Physics 2 Quiz: Interpreting Rc Rl Circuit Graphs

What this quiz covers

This quiz focuses on Interpreting Rc Rl Circuit Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The slope $m = -1/(RC)$ and the y-intercept $b = 0$, since $V_C(0) = \mathcal{E}$ for a fully charged capacitor; a nonzero $b$ would indicate the capacitor was not fully charged to $\mathcal{E}$, corresponding to $V_{C,0} = \mathcal{E}\,e^b$, which is less than $\mathcal{E}$ if $b < 0$.
2. The slope $m = -RC$ and the y-intercept $b = \ln(\mathcal{E})$, because the linearized discharge equation gives $\ln V_C = -RCt + \ln \mathcal{E}$; a nonzero intercept $b \neq \ln \mathcal{E}$ would indicate a measurement error in the reference voltage.
3. 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 $V_{C,0} = \mathcal{E}\,e^b$. (correct answer)
4. The slope $m = -1/\tau$ where $\tau = R/C$, and the y-intercept $b = -1$ for a fully charged capacitor; a value $b \neq -1$ indicates that the resistance or capacitance differs from the assumed values used in computing the normalization.

Question 2

1. 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. (correct answer)
2. The expression for $R$ is incorrect because at $t = 0$ the capacitor acts as a short circuit, making the initial current infinite; the correct initial current must account for internal resistance of the battery, rendering $R = \mathcal{E}/I_0$ an overestimate of the true resistance.
3. The expression for $C$ is incorrect because $\tau$ is defined as the time for the current to fall to $1/e$ of its initial value only for RL circuits; in RC circuits the time constant governs voltage, not current, so $\tau \neq RC$ for the current graph.
4. The expression for $R$ is correct, but the expression for $C$ is wrong because $\tau = RC$ applies only when both $R$ and $C$ are measured in SI base units and the graph's time axis must be verified to be in seconds before computing $C = \tau/R$; otherwise the formula gives an incorrect capacitance.

Question 3

1. The student's conclusion is partially correct: $RC = L/R$ is necessary for the current graphs to overlap, but the normalized energy graphs are always identical because the squaring operation removes the distinction between time constants, making the curves degenerate regardless of circuit parameters.
2. The student's conclusion is incorrect: the energy curves are always identical regardless of whether $RC = L/R$, because both $U_C$ and $U_L$ are proportional to the square of their respective state variables, which always decay with the same normalized shape $e^{-2t/\tau}$ scaled to unity at $t=0$.
3. 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. (correct answer)
4. The student's conclusion is incorrect: the energy curves are never identical because $U_C$ decays as $e^{-t/(RC)}$ while $U_L$ decays as $e^{-2t/(L/R)}$; the factor of 2 in the RL exponent arises because the inductor dissipates energy at twice the rate of the capacitor for equal stored energies.

Question 4

1. The decay is faster because the inductor now dissipates energy through both $R$ and $R'$ in series, giving $\tau_{\text{decay}} = L/(R + R')$, while the growth time constant is $\tau_{\text{growth}} = L/R$; the ratio is $\tau_{\text{decay}}/\tau_{\text{growth}} = R/(R+R') < 1$.
2. The decay is faster because the effective inductance decreases after the switch opens due to back-EMF opposition, giving $\tau_{\text{decay}} = L'/(R')$ where $L' < L$; the ratio of time constants depends on how much inductance is lost and cannot be determined without additional information.
3. The decay is faster because the battery EMF is no longer present to sustain the current, reducing the effective driving voltage; the time constants are equal ($\tau_{\text{decay}} = \tau_{\text{growth}} = L/R'$) but the decay appears faster on the graph due to the lower initial amplitude after $t = T$.
4. 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 $R' > R$. (correct answer)

Question 5

1. The student is correct: the slope at $t = \tau$ equals $I_{\max}/\tau$ because the time constant is defined as the time at which the tangent to the exponential at $t=0$ intersects the asymptote, and this geometric property guarantees the slope is constant throughout the growth.
2. 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 $e$. (correct answer)
3. The student is incorrect: the slope at $t = \tau$ is $I_{\max}(1-e^{-1})/\tau \approx 0.632\,I_{\max}/\tau$, because the instantaneous slope at any time equals the ratio of the current value to the time constant, not the ratio of the maximum current to the time constant.
4. The student is incorrect: the slope at $t = \tau$ is zero, because the current has reached 63.2% of its maximum and the derivative of an exponential growth function passes through an inflection point at $t = \tau$, after which the rate of change begins to accelerate before decelerating again.

Question 6

1. $\tau_{\text{reported}} = RC$ 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 $\mathcal{E}$. (correct answer)
2. $\tau_{\text{reported}} = 2RC$ and $V_0 = \mathcal{E}/2$, because the resistor voltage is the complement of the capacitor voltage, and summing two exponentials of the same rate effectively doubles the apparent time constant when fit to a single exponential form.
3. $\tau_{\text{reported}} = RC$ and $V_0 = \mathcal{E}/2$, because the resistor voltage begins at $\mathcal{E}/2$ when the capacitor is half-charged at $t = 0$, and the decay rate is still governed by $\tau = RC$.
4. $\tau_{\text{reported}} = RC/2$ and $V_0 = \mathcal{E}$, because the resistor dissipates energy twice as fast as the capacitor stores it, effectively halving the observed time constant, while the initial voltage across the resistor equals the full EMF at $t = 0$.

Question 7

1. The student correctly identifies $\tau = R_1 C$, but the asymptote is wrong: $V_C(\infty) = \mathcal{E}\cdot R_2/(R_1 + R_2) = 5\,\text{V}$ because $R_2$ forms a voltage divider with $R_1$ at steady state, and the time constant is unaffected by $R_2$ since it only loads the final voltage.
2. 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 $R_1 \| R_2$. (correct answer)
3. Both predictions are wrong: the asymptote is $V_C(\infty) = 10\,\text{V}$ (correct, since $R_2$ draws no current at steady state when $V_C$ is constant), and the time constant is $\tau = (R_1 + R_2)\cdot C = 2\,\text{s}$ because the two resistors are effectively in series during the transient.
4. Only the time constant is wrong: $V_C(\infty) = 10\,\text{V}$ is correct because at steady state the capacitor is fully charged to the source voltage, but $\tau = (R_1 + R_2)C = 2\,\text{s}$ because the charging current must flow through both resistors before reaching the capacitor.

Question 8

1. $V_C(\tau) \approx 5.06\,\text{V}$, because at one time constant the capacitor has charged to 63.2% of its final value, giving $0.632 \times 8 = 5.06\,\text{V}$; the general formula is $V_C(\tau) = (1 - e^{-1})\,V_f$.
2. $V_C(\tau) \approx 6.59\,\text{V}$, because the time constant marks the point where the remaining gap to the final value has shrunk to $1/e$ of the total gap measured from the final value, giving $8 - 8/e \approx 8 - 2.94 = 5.06\,\text{V}$; the generalization is $V_C(\tau) = V_f(1 - e^{-1})$.
3. $V_C(\tau) = 5\,\text{V}$, because at one time constant the capacitor voltage is exactly halfway between the initial and final values by definition of the exponential midpoint, giving $(2 + 8)/2 = 5\,\text{V}$; the generalization is $V_C(\tau) = (V_i + V_f)/2$.
4. $V_C(\tau) \approx 5.79\,\text{V}$, 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. (correct answer)