Physics 2 Quiz: Rl Circuits Current Growth And Decay

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Rl Circuits Current Growth And DecayQuestion 1 of 10

A student measures the current in a series RL circuit at two specific times after the switch is closed: at $t_1 = 2\,\text{ms}$ the current is $I_1 = 0.432\,\text{A}$ , and at $t_2 = 6\,\text{ms}$ the current is $I_2 = 0.865\,\text{A}$ . The student knows the battery EMF is $\mathcal{E} = 1.00\,\text{V}$ and wishes to determine $L$ and $R$ from these measurements alone.

Using only the given data, which of the following correctly identifies the time constant $\tau$ of this circuit?

\tau = 4\,\text{ms}

, found by taking the difference

t_2 - t_1

and recognizing that the current nearly doubles over this interval, which corresponds to one time constant in exponential growth.

\tau \approx 2\,\text{ms}

, found by noting that

I_1 \approx 0.432\,\text{A}

is close to

\mathcal{E}(1-e^{-1})/R

, implying

t_1 \approx \tau

, and inferring

R

from the approximate steady-state limit of

I_2

\tau \approx 2.89\,\text{ms}

, found by exploiting the fact that

t_2 = 3t_1

to write

I_2/I_1 = 1 + x + x^2

where

x = e^{-t_1/\tau}

, solving for

x

, and then computing

\tau = -t_1/\ln x

\tau \approx 1.45\,\text{ms}

, found by noting that

I_2 \approx 0.865\,\text{A}

suggests the circuit is near steady state, estimating

I_f \approx I_2

, then using

I_1 = I_f(1-e^{-t_1/\tau})

to solve for

\tau

directly.

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Physics 2 Quiz

Physics 2 Quiz: Rl Circuits Current Growth And Decay

Practice Rl Circuits Current Growth And Decay 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 Rl Circuits Current Growth And Decay, 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

Using only the given data, which of the following correctly identifies the time constant $\tau$ of this circuit?

$\tau = 4\,\text{ms}$ , found by taking the difference $t_2 - t_1$ and recognizing that the current nearly doubles over this interval, which corresponds to one time constant in exponential growth.
$\tau \approx 2\,\text{ms}$ , found by noting that $I_1 \approx 0.432\,\text{A}$ is close to $\mathcal{E}(1-e^{-1})/R$ , implying $t_1 \approx \tau$ , and inferring $R$ from the approximate steady-state limit of $I_2$ .
$\tau \approx 2.89\,\text{ms}$ , found by exploiting the fact that $t_2 = 3t_1$ to write $I_2/I_1 = 1 + x + x^2$ where $x = e^{-t_1/\tau}$ , solving for $x$ , and then computing $\tau = -t_1/\ln x$ . (correct answer)
$\tau \approx 1.45\,\text{ms}$ , found by noting that $I_2 \approx 0.865\,\text{A}$ suggests the circuit is near steady state, estimating $I_f \approx I_2$ , then using $I_1 = I_f(1-e^{-t_1/\tau})$ to solve for $\tau$ directly.

Explanation: When you have two current measurements from a charging RL circuit and need to extract

\tau

, your tool is the equation

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

, where

I_f = \mathcal{E}/R

. The challenge is that you have two unknowns (

I_f

and

\tau

), so you need to eliminate one algebraically using both data points together. Notice that

t_2 = 6\,\text{ms} = 3 \times 2\,\text{ms} = 3t_1

. Let

x = e^{-t_1/\tau}

, so

e^{-t_2/\tau} = x^3

. Writing both current equations and dividing eliminates

I_f

\frac{I_2}{I_1} = \frac{1-x^3}{1-x} = 1 + x + x^2

. Plugging in

I_2/I_1 = 0.865/0.432 \approx 2.002

gives

x^2 + x - 1.002 \approx 0

, yielding

x \approx 0.501

. Then

\tau = -t_1/\ln(0.501) \approx 2/(0.691) \approx 2.89\,\text{ms}

. This is exactly what C describes — a clean algebraic elimination that uses the special ratio

t_2/t_1 = 3

. A is wrong because "doubling over one time constant" is not a property of exponential growth toward a limit — it misapplies intuition from pure exponential decay. B makes a circular assumption: you cannot infer

\tau \approx t_1

simply because

I_1

looks like it might be near

63\%

of steady state without first knowing

I_f

, which itself requires

R

. D commits the same error more explicitly — it assumes

I_2 \approx I_f

, but at roughly

t \approx 2\tau

the current is only about

86\%

of steady state, so this approximation introduces significant error. Your strategy: when two measurements share a clean integer time ratio, use substitution to eliminate unknowns algebraically rather than approximating steady-state values you haven't verified.

