so the loop rule reads ε − IR − L dI/dt = 0.
Students often think The ε in ε = −L dI/dt is the emf of the battery in the circuit, so the rate of change of current is −ε/L whatever the current. In fact No. In ε = −L dI/dt, ε is the emf induced in the inductor by its own changing current. The battery's emf is a separate quantity; in a series LR circuit the loop rule connects them: εbattery − IR − L dI/dt = 0.
13.5.A.3 Time constant, τ Fix
Time constant, τ
The characteristic time of the exponential changes in an LR circuit after a switch is closed or opened: in each interval τ, the difference between a current or potential difference and its final value decreases by the same factor, e⁻¹ ≈ 0.37. Unit: second (s).
τ = L/Req
For an LR circuit, τ = L/Req, where Req is the equivalent resistance of the loop that the inductor's current follows. A larger inductance makes τ longer; a larger resistance makes it shorter. The units check: H/Ω = s.
τ and the initial rate of change
Just after a switch changes, the current changes at its greatest rate. If it continued at that rate, it would reach its final value after one time constant: for growth from zero in a series circuit the initial rate is ε/L and the final current ε/R, and (ε/R)/(ε/L) = L/R = τ. On an I(t) graph, the tangent at t = 0 meets the final-value line at t = τ.
Growth to approximately 63 percent
For an inductor with zero initial current, I(t) = If(1 − e−t/τ), where If is the final current, so at t = τ the current is (1 − e−1) ≈ 63 percent of If.
Decay to approximately 37 percent
For an inductor with initial current I₀ whose current decays through resistors, I(t) = I₀e−t/τ, so at t = τ the current is e−1 ≈ 37 percent of I₀, and in each further interval τ it falls by the same factor.
Students often think After one time constant a decaying current keeps 63 percent of its initial value, and a growing current has reached 37 percent of its final value: the two percentages are exchanged. In fact No. A decaying current keeps approximately 37 percent of its initial value after one time constant (it has lost about 63 percent), and a current growing from zero has reached approximately 63 percent of its final value.
Students often think The time constant is the time the current takes to reach its final, steady value, after which nothing in the circuit changes. In fact No. After one time constant a current growing from zero has reached approximately 63 percent of its final value, and a decaying current has fallen to approximately 37 percent of its initial value. The current approaches its steady value more and more slowly and is steady, for practical purposes, only after a time much greater than τ.
13.5.A.4 Transient and steady state Fix
Transient and steady state
While the current in an inductor is changing, its induced emf and the currents and potential differences in the circuit change with time. After a time much greater than τ, the currents no longer change, and the circuit is in a steady state.
Inductor at the instant of switching
An inductor's current cannot change instantaneously. When a switch is closed or opened, the induced emf is equal in magnitude and opposite in direction to the potential difference applied across the branch containing the inductor: an inductor with zero current just before a switch closes still has zero current just after, and an inductor carrying a current keeps that current, for that instant, in whatever loop remains.
Exponential dependence on time
For growth from zero in a series LR circuit, I(t) = (ε/R)(1 − e−t/τ), |ΔVL(t)| = εe−t/τ and UL(t) = (1/2)L(ε/R)²(1 − e−t/τ)². For decay from a current I₀, I(t) = I₀e−t/τ and UL(t) = (1/2)LI₀²e−2t/τ. Each quantity approaches an asymptote set by the circuit's conditions.
Energy stored in an inductor, UL
UL = (1/2)LI², the energy stored in the magnetic field of an inductor of inductance L carrying current I. Because it depends on I², it rises and falls with a different time dependence from the current. Unit: joule (J).
Inductor in the steady state
After a time much greater than τ, dI/dt = 0, so an ideal inductor has zero potential difference across it and behaves as a conducting wire with zero resistance; it still carries current and stores energy (1/2)LI².
