Physics 2 Quiz: Rc Circuit Time Constant

What this quiz covers

This quiz focuses on Rc Circuit Time Constant, 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. Slope $= -1/RC$, vertical intercept $= 0$, because at $t = 0$ the capacitor is uncharged so $V_C = 0$ and the quantity $\mathcal{E} - V_C = \mathcal{E}$; however, $\ln(\mathcal{E} - 0) = 0$ only when $\mathcal{E} = 1\,\text{V}$, which is taken as the reference condition.
2. Slope $= -RC$, vertical intercept $= \ln(\mathcal{E})$, because the time constant $\tau = RC$ appears in the denominator of the exponent, and linearizing the equation brings $RC$ into the coefficient of $t$.
3. Slope $= -1/RC$, vertical intercept $= \ln(\mathcal{E})$, because $\mathcal{E} - V_C = \mathcal{E}\,e^{-t/RC}$ and taking the natural log gives a linear function of $t$ with these parameters. (correct answer)
4. Slope $= +1/RC$, vertical intercept $= \ln(\mathcal{E})$, because $\mathcal{E} - V_C$ decreases over time, making the slope of its logarithm positive to compensate for the decreasing argument.

Question 2

1. $W_R = C\mathcal{E}^2$, because the battery delivers a total energy of $C\mathcal{E}^2$ and all of it is eventually dissipated in the resistor once the capacitor fully charges and current drops to zero.
2. $W_R = \frac{\mathcal{E}^2}{2R}\tau$, where $\tau = RC$, because the average power dissipated is $\mathcal{E}^2/(2R)$ and this persists for one time constant.
3. $W_R = \frac{1}{2}C\mathcal{E}^2$, because the energy stored in the capacitor at full charge equals $\frac{1}{2}C\mathcal{E}^2$, and by energy conservation the battery supplies $C\mathcal{E}^2$ total, so the resistor dissipates exactly half regardless of the value of $R$. (correct answer)
4. $W_R = \frac{1}{2}C\mathcal{E}^2 \cdot (1 - e^{-2})$, because the resistor only dissipates energy during the first two time constants when significant current flows, and the remainder is stored in the capacitor's electric field.

Question 3

1. $V_0 e^{-2} \approx 1.62\,\text{V}$, because the effective resistance is $5\,\text{k}\Omega$, giving $\tau = 20\,\text{ms}$, and the voltage decays as $V_0 e^{-t/\tau}$.
2. $V_0 e^{-2} \approx 1.62\,\text{V}$, because the effective resistance is $6\,\text{k}\Omega$, giving $\tau = 24\,\text{ms}$, and the voltage decays as $V_0 e^{-t/\tau}$. (correct answer)
3. $V_0 e^{-2} \approx 1.62\,\text{V}$, because the effective resistance is $9\,\text{k}\Omega$, giving $\tau = 36\,\text{ms}$, and the voltage decays as $V_0 e^{-t/\tau}$.
4. $V_0 e^{-2} \approx 1.62\,\text{V}$, because the effective resistance is $15\,\text{k}\Omega$, giving $\tau = 60\,\text{ms}$, and the voltage decays as $V_0 e^{-t/\tau}$.

Question 4

1. The claim is correct, because after three time constants the voltage has fallen to $e^{-3} V_0 \approx 0.05 V_0$, meaning $95\%$ of the voltage — and therefore $95\%$ of the energy — has been lost.
2. The claim is incorrect, because after three time constants the voltage is $e^{-3} V_0$, and since energy scales as $V^2$, the remaining energy fraction is $e^{-6} \approx 0.0025$, so approximately $99.75\%$ of the energy has been lost, not $95\%$. (correct answer)
3. The claim is incorrect, because the energy lost after three time constants is $1 - e^{-3} \approx 0.950$, but this expression gives the fraction of charge lost, not energy lost; the energy lost is $1 - e^{-3/2} \approx 0.777$, which differs from $95\%$.
4. The claim is correct, because the power dissipated in $R$ integrates to exactly $95\%$ of $\frac{1}{2}CV_0^2$ over the interval $0$ to $3\tau$, a result that follows directly from the exponential decay of current.