Question 2

In a series RL circuit connected to a battery, the current as a function of time during growth is $I(t) = I_f\left(1 - e^{-t/\tau}\right)$ . A student argues that the voltage across the inductor at $t = 0$ must be zero because no current is flowing and $V_L = L\,dI/dt$ . Which of the following best identifies the flaw in this reasoning?

The student is correct that $V_L = 0$ at $t = 0$ ; the flaw is in the formula — at $t = 0$ the correct expression is $V_L = \mathcal{E} - I_f R$ , which evaluates to zero since $I = 0$ makes $I_f R = \mathcal{E}$ .
The student conflates the instantaneous current with the rate of change of current. At $t = 0$ , $I = 0$ but $dI/dt = I_f/\tau = \mathcal{E}/L \neq 0$ , so $V_L = L(\mathcal{E}/L) = \mathcal{E}$ , not zero. (correct answer)
The student's error is dimensional: the formula $V_L = L\,dI/dt$ applies only after the current has been established, so a separate initial-condition formula $V_L = \mathcal{E}$ must be invoked for $t = 0$ .
The student confuses the inductor with a capacitor. For a capacitor, zero current implies zero voltage; for an inductor, zero current implies maximum voltage, which equals $\mathcal{E}/2$ at $t = 0$ due to energy partitioning.

Explanation: Whenever you see a question about inductors at the moment a circuit is switched on, your first instinct should be to think carefully about what the inductor formula actually depends on — and that's rate of change, not current itself. The formula

V_L = L\,\frac{dI}{dt}

is always valid, including at

t = 0

. The student's mistake is assuming that because

I(0) = 0

, the derivative must also be zero. These are independent quantities. Differentiating the growth equation gives

\frac{dI}{dt} = \frac{I_f}{\tau}e^{-t/\tau}

. At

t = 0

, this is at its maximum:

\frac{dI}{dt}\big|_{t=0} = \frac{I_f}{\tau}

. Since

I_f = \mathcal{E}/R

and

\tau = L/R

, we get

\frac{dI}{dt}\big|_{t=0} = \frac{\mathcal{E}}{L}

, so

V_L = L \cdot \frac{\mathcal{E}}{L} = \mathcal{E}

. At the instant of switch-on, the inductor carries the full EMF — consistent with Kirchhoff's voltage law, since

IR = 0

when

I = 0

. Choice B captures this exactly. A is wrong because

V_L

t=0

is not zero, and the claim that

I_f R = \mathcal{E}

only holds at steady state (

t \to \infty

), not at

t = 0

. C invents a fictional "separate formula" for initial conditions —

V_L = L\,dI/dt

is universal and requires no replacement. D is a false analogy; for an inductor, zero current means maximum rate of change, not maximum voltage from some energy-partitioning rule, and

\mathcal{E}/2

is simply incorrect. Remember: for inductors, current cannot change instantaneously, but voltage can be anything. Confusing

I

with

dI/dt

is the classic trap here.

Question 3

An RL circuit consists of a resistor $R = 40\,\Omega$ , an inductor $L = 200\,\text{mH}$ , and an ideal battery of EMF $\mathcal{E} = 12\,\text{V}$ . The switch is closed at $t = 0$ .

At the instant when the current in the circuit equals exactly half its final steady-state value, what is the rate of change of current, $dI/dt$ ?