Students often think The energy stored in an inductor changes with time in the same way as the current, so it reaches the same fraction of its final value at the same time (63 percent after one time constant for growth, 37 percent for decay… In fact No. The stored energy is UL = (1/2)LI², proportional to the square of the current, so it has a different time dependence from the current. When a growing current is at 63 percent of its final value, the energy is about (0.63)² ≈ 40 percent of its final value; when a decaying current is at 37 percent of its initial value, the energy is about (0.37)² ≈ 14 percent of its initial value.
Students often think The inductor's emf stays equal to the battery's emf for as long as the current is changing, and drops to zero only when the current becomes steady. In fact No. In a series LR circuit, the inductor's emf equals the battery's emf in magnitude only at the instant the switch is closed, when the current is still zero. As the current grows, the potential difference across the resistor grows, and the inductor's emf decreases toward zero.
17 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 17
An ideal inductor of inductance 0.40 H carries a steady current of 4.0 A. At t = 0 a switch disconnects the battery and, at the same instant, connects the inductor across a 12 Ω resistor, so the current decays in the loop formed by the inductor and the resistor. How much energy is dissipated in the resistor between t = 0 and t = τ, where τ is the time constant of that loop?
Answer and reasoning
A2.8 JCorrect The initial stored energy is (1/2)(0.40 H)(4.0 A)² = 3.2 J. At t = τ the current is I₀e−1, so the stored energy, proportional to I², is (3.2 J)e−2 ≈ 0.43 J. The resistor has dissipated the difference: 3.2 J − 0.43 J ≈ 2.8 J.
B2.0 J A student who applies the 63 percent value to the energy instead of the current picks this: 0.63 × 3.2 J ≈ 2.0 J. The current falls to about 37 percent of I₀, so the energy falls to about (0.37)² ≈ 14 percent of 3.2 J, and about 86 percent has been dissipated.
C1.9 J A student who thinks a decaying current keeps 63 percent of its initial value after one time constant finds (0.63)² × 3.2 J ≈ 1.3 J still stored and 1.9 J dissipated. The current keeps about 37 percent, not 63 percent.
D3.2 J A student who thinks the current has decayed completely after one time constant picks all 3.2 J. At t = τ the current is still about 37 percent of I₀, and about 0.43 J is still stored.
Working U₀ = (1/2)LI₀² = (1/2)(0.40 H)(4.0 A)² = 3.2 J. The current decays as I = I₀e−t/τ, so the stored energy decays as U = U₀e−2t/τ. At t = τ: U = (3.2 J)e−2 = 0.43 J. Energy dissipated in the resistor = 3.2 J − 0.43 J = 2.77 J ≈ 2.8 J. (τ = L/R = 0.40 H/12 Ω ≈ 0.033 s; its value is not needed for the fraction.)
A series circuit contains an ideal 9.0 V battery, a 25 Ω resistor, an ideal inductor and a switch. The switch is closed at t = 0. At a later instant, the current in the circuit is 0.24 A. What is the magnitude of the potential difference across the inductor at that instant?
Answer and reasoning
A0.0 V A student who thinks an ideal inductor, having no resistance, has no potential difference across it picks this. Its potential difference is L dI/dt, which is not zero while the current is still increasing.
B9.0 V A student who thinks the inductor's emf stays equal to the battery's emf while the current changes picks this. That is true only at t = 0, when I = 0; now 6.0 V is across the resistor, leaving 3.0 V.
C6.0 V A student who thinks elements carrying the same current have the same potential difference gives the inductor the resistor's 6.0 V. The inductor's potential difference is L dI/dt, found here from the loop rule: 9.0 V − 6.0 V.
D3.0 VCorrect The potential difference across the resistor is IR = (0.24 A)(25 Ω) = 6.0 V. The loop rule, ε = IR + L dI/dt, leaves 9.0 V − 6.0 V = 3.0 V across the inductor.
Working ΔVR = IR = (0.24 A)(25 Ω) = 6.0 V. Loop rule: ε = IR + L dI/dt, so |ΔVL| = L dI/dt = 9.0 V − 6.0 V = 3.0 V.
A series circuit containing an ideal battery, a resistor of resistance R and an ideal inductor of inductance L has time constant τ₁. The inductor is replaced by one of inductance 2L, and the resistor by one of resistance R/2. What is the new time constant of the circuit?
