Interpreting RC/RL Circuit Graphs - Physics 2

1. From the graph described, a student extracts the time constant $\tau = t_1$ and the initial current $I_0$. The student then claims: *'The resistance in the circuit is $R = \mathcal{E}/I_0$ and the capacitance is $C = \tau/R$.'* Which of the following best evaluates this claim?
   Answer: Both expressions are correct: $I_0 = \mathcal{E}/R$ follows from Kirchhoff's voltage law at $t = 0$ when the capacitor voltage is zero, and $C = \tau/R$ follows from the definition $\tau = RC$, so both relations are valid. When analyzing RC charging circuits, your anchor should be Kirchhoff's Voltage Law (KVL) applied at two key moments: $t = 0$ and as $t \to \infty$. This question tests whether you understand both the physical meaning of initial conditions and the definition of the RC time constant. At $t = 0$, the capacitor is completely uncharged, so it contributes zero voltage — it behaves like a plain wire. KVL gives $\mathcal{E} = I_0 R$, which immediately yields $I_0 = \mathcal{E}/R$, or equivalently $R = \mathcal{E}/I_0$. The student's first expression is correct. For the time constant, the current in an RC circuit follows $I(t) = I_0 e^{-t/\tau}$ where $\tau = RC$. The graph confirms $\tau = t_1$ (the time to reach 37%, i.e., $1/e$, of $I_0$), so solving gives $C = \tau/R$ — the student's second expression is also correct. Answer A captures both of these valid relationships precisely. B is wrong because it invents a problem that doesn't exist. Nothing in the problem introduces internal resistance, and the formula $R = \mathcal{E}/I_0$ is perfectly valid for the given ideal circuit. The capacitor at $t = 0\$ does *not* make current infinite — the resistor$R$is still present and limits current. **C** is wrong because the exponential decay$$e^{-t/RC}$$governs *both* the current *and* the voltage in an RC circuit (current decays, voltage rises). The time constant$\tau = RC$appears in both expressions — it is not exclusive to RL circuits or to voltage alone. **D** is a red herring. Using consistent SI units is always assumed in physics problems unless told otherwise; it's not a flaw in the student's reasoning. **Study tip:** Memorize the two boundary conditions for RC circuits — at$t = 0$, the capacitor is a wire (zero voltage); as$t \to \infty$$, it's an open circuit (zero current). These two snapshots solve most RC problems quickly.
2. If the student fits the mistakenly labeled graph to the function $V_0 e^{-t/\tau_{\text{reported}}}$ and extracts $\tau_{\text{reported}}$, how does $\tau_{\text{reported}}$ compare to the true time constant $\tau = RC$, and what is $V_0$?
   Answer: $\tau_{\text{reported}} = RC$ and $V_0 = \mathcal{E}$, because $V_R(t) = \mathcal{E}\,e^{-t/RC}$ is itself a decaying exponential with the same time constant as $V_C(t)$; fitting it to $V_0 e^{-t/\tau}$ correctly recovers $\tau = RC$ and initial amplitude $\mathcal{E}$. When analyzing RC circuits, always start from Kirchhoff's voltage law: the EMF must equal the sum of voltages across each element at every moment. This means $V_R(t) + V_C(t) = \mathcal{E}$ at all times. Since the capacitor starts uncharged, $V_C(t) = \mathcal{E}(1 - e^{-t/RC})$. Substituting into Kirchhoff's law gives $V_R(t) = \mathcal{E} - \mathcal{E}(1 - e^{-t/RC}) = \mathcal{E}\,e^{-t/RC}$. This is already a perfect decaying exponential with initial value $\mathcal{E}$ and time constant $RC$. So when the student fits the resistor voltage data to $V_0 e^{-t/\tau_{\text{reported}}}$, the fit returns exactly $\tau_{\text{reported}} = RC$ and $V_0 = \mathcal{E}$. The label on the graph is wrong, but the extracted parameters are mathematically correct — making A the right answer. Choice B is wrong because no "doubling" occurs. $V_R(t)$ is a single clean exponential, not a sum of two separate exponentials requiring a combined fit. Choice C is wrong on two counts: the capacitor is uncharged at $t = 0$, so $V_R(0) = \mathcal{E}$, not $\mathcal{E}/2$. There is no half-charging condition at the start. Choice D is wrong because energy dissipation rate does not alter the circuit's time constant — $\tau = RC$ is fixed by the circuit topology, regardless of power considerations. A useful habit: always write out $V_R(t) = \mathcal{E} - V_C(t)$ explicitly before drawing conclusions. Many RC circuit traps disappear once you see the full exponential form of each voltage.
