Charging & Discharging Capacitors - Physics 2

1. Which of the following correctly describes how the time constant for this discharge compares to the time constant if only $R_1$ were connected across the capacitor, and how the initial discharge current compares?
   Answer: The time constant with both resistors is $C R_1 R_2/(R_1 + R_2)$, shorter than $CR_1$ alone, and the initial total current from the capacitor is $V_0(1/R_1 + 1/R_2)$, larger than $V_0/R_1$ alone. When a capacitor discharges through resistors, two key quantities determine the behavior: the time constant $\tau = RC_{eq}$ and the initial current $I_0 = V_0/R_{eq}$. The critical skill here is correctly identifying the equivalent resistance when resistors are in parallel. For two resistors in parallel, the equivalent resistance is $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$, which is always less than either resistor alone. This means the time constant with both resistors is $\tau = \frac{CR_1R_2}{R_1+R_2}$, which is shorter than $CR_1$ alone — the capacitor drains faster because it now has two discharge paths. For the initial current, each resistor independently "sees" the full voltage $V_0$, so the total current is $I_0 = \frac{V_0}{R_1} + \frac{V_0}{R_2} = V_0\!\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$, which is larger than $V_0/R_1$ alone. This confirms D is correct. Choice A uses $R_1 + R_2$ (series formula) for both quantities and gets the current direction wrong. Choice B also uses the series formula for the time constant, so the time constant comparison is backwards — though it correctly identifies the current expression. Choice C gets the equivalent resistance right but then incorrectly computes the initial current as $V_0/(R_1+R_2)$, confusing the parallel current rule with a series current calculation. A useful memory anchor: parallel paths always lower resistance, shorten the time constant, and increase total current. If you see resistors in parallel on a discharge problem, expect a faster decay and a larger initial current than any single resistor alone.
2. Which of the following correctly evaluates the student's claim?
   Answer: The student is incorrect: doubling $R$ increases the time constant $\tau = RC$, so the capacitor takes longer to reach steady state, but the final charge $Q = C\mathcal{E}$ is determined only by $C$ and $\mathcal{E}$ and is completely independent of $R$. When analyzing RC circuits, you need to distinguish between two separate questions: how fast does the capacitor charge, and how much charge does it ultimately store? These are governed by different quantities entirely. The time constant $\tau = RC$ controls the charging rate — it tells you how quickly the capacitor approaches its final voltage. The final (steady-state) charge, however, is determined by $Q = C\mathcal{E}$. At steady state, no current flows, meaning the resistor has zero voltage drop across it (since $V = IR = 0$). The full EMF therefore appears across the capacitor, giving $Q = C\mathcal{E}$ regardless of what $R$ is. Doubling $R$ doubles $\tau$, so the capacitor charges more slowly — but it still reaches the same final charge. B captures this correctly. A is wrong because it conflates charging rate with final charge. A longer time constant means slower charging, not less total charge stored at steady state. C contains a subtle logical error. It correctly notes that the initial current drops from $\mathcal{E}/R$ to $\mathcal{E}/(2R)$, but then incorrectly concludes this means less total charge transfers. The lower current flows for a proportionally longer time, and the integral of current over time — which equals total charge — remains exactly $C\mathcal{E}$. D reverses the relationship entirely. A larger resistance increases $\tau = RC$, it does not decrease it. Greater resistance slows the charging process. Study tip: Whenever a question involves RC steady state, immediately ask yourself: does current flow at steady state? If not, $R$ is irrelevant to the final charge — only $C$ and $\mathcal{E}$ matter.
3. Long after the switch is closed ($t \to \infty$), which of the following correctly describes the current through $R_1$, the current through $R_2$, and the voltage across the capacitor $C$?
