RC Circuit Time Constant - Physics 2

1. Using only these two data points, what is the best estimate of the time constant $\tau$ of the circuit, and what voltage would be expected at $t_3 = 90\,\text{ms}$?
   Answer: $\tau = \frac{60}{\ln 4} \approx 43.3\,\text{ms}$; the expected voltage at $t_3 = 90\,\text{ms}$ is approximately $V_3 = 8.0\,e^{-90/43.3} \approx 1.0\,\text{V}$. When a capacitor discharges through a resistor, the voltage follows $V(t) = V_0\,e^{-t/\tau}$. Your job with two data points is to extract $\tau$ by comparing them — this eliminates $V_0$ and isolates the exponential decay rate. Dividing the two measurements: $\frac{V_2}{V_1} = \frac{2.0}{8.0} = \frac{1}{4} = e^{-(t_2 - t_1)/\tau}$. Taking the natural log of both sides gives $-\ln 4 = -\frac{60}{\tau}$, so $\tau = \frac{60}{\ln 4} \approx 43.3\,\text{ms}$. Plugging into the full equation: $V_3 = 8.0\,e^{-90/43.3} \approx 1.0\,\text{V}$. This is exactly answer A, and it correctly anchors the exponential at $t = 0$ using $V_0 = 8.0\,\text{V}$. Answer B is wrong because it uses $\ln 2$ instead of $\ln 4$. This would only be valid if the voltage had halved — it actually dropped to one-quarter, so you need $\ln 4 = 2\ln 2$. Answer C confuses the half-life with $\tau$. Noticing that voltage quarters in 60 ms correctly implies a half-life of 30 ms, but $\tau$ and half-life are related by $t_{1/2} = \tau \ln 2 \approx 0.693\,\tau$ — they are not the same value. Answer D uses the correct $\tau$ but makes an error by starting the exponential from the second data point $(t_2, V_2)$ rather than from the initial condition. The formula $V(t) = V_0\,e^{-t/\tau}$ requires $t$ measured from $t = 0$, not from an arbitrary reference. A reliable strategy: always divide two voltage measurements to cancel $V_0$, solve for $\tau$, then return to the original $V_0$ at $t = 0$ when predicting future voltages.
2. After many cycles, the capacitor voltage oscillates between a minimum value $V_{\min}$ and a maximum value $V_{\max}$. Which of the following best characterizes $V_{\min}$ and $V_{\max}$?
   Answer: $V_{\min} > 0\,\text{V}$ 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 $5\,\text{V}$. When an RC circuit is driven by a periodic square wave long enough to reach steady state, the capacitor voltage settles into a repeating pattern — it neither fully charges nor fully discharges unless the half-period is much larger than the time constant $\tau = RC$. Here, $\tau = (2\,\text{k}\Omega)(50\,\mu\text{F}) = 0.1\,\text{s}$, and the half-period is $T/2 = 0.25\,\text{s} = 2.5\tau$. That's comparable to — not much greater than — $\tau$, so the capacitor partially charges and partially discharges each half-cycle. In steady state, whatever voltage is gained during the high phase must be lost during the low phase. Because charging targets $10\,\text{V}$ and discharging targets $0\,\text{V}$, and both half-cycles share the same $\tau$ and same duration, the system is perfectly symmetric about $5\,\text{V}$. Solving the steady-state equations confirms $V_{\min} \approx 1.60\,\text{V}$ and $V_{\max} \approx 8.40\,\text{V}$, equidistant from $5\,\text{V}$. D is correct. A assumes $T/2 \gg \tau$, but $2.5\tau$ is nowhere near large enough for complete charge/discharge. B makes the same error about full discharge, and also incorrectly treats each cycle as independent rather than recognizing that in steady state the capacitor starts each half-cycle from a nonzero residual voltage. C is tempting but wrong — the asymmetry argument fails because the same exponential behavior applies symmetrically in both half-cycles around the midpoint $5\,\text{V}$. When you see RC circuits driven by square waves, always compare $T/2$ to $\tau$ first, then ask whether steady-state symmetry applies — it does whenever the circuit and drive are symmetric about the signal's midpoint.
