Physics 2 Quiz: Energy Stored In Inductors

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Energy Stored In InductorsQuestion 1 of 7

A battery of EMF $\mathcal{E}$ and negligible internal resistance is connected in series with resistance $R$ and inductance $L$ . After the circuit reaches steady state, the battery is suddenly removed (replaced by a wire) at $t = 0$ .

What fraction of the energy that was stored in the inductor at $t = 0$ has been dissipated in the resistor by time $t = 2\tau$ , where $\tau = L/R$ ?

e^{-2}

, because the fraction dissipated equals the fraction by which the current has decayed, and at

t = 2\tau

the current is

e^{-2}

of its initial value.

1 - e^{-2}

, because the energy stored decays with time constant

\tau

, not

\tau/2

, so the remaining energy fraction at

t = 2\tau

e^{-2}

\frac{1}{2}(1 - e^{-4})

, because only half the initial stored energy can ever be recovered as heat due to the reactive nature of the inductor.

1 - e^{-4}

, because the current decays as

e^{-t/\tau}

, the energy decays as

e^{-2t/\tau}

, and the fraction dissipated is

1

minus the remaining energy fraction

e^{-2(2\tau)/\tau} = e^{-4}

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Physics 2 Quiz

Physics 2 Quiz: Energy Stored In Inductors

Practice Energy Stored In Inductors in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Energy Stored In Inductors, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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Question 1

What fraction of the energy that was stored in the inductor at $t = 0$ has been dissipated in the resistor by time $t = 2\tau$ , where $\tau = L/R$ ?

$e^{-2}$ , because the fraction dissipated equals the fraction by which the current has decayed, and at $t = 2\tau$ the current is $e^{-2}$ of its initial value.
$1 - e^{-2}$ , because the energy stored decays with time constant $\tau$ , not $\tau/2$ , so the remaining energy fraction at $t = 2\tau$ is $e^{-2}$ .
$\frac{1}{2}(1 - e^{-4})$ , because only half the initial stored energy can ever be recovered as heat due to the reactive nature of the inductor.
$1 - e^{-4}$ , because the current decays as $e^{-t/\tau}$ , the energy decays as $e^{-2t/\tau}$ , and the fraction dissipated is $1$ minus the remaining energy fraction $e^{-2(2\tau)/\tau} = e^{-4}$ . (correct answer)

Explanation: When an inductor discharges through a resistor, you need to track energy, not just current — and these decay at different rates. That distinction is what this question is testing. At steady state, the initial current is

I_0 = \mathcal{E}/R

, and the energy stored is

U_0 = \frac{1}{2}LI_0^2

. After the battery is removed, the current decays as

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

. The energy remaining in the inductor at any time is

U(t) = \frac{1}{2}L I(t)^2 = \frac{1}{2}L I_0^2\, e^{-2t/\tau} = U_0\, e^{-2t/\tau}

. Notice the exponent is

-2t/\tau

, not

-t/\tau

, because energy goes as current squared. At

t = 2\tau

, the remaining energy fraction is

e^{-2(2\tau)/\tau} = e^{-4}

. Since all energy lost by the inductor is dissipated in the resistor, the fraction dissipated is

1 - e^{-4}

, confirming D. Choice A confuses the current decay factor

e^{-2}

with an energy fraction — current and energy decay on different timescales. Choice B correctly identifies that energy (not current) governs the answer, but then applies a time constant of

\tau

instead of

\tau/2

, arriving at the remaining fraction rather than accounting for the squared relationship properly. Choice C invents a false physical rule — all stored energy is eventually dissipated as heat; the inductor has no mechanism to "keep" half of it. Study tip: Whenever you see energy and exponential decay together, ask yourself: is the exponent coming from a quantity, or from that quantity squared? Energy always decays twice as fast as the current in an RL circuit.

Question 2

An ideal LC circuit consists of an inductor $L = 10\text{ mH}$ and a capacitor $C = 40\text{ μF}$ . At $t = 0$ , the capacitor is fully charged to voltage $V_0 = 50\text{ V}$ and the current through the inductor is zero.

At the instant when the energy stored in the inductor equals three times the energy stored in the capacitor, what is the magnitude of the current through the inductor?