   Answer: Current through $R_1$ is $\mathcal{E}/(R_1 + R_2)$, current through $R_2$ is $\mathcal{E} R_2/(R_1+R_2) \cdot 1/R_2 = \mathcal{E}/(R_1+R_2)$, and voltage across $C$ equals $\mathcal{E} R_2/(R_1 + R_2)$, because at steady state the fully charged capacitor blocks DC and $R_1$ and $R_2$ carry the same series current. When analyzing RC circuits at steady state ($t \to \infty$), the single most important principle is that a fully charged capacitor acts as an open circuit — no current flows through it. This immediately tells you how to redraw the circuit for analysis. In this circuit, $R_2$ and $C$ are in parallel. At steady state, the capacitor branch is open, so all current that flows must pass through $R_2$ instead. That means $R_1$ and $R_2$ are effectively in series, carrying the same current: $I = \mathcal{E}/(R_1 + R_2)$. The voltage across the parallel combination (and therefore across the capacitor) is the voltage drop across $R_2$: $V_C = IR_2 = \mathcal{E}R_2/(R_1 + R_2)$. This is exactly what answer A describes — making it correct. Answer B claims all current stops and $V_C = \mathcal{E}$. This would only be true if $R_2$ weren't present — but $R_2$ remains a conducting path even when $C$ is fully charged, so current continues flowing through $R_1$ and $R_2$. Answer C claims $I_{R_1} = \mathcal{E}/R_1$, implying no voltage drop across $R_1$. This contradicts Kirchhoff's voltage law — current through $R_1$ must produce a drop, so the full EMF cannot appear across the parallel combination. Answer D says the capacitor and $R_2$ "share current equally," which misunderstands steady state entirely. At $t \to \infty$, zero current flows through the capacitor branch — there's no sharing. Study tip: For any steady-state RC problem, replace every capacitor with an open circuit and solve the resulting resistor network using standard series/parallel rules.
4. After the switch is opened at $t = \tau$, which of the following correctly describes the subsequent behavior of the charge stored on the capacitor?
   Answer: The charge remains constant at approximately $0.632\, C\mathcal{E}$ indefinitely, because once the circuit is open there is no path for current to flow onto or off of the capacitor's plates. When analyzing RC circuits, always ask yourself: what does the current path look like right now? Current — and therefore any change in charge — requires a closed conducting loop. That single question resolves this entire problem. At $t = \tau = RC$, the capacitor has charged to $Q = C\mathcal{E}(1 - e^{-1}) \approx 0.632\, C\mathcal{E}$. The moment the switch opens, the conducting path is completely broken. With no closed loop, no current can flow — and since $I = dQ/dt$, a zero current means zero rate of change of charge. The charge simply freezes at $0.632\, C\mathcal{E}$ and stays there indefinitely. That's why C is correct. A is wrong because charging doesn't have "momentum." The capacitor charges only because current flows through the resistor from the battery. Remove the path, remove the current, and the process stops instantly — there is no residual effect carrying it forward. B describes real RC discharge, but discharge requires a closed loop through the resistor. Here, the switch disconnects both the battery and the resistor simultaneously, so there is no discharge path. If only the battery were removed while the resistor remained in the loop, B would be correct — but that's not what the problem states. D confuses the battery's role. The battery drove the charging, but the charge that accumulated is stored in the electric field between the capacitor plates. Removing the source doesn't erase what's already stored, just as unplugging a lamp doesn't drain a nearby charged battery. Key strategy: On capacitor questions, always identify the current path before predicting behavior. No path = no current = no change in charge.
5. After the capacitor begins discharging through $R/2$, approximately how many time constants of the new discharge circuit does it take for the capacitor to lose approximately 95% of the charge it had at the start of the discharge, and what is the actual elapsed clock time for this process?