$30\,\text{A/s}$ , because at half the steady-state current the resistor drops half the battery voltage, leaving a net EMF of 6 V across the inductor, and $dI/dt = (\mathcal{E} - IR)/L = 6/0.2$ . (correct answer)
$60\,\text{A/s}$ , because at half the steady-state current the inductor's back-EMF equals half the battery EMF, so the full battery voltage divided by the inductance gives the rate of change.
$15\,\text{A/s}$ , because the rate of change at any moment is the initial rate scaled by the fraction of steady-state current not yet reached, divided by two.
$120\,\text{A/s}$ , because at half the steady-state current the inductor stores half its final energy, so the EMF available to change the current is doubled relative to the final state.

Explanation: When analyzing an RL circuit at any specific moment, your go-to equation is Kirchhoff's voltage law applied instantaneously:

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

, which rearranges to

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

. This single equation tells you everything about the current's rate of change at any snapshot in time. The steady-state current is

I_f = \mathcal{E}/R = 12/40 = 0.3\,\text{A}

. At half that value,

I = 0.15\,\text{A}

. Plugging into the formula:

\frac{dI}{dt} = \frac{12 - (0.15)(40)}{0.2} = \frac{12 - 6}{0.2} = \frac{6}{0.2} = 30\,\text{A/s}

. That confirms A is correct — the resistor drops 6 V, leaving 6 V across the inductor to drive further change. Choice B claims the full 12 V drives the rate of change, confusing the initial condition (where

I = 0

and the full EMF appears across the inductor) with the halfway point. Choice C invents a "divide by two" scaling rule that has no basis in the governing equation —

dI/dt

does scale with remaining EMF, but the factor isn't simply halved again. Choice D brings in energy stored in the inductor (

\frac{1}{2}LI^2

), which is irrelevant here; energy storage doesn't amplify the available EMF — it's a completely separate quantity. A useful habit: whenever an RL question asks about a specific moment, write

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

immediately and substitute the given current. That one equation resolves nearly every instantaneous-rate question you'll encounter.

Question 4

A series RL circuit has $R = 20\,\Omega$ , $L = 100\,\text{mH}$ , and is driven by a battery $\mathcal{E} = 10\,\text{V}$ . After reaching steady state, the battery is disconnected and the inductor is simultaneously connected across a second resistor $R_2 = 30\,\Omega$ , forming a new closed loop containing only $L$ and $R_2$ .

How does the time constant for the decay phase compare to the time constant for the growth phase, and what is the initial current at the start of the decay?

The decay time constant is larger than the growth time constant ( $\tau_{\text{decay}} = L/R_2 > L/R$ ), and the initial decay current equals $\mathcal{E}/R = 0.5\,\text{A}$ .
The decay time constant is smaller than the growth time constant ( $\tau_{\text{decay}} = L/R_2 < L/R$ ), and the initial decay current equals $\mathcal{E}/R = 0.5\,\text{A}$ . (correct answer)
The decay time constant is smaller than the growth time constant ( $\tau_{\text{decay}} = L/R_2 < L/R$ ), and the initial decay current equals $\mathcal{E}/(R + R_2) = 0.2\,\text{A}$ because both resistors are momentarily in series at the instant of switching.
The decay time constant equals the growth time constant (both equal $L/(R + R_2)$ ), and the initial decay current equals $\mathcal{E}/R = 0.5\,\text{A}$ because the inductor enforces current continuity.

Explanation: When a circuit switches configurations, you need to separately analyze each phase using only the components active in that phase — a common source of confusion on RL circuit problems. During the growth phase, the battery,

R

, and

L

form a series loop. The time constant is

\tau_{\text{growth}} = L/R = 0.1/20 = 5\,\text{ms}

, and current grows toward a steady-state value of

\mathcal{E}/R = 10/20 = 0.5\,\text{A}

. At the moment of switching, the inductor enforces current continuity — it cannot change its current instantaneously. So the current at the start of the decay phase is exactly the steady-state current from the growth phase:

I_0 = 0.5\,\text{A}

. The battery and

R

are now disconnected; the only components in the new loop are

L

and

R_2

. Therefore

\tau_{\text{decay}} = L/R_2 = 0.1/30 \approx 3.33\,\text{ms}

. Since

R_2 > R

, the decay time constant is smaller than the growth time constant — the current collapses faster than it rose. This confirms answer B. Answer A gets the initial current right but reverses the comparison, incorrectly claiming

R_2 > R

produces a larger time constant. Answer C treats both resistors as momentarily in series at switching, which would only apply if both remained in the circuit — they don't. The battery and

R

are disconnected. Answer D wrongly applies

L/(R+R_2)

to both phases, ignoring that each phase has a distinct circuit topology. As a general strategy: always redraw the circuit for each phase and identify only the active components. The inductor's current is continuous across switching, but the time constant is not.