Answer and reasoning
AExactly as long as before (τ₁) A student who uses τ = RL, by analogy with RC, picks this: (2L)(R/2) = RL. For an LR circuit τ = L/R, so the two changes multiply instead of cancelling.
BTwice as long as before (2τ₁) A student who thinks the inductance alone sets the time constant picks this: 2L gives 2τ₁. The resistance matters too: halving R doubles τ again.
CFour times as long as before (4τ₁)Correct τ = L/Req. Doubling the inductance doubles τ, and halving the resistance doubles it again: τ = (2L)/(R/2) = 4L/R = 4τ₁.
DHalf as long as originally (τ₁/2) A student who thinks the resistance alone sets how long the current takes to change, as for a capacitor charging through a resistor, picks this: R/2 gives τ₁/2. In τ = L/R a smaller R makes τ larger, and the larger L increases it further.
The diagram shows three circuits, each with an identical ideal battery. Every resistor has resistance R, and the ideal inductor in each circuit has inductance L. τ₁, τ₂ and τ₃ are the time constants of Circuits 1, 2 and 3. Which ranking of the time constants is correct?
Answer and reasoning
Aτ₃ > τ₁ > τ₂Correct τ = L/Req. Circuit 1: Req = R, so τ₁ = L/R. Circuit 2: two resistors in series, Req = 2R, so τ₂ = L/(2R). Circuit 3: two resistors in parallel, Req = R/2, so τ₃ = 2L/R. Hence τ₃ > τ₁ > τ₂.
Bτ₂ > τ₁ > τ₃ A student who uses τ = RL picks this: RL, 2RL and RL/2. For an LR circuit τ = L/Req, so the circuit with the largest equivalent resistance has the smallest time constant.
Cτ₁ > τ₂ = τ₃ A student who thinks any added resistor increases the resistance gives Circuit 3 Req = 2R, so τ₃ = τ₂ = L/(2R). Two resistors in parallel give Req = R/2, so τ₃ = 2L/R is the largest.
Dτ₁ = τ₂ = τ₃ A student who thinks the inductance alone sets the time constant picks this, since each circuit has inductance L. τ = L/Req also depends on the equivalent resistance, which differs among the circuits.
In the circuit shown, the battery and the inductor are ideal. Switch S has been closed for a long time and is then opened. What is the time constant for the decrease of the current in the inductor after S is opened?
Answer and reasoning
AL/R A student who uses only the resistance that was in series with the inductor while the battery was connected picks this. After S opens, the inductor's current must also pass through 2R, so Req = 3R.
B(3R)L A student who uses τ = RL, by analogy with RC, picks this. For an LR circuit τ = L/Req; the product (3R)L does not even have units of time.
C3L/(2R) A student who treats R and 2R as parallel because they are drawn in separate branches picks this: Req = (R)(2R)/(3R) = 2R/3. With S open, the same current passes through one and then the other, so they are in series.
DL/(3R)Correct After S opens, the battery's branch carries no current, and the only closed loop through the inductor passes through R, the inductor and 2R one after another. The resistors are in series, Req = 3R, so τ = L/Req = L/(3R).
Working With S open, the loop containing the inductor is: inductor → R → top wire → 2R → bottom wire → inductor. Same current, no junction in use: series, Req = R + 2R = 3R. τ = L/Req = L/(3R). (While S was closed, the growth time constant was L/R, because the ideal battery holds its branch at ε; that is not the question.)
In Circuit 1, an ideal battery of emf ε, a resistor of resistance R, an ideal inductor of inductance L and a switch are connected in series. When the switch is closed, the current begins to increase at a rate r₁. Circuit 2 is identical except that its resistor has resistance 2R. At what rate does the current in Circuit 2 begin to increase when its switch is closed?
Answer and reasoning
AAt exactly the same rate as beforeCorrect Just after closing, I = 0, so there is no potential difference across the resistor, and the loop rule gives ε = L dI/dt: the initial rate is ε/L whatever the resistance. Equivalently, doubling R halves both the final current and τ, and final current/τ = (ε/(2R))/(L/(2R)) = ε/L = r₁.