   Answer: Approximately 3 time constants of the new circuit, and the elapsed clock time is $3 \times CR/2 = 1.5\,CR$, which is half the actual elapsed time the original charging process took, because the new time constant $CR/2$ is half the original $CR$. When dealing with RC circuit timing questions, always keep two quantities distinct: the number of time constants needed for a given percentage change, and the actual elapsed clock time, which depends on the specific time constant of that circuit. The fraction of charge remaining on a discharging capacitor follows $q(t) = q_0 e^{-t/\tau}$. To lose 95% of charge means retaining only 5%, so you need $e^{-t/\tau} = 0.05$, giving $t/\tau \approx 3$ time constants (since $\ln 20 \approx 3$). This number — roughly 3 — depends only on the percentage discharged, not on the resistance. So the new discharge circuit through $R/2$ still requires approximately 3 time constants to reach 95% discharge. The new time constant, however, is $\tau' = C \cdot (R/2) = CR/2$. Therefore, the actual elapsed clock time is $3 \times (CR/2) = 1.5\,CR$. This confirms D as correct. Choice A wrongly claims that changing the resistance changes how many time constants are needed — it doesn't. The number of time constants for a given percentage is fixed by the math of the exponential, not by $R$. Choice B makes the opposite arithmetic error: it doubles the number of time constants (to 6) while also halving them via the new $\tau'$, arriving at a correct clock time by accident but for completely wrong reasoning. Choice C correctly identifies that 3 time constants are needed but then incorrectly claims the elapsed time is unchanged — it forgets that clock time equals the number of time constants multiplied by the value of each one. Your takeaway: always separate "how many $\tau$'s?" (determined by the target percentage) from "how long is each $\tau$?" (determined by $RC$). Confusing these two is the central trap in multi-step RC timing problems.
6. After a time equal to one time constant $\tau = RC$ has elapsed in both circuits, which of the following correctly compares the charge on the capacitors in the two circuits?
   Answer: Circuit 2 has exactly twice the charge of Circuit 1, and Circuit 1's capacitor has reached approximately 63% of its final charge $C\mathcal{E}$, while Circuit 2's capacitor has reached approximately 63% of its final charge $2C\mathcal{E}$. When you see an RC charging question, your anchor equation should be $Q(t) = Q_f\left(1 - e^{-t/\tau}\right)$, where $Q_f = C\mathcal{E}$ is the final (maximum) charge and $\tau = RC$ is the time constant. Notice that the shape of the charging curve — what fraction of the final charge is reached at any time — is universal for all RC circuits with the same $\tau$, regardless of the battery voltage. At $t = \tau$, the factor $(1 - e^{-1}) \approx 0.632$, so every RC circuit reaches about 63% of its own final charge after one time constant. For Circuit 1, $Q_1 = C\mathcal{E}(1-e^{-1}) \approx 0.63\,C\mathcal{E}$. For Circuit 2, $Q_2 = 2C\mathcal{E}(1-e^{-1}) \approx 1.26\,C\mathcal{E}$. Since both circuits share the same $\tau = RC$, they reach 63% of their respective finals simultaneously, and Circuit 2 carries exactly twice the charge — confirming A is correct. Choice B is tempting but wrong: a shared time constant means the circuits evolve at the same rate fractionally, not that they store equal absolute charge. The final charges differ, so the absolute charges differ too. Choice C correctly identifies the 2:1 ratio but confuses the one-time-constant milestone with a half-life; 50% is reached at $t = \tau\ln 2 \approx 0.69\tau$, not $t = \tau$. Choice D incorrectly claims Circuit 1 charges faster — both circuits have identical $\tau$, so neither is faster; the rate of fractional charging is the same for both. A useful reminder: $\tau$ controls how fast you approach the final value as a fraction, while $\mathcal{E}$ controls what that final value is. Keep those two ideas separate and these questions become straightforward.
7. At the instant the circuit is closed ($t = 0^+$), which of the following correctly describes the voltage across the capacitor $V_C$, the voltage across the resistor $V_R$, and the current $I$ in the circuit?