$I = \sqrt{\frac{3CV_0^2}{2L}} = \sqrt{\frac{3(40\times10^{-6})(2500)}{2(0.01)}} \approx 4.24\text{ A}$ , found by setting $\frac{1}{2}LI^2 = 3 \cdot \frac{1}{2}CV_C^2$ and using energy conservation to find $V_C$ .
$I = V_0\sqrt{\frac{3C}{4L}} = 50\sqrt{\frac{3(40\times10^{-6})}{4(0.01)}} \approx 2.74\text{ A}$ , found by setting the inductor energy equal to three-fourths of the total initial energy. (correct answer)
$I = V_0\sqrt{\frac{C}{L}} = 50\sqrt{\frac{40\times10^{-6}}{0.01}} = 1\text{ A}$ , which is the maximum current in the LC circuit and serves as an upper bound that the given condition cannot exceed.
$I = V_0\sqrt{\frac{3C}{L}} = 50\sqrt{\frac{3(40\times10^{-6})}{0.01}} \approx 1.73\text{ A}$ , found by applying energy conservation with the assumption that the capacitor energy has dropped to one-third of its initial value.

Explanation: When you see an LC circuit energy problem, your instinct should be to write two equations: the energy condition given, and total energy conservation. Together, they fully determine the unknowns. Here's the logic for choice B. The total initial energy is stored entirely in the capacitor:

U_{total} = \frac{1}{2}CV_0^2

. At any later moment, this energy splits between the inductor and capacitor. If the inductor holds three times the capacitor's energy, and those two portions must sum to the total, then

U_L + U_C = U_{total}

becomes

3U_C + U_C = \frac{1}{2}CV_0^2

, giving

U_C = \frac{1}{8}CV_0^2

and

U_L = \frac{3}{4} \cdot \frac{1}{2}CV_0^2

. Setting

\frac{1}{2}LI^2 = \frac{3}{4} \cdot \frac{1}{2}CV_0^2

and solving:

I = V_0\sqrt{\frac{3C}{4L}} \approx 2.74\text{ A}

. That's B, and it's correct. Choice A makes a critical algebra error: it sets

U_L = 3 \cdot \frac{1}{2}CV_0^2

, treating the initial capacitor energy as the benchmark instead of the current capacitor energy. This ignores the constraint that both energies must sum to

U_{total}

, badly overcounting. Choice C computes the maximum possible current (when all energy is in the inductor), which is a useful reference but doesn't satisfy the stated condition — it's an upper bound, not the answer. Choice D assumes the capacitor's energy dropped to one-third of its initial value, which misreads the condition; the problem says the inductor energy is three times the current capacitor energy, not one-third of the original. The key strategy: whenever a problem says one energy is a multiple of the other, immediately write that relationship and the conservation equation simultaneously. Two unknowns, two equations — don't skip either step.

Question 3

An engineer stores energy in two inductors by connecting them in series to a current source delivering $I = 5\text{ A}$ . Inductor 1 has $L_1 = 3\text{ H}$ and Inductor 2 has $L_2 = 7\text{ H}$ with a mutual inductance $M = 2\text{ H}$ between them (the coils are wound so their fields add). What is the total magnetic energy stored in this coupled system?

$U = \frac{1}{2}(L_1 + L_2)I^2 = \frac{1}{2}(10)(25) = 125\text{ J}$ , because mutual inductance does not contribute to stored energy when the inductors are in series.
$U = \frac{1}{2}(L_1 + L_2 + M)I^2 = \frac{1}{2}(12)(25) = 150\text{ J}$ , because the mutual inductance term $M$ adds once to the effective inductance when fields are aiding.
$U = \frac{1}{2}(L_1 + L_2 + 2M)I^2 = \frac{1}{2}(14)(25) = 175\text{ J}$ , because for magnetically coupled series inductors with aiding fields, the effective inductance is $L_1 + L_2 + 2M$ . (correct answer)
$U = \frac{1}{2}L_1 I^2 + \frac{1}{2}L_2 I^2 + \frac{1}{2}MI^2 = 37.5 + 87.5 + 25 = 150\text{ J}$ , found by treating the mutual energy term symmetrically with the self-energy terms, each weighted by $\frac{1}{2}$ .

