RC Circuits: Voltage, Current vs. Time - Physics 2

1. A resistor $R_1 = 10\,\text{k}\Omega$ and a capacitor $C = 1\,\mu\text{F}$ are connected in series with an ideal battery of EMF $\mathcal{E}$. A second resistor $R_2 = 10\,\text{k}\Omega$ is connected in parallel with the capacitor only (not with $R_1$). The circuit is assembled at $t = 0$ with the capacitor initially uncharged. What is the time constant for the charging transient?
   Answer: $\tau = \frac{R_1 R_2}{R_1 + R_2}C = 5\,\text{ms}$, because the Thévenin resistance seen by the capacitor is $R_1 \parallel R_2$ when the battery is replaced by a short circuit. Whenever a circuit has a capacitor mixed with multiple resistors and a source, your go-to tool is the Thévenin equivalent seen from the capacitor's terminals. Replace the independent source with its ideal equivalent (a short circuit for an ideal battery), then find the resistance looking into those terminals — that's your $R_{th}$, and the time constant is simply $\tau = R_{th} C$. With the battery shorted, trace what the capacitor "sees": $R_1$ is connected from the positive terminal node to the junction, and $R_2$ is connected across the capacitor's terminals. From the capacitor's perspective, $R_1$ and $R_2$ share both terminal nodes, placing them in parallel. So $R_{th} = R_1 \parallel R_2 = \frac{(10)(10)}{10+10} = 5\,\text{k}\Omega$, giving $\tau = 5\,\text{k}\Omega \times 1\,\mu\text{F} = 5\,\text{ms}$. That confirms C is correct. Choice A wrongly ignores $R_1$ entirely — even though $R_2$ is directly across the capacitor, current must still flow through $R_1$ to reach the circuit, so $R_1$ absolutely influences the time constant. Choice B treats the resistors as series from the capacitor's view, but series would require the battery to remain active; once you short the battery for Thévenin analysis, $R_1$ and $R_2$ are clearly in parallel, not series. Choice D makes the opposite error from A — it keeps $R_1$ but discards $R_2$, reasoning that a parallel element "carries no net current." In reality, $R_2$ provides an additional discharge/charge path that actively reduces the effective resistance. Study tip: Always find $\tau$ by computing Thévenin resistance at the capacitor's terminals with sources zeroed — never just pick whichever resistor "looks closest" to the capacitor.
2. Which of the following correctly compares how the current through the resistor varies with time after the switch is closed at $t = 0$ in each circuit, assuming capacitors are initially uncharged?
   Answer: In Circuit 1 the resistor current decays exponentially from $\mathcal{E}/R$ toward zero; in Circuit 2 the resistor current is constant at $\mathcal{E}/R$ at all times $t \geq 0$. When analyzing RC circuits, your first move should always be to identify what each element is connected to — specifically, whether components share a loop or share terminals (i.e., series vs. parallel with the source). In Circuit 1 (series RC), the battery, resistor, and capacitor share a single loop. At $t = 0$, the uncharged capacitor acts like a short circuit, so all the voltage drops across R, giving an initial current of $\mathcal{E}/R$. As the capacitor charges, it opposes the battery, and the current decays exponentially: $i(t) = \frac{\mathcal{E}}{R}e^{-t/RC}$, approaching zero as the capacitor reaches full voltage. In Circuit 2 (parallel), the resistor and capacitor are each connected directly across the ideal battery — independently. The resistor always sees the full EMF $\mathcal{E}$ across its terminals, so by Ohm's law it always carries $\mathcal{E}/R$, regardless of what the capacitor is doing. The capacitor charges separately but doesn't influence the resistor's current at all. This makes the resistor current constant for all $t \geq 0$, confirming answer A. Answer B is wrong because it describes the resistor current in Circuit 2 as growing from zero — that's actually the behavior of the capacitor's current in a series circuit, not a parallel resistor. Answer C incorrectly assumes both circuits produce identical resistor behavior; the topology completely changes the physics. Answer D wrongly claims no current flows through the resistor in Circuit 2 — the capacitor is in a separate parallel branch and cannot block current through R. The key study tip: topology determines behavior. Always ask "what voltage appears across this element?" A component directly across an ideal battery always sees $\mathcal{E}$ — full stop.