BAt half the rate it had before A student who thinks the inductance alone sets the time constant keeps τ fixed while the final current halves, and picks this. τ = L/R also halves, so the initial rate, final current/τ, is unchanged.
CAt a quarter of the rate before A student who uses τ = RL finds τ doubled and the final current halved, giving (1/2)/2 = 1/4 of the rate. With τ = L/R, τ halves and the initial rate is unchanged.
DAt double the rate it had before A student who thinks a smaller time constant always means a steeper start picks this, since τ halves. The final current also halves, so the initial slope, final current/τ, stays ε/L.
Working At t = 0, I = 0, so ε − 0·R − L dI/dt = 0 and dI/dt = ε/L for both circuits: r₂ = r₁. Check with the time constant: initial rate = If/τ; Circuit 1: (ε/R)/(L/R) = ε/L; Circuit 2: (ε/(2R))/(L/(2R)) = ε/L.
The graph shows the current I in a series circuit containing an ideal battery, a resistor and an ideal inductor, as a function of time t after the switch is closed at t = 0. The horizontal dashed line shows the final current, and the sloping dashed line is the tangent to the curve at t = 0. A student concludes that the time constant is 4.0 ms, the time at which the tangent meets the final-current line. Which reasoning correctly supports the student's method?
Answer and reasoning
AThe current rises at the constant rate the tangent shows, until it reaches its final value at τ. A student who thinks the current rises at a constant rate picks this. The curve bends away from the tangent: the rate decreases continuously, and at 4.0 ms the current is only about 0.32 A, not 0.50 A.
BThe tangent meets the final-current line when the curve is at half its final value, at t = τ. A student who thinks τ is the time to reach half the final value picks this. At 4.0 ms the curve is at about 0.32 A, about 63 percent of 0.50 A; it passes 0.25 A earlier, at about 2.8 ms.
CThe tangent meets the final-current line just as the curve reaches 37 percent of its final value. A student who exchanges the 63 and 37 percent values picks this. For a current growing from zero, the curve is at about 63 percent of its final value at t = τ (about 0.32 A here), not 37 percent.
DIf the current kept increasing at its initial rate, it would reach its final value at t = τ.Correct At t = 0 the loop rule gives dI/dt = ε/L, the slope of the tangent. At that rate the current would reach its final value, ε/R, after (ε/R)/(ε/L) = L/R = τ, so the tangent meets the final-current line at t = τ. The curve itself has reached only about 63 percent of its final value then, about 0.32 A.
Working Initial slope from the loop rule at I = 0: dI/dt = ε/L. Time to reach ε/R at that slope: (ε/R)/(ε/L) = L/R = τ. From the graph: tangent reaches 0.50 A at 4.0 ms; the curve is at 0.50(1 − e−1) ≈ 0.32 A there, consistent with τ = 4.0 ms.
The graph shows the current I in a series circuit as a function of time t after the switch is closed at t = 0. The circuit contains an ideal battery, a 12 Ω resistor and an ideal inductor. The dashed line shows the final value of the current. What is the inductance of the inductor?
Answer and reasoning
A1.1 × 10⁻¹ H A student who takes τ as the time to reach half the final value reads t ≈ 9.0 ms at 0.25 A and gets (9.0 × 10⁻³ s)(12 Ω) ≈ 0.11 H. At t = τ the current is about 63 percent of its final value, which the curve reaches at 13 ms.
B7.2 × 10⁻² H A student who uses 37 percent for a growing current reads t ≈ 6.0 ms at about 0.19 A and gets (6.0 × 10⁻³ s)(12 Ω) = 0.072 H. For growth from zero, τ is the time to reach about 63 percent of the final value.
C1.6 × 10⁻¹ HCorrect The final current is 0.50 A, and 63 percent of it is about 0.32 A. The curve reaches 0.32 A at t ≈ 13 ms, so τ = 13 ms. τ = L/R gives L = τR = (13 × 10⁻³ s)(12 Ω) ≈ 0.16 H.