Explanation: When two inductors are magnetically coupled and carry the same current (series connection), you cannot treat them as isolated — their magnetic fields interact, and that interaction contributes real, physical energy to the system. The total energy stored is:

U = \frac{1}{2}L_1 I^2 + \frac{1}{2}L_2 I^2 \pm MI^2

where the

\pm

depends on whether the fields aid (+) or oppose (−) each other. Notice the mutual energy term carries no

\frac{1}{2}

— this is because the coupling energy between coil 1 and coil 2 is counted once from each coil's perspective, and when you derive it from first principles (integrating power), those two contributions combine to give a single

MI^2

. Factoring out, the effective inductance for aiding fields is

L_{eff} = L_1 + L_2 + 2M

, making the total energy

\frac{1}{2}(3 + 7 + 4)(25) = \frac{1}{2}(14)(25) = \mathbf{175\text{ J}}

, confirming C is correct. A is wrong because it ignores mutual inductance entirely — a critical error whenever the problem states the coils are magnetically coupled. B adds

M

only once, as if the coupling appears in just one coil's equation, but physically both coils experience the mutual flux. D makes the subtle mistake of writing the mutual energy as

\frac{1}{2}MI^2

, applying the

\frac{1}{2}

factor that belongs only to self-inductance terms — the mutual energy term is

MI^2

, not

\frac{1}{2}MI^2

. A reliable memory rule: self-energy terms get

\frac{1}{2}

, mutual energy terms do not. For aiding series inductors, always add

2M

to the inductances before computing energy.

Question 4

A student argues that because $U = \frac{1}{2}LI^2$ , an inductor with larger inductance always stores more energy than one with smaller inductance when both are in the same circuit.

Which of the following scenarios most directly refutes the student's claim by demonstrating a case where the inductor with smaller inductance stores more energy?

Two inductors $L_1 = 1\text{ H}$ and $L_2 = 4\text{ H}$ are connected in parallel directly across an ideal DC voltage source. Because ideal inductors have zero resistance, the current in each branch grows without bound and no finite steady-state energy comparison is possible.
Two inductors $L_1 = 1\text{ H}$ and $L_2 = 4\text{ H}$ are connected in series to a current source delivering $I = 5\text{ A}$ . Both carry the same current, so $U_1 = 12.5\text{ J}$ and $U_2 = 50\text{ J}$ , meaning the larger inductor stores more — this confirms, rather than refutes, the student's claim.
A single inductor $L_1 = 1\text{ H}$ in one independent circuit carries $I = 4\text{ A}$ and stores $U_1 = 8\text{ J}$ , while a separate inductor $L_2 = 4\text{ H}$ in a different circuit carries $I = 1\text{ A}$ and stores $U_2 = 2\text{ J}$ . Although the smaller inductor stores more here, the two inductors are not in the same circuit, so this does not address the student's claim.
Two inductors $L_1 = 1\text{ H}$ and $L_2 = 4\text{ H}$ are connected in parallel across a constant-current source of $I_{\text{total}} = 5\text{ A}$ . The current divides inversely with inductance, giving $I_1 = 4\text{ A}$ and $I_2 = 1\text{ A}$ , so $U_1 = 8\text{ J}$ and $U_2 = 2\text{ J}$ : the smaller inductor stores more. (correct answer)

Explanation: When evaluating a claim about inductors, always ask: what constraint does the circuit impose? The student assumes larger

L

always wins, but that ignores how current distributes depending on circuit configuration. The key insight in D is how inductors share current in a parallel configuration driven by a constant-current source. Unlike resistors, inductors in parallel divide current inversely proportional to their inductance (because equal voltage across both means

V = L\frac{dI}{dt}

, so the smaller inductor ramps up current faster). With

I_{\text{total}} = 5\text{ A}

, the split gives

I_1 = 4\text{ A}

(smaller

L_1 = 1\text{ H}

) and

I_2 = 1\text{ A}

(larger

L_2 = 4\text{ H}

). Plugging into

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

U_1 = \frac{1}{2}(1)(16) = 8\text{ J}

and

U_2 = \frac{1}{2}(4)(1) = 2\text{ J}

. The smaller inductor stores more energy — directly refuting the student's claim within the same circuit. A is a trap: it correctly identifies that ideal inductors in parallel across a voltage source produce unbounded current, but this means no valid comparison exists — it doesn't refute anything. B actually confirms the student's claim because series inductors share the same current, so larger

L

always wins. C presents a valid counterexample mathematically but fails the "same circuit" requirement the student's claim specifies, making it logically insufficient as a refutation. Your takeaway: whenever a claim says "always," look for the configuration that breaks the assumption. Parallel inductors with a current source are a classic case where current distribution flips the expected energy ranking.