3. Which expression correctly gives the current through the capacitor as a function of time for $t > 0$?
   Answer: $I(t) = \frac{\mathcal{E}}{3R}\,e^{-t/(3RC)}$, because both $R$ and $2R$ are now in series in the discharge loop, giving total resistance $3R$ and time constant $3RC$. When a capacitor discharges through resistors, two things determine the current: the initial voltage on the capacitor and the total resistance in the discharge loop. Both matter — and this question is specifically designed to test whether you treat the discharge circuit independently from the charging circuit. At $t = 0^+$, the capacitor holds voltage $\mathcal{E}$ (fully charged). The moment the switch moves to position B, the capacitor drives current through a closed loop containing both $R$ and $2R$ in series. The total resistance is $R + 2R = 3R$. Applying Ohm's law at the initial moment gives $I(0) = \mathcal{E}/3R$. The time constant for any RC discharge is $\tau = R_{\text{total}} \cdot C = 3RC$. Putting it together: $I(t) = \frac{\mathcal{E}}{3R}\,e^{-t/(3RC)}$, confirming D. Choice A ignores $2R$ entirely — it uses only the original charging resistor, as if the new resistor doesn't exist. The discharge loop has changed; you must use the new circuit. Choice B correctly identifies the initial current $\mathcal{E}/3R$ but then incorrectly uses $RC$ as the time constant, as if $2R$ only affects amplitude but not the rate of decay. Both resistors in series slow the discharge — $2R$ absolutely affects $\tau$. Choice C arbitrarily elevates $2R$ to sole importance, which has no physical justification; series resistors add together. Study tip: Always redraw the circuit for each phase of a switching problem. The time constant $\tau = RC$ uses the total resistance seen by the capacitor in the current loop — not the resistance from a previous phase.
4. A student claims: *'The final voltage across the capacitor equals $\mathcal{E} \cdot \frac{R}{R+r}$ because the internal resistance $r$ permanently reduces the voltage available to the capacitor, just as in a resistive voltage divider.'* Which of the following correctly evaluates this claim?
   Answer: The student's conclusion is wrong. At steady state the capacitor is an open circuit, so current is zero, voltage drops across both $R$ and $r$ are zero, and the full EMF $\mathcal{E}$ appears across the capacitor. The internal resistance $r$ only affects the time constant $(R+r)C$, not the final voltage. Whenever you see an RC circuit question involving steady-state behavior, your first move should be to ask: what happens to current as time goes to infinity? A capacitor in DC steady state acts as an open circuit — no current flows through the branch containing it. Because the capacitor and resistors are in series, zero current through the capacitor means zero current everywhere in the loop. With no current flowing, the voltage drop across $R$ is $V_R = IR = 0$, and the voltage drop across the internal resistance is $V_r = Ir = 0$. By Kirchhoff's voltage law, all of the EMF must appear across the capacitor: $V_C = \mathcal{E}$. The internal resistance $r$ is irrelevant to the final voltage — it only affects how quickly the capacitor charges, giving a time constant $\tau = (R + r)C$. Answer C is correct. Answer A contains a genuine insight — it correctly identifies that the voltage-divider formula requires nonzero current — but then wrongly concludes the formula gives the "wrong answer" without stating the right answer, leaving the reasoning incomplete and misleading. Answer B commits the classic trap: applying a resistive voltage-divider analogy to a capacitor at steady state. A voltage divider works only when current is continuously flowing through both resistors. At steady state, current is zero, so the divider analogy breaks down entirely. Answer D simply inverts the fraction from B — it's equally wrong for the same reason, and also mischaracterizes which component "charges" the capacitor. Study tip: On any steady-state DC circuit question involving a capacitor, immediately replace the capacitor with an open circuit. This single substitution resolves most voltage and current questions at $t = \infty$.
5. In steady-state operation (after many cycles), which of the following best describes the voltage waveform across the capacitor?