3. Which of the following correctly interprets both the slope $m$ and y-intercept $b$ of this semi-log plot, and what does a non-zero y-intercept physically indicate?
   Answer: The slope $m = -1/(RC)$ and the y-intercept $b = \ln(V_{C,0}/\mathcal{E})$, where $V_{C,0}$ is the initial capacitor voltage; a nonzero $b$ indicates the capacitor's initial voltage differed from $\mathcal{E}$, with $V_{C,0} = \mathcal{E}\,e^b$. When you see a semi-log plot of a decaying exponential, your first move should be to linearize the governing equation and match its form to $y = mt + b$. For RC discharge, the voltage follows $V_C(t) = V_{C,0}\,e^{-t/RC}$. Dividing by the reference voltage $\mathcal{E}$ and taking the natural log gives: $\ln\!\left(\frac{V_C(t)}{\mathcal{E}}\right) = \ln\!\left(\frac{V_{C,0}}{\mathcal{E}}\right) - \frac{1}{RC}\,t$ This is exactly $y = b + mt$, where the slope is $m = -1/(RC)$ and the y-intercept is $b = \ln(V_{C,0}/\mathcal{E})$. A nonzero $b$ simply means $V_{C,0} \neq \mathcal{E}$—the capacitor wasn't charged to exactly $\mathcal{E}$—and you can recover the actual initial voltage as $V_{C,0} = \mathcal{E}\,e^b$. This makes C correct. A is almost right on the slope but incorrectly assumes $b = 0$ as the default, then contradicts itself—if $b=0$ requires full charging, a nonzero $b$ can't simultaneously be the definition of the intercept. B gets the slope completely wrong: $m = -RC$ has units of $\Omega \cdot F = \text{s}^2/\text{s} \cdot \text{s}$, which is dimensionally inconsistent as a rate. The correct slope must carry units of $\text{s}^{-1}$. D invents an incorrect time constant $\tau = R/C$ (it should be $\tau = RC$) and fabricates a required intercept of $-1$, which has no physical basis from the linearized equation. Study tip: On semi-log problems, always linearize first, then read off slope and intercept by direct comparison—never guess the form.
4. Based solely on the graph information provided, what is $V_C(\tau)$, and which of the following expressions correctly generalizes the result for arbitrary initial and final voltages?
   Answer: $V_C(\tau) \approx 5.79\,\text{V}$, because the general formula is $V_C(\tau) = V_i + (V_f - V_i)(1 - e^{-1})$, which gives $2 + 6(0.632) \approx 2 + 3.79 = 5.79\,\text{V}$; at one time constant the capacitor has traversed 63.2% of the gap between its initial and final values. Whenever an RC circuit problem gives you both an initial and final voltage, your first instinct should be to use the general charging equation, not a simplified version that assumes the capacitor starts from zero. The complete formula for capacitor voltage at any time is $V_C(t) = V_f + (V_i - V_f)e^{-t/\tau}$. At exactly one time constant ($t = \tau$), the exponential term becomes $e^{-1} \approx 0.368$, so the gap remaining between the capacitor and its final value shrinks to 36.8% of the original gap — meaning the capacitor has closed 63.2% of that gap. Plugging in $V_i = 2\,\text{V}$ and $V_f = 8\,\text{V}$: $V_C(\tau) = 2 + (8 - 2)(1 - e^{-1}) \approx 2 + 6(0.632) \approx 5.79\,\text{V}$. This confirms D is correct. Choice A applies the 63.2% rule directly to the final voltage ($0.632 \times 8$), which only works when $V_i = 0$. Since the capacitor starts at 2 V, this ignores the initial condition entirely. Choice B makes the same conceptual error — it computes $8 - 8/e$, again treating the initial voltage as zero — and compounds it by claiming the answer is 6.59 V while actually calculating 5.06 V, making it doubly wrong. Choice C confuses the time constant with a geometric midpoint; exponential functions are not linear, and the halfway point occurs at a different time than $\tau$. The key strategy: always apply the full formula $V_C(\tau) = V_i + (V_f - V_i)(1 - e^{-1})$. The 63.2% figure describes the fraction of the gap traveled, never a fraction of $V_f$ alone.