Question 5

1. Both time constants are equal at $\tau = RC = 1.0\,\text{s}$, because the same $R$ and $C$ govern the exponential behavior in both phases regardless of whether the battery is present. (correct answer)
2. The charging time constant is $\tau_{\text{charge}} = RC = 1.0\,\text{s}$, but the discharging time constant is $\tau_{\text{discharge}} = RC/2 = 0.5\,\text{s}$, because without the battery the effective resistance driving the discharge is halved.
3. The charging time constant is $\tau_{\text{charge}} = RC/\mathcal{E} = 1.0\,\text{s}$ (independent of $\mathcal{E}$), but the discharging time constant is $\tau_{\text{discharge}} = 2RC = 2.0\,\text{s}$, because the capacitor must discharge through twice the path length without the battery maintaining current flow.
4. The charging time constant is $\tau_{\text{charge}} = 2RC = 2.0\,\text{s}$, because the battery and the capacitor share the driving role during charging, effectively doubling the time scale; the discharging time constant is $\tau_{\text{discharge}} = RC = 1.0\,\text{s}$.

Question 6

1. $V_{\min} = 0\,\text{V}$ and $V_{\max} = 10\,\text{V}$, because after many cycles the capacitor fully charges to $10\,\text{V}$ during each high phase and fully discharges to $0\,\text{V}$ during each low phase, since $T/2 \gg \tau$.
2. $V_{\min} = 0\,\text{V}$ and $V_{\max} = 10(1 - e^{-2.5}) \approx 9.18\,\text{V}$, because the capacitor fully discharges to $0\,\text{V}$ during each low phase and then charges for exactly $T/2 = 2.5\tau$ during the high phase, as if starting from zero each cycle.
3. $V_{\min} > 0\,\text{V}$ and $V_{\max} < 10\,\text{V}$, but $V_{\min}$ and $V_{\max}$ are not symmetric about $5\,\text{V}$; because charging occurs toward $10\,\text{V}$ while discharging occurs toward $0\,\text{V}$, the exponential approach is inherently asymmetric, pushing $V_{\max}$ significantly closer to $10\,\text{V}$ than $V_{\min}$ is to $0\,\text{V}$.
4. $V_{\min} > 0\,\text{V}$ and $V_{\max} < 10\,\text{V}$, with $V_{\min}$ and $V_{\max}$ symmetric about $5\,\text{V}$, because the half-period $T/2 = 0.25\,\text{s}$ is only about $2.5$ time constants ($\tau = 0.1\,\text{s}$), so the capacitor neither fully charges nor fully discharges, and the equal time constant in both half-cycles forces symmetry about $5\,\text{V}$. (correct answer)

Question 7

1. $t^*$ decreases, but not by exactly a factor of 2; the parallel resistance lowers $\tau$ while simultaneously reducing the steady-state voltage the capacitor approaches, so the capacitor may never reach $8\,\text{V}$ depending on the new Thévenin equivalent.
2. $t^*$ increases by a factor of 2, because adding a parallel resistor doubles the current paths, effectively doubling the time constant and slowing the rate at which the capacitor charges.
3. $t^*$ is unchanged, because the time to reach a given fraction of $V_s$ depends only on the ratio $V_C/V_s$, which is independent of $R$; the resistor affects only the maximum current, not the voltage threshold.
4. $t^*$ decreases by a factor of 2, because the parallel combination halves the effective resistance to $2.5\,\text{k}\Omega$, halving $\tau$; since $t^* = \tau \ln 5$, the time is halved accordingly. (correct answer)

Question 8

1. $\tau = \frac{60}{\ln 4} \approx 43.3\,\text{ms}$; the expected voltage at $t_3 = 90\,\text{ms}$ is approximately $V_3 = 8.0\,e^{-90/43.3} \approx 1.0\,\text{V}$. (correct answer)
2. $\tau = \frac{60}{\ln 2} \approx 86.6\,\text{ms}$; the expected voltage at $t_3 = 90\,\text{ms}$ is approximately $V_3 = 8.0\,e^{-90/86.6} \approx 3.0\,\text{V}$.
3. $\tau = 30\,\text{ms}$; the expected voltage at $t_3 = 90\,\text{ms}$ is approximately $V_3 = 8.0\,e^{-3} \approx 0.40\,\text{V}$, since the voltage quarters every $60\,\text{ms}$, implying a half-life of $30\,\text{ms}$, which is mistakenly used as $\tau$.
4. $\tau = \frac{60}{\ln 4} \approx 43.3\,\text{ms}$; the expected voltage at $t_3 = 90\,\text{ms}$ is approximately $V_3 = 2.0\,e^{-30/43.3} \approx 1.27\,\text{V}$, extrapolating forward from the second data point.