Question 5

In a series RL circuit driven by a battery of EMF $\mathcal{E}$ , resistance $R$ , and inductance $L$ , the energy stored in the inductor at time $t$ is $U_L(t) = \frac{1}{2}L I^2(t)$ . At what time $t^*$ does the energy stored in the inductor equal exactly half of its maximum (steady-state) stored energy?

$t^* = \frac{\tau}{2}$ , because energy grows linearly in time during the initial transient and reaches half its maximum value at half the time constant.
$t^* = \tau \ln 2$ , because the current reaches half its maximum at $t = \tau \ln 2$ , and since energy is proportional to current, the energy also reaches half its maximum at the same time.
$t^* = \tau \ln\!\left(\frac{1}{1 - 1/\sqrt{2}}\right)$ , because the energy is proportional to current squared, so the current must reach $1/\sqrt{2}$ of its maximum for the energy to be at half maximum, giving this logarithmic expression. (correct answer)
$t^* = \frac{\tau}{2}\ln 2$ , because the energy is proportional to the square of the current, which introduces a factor of 2 in the exponent that halves the time constant governing energy growth.

Explanation: When analyzing energy storage in an RL circuit, remember that you must work with the right quantity — energy, not current — when setting up your equation. The current in a series RL circuit grows as

I(t) = I_{max}(1 - e^{-t/\tau})

, where

\tau = L/R

and

I_{max} = \mathcal{E}/R

. Since

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

, the maximum energy is

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

. Setting

U_L(t^*) = \frac{1}{2}U_{max}

gives

\frac{1}{2}LI^2(t^*) = \frac{1}{2}\cdot\frac{1}{2}LI_{max}^2

, which simplifies to

I(t^*) = \frac{I_{max}}{\sqrt{2}}

. Substituting the current expression:

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

, so

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

, giving

t^* = \tau\ln\!\left(\frac{1}{1 - 1/\sqrt{2}}\right)

. That confirms C is correct. Choice A is wrong because current — and therefore energy — grows exponentially, not linearly. Halving the time does not halve the energy. Choice B confuses the condition for half-current with half-energy. Finding when

I = \frac{1}{2}I_{max}

gives

t = \tau\ln 2

, but half-current means only one-quarter maximum energy (since

U \propto I^2

). Choice D invents a nonexistent rule; there is no factor of 2 that simply halves the time constant for energy growth. The key study tip: whenever a question shifts from current to energy in an inductor or capacitor, remember to square the relevant quantity before setting up your equation. That squaring step changes the threshold you're solving for and is the most common trap on these circuit-energy problems.

Question 6

A series RL circuit (resistance $R$ , inductance $L$ , ideal battery EMF $\mathcal{E}$ ) reaches steady state. The battery is then removed and replaced by a short circuit, allowing the current to decay. Which of the following statements about the voltage across the resistor and the voltage across the inductor during decay is correct?

The voltage across the resistor decays exponentially from $\mathcal{E}$ to zero, while the voltage across the inductor remains zero throughout because the inductor behaves as a short circuit once current is established.
The resistor voltage decays from $\mathcal{E}$ and the inductor voltage grows from zero, with their sum always equaling $\mathcal{E}$ , because the total EMF in the circuit is conserved throughout the decay.
The resistor voltage decays exponentially from $\mathcal{E}$ toward zero, while the inductor voltage is always zero because the inductor's back-EMF exactly cancels its own forward-EMF, leaving no net voltage across it.
Both the resistor voltage and the inductor voltage decay exponentially from $\mathcal{E}$ to zero, but they are always equal in magnitude and opposite in sign, summing to zero at every instant in agreement with KVL around the closed loop. (correct answer)