D1.1 × 10⁻³ H A student who uses τ = RL reads τ = 13 ms correctly but computes L = τ/R = (13 × 10⁻³ s)/(12 Ω) ≈ 1.1 × 10⁻³ H. For an LR circuit τ = L/R, so L = τR.
Working From the graph: final current 0.50 A; 0.63 × 0.50 A ≈ 0.32 A is reached at t ≈ 13 ms, so τ = 13 ms. L = τR = (0.013 s)(12 Ω) = 0.156 H ≈ 0.16 H. (Curve drawn with τ = 13 ms exactly: 0.25 A at 9.0 ms, 0.185 A at 6.0 ms.)
An ideal inductor of inductance 0.30 H carries a steady current of 4.0 A. At t = 0 a switch disconnects the battery and, at the same instant, connects the inductor across a 75 Ω resistor. What is the current in the inductor at t = 4.0 ms?
Answer and reasoning
A2.5 A A student who thinks a decaying current keeps 63 percent of its initial value after one time constant picks this: 0.63 × 4.0 A ≈ 2.5 A. About 63 percent is lost, leaving about 37 percent, 1.5 A.
B1.5 ACorrect τ = L/R = (0.30 H)/(75 Ω) = 4.0 × 10⁻³ s, so t = τ. A decaying current keeps approximately 37 percent of its initial value after one time constant: 0.37 × 4.0 A ≈ 1.5 A.
C2.0 A A student who takes τ as the time for the current to halve picks this: 4.0 A/2 = 2.0 A. After one time constant about 37 percent remains, not 50 percent.
D4.0 A A student who uses τ = RL gets τ = (75 Ω)(0.30 H) = 22.5 s, so 4.0 ms seems negligible and the current is still about 4.0 A. τ = L/R = 4.0 ms, so t is a full time constant.
Working τ = L/R = 0.30 H/75 Ω = 4.0 × 10⁻³ s = 4.0 ms, so t = τ. I = I₀e−t/τ = (4.0 A)e−1 = 1.47 A ≈ 1.5 A (0.37 × 4.0 A = 1.48 A).
In the circuit shown, the battery and the inductor are ideal, and switch S is initially open. S is closed at t = 0. I₀ is the current in resistor R₂ just after S is closed, and If is the current in R₂ a long time after S is closed. Which statement is correct?
Answer and reasoning
AI₀ = If = 0 A student who thinks the inductor acts like a wire as soon as S closes, as an uncharged capacitor does, gives R₂ no current at first. The inductor's current starts at zero and cannot jump, so at first the whole current passes through R₂: I₀ = ε/(R₁ + R₂).
BI₀ = If > 0 A student who thinks the inductor acts like an open circuit after a long time, as a charged capacitor does, keeps the current ε/(R₁ + R₂) in R₂. After a long time the inductor is a wire with zero resistance, so the potential difference across R₂ is zero and If = 0.
CI₀ > If > 0 A student who thinks the inductor still opposes the steady current, like an extra resistance, has the steady current shared between the inductor and R₂. With a steady current, dI/dt = 0, the ideal inductor has zero potential difference across it, and R₂, in parallel with it, carries no current.
DI₀ > If = 0Correct Just after S closes, the inductor's current is still zero, because it cannot change instantaneously, so the battery's current passes through R₁ and R₂ in series: I₀ = ε/(R₁ + R₂). A long time later the current is steady, and the ideal inductor behaves as a wire with zero resistance in parallel with R₂; the potential difference across R₂ is then zero, so If = 0, and the whole current, ε/R₁, passes through the inductor.
Working t = 0⁺: IL = 0 (current in an inductor is continuous), so R₁ and R₂ are in series: I₀ = ε/(R₁ + R₂) > 0. t ≫ τ: dI/dt = 0, inductor is a zero-resistance wire across R₂, ΔVR₂ = 0, so If = 0; IL = ε/R₁.