   Answer: $V_C = 0$, $V_R = \mathcal{E}$, and $I = \mathcal{E}/R$, because an uncharged capacitor acts like a short circuit, placing the full EMF across the resistor at $t = 0$. When analyzing RC circuits at specific moments in time, the key is understanding how a capacitor behaves at $t = 0$ versus at steady state. Think of the capacitor's voltage as its "memory" — it cannot change instantaneously, because that would require infinite current. Since the capacitor starts uncharged, its voltage must be zero at $t = 0^+$. With $V_C = 0$, the capacitor behaves exactly like a wire (a short circuit) at that instant. Applying Kirchhoff's voltage law around the loop: $\mathcal{E} = V_R + V_C = V_R + 0 \implies V_R = \mathcal{E}$ The current is then simply Ohm's Law across the resistor: $I = \frac{V_R}{R} = \frac{\mathcal{E}}{R}$ This confirms B as correct. A has it exactly backwards — $V_C = \mathcal{E}$ describes the steady-state condition (as $t \to \infty$), not $t = 0$. At steady state, the capacitor is fully charged, current drops to zero, and all EMF appears across the capacitor. Confusing initial and final conditions is one of the most common RC circuit mistakes. C is a fabricated middle-ground with no physical justification. There is no rule that voltage splits equally at $t = 0$ — voltage division depends on impedance, not time alone. D confuses the long-term behavior of a capacitor (blocking DC at steady state) with its initial behavior. At $t = 0$, a fully uncharged capacitor allows current to flow freely. Study tip: Memorize the two boundary conditions — at $t = 0$, a capacitor acts like a short circuit ($V_C = 0$); at $t = \infty$, it acts like an open circuit ($I = 0$). These two rules will unlock almost every RC circuit question.
8. As the capacitor discharges, which of the following correctly describes how the rate of energy dissipation in the resistor changes over time, and what ultimately happens to the total energy initially stored in the capacitor?
   Answer: The rate of energy dissipation decreases over time as the current decreases, and the total energy dissipated in the resistor equals $\frac{1}{2}CV_0^2$, the full initial stored energy. When a capacitor discharges through a resistor, think about what's driving the current: the voltage across the capacitor. Since that voltage decays exponentially as $V(t) = V_0 e^{-t/RC}$, the current follows the same pattern: $I(t) = \frac{V_0}{R}e^{-t/RC}$. Both start at their maximum values and shrink toward zero — never increase. The rate of energy dissipation is the power in the resistor, $P(t) = I^2 R = \frac{V_0^2}{R}e^{-2t/RC}$. This is largest at $t = 0$ and decreases continuously — so the rate of dissipation is definitely falling over time. For total energy, you integrate this power over all time, and the result is exactly $\frac{1}{2}CV_0^2$ — the full initial energy stored in the capacitor. Every joule stored in the electric field ends up as heat in the resistor. This confirms B as correct. A is wrong because it claims power stays constant. Power is proportional to $I^2$, and since current is exponentially decaying, power must also decay — it's never constant during RC discharge. C contains a tempting but false idea: that some energy "escapes" as electromagnetic radiation. In an ideal RC circuit, there is no such loss mechanism. All stored energy is accounted for in resistive heating, and the total is the full $\frac{1}{2}CV_0^2$, not half of it. D inverts the physics entirely — current decreases, not increases, and energy dissipated cannot exceed what was initially stored. A useful habit: whenever you see RC discharge questions, immediately sketch the exponential decay of current in your mind. Everything — voltage, current, and power — follows that same decaying shape.
9. Immediately after the dielectric is inserted but before the circuit reaches its new steady state, which of the following correctly describes what happens to the charge on the capacitor, the current in the circuit, and the voltage across the capacitor?