Explanation: When analyzing an RL decay circuit, your most powerful tool is Kirchhoff's Voltage Law (KVL). Once the battery is removed and replaced by a short, the only two elements in the closed loop are the resistor and inductor. KVL demands their voltages sum to zero at every instant. During decay, the current follows

i(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}

, where

\tau = L/R

. The resistor voltage is simply

V_R = iR = \mathcal{E}\,e^{-t/\tau}

, which starts at

\mathcal{E}

and decays exponentially to zero. Now apply KVL:

V_R + V_L = 0

, so

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

. The inductor voltage has the same exponential magnitude but opposite sign — it's the back-EMF driving the current against the resistor's dissipation. Both decay from

\mathcal{E}

in magnitude, and their sum is zero at every moment. That's exactly what D describes. A is wrong because the inductor is not a short circuit during decay — it actively maintains current by generating a back-EMF. B contains two errors: the inductor voltage doesn't grow from zero, and their sum must equal zero (not

\mathcal{E}

), since the battery is gone. C invents a fictional "cancellation" between forward and back-EMF within the inductor itself — no such mechanism exists; the inductor simply has a real, nonzero terminal voltage equal to

L\,di/dt

. Your study tip: in any RL or RC decay problem, remove the source first, then write KVL with only the passive elements. The voltages must sum to zero — this alone eliminates B and confirms D immediately.

Question 7

A series RL circuit with resistance $R$ and inductance $L$ is connected to a battery of EMF $\mathcal{E}$ at $t=0$ . A student claims: "Doubling both $R$ and $L$ simultaneously leaves the time constant unchanged and also leaves the steady-state current unchanged." Which of the following correctly evaluates this claim?

The claim is entirely correct: the time constant $\tau = L/R$ is unchanged because both numerator and denominator double, and the steady-state current $I_f = \mathcal{E}/R$ is unchanged because $R$ does not change.
The claim is half correct: the time constant $\tau = L/R$ is indeed unchanged because both $L$ and $R$ double, but the steady-state current $I_f = \mathcal{E}/R$ is halved because only $R$ determines it and $R$ has doubled. (correct answer)
The claim is entirely incorrect: the time constant doubles because $L$ doubles while $R$ remains the dominant factor in transient behavior, and the steady-state current is halved because $R$ has doubled.
The claim is half correct: the steady-state current is indeed unchanged because the ratio $L/R$ is preserved, which also governs the DC response, but the time constant doubles because inductance dominates the transient.

Explanation: When analyzing a series RL circuit, you need to track two independent quantities separately: the time constant

\tau = L/R

and the steady-state current

I_f = \mathcal{E}/R

. The key insight is that these formulas share

R

but depend on it differently. If you double both

L

and

R

, the time constant becomes

\tau' = (2L)/(2R) = L/R

, so

\tau

is indeed unchanged — the student's first claim holds. However, the steady-state current becomes

I_f' = \mathcal{E}/(2R)

, which is exactly half the original value. The steady-state current has no dependence on

L

at all — inductors are just wires in DC steady state — so doubling

L

does nothing to

I_f

, while doubling

R

cuts it in half. This makes B correct. Answer A is the trap most students fall into. It correctly identifies that

\tau

is unchanged but then incorrectly claims

R

"does not change" —

R$ absolutely doubled; the student is confusing the *ratio*

L/R

being preserved with

R$$ itself being preserved. Answer C is doubly wrong: it invents a false rule that "inductance dominates" the time constant and incorrectly says

\tau

doubles, while it does get

I_f

halved correctly. Answer D reverses the errors of A — it correctly identifies

I_f

behavior for the wrong reason (claiming

L/R

governs DC response, which it does not) and incorrectly doubles

\tau

. Study tip: Always evaluate

\tau = L/R

and

I_f = \mathcal{E}/R

as completely separate formulas — changes to

L

affect only

\tau

, while changes to

R

affect both.

Question 8

Two RL circuits share the same ideal battery (EMF $\mathcal{E}$ ). Circuit 1 has resistance $R_1 = 10\,\Omega$ and inductance $L_1 = 50\,\text{mH}$ . Circuit 2 has resistance $R_2 = 100\,\Omega$ and inductance $L_2 = 50\,\text{mH}$ . Both switches are closed simultaneously at $t = 0$ .