In the circuit shown, the battery and the inductor are ideal. Switch S has been closed for a long time and is then opened. What is the magnitude of the emf induced in the inductor just after S is opened?
Answer and reasoning
A3.0 V A student who thinks no potential difference can be larger than the battery's emf picks this. The inductor keeps its 1.5 A for an instant and forces it through the 4.0 Ω resistor, so its emf is 6.0 V, twice the battery's emf.
B6.0 VCorrect Before S opens, the inductor acts as a wire, so R₂ carries no current and the inductor carries ε/R₁ = (3.0 V)/(2.0 Ω) = 1.5 A. The inductor's current cannot change instantly, so just after S opens, 1.5 A passes around the loop formed by the inductor and R₂: the potential difference across R₂ is (1.5 A)(4.0 Ω) = 6.0 V, and the inductor's emf has the same magnitude.
C9.0 V A student who sends the inductor's current through every resistor, including R₁ in the branch the open switch has disconnected, gets (1.5 A)(2.0 Ω + 4.0 Ω) = 9.0 V. With S open, R₁ carries no current; the loop contains only the inductor and R₂.
D0.0 V A student who thinks the inductor acts as an open circuit after a long time gives it no current before S opens, and so no emf after. In the steady state the inductor is a wire carrying 1.5 A, and that current continues through R₂.
Working Steady state (S closed): inductor = wire, so R₂ is bypassed; IL = ε/R₁ = 3.0 V/2.0 Ω = 1.5 A. Just after opening: IL is still 1.5 A, and the only closed loop is inductor + R₂, so |ΔVR₂| = (1.5 A)(4.0 Ω) = 6.0 V; loop rule: |εL| = 6.0 V.
A series circuit contains an ideal battery of emf ε, a resistor of resistance R, an ideal inductor of inductance L and a switch. The switch is closed at t = 0. At what time t is the energy stored in the inductor one-quarter of its final value?
Answer and reasoning
A0.29 L/R A student who thinks the energy has the same time dependence as the current sets 1 − e−Rt/L = 1/4, giving t = (L/R) ln(4/3) ≈ 0.29 L/R. The energy depends on I², so one-quarter of the final energy needs one-half of the final current.
B0.50 L/R A student who thinks the current rises at its constant initial rate, ε/L, finds half the final current at (ε/(2R))/(ε/L) = L/(2R) = 0.50 L/R. The rate decreases as the current grows, so half the final current is reached later, at (L/R) ln 2.
C0.69 L/RCorrect The stored energy is (1/2)LI², so it is one-quarter of its final value when the current is one-half of its final value. From I = (ε/R)(1 − e−Rt/L) = ε/(2R), e−Rt/L = 1/2, so t = (L/R) ln 2 ≈ 0.69L/R.
D1.00 L/R A student who takes the time constant, L/R, as the time for the current to reach half its final value picks t = 1.00 L/R. At that time the current is about 63 percent of its final value and the energy about 40 percent.
Working UL = (1/2)LI², Ufinal = (1/2)L(ε/R)². UL = Ufinal/4 ⇔ I = ε/(2R). I(t) = (ε/R)(1 − e−Rt/L) (solution of ε = IR + L dI/dt, I(0) = 0). 1 − e−Rt/L = 1/2 ⇒ e−Rt/L = 1/2 ⇒ t = (L/R) ln 2 = 0.693 L/R ≈ 0.69 L/R.
A series circuit contains an ideal battery of emf ε, a resistor of resistance R, an ideal inductor of inductance L and a switch. The switch is closed at t = 0. Which expression gives the magnitude of the potential difference across the inductor as a function of time t, for t > 0?
Answer and reasoning
A|ΔVL| = εe−Rt/LCorrect Solving ε = IR + L dI/dt with I(0) = 0 gives I = (ε/R)(1 − e−Rt/L). The potential difference across the inductor is ε − IR = εe−Rt/L: equal to ε at t = 0, and decreasing exponentially toward zero as the current becomes steady.