   Answer: The charge on the capacitor immediately increases above $C\mathcal{E}$, a transient current flows from the battery to the capacitor, and the voltage across the capacitor momentarily drops below $\mathcal{E}$ before returning to $\mathcal{E}$ at the new steady state. When a battery-connected capacitor reaches steady state, the voltage across it equals $\mathcal{E}$ and the charge is $Q = C\mathcal{E}$. Inserting a dielectric changes the capacitance to $\kappa C$, and this is where careful reasoning matters: since the battery remains connected, it enforces a fixed terminal voltage — but not instantaneously across the capacitor. Here's the key insight: charge cannot redistribute instantaneously. At the very moment of insertion, the charge on the plates hasn't changed yet, but the capacitance has jumped to $\kappa C$. Using $V = Q/C$, the voltage across the capacitor momentarily drops to $\frac{C\mathcal{E}}{\kappa C} = \frac{\mathcal{E}}{\kappa} < \mathcal{E}$. This creates a voltage imbalance — the battery now "sees" a capacitor voltage below $\mathcal{E}$, so it drives a transient current from battery to capacitor. This current deposits additional charge until $Q_{new} = \kappa C \cdot \mathcal{E}$, restoring the voltage to $\mathcal{E}$ at the new steady state. This is exactly what A describes, making it correct. B is wrong because it claims charge decreases and current flows backward — the opposite of what happens when a larger capacitor needs more charge to reach the same voltage. C is wrong on two counts: charge doesn't freeze at $C\mathcal{E}$, and the voltage never rises to $\kappa\mathcal{E}$ — the battery prevents overvoltage. D is the most tempting trap: yes, the battery maintains $\mathcal{E}$ at steady state, but during the transient, the capacitor voltage genuinely dips below $\mathcal{E}$. Remember: a connected battery fixes the final voltage, not the instantaneous voltage. Transient behavior is governed by the RC time constant, and the circuit needs time to recharge.
10. After a very long time, what is the final voltage across the original capacitor $C$, and how does the total energy stored in the system compare to the initial energy $\frac{1}{2}CV_0^2$?
    Answer: The final voltage across $C$ is $V_0/3$, the final voltage across $2C$ is also $V_0/3$, and the total final energy is less than $\frac{1}{2}CV_0^2$ because energy is dissipated in $R$ during charge redistribution. When two capacitors share charge through a resistor, you need two tools: charge conservation and the series voltage constraint at equilibrium. At steady state, current stops flowing, so there's no voltage drop across $R$. This means the two capacitors must have equal voltage across them (they're in series forming a single loop, and with no resistor drop, $V_C = V_{2C}$). Start with charge conservation. Initially, only $C$ holds charge: $Q_0 = CV_0$. After redistribution, that charge splits between $C$ and $2C$. Since they're in series, the same charge $Q_f$ lands on each capacitor (charge on one plate of each must be equal in a series loop). So $Q_f + Q_f = Q_0$... wait — actually charge conservation gives $Q_f$ on $C$ and $Q_f$ on $2C$, with $Q_f = CV_0/3 \cdot ...$. Let's be direct: setting voltages equal, $Q_f/C = Q_f/(2C)$ is impossible unless $Q_f = 0$, so instead use $Q_f = CV_0 - q$ on $C$ and $q$ on $2C$, with $V_C = V_{2C}$: $(CV_0 - q)/C = q/(2C)$, giving $q = 2CV_0/3$. So $V_{2C} = q/(2C) = V_0/3$ and $V_C = V_0/3$. Both capacitors settle at $V_0/3$, confirming C. Final energy: $\frac{1}{2}C(V_0/3)^2 + \frac{1}{2}(2C)(V_0/3)^2 = \frac{1}{6}CV_0^2$, which is less than $\frac{1}{2}CV_0^2$ — the difference is lost to heat in $R$. A is wrong because it assigns different voltages ($V_0/3$ and $2V_0/3$) to the two capacitors, violating the equilibrium condition that $V_C = V_{2C}$. B claims equal voltages of $V_0/2$ each, which violates charge conservation, and incorrectly states energy is conserved — charge conservation never guarantees energy conservation. D gets the voltage fractions backwards and wrongly claims energy is conserved. Your key takeaway: charge conservation ≠ energy conservation. Whenever charge redistributes through a resistor, energy is always lost as heat — even if $R \to 0$. Memorize this: resistive charge redistribution is inherently dissipative.

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