5. Is the student's claim about the slope at $t = \tau$ correct, and what is the actual slope of $I(t)$ at $t = \tau$?
   Answer: The student is incorrect: the slope at $t = \tau$ is $I_{\max}/(e\tau) \approx 0.368\,I_{\max}/\tau$, because $dI/dt = (I_{\max}/\tau)e^{-t/\tau}$, which at $t = \tau$ gives $(I_{\max}/\tau)e^{-1}$, smaller than the initial slope by a factor of $e$. When analyzing RL circuit behavior, always go back to the mathematical definition of the current function rather than relying on intuition about geometric properties. The current in an RL circuit follows $I(t) = I_{\max}(1 - e^{-t/\tau})$. To find the slope at any moment, take the derivative: $\frac{dI}{dt} = \frac{I_{\max}}{\tau}e^{-t/\tau}$. At $t = 0$, this gives the initial slope $I_{\max}/\tau$. At $t = \tau$, substituting directly yields $\frac{dI}{dt}\big|_{t=\tau} = \frac{I_{\max}}{\tau}e^{-1} \approx \frac{0.368\,I_{\max}}{\tau}$. The slope has decreased by a factor of $e$ — confirming B is correct and the student's claim is wrong. A contains a true geometric fact — the tangent line at $t = 0\$ does intersect the asymptote at$t = \tau$— but draws a false conclusion from it. That intersection defines$\tau$; it does not mean the slope stays constant. The slope is continuously decreasing as the exponential decays. **C** confuses the *value* of the current with the *derivative* of the current. While$I(\tau) \approx 0.632\,I_{\max}$, the instantaneous rate of change is not the ratio of current value to$\tau$. Those are entirely different quantities. **D** is completely wrong on two counts: the derivative of a pure exponential growth-toward-asymptote function has no inflection point, and the slope never reaches zero until$t \to \infty$. A reliable strategy: whenever a question involves exponential functions, differentiate explicitly. Don't rely on verbal descriptions of the graph — the derivative of$e^{-t/\tau}$carries its own factor of$e^{-t/\tau}$$, which always reduces the slope over time.
6. The graph shows that the decay after $t = T$ is significantly faster than the growth before $t = T$. Which of the following correctly identifies both the cause of this difference and the ratio of the two time constants?
   Answer: The decay is faster because the inductor drives current through $R'$ alone (the battery branch is open), giving $\tau_{\text{decay}} = L/R'$, while the growth time constant is $\tau_{\text{growth}} = L/R$; the ratio is $\tau_{\text{decay}}/\tau_{\text{growth}} = R/R'$, which is less than 1 only if $R' > R$. When analyzing RL circuits with switching events, your key tool is the time constant formula $\tau = L/R_{\text{eff}}$, where $R_{\text{eff}}$ is whatever resistance the inductor actually "sees" in its current loop. The circuit topology changes when the switch opens, so you must redraw the circuit for each phase separately. During growth ($0 < t < T$), the battery drives current through the inductor and resistance $R$. The time constant is $\tau_{\text{growth}} = L/R$. After the switch opens at $t = T$, the battery branch is disconnected entirely. The inductor now drives its stored current through the only available closed path: the freewheeling resistor $R'$. This gives $\tau_{\text{decay}} = L/R'$. If $R' > R$, then $\tau_{\text{decay}} < \tau_{\text{growth}}$, which is exactly what the faster decay on the graph tells you. The ratio is $\tau_{\text{decay}}/\tau_{\text{growth}} = R/R'$, confirming D is correct. Choice A is tempting but wrong — it claims $R$ and $R'$ are in series during decay. They are not; the switch is open, so $R$ is disconnected from the loop entirely. Only $R'$ carries current after $t = T$. Choice B incorrectly invents a changing inductance — inductance is a fixed physical property of the coil and does not change when the switch opens. Choice C makes two errors: it claims both time constants equal $L/R'$ (wrong for the growth phase) and misattributes the faster decay to amplitude rather than $\tau$. A reliable strategy: whenever a switch opens or closes, redraw the circuit and identify the new current loop. The inductor always "sees" only the resistance in its closed path — nothing more, nothing less.