B|ΔVL| = ε(1−e−Rt/L) A student who treats the inductor like a charging capacitor, whose potential difference rises toward ε as it blocks the current, picks this. The inductor's potential difference is L dI/dt, largest at t = 0 and approaching zero; this expression is the resistor's potential difference, IR.
C|ΔVL| = εe−t/(RL) A student who uses τ = RL, by analogy with RC, picks this. The loop rule gives τ = L/R, so the exponent is −Rt/L; the exponent here is not even dimensionless.
D|ΔVL| = ε(1−Rt/L) A student who thinks the potential difference keeps decreasing at its initial rate picks this straight line, which reaches zero at t = L/R and would then become negative. The potential difference decreases more and more slowly and approaches zero without reaching it.
Working ε = IR + L dI/dt, I(0) = 0 ⇒ I = (ε/R)(1 − e−Rt/L). |ΔVL| = L dI/dt = ε − IR = εe−Rt/L. Initial slope −εR/L; asymptote 0.
A series circuit contains an ideal battery, a resistor, an ideal inductor and a switch. The switch is closed at t = 0. The graphs, labeled (1) to (4), show four predictions of the energy U stored in the inductor as a function of time t. Which graph best represents the energy?
Answer and reasoning
AGraph (1) A student who thinks the energy has the same time dependence as the current picks this curve, which rises most steeply at t = 0. Since U is proportional to I², the energy rises slowly at first, while the current is small: the graph starts flat.
BGraph (2)Correct U = (1/2)LI², with I = If(1 − e−t/τ). At t = 0 the current is zero, so dU/dt = LI dI/dt is zero and the graph starts flat. The energy then rises and levels off at (1/2)LIf² as the current becomes steady.
CGraph (3) A student who thinks an inductor stores energy only while its current is changing picks this curve, which falls back to zero. A steady current keeps a steady magnetic field, so the stored energy stays at (1/2)LIf².
DGraph (4) A student who thinks the stored energy keeps increasing while the battery drives a current picks this straight line. U = (1/2)LI² depends only on the present current, which levels off, so the energy levels off too.
Working U = (1/2)LIf²(1 − e−t/τ)². dU/dt = LI dI/dt = 0 at t = 0 (I = 0), maximum slope later, asymptote (1/2)LIf²: an S-shaped rise.
A series circuit contains an ideal battery, a resistor, an ideal inductor and a switch. A long time after the switch is closed, a student claims that the potential difference across the inductor is zero. Which reasoning correctly supports the claim?
Answer and reasoning
AThe inductor has zero resistance, so by ΔV = IR there is no potential difference across it. A student who applies ΔV = IR to the inductor picks this. That reasoning would make the potential difference zero at every instant, yet just after the switch closes the potential difference across the ideal inductor is ε, although its resistance is zero. What makes it zero now is that the current has stopped changing.
BThe current reached its final value after one time constant, and nothing has changed since then. A student who thinks τ is the time to reach the steady state picks this. After one time constant the current has reached only about 63 percent of its final value and is still changing; it is steady only after a time much greater than τ.
CThe inductor's magnetic field has disappeared, so there is no magnetic flux through it. A student who thinks an inductor has a magnetic field only while its current changes picks this. The steady current keeps a steady field and flux in the inductor; the emf is zero because the flux is not changing, not because there is none.
DThe current has become constant, so dI/dt and the inductor's emf, −L dI/dt, are now both zero.Correct After a time much greater than τ the current is steady, so dI/dt is zero. The inductor's emf, −L dI/dt, is then zero, so there is no potential difference across the ideal inductor: it behaves as a wire with zero resistance.
An ideal battery of emf ε and a switch S are connected across two parallel branches. One branch contains a resistor of resistance R in series with an ideal inductor of inductance L; the other branch contains a resistor of resistance 3R. S has been closed for a long time and is then opened. How much energy is dissipated in the resistor of resistance 3R after S is opened?
Answer and reasoning
A(1/8)(Lε²/R²) A student who thinks the larger resistance dissipates less, as it would in parallel, gives 3R only 1/4 of the stored energy (1/2)L(ε/R)². In series the current is shared, so P = I²R and the larger resistance dissipates more.
B(1/4)(Lε²/R²) A student who thinks elements carrying the same current have the same potential difference splits the stored energy equally between R and 3R. The same current gives potential differences IR and 3IR, so 3R dissipates three times as much as R.
C(3/8)(Lε²/R²)Correct In the steady state the inductor is a wire, so the inductor's branch has the full emf across R: I₀ = ε/R, and the stored energy is (1/2)L(ε/R)². After S opens, R and 3R are in series in the loop with the inductor, so at every instant they carry the same current and dissipate power in the ratio 1 : 3. The 3R resistor receives 3/4 of the stored energy: (3/8)(Lε²/R²).
D(2/3)(Lε²/R²) A student who takes the inductor's steady current to be the battery's total current, ε/Req with Req = 3R/4, gets I = 4ε/(3R), a stored energy of (8/9)(Lε²/R²), and 3/4 of that. The inductor's branch has ε across R alone, so its current is ε/R.
Working Steady state: inductor = wire; its branch is directly across the ideal battery, so I₀ = ε/R and U₀ = (1/2)L(ε/R)² = Lε²/(2R²). After S opens: loop inductor–R–3R, series, same current i(t); energy to 3R = ∫i²(3R)dt = (3R/4R) × U₀ (all of U₀ is dissipated in R + 3R). E₃R = (3/4)(Lε²/(2R²)) = (3/8)(Lε²/R²).
An ideal battery of emf ε, an ideal inductor of inductance L and two resistors, each of resistance R, are connected in series. A switch S is connected across one of the resistors, so that while S is closed that resistor carries no current. S has been closed for a long time. At t = 0, S is opened. At what time t has the current in the inductor decreased to three-quarters of its value at t = 0?
Answer and reasoning
A0.69 L/R A student who keeps the time constant L/R, set by the single resistor in series with the inductor before S was opened, solves e−Rt/L = 1/2 and picks this. Once S is open, both resistors are in the inductor’s loop, so τ = L/(2R).
B0.25 L/R A student who assumes the current keeps falling at its initial rate, ε/L (from ε − (ε/R)(2R) = L dI/dt), finds that it drops by ε/(4R) in L/(4R) and picks this. The rate of decrease shrinks as the current approaches ε/(2R), so the drop takes longer.
C0.35 L/RCorrect After S opens, the inductor’s loop has resistance 2R, so τ = L/(2R), and the current falls from ε/R toward ε/(2R): I = ε/(2R) + (ε/(2R))e−2Rt/L. It reaches 3ε/(4R) when e−2Rt/L = 1/2, at t = (L/(2R)) ln 2 ≈ 0.35 L/R.
D0.14 L/R A student who assumes the current decays exponentially toward zero, I = (ε/R)e−2Rt/L, solves e−2Rt/L = 3/4 and picks this. With the battery still in the circuit the current approaches ε/(2R), not zero, and only the part above ε/(2R) decays exponentially.
Working Before t = 0 the current is steady and the inductor acts as a wire: I₀ = ε/R. After S opens, both resistors are in series with the inductor: ε = I(2R) + L dI/dt, with I(0) = ε/R (the inductor’s current cannot change instantly). Final value If = ε/(2R); τ = L/(2R). Solution: I = ε/(2R) + (ε/(2R))e−2Rt/L. I = (3/4)(ε/R) ⇔ (ε/(2R))e−2Rt/L = ε/(4R) ⇔ e−2Rt/L = 1/2 ⇔ t = (L/(2R)) ln 2 = 0.347 L/R ≈ 0.35 L/R. Distractors (sympy): τ kept at L/R, e−Rt/L = 1/2, t = (L/R) ln 2 = 0.69 L/R; initial rate dI/dt = (ε − 2ε)/L = −ε/L held constant, drop of ε/(4R) in L/(4R) = 0.25 L/R; decay toward zero, e−2Rt/L = 3/4, t = (L/(2R)) ln(4/3) = 0.14 L/R.
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