Which circuit reaches 90% of its steady-state current first, and approximately how much sooner does it do so compared to the other circuit?

Circuit 2 reaches 90% first, approximately $0.46\,\text{ms}$ sooner than Circuit 1, because its larger resistance gives a smaller time constant $\tau_2 = L/R_2 \ll \tau_1$ .
Circuit 1 reaches 90% first, approximately $0.46\,\text{ms}$ sooner than Circuit 2, because its smaller resistance allows current to build more quickly in absolute terms, even though its time constant is larger.
Both circuits reach 90% of their respective steady-state currents at exactly the same time, because they share the same inductance $L$ and the time to reach any given fraction of steady-state current depends only on $L$ .
Circuit 2 reaches 90% first, approximately $10.3\,\text{ms}$ sooner than Circuit 1, because its larger resistance gives a smaller time constant, and the time to reach any fixed fraction of steady-state current scales directly with $\tau$ . (correct answer)

Explanation: When analyzing RL circuits, the key concept is the time constant

\tau = L/R

, which controls how quickly current builds toward its steady-state value. The current follows

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

, and the time to reach any fixed fraction of steady-state current depends entirely on

\tau

— not on

R

\mathcal{E}

individually. To find when a circuit hits 90% of steady-state, set

1 - e^{-t/\tau} = 0.90

, which gives

t_{90} = \tau \ln(10) \approx 2.303\tau

. For Circuit 1:

\tau_1 = 50\,\text{mH}/10\,\Omega = 5\,\text{ms}

, so

t_{90,1} \approx 2.303 \times 5 = 11.5\,\text{ms}

. For Circuit 2:

\tau_2 = 50\,\text{mH}/100\,\Omega = 0.5\,\text{ms}

, so

t_{90,2} \approx 2.303 \times 0.5 = 1.15\,\text{ms}

. Circuit 2 reaches 90% first, and the difference is

11.5 - 1.15 \approx 10.3\,\text{ms}

, confirming D is correct. Choice A correctly identifies that Circuit 2 wins due to its smaller

\tau

, but the time difference of

0.46\,\text{ms}

is badly miscalculated — likely from forgetting to multiply by

\ln(10)

. Choice B is backwards: smaller resistance means slower approach to steady state (larger

\tau

), even though the final current is larger. Choice C is a classic trap — same

L

does not mean same timing;

\tau

depends on both

L

and

R

. Study tip: Memorize

t_{90} \approx 2.303\tau

. Any question asking about "time to reach X% of steady-state" is really asking you to compare time constants — calculate

\tau

first, always.

Question 9

In a series RL circuit, the switch is closed at $t = 0$ , connecting a battery (EMF $\mathcal{E}$ , internal resistance $r$ ) to an external resistor $R$ and inductor $L$ in series. After the circuit reaches steady state, the EMF source alone is suddenly removed and replaced by a wire at time $t = t_1$ , so that the internal resistance $r$ remains in the loop along with $R$ and $L$ .

Immediately after this change is applied at $t = t_1$ , which expression correctly gives the initial rate of current decay, $|dI/dt|$ , in the now-decaying circuit?

$\dfrac{\mathcal{E}}{L}$ , because the inductor maintains the steady-state current and the only EMF driving change is the original battery voltage divided by the inductance, independent of resistance. (correct answer)
$\dfrac{\mathcal{E}(R+r)}{RL}$ , because the total resistance in the loop at the moment of switching is $R+r$ , and this must be multiplied by the steady-state current and divided by $L$ .
$\dfrac{\mathcal{E}}{(R+r)L}$ , because the decay rate equals the battery EMF divided by the product of total resistance and inductance.
$\dfrac{\mathcal{E}}{rL}$ , because after the EMF is removed, only the internal resistance $r$ remains active in driving the transient, making it the sole factor alongside $L$ .

Explanation: When analyzing RL circuit switching problems, your first instinct should be to identify what quantities are continuous across the switching moment and what the governing differential equation looks like in the new configuration. The key physics principle here is that inductor current cannot change instantaneously. At steady state (just before

t = t_1

), the current is

I_0 = \mathcal{E}/(R+r)

, and this same current flows immediately after the switch. Once the EMF is removed and replaced by a wire, Kirchhoff's voltage law around the loop gives

L\,dI/dt + (R+r)I = 0

, so the rate of change is:

\left|\frac{dI}{dt}\right| = \frac{(R+r)}{L} \cdot I_0 = \frac{(R+r)}{L} \cdot \frac{\mathcal{E}}{R+r} = \frac{\mathcal{E}}{L}

The

(R+r)

factors cancel cleanly, making A correct — though for a subtle reason that's easy to miss. Choice B almost gets there but stops one step short: it correctly identifies

(R+r)I_0/L

but then substitutes

I_0 = \mathcal{E}/R

instead of

\mathcal{E}/(R+r)

, yielding a wrong expression with an extra

(R+r)/R

factor. Choice C incorrectly divides

\mathcal{E}

by both

(R+r)

and

L

without accounting for the current — it confuses the formula structure entirely. Choice D incorrectly assumes only

r

remains in the circuit after removal of the source, forgetting that

R

is still in series. Study tip: Whenever a source is switched out of an RL circuit, always write

|dI/dt| = (R_{\text{total}}/L) \cdot I_0

first, then substitute

I_0

— the cancellations often surprise you and reveal a simpler final form.

Question 10

An inductor of inductance $L$ and a resistor of resistance $R$ are connected in series with a switch and ideal battery of EMF $\mathcal{E}$ . The switch is closed at $t = 0$ . At time $t = \tau$ (one time constant), the power dissipated in the resistor is closest to which of the following?

$\frac{\mathcal{E}^2}{R}\left(1 - e^{-1}\right)^2 \approx 0.400\,\frac{\mathcal{E}^2}{R}$ , because the current at one time constant is $I_f(1-e^{-1})$ and power scales as current squared. (correct answer)
$\frac{\mathcal{E}^2}{R}\left(1 - e^{-2}\right) \approx 0.865\,\frac{\mathcal{E}^2}{R}$ , because the power involves the square of the exponential factor, which transforms the argument from $-1$ to $-2$ .
$\frac{\mathcal{E}^2}{R}\left(1 - e^{-1}\right) \approx 0.632\,\frac{\mathcal{E}^2}{R}$ , because at one time constant the current reaches $63.2\%$ of its final value, and power is proportional to current, not current squared.
$\frac{\mathcal{E}^2}{2R} \approx 0.500\,\frac{\mathcal{E}^2}{R}$ , because at one time constant the inductor has stored half its maximum energy, leaving the other half to be dissipated in the resistor at that instant.

Explanation: When analyzing an RL circuit at a specific moment in time, always anchor your reasoning to the current equation first, then derive other quantities like power from it — never work backwards from energy or approximations. For a series RL circuit, the current grows as

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

, where

\tau = L/R

. At

t = \tau

, this gives

I(\tau) = \frac{\mathcal{E}}{R}(1 - e^{-1})

. Power dissipated in the resistor is

P = I^2 R

, so you square the current and multiply by

R

P = \left[\frac{\mathcal{E}}{R}(1-e^{-1})\right]^2 R = \frac{\mathcal{E}^2}{R}(1-e^{-1})^2 \approx 0.400\,\frac{\mathcal{E}^2}{R}

This confirms A is correct. B is a subtle algebra error — squaring

(1 - e^{-1})^2

does not equal

(1 - e^{-2})

. Those are genuinely different numbers (≈0.400 vs. ≈0.865), and conflating them is a common algebraic trap. C applies the wrong power law entirely. Power scales as

I^2

, not

I

, so you cannot simply say "63.2% of maximum current means 63.2% of maximum power." D confuses instantaneous power with energy storage. The inductor's stored energy at

t = \tau

tells you nothing directly about the resistor's instantaneous dissipation rate at that moment — these are fundamentally different quantities. Study tip: On circuit problems, write down

I(t)

first, then calculate

P = I^2R

explicitly. Never shortcut by assuming power and current share the same proportionality — they don't.