3. How does adding $R'$ in parallel with $R$ affect the time $t^*$ required for the capacitor to reach $8\,\text{V}$?
   Answer: $t^*$ 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. When you see a series RC circuit with a step input, your first instinct should be to identify two things: the time constant $\tau = RC$ and the steady-state (Thévenin) voltage the capacitor ultimately charges toward. Both determine how quickly $V_C$ reaches any target value. Here, the capacitor charges according to $V_C(t) = V_s(1 - e^{-t/\tau})$. To find $t^*$ when $V_C = 8\,\text{V}$: solving $8 = 10(1 - e^{-t^*/\tau})$ gives $t^* = \tau \ln 5$. The original time constant is $\tau = (5\,\text{k}\Omega)(20\,\mu\text{F}) = 0.1\,\text{s}$, so $t^* = 0.1\ln 5$. Adding $R' = 5\,\text{k}\Omega$ in parallel with $R$ gives an equivalent resistance of $R_{eq} = 2.5\,\text{k}\Omega$, halving $\tau$ to $0.05\,\text{s}$. Crucially, both resistors connect between the source and the same nodes — the Thévenin voltage seen by the capacitor remains $10\,\text{V}$. So the same formula applies: $t^* = 0.05\ln 5$, exactly half the original. Answer D is correct. Answer A incorrectly claims the Thévenin voltage changes. It would change only if $R'$ were placed in a voltage-divider configuration (e.g., in series), not in parallel across $R$. Answer B has the direction of the effect exactly backwards — parallel resistance reduces $R_{eq}$, which decreases $\tau$, not increases it. Answer C wrongly treats the resistor as irrelevant; while the target fraction $8/10$ doesn't change, $\tau$ itself depends directly on $R$, so $t^*$ absolutely changes. Study tip: Always separately check (1) how a circuit modification changes $\tau$ and (2) whether it changes the Thévenin voltage. Confusing these two effects is the most common trap in RC transient problems.
4. How does the time constant for the discharging phase compare to the time constant for the charging phase, and what is the numerical value of each?
   Answer: 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. When analyzing RC circuits, the key question to ask is: what elements are in the loop when current flows? The time constant $\tau = RC$ depends only on the resistance and capacitance in the active loop — not on the EMF source, the initial charge, or which direction current flows. During charging, the battery drives current through resistor $R$ into capacitor $C$. The governing equation yields a time constant $\tau = RC = (10{,}000\,\Omega)(100 \times 10^{-6}\,\text{F}) = 1.0\,\text{s}$. During discharging, the battery is removed, and the capacitor drives current through the same resistor $R$ in the same loop. The governing equation is identical in form, giving $\tau = RC = 1.0\,\text{s}$ again. The battery's EMF affects the final voltage the capacitor reaches, not the time scale of the exponential process. Answer A is correct. Answer B is wrong because it invents a "halved resistance" during discharge — removing the battery doesn't split or reduce $R$; the same resistor carries the discharge current. Answer C is wrong on two counts: it incorrectly ties the charging time constant to $\mathcal{E}$ (EMF never appears in $\tau$), and it fabricates a "doubled path length" during discharge with no physical basis. Answer D is wrong because it claims the battery and capacitor share the driving role during charging in a way that doubles the time constant — in reality, both the battery and capacitor act through the same single resistor $R$, so $\tau$ remains $RC$. A useful rule of thumb: whenever a question tries to connect the time constant to voltage, EMF, or which device is "driving" the circuit, treat that as a red flag. $\tau = RC$ — full stop.
5. A resistor $R$ and capacitor $C$ are in series with an ideal battery $\mathcal{E}$. After many time constants, the switch is opened, disconnecting the battery but leaving $R$ and $C$ in a closed loop. Which of the following expressions correctly gives the total energy dissipated in the resistor during the charging phase only (i.e., from $t = 0$ until the capacitor is fully charged)?
   Answer: $W_R = \frac{1}{2}C\mathcal{E}^2$, 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 $R$. Whenever you see an RC charging problem asking about energy, your instinct should be to apply energy conservation across the entire process rather than integrating complicated time-dependent functions. When a capacitor charges through a resistor from a battery of EMF $\mathcal{E}$, the battery delivers a total charge $Q = C\mathcal{E}$ at voltage $\mathcal{E}$, so the total energy supplied by the battery is $W_{\text{battery}} = Q\mathcal{E} = C\mathcal{E}^2$. At full charge, the capacitor stores $U_C = \frac{1}{2}C\mathcal{E}^2$. Since energy must be conserved, the resistor dissipates the difference: $W_R = C\mathcal{E}^2 - \frac{1}{2}C\mathcal{E}^2 = \frac{1}{2}C\mathcal{E}^2$. This result is independent of R — a surprising but important fact. Answer C is correct. Answer A is wrong because it claims the battery supplies $C\mathcal{E}^2$ and the resistor dissipates all of it — but this ignores the $\frac{1}{2}C\mathcal{E}^2$ stored in the capacitor's electric field. That energy isn't dissipated; it's stored. Answer B constructs a plausible-looking formula, but the average power during charging is not simply $\mathcal{E}^2/(2R)$, and multiplying by one time constant doesn't correctly integrate the actual power dissipation curve. It's dimensional coincidence dressed as physics. Answer D incorrectly treats the charging process as if it terminates after two time constants. In reality, current flows (and energy dissipates) for all time from $t = 0$ to $t \to \infty$; cutting it off at $2\tau$ underestimates the total dissipation. Study tip: When energy problems involve RC circuits, always use conservation: battery energy in = capacitor energy stored + resistor heat. The 50/50 split is a classic result worth memorizing.
6. A capacitor $C$ is charged to voltage $V_0$ and discharges through resistance $R$. A student claims: *'After three time constants, the capacitor has lost exactly $95\%$ of its initial stored energy.'* Which of the following best evaluates this claim?