Question 5

A superconducting toroidal inductor with self-inductance $L = 4 \text{ H}$ carries a steady current $I_0 = 3 \text{ A}$ . The superconducting loop is then broken by a tiny resistive segment with resistance $R = 100 \ \Omega$ , and the current decays exponentially with time constant $\tau = L/R$ .

How does the energy stored in the inductor at time $t = \tau$ compare to the initial stored energy $U_0$ ?

$U(\tau) = U_0 / e^2$ , because the current has decayed to $I_0/e$ and energy depends on the square of the current. (correct answer)
$U(\tau) = U_0 / e$ , because energy is proportional to current and the current has fallen by a factor of $e$ after one time constant.
$U(\tau) = U_0 (1 - 1/e)$ , because the energy dissipated in the resistor after one time constant equals a fraction $(1-1/e)$ of $U_0$ , leaving the remainder stored.
$U(\tau) = U_0 / (2e)$ , because one time constant corresponds to the half-life of the current decay, and the energy is halved relative to the $1/e$ reduction.

Explanation: When a question links exponential decay to stored energy, your first instinct should be to track the current carefully, then apply the energy formula — don't jump straight to the energy decay without thinking about the exponent. Here's the chain of reasoning: when the resistive segment is introduced, the current decays as

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

. At

t = \tau

, the current is

I(\tau) = I_0 e^{-1} = I_0/e

. Now apply the energy formula for an inductor:

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

. Substituting the decayed current gives

U(\tau) = \frac{1}{2}L\left(\frac{I_0}{e}\right)^2 = \frac{1}{2}LI_0^2 \cdot \frac{1}{e^2} = \frac{U_0}{e^2}

. The squaring of the exponential factor is the key step, confirming answer A. Answer B claims energy is proportional to current (not current squared), so it only divides by

e

once. This confuses the linear relationship between current and, say, flux, with the quadratic relationship between current and energy — a classic trap. Answer C describes the energy dissipated, not the energy remaining. It's true that roughly

(1 - 1/e)

of the energy has been lost by

t = \tau

, but the question asks what's still stored, which is

U_0/e^2

, not

U_0(1-1/e)

. Answer D introduces a false concept: one time constant is not equivalent to a half-life. The half-life would require

e^{-t/\tau} = 1/2

, which doesn't occur at

t = \tau

. Study tip: Whenever energy depends on a decaying quantity, always square the decay factor. Energy goes as

e^{-2t/\tau}

, not

e^{-t/\tau}

Question 6

A long solenoid of length $\ell$ , cross-sectional area $A$ , and $n$ turns per unit length carries current $I$ . The energy stored per unit volume in the magnetic field inside the solenoid is $u_B$ . If both the number of turns per unit length and the current are simultaneously doubled while $\ell$ and $A$ remain fixed, which of the following correctly describes the changes in total stored energy $U$ and energy density $u_B$ ?

Both $U$ and $u_B$ increase by a factor of 16, because the magnetic field doubles due to doubling $n$ and doubles again due to doubling $I$ , so $B$ increases by 4, and both energy and energy density scale as $B^2$ . (correct answer)
$U$ increases by a factor of 16 but $u_B$ increases by a factor of 4, because total energy depends on inductance (which scales as $n^2$ ) and $I^2$ , while energy density depends only on field strength squared.
Both $U$ and $u_B$ increase by a factor of 4, because the inductance doubles when $n$ doubles and the energy scales as $LI^2$ , giving an overall factor of $2 \times 2 = 4$ .
$U$ increases by a factor of 16 and $u_B$ also increases by a factor of 16, but the mechanism differs: $U$ grows because $L \propto n^2$ (factor of 4) and $I^2$ (factor of 4), while $u_B$ grows because $B = \mu_0 nI$ doubles twice so $B^2$ quadruples twice.