   Answer: The capacitor voltage is nearly constant, fluctuating only slightly above and below an approximately constant DC level near $V_0/2$, because the half-period is much shorter than $\tau$ and the capacitor barely charges or discharges during each half-cycle. When an RC circuit is driven by a periodic signal, the key question is how the half-period compares to the time constant $\tau = RC$. Here, $\tau = 1\,\text{s}$ but the half-period is only $0.1\,\text{s}$ — just one-tenth of $\tau$. This ratio tells you everything. During each half-cycle, the capacitor tries to charge or discharge, but the exponential approach only progresses by roughly $1 - e^{-0.1/1} \approx 9.5\%$ of the remaining gap. In other words, the capacitor barely moves toward its target voltage before the source switches again. Over many cycles, this process reaches a steady state where the tiny upward charges during the $+V_0$ phase exactly balance the tiny downward discharges during the $0$ phase. The equilibrium settles near the average value of the square wave, which is $V_0/2$, with only small ripples around it. That makes D correct. A has the logic backwards — a short $RC$ relative to the period would allow nearly complete charging/discharging. Here $RC$ is long, so the capacitor is nearly frozen each half-cycle. B confuses "small fluctuations around a DC level" with "exactly zero." The average of the waveform is $V_0/2$, not zero, so the capacitor settles near $V_0/2$, not at ground. C describes an unstable, ever-growing voltage, which is physically impossible in a passive RC circuit — energy dissipation in the resistor guarantees a bounded steady state. Study tip: Always compare the switching period to $\tau$. If period $\ll \tau$, the capacitor acts like a nearly constant voltage source sitting at the waveform's DC average — this is the principle behind power-supply filtering capacitors.
6. Which of the following correctly describes the final voltage across the originally charged capacitor (capacitance $C$) as $t \to \infty$?
   Answer: $V_f = \frac{V_0}{3}$, because charge is conserved and redistributes so that both capacitors reach the same potential, with the larger capacitor holding twice the charge. When two capacitors are connected in a circuit with no battery, the key principle to apply is conservation of charge combined with the condition that, at steady state, current stops flowing — meaning both capacitors must reach the same voltage (otherwise current would still flow). Initially, capacitor $C$ holds charge $Q_0 = CV_0$, and the uncharged capacitor $2C$ holds zero charge. Once connected, charge redistributes until both capacitors sit at the same final voltage $V_f$. Setting up the conservation equation: the total charge is preserved, so $Q_0 = CV_f + 2C \cdot V_f = 3CV_f$. Solving gives $V_f = \frac{Q_0}{3C} = \frac{V_0}{3}$, confirming that A is correct. Choice B claims the voltage splits equally, as if both capacitors had the same capacitance. Equal voltage would require equal capacitance — but here the capacitors differ by a factor of two, so they hold different amounts of charge at the same voltage, and the larger one draws proportionally more charge from the system. Choice C is a common trap: a capacitor discharging through a resistor alone would go to zero, but here the second capacitor acts as a "voltage backstop." Once both are at the same potential, current stops — the system reaches equilibrium before full discharge. Choice D inverts the logic; because $2C$ is larger, it stores more charge, which actually lowers the final voltage rather than preserving more of it for the smaller capacitor. Your study tip: whenever you see two capacitors reaching steady state with no battery, use $Q_{total} = (C_1 + C_2)V_f$ to find the final voltage — conservation of charge is always your starting point.
7. A fully charged capacitor (capacitance $C$, initial voltage $V_0$) discharges through two resistors: $R_1$ and $R_2$ connected in series with the capacitor. A voltmeter with very high (but finite) internal resistance $R_V \gg R_1, R_2$ is connected directly across $R_2$. Which expression best represents the voltage the voltmeter reads at $t = 0^+$ (immediately after discharge begins)?
   Answer: $V_{\text{meter}} = V_0 \cdot \frac{R_2}{R_1 + R_2}$, because the voltmeter resistance is very large so the parallel combination $R_2 \parallel R_V \approx R_2$, and the voltage divides between $R_1$ and $R_2$. When a capacitor discharges through resistors, treat the capacitor as a voltage source equal to its current charge. At $t = 0^+$, the capacitor still holds $V_0$, so current immediately begins flowing — this is the key insight that eliminates wrong answers right away. The circuit at $t = 0^+$ looks like: a voltage source $V_0$ driving current through $R_1$ in series with the parallel combination of $R_2$ and $R_V$. Since $R_V \gg R_2$, the parallel combination $R_2 \parallel R_V \approx R_2$. The voltage across $R_2$ (which is what the voltmeter reads) is simