RL Circuits: Current Growth & Decay - Physics 2

1. 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?
   Answer: $t^* = \tau \ln\!\left(\frac{1}{1 - 1/\sqrt{2}}\right)$, 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. 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.
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?
   Answer: 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. 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 $V_L$ at $t=0$ 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.
3. 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?
   Answer: $\frac{\mathcal{E}^2}{R}\left(1 - e^{-1}\right)^2 \approx 0.400\,\frac{\mathcal{E}^2}{R}$, because the current at one time constant is $I_f(1-e^{-1})$ and power scales as current squared. 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.
4. Using only the given data, which of the following correctly identifies the time constant $\tau$ of this circuit?
   Answer: $\tau \approx 2.89\,\text{ms}$, 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 $\tau = -t_1/\ln x$. 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.
5. Which circuit reaches 90% of its steady-state current first, and approximately how much sooner does it do so compared to the other circuit?
   Answer: 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 $\tau$. 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 $R$ or $\mathcal{E}$ 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.
6. 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?
   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 $\mathcal{E}/R = 0.5\,\text{A}$. 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.
7. 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?
   Answer: $\dfrac{\mathcal{E}}{L}$, 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. 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.
8. 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, $dI/dt$?
   Answer: $30\,\text{A/s}$, 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 $dI/dt = (\mathcal{E} - IR)/L = 6/0.2$. 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.
9. 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?
   Answer: 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. 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.
10. 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?
    Answer: 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. 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.

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