Explanation: Whenever you see a question about energy stored in a magnetic field, your two essential formulas are the magnetic field inside a solenoid,

B = \mu_0 n I

, and the energy density,

u_B = \frac{B^2}{2\mu_0}

. Everything else follows from these. When you double both

n

and

I

, the field becomes

B' = \mu_0 (2n)(2I) = 4\mu_0 nI = 4B

. Since energy density scales as

B^2

, you get

u_B' = \frac{(4B)^2}{2\mu_0} = 16\,u_B

— a factor of 16. For total stored energy, since the geometry (

\ell

and

A

) is unchanged, the volume

V = A\ell

is constant, so

U = u_B \cdot V

also scales by exactly the same factor of 16. Answer A is correct. Answer B claims

u_B

only increases by 4, confusing energy density (which depends purely on

B

) with inductance arguments — but

u_B = B^2/2\mu_0

has nothing to do with

L

directly. Answer C incorrectly states inductance merely doubles when

n

doubles; in fact

L = \mu_0 n^2 A \ell

, so doubling

n

quadruples

L

, and the full energy

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

gives

4 \times 4 = 16

, not 4. Answer D states the same final answer as A but invents a flawed "doubles twice" mechanism —

B

does not "double twice"; it quadruples in a single step because both

n

and

I

double simultaneously. A reliable strategy: always go back to

B = \mu_0 nI

first, compute the new

B

, then apply

u_B \propto B^2

. This single chain of reasoning handles both

u_B

and

U

without needing inductance formulas.

Question 7

Two inductors $L_1 = 2\text{ H}$ and $L_2 = 8\text{ H}$ are connected in series (no mutual inductance) and carry a common current of $I = 3\text{ A}$ .

A student claims: 'The inductor with the larger inductance stores more energy, so the ratio of energy stored in $L_2$ to energy stored in $L_1$ equals the ratio of their inductances, which is 4:1.' A second student counters: 'Because they are in series, the voltage across each inductor is proportional to its inductance, so the energy ratio must equal the square of the inductance ratio, giving 16:1.' Which student, if either, is correct, and what is the actual energy ratio $U_2 : U_1$ ?

The first student is correct: $U_2 : U_1 = 4:1$ , because $U = \frac{1}{2}LI^2$ and the current $I$ is the same for both, so the energy ratio equals the inductance ratio directly. (correct answer)
The second student is correct: $U_2 : U_1 = 16:1$ , because the voltage across each inductor is proportional to its inductance and energy is proportional to voltage squared times inductance.
Neither student is correct: $U_2 : U_1 = 2:1$ , because in a series circuit energy distributes in proportion to the square root of the inductance ratio.
Neither student is correct: $U_2 : U_1 = 1:4$ , because the larger inductor opposes current more strongly, reducing its contribution to the total stored energy relative to the smaller inductor.

Explanation: When inductors are connected in series with no mutual inductance, every element shares the same current at every instant. This is the critical insight that unlocks the entire question — start there whenever you see energy storage in series inductors. The energy stored in any inductor is

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

. Since both inductors carry the same current

I = 3\text{ A}

, you can write the ratio directly:

\frac{U_2}{U_1} = \frac{\frac{1}{2}L_2 I^2}{\frac{1}{2}L_1 I^2} = \frac{L_2}{L_1} = \frac{8}{2} = 4

U_2 : U_1 = 4:1

, confirming that answer A is correct. The first student's reasoning is sound. The second student (answer B) confuses two separate ideas. Yes, the voltage across each inductor is proportional to its inductance (

V = L\frac{dI}{dt}

), but energy stored in an inductor is not

\frac{1}{2}\frac{V^2}{L}

in the way resistive power is. That formula applies to resistors with steady-state voltages — not to inductors storing magnetic energy. Squaring the inductance ratio to get 16:1 has no physical basis here. Answer C is a fabricated rule — energy in series inductors does not scale with the square root of inductance. No such relationship exists. Answer D reverses the logic entirely; a larger inductance doesn't reduce stored energy when current is identical — it increases it. Study tip: In series circuits, current is always the equalizer. When a question asks you to compare energy in series inductors, immediately recognize that

I

cancels in the ratio, leaving only

L_2/L_1