7. The student concludes that $\tau_{RC} = \tau_{RL}$. Assuming $\tau_{RC} = RC$ and $\tau_{RL} = L/R$, and given that the energy curves are identical, which of the following correctly identifies what must be true and why the energy curves match?
   Answer: The student's conclusion is correct: $RC = L/R$ must hold; the energy curves are identical because both $U_C(t)/U_{C,0} = e^{-2t/(RC)}$ and $U_L(t)/U_{L,0} = e^{-2t/(L/R)}$ decay at the same rate when $RC = L/R$, confirming the time constants are equal. When analyzing RC and RL discharge circuits, the key is carefully deriving how energy—not just voltage or current—decays over time, then comparing those expressions directly. In an RC circuit, voltage decays as $V(t) = V_0 e^{-t/(RC)}$, so normalized stored energy is $U_C/U_{C,0} = (V/V_0)^2 = e^{-2t/(RC)}$. In an RL circuit, current decays as $I(t) = I_0 e^{-t/(L/R)}$, so normalized stored energy is $U_L/U_{L,0} = (I/I_0)^2 = e^{-2t/(L/R)}$. Both curves share the form $e^{-2t/\tau}$, but with their own respective time constants. For these two curves to be identical, you need $RC = L/R$—confirming the student's conclusion in answer C is exactly right. Answer A is wrong because it claims squaring "removes the distinction" between time constants—it doesn't. Squaring changes $e^{-t/\tau}$ to $e^{-2t/\tau}$, but the time constant $\tau$ remains fully present and determines the decay rate. The curves only overlap if $RC = L/R$. Answer B makes the same error as A: both normalized energy curves do have the same functional form $e^{-2t/\tau}$, but "same form" doesn't mean "same curve." They match only when the time constants are equal, not universally. Answer D introduces a false asymmetry, incorrectly claiming $U_C$ decays as $e^{-t/(RC)}$ (missing the factor of 2) and that RL energy decays faster by an additional factor. Both energy expressions carry the factor of 2 symmetrically. Study tip: Always distinguish between the state variable ($V$ or $I$, decaying as $e^{-t/\tau}$) and stored energy (proportional to the square, decaying as $e^{-2t/\tau}$). Mixing these up is the most common trap on transient circuit questions.
8. Which of the following correctly identifies the errors in the student's prediction?
   Answer: Both predictions are wrong: the asymptote is $V_C(\infty) = 5\,\text{V}$ (voltage divider between $R_1$ and $R_2$), and the time constant is $\tau = (R_1 \| R_2)\cdot C = 0.5\,\text{s}$, because from the capacitor's perspective the Thévenin resistance is $R_1 \| R_2$. Whenever you analyze an RC circuit with resistors in parallel with the capacitor, your first move should be to find the Thévenin equivalent seen by the capacitor — this gives you both the steady-state voltage and the time constant in one step. To find the Thévenin equivalent, you "look into" the capacitor's terminals with the source active. The Thévenin voltage is determined by the voltage divider that $R_1$ and $R_2$ form: $V_{th} = \mathcal{E} \cdot \frac{R_2}{R_1 + R_2} = 10 \cdot \frac{10}{20} = 5\,\text{V}$. This is the asymptote $V_C(\infty)$, because at steady state the capacitor stops charging and acts like an open circuit — but current still flows through $R_1$ and $R_2$, so the source voltage is divided between them. The Thévenin resistance (with the voltage source replaced by a short) is $R_{th} = R_1 \| R_2 = 5\,\text{k}\Omega$, giving $\tau = R_{th} \cdot C = 5{,}000 \times 100 \times 10^{-6} = 0.5\,\text{s}$. Answer B captures both corrections exactly. Answer A correctly finds the asymptote but wrongly claims $R_2$ doesn't affect the time constant — it absolutely does, since it appears in the Thévenin resistance. Answer C gets the asymptote wrong (the steady-state current through $R_2$ means $V_C \neq \mathcal{E}$) and incorrectly adds the resistors in series for $\tau$. Answer D makes the same asymptote error as C; $R_2$ is in parallel with $C$, not in series with $R_1$ in the charging path. Study tip: For any RC transient problem, always apply Thévenin's theorem at the capacitor's terminals first — it immediately gives you both $V_C(\infty)$ and $\tau$, preventing both errors the student made here.

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