   Answer: 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 $95\%$. When analyzing RC discharge problems, the key is to track which quantity is decaying exponentially and remember that energy depends on the square of voltage — a distinction that trips up many students. In an RC circuit discharging from $V_0$, the voltage decays as $V(t) = V_0 e^{-t/\tau}$, where $\tau = RC$. After three time constants, $V = V_0 e^{-3} \approx 0.050 V_0$. So far, so good. But since stored energy is $U = \frac{1}{2}CV^2$, the remaining energy fraction is: $\frac{U(3\tau)}{U_0} = \frac{V^2}{V_0^2} = e^{-6} \approx 0.0025$ That means only about $0.25\%$ of the energy remains, so roughly $99.75\%$ has been lost — far more than the claimed $95\%$. This makes B correct. A commits the critical error of assuming energy scales linearly with voltage. It doesn't — energy scales as $V^2$, so losing $95\%$ of the voltage does not mean losing $95\%$ of the energy. C introduces a plausible-sounding but fabricated correction. The expression $1 - e^{-3/2}$ has no physical basis here; it neither correctly represents energy nor charge loss in this context. D sounds rigorous by invoking integration of power, but its conclusion is wrong. Integrating the dissipated power correctly yields the same result as B — confirming that $\approx 99.75\%$ of energy is lost, not $95\%$. Study tip: Whenever a problem involves energy in an RC circuit, immediately square the exponential decay factor. Energy goes as $e^{-2t/\tau}$, which decays much faster than voltage alone.
7. Which of the following correctly identifies both the slope and the vertical intercept of this graph?
   Answer: 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. Whenever you see a question involving a logarithmic graph of a circuit quantity, your first move should be to write down the original equation and apply the natural log algebraically — let the math tell you the slope and intercept directly. For an RC charging circuit, the voltage across the capacitor is $V_C(t) = \mathcal{E}\left(1 - e^{-t/RC}\right)$. Subtracting from $\mathcal{E}$ gives $\mathcal{E} - V_C(t) = \mathcal{E}\,e^{-t/RC}$. Taking the natural log of both sides: $\ln(\mathcal{E} - V_C) = \ln(\mathcal{E}) + \ln\!\left(e^{-t/RC}\right) = \ln(\mathcal{E}) - \dfrac{1}{RC}\,t$. This is exactly the form $y = b + mx$, where the slope is $-1/RC$ and the vertical intercept is $\ln(\mathcal{E})$. That's answer C, and it follows directly from clean algebra — no guessing required. Answer A correctly identifies the slope but incorrectly claims the intercept is zero. At $t = 0$, $\ln(\mathcal{E} - V_C) = \ln(\mathcal{E})$, which equals zero only if $\mathcal{E} = 1\,\text{V}$ — that's a special numerical coincidence, not a general result. Don't confuse "the capacitor is uncharged" with "the intercept is zero." Answer B flips the slope: the exponent is $-t/RC$, so the coefficient of $t$ is $-1/RC$, not $-RC$. Mixing up $\tau$ with $1/\tau$ is a classic error. Answer D inverts the sign of the slope. Because $\mathcal{E} - V_C$ is decreasing, its logarithm is also decreasing — the slope is negative, not positive. Study tip: When linearizing exponential equations, always take the log first and match the result to $y = mx + b$ term by term. Never guess the sign of the slope from physical intuition alone.
8. What is the voltage across the capacitor at time $t = 2\tau$, where $\tau$ is the time constant of the discharge circuit?
   Answer: $V_0 e^{-2} \approx 1.62\,\text{V}$, because the effective resistance is $6\,\text{k}\Omega$, giving $\tau = 24\,\text{ms}$, and the voltage decays as $V_0 e^{-t/\tau}$. When a capacitor discharges through a resistor network, the voltage decays as $V(t) = V_0 e^{-t/\tau}$, where $\tau = R_{\text{eff}} \cdot C$. The critical skill here is correctly combining the resistors to find $R_{\text{eff}}$ — and remember, you find the effective resistance from the capacitor's "perspective," looking into the network with the capacitor removed. Here, $R_2$ and $R_3$ are in parallel: $R_{23} = \frac{6 \times 6}{6 + 6} = 3\,\text{k}\Omega$. This parallel combination is in series with $R_1$, so $R_{\text{eff}} = 3 + 3 = 6\,\text{k}\Omega$. The time constant is then $\tau = R_{\text{eff}} \cdot C = 6000 \times 4 \times 10^{-6} = 24\,\text{ms}$. At $t = 2\tau$, the voltage is $V(2\tau) = 12 e^{-2} \approx 1.62\,\text{V}$, confirming B is correct. Choice A uses $R_{\text{eff}} = 5\,\text{k}\Omega$, which has no valid circuit justification — it likely comes from adding resistors incorrectly or misreading the network. Choice C uses $9\,\text{k}\Omega$, the mistake of adding all three resistors in series ($3 + 6 + 6$) without recognizing that $R_2$ and $R_3$ are parallel. Choice D uses $15\,\text{k}\Omega$, which comes from treating all resistors as purely in series ($3 + 6 + 6$) and double-counting or misapplying the parallel rule entirely. A reliable strategy: always sketch the circuit and reduce it step by step — parallel branches first, then series combinations. Never add all resistors together before checking the topology.

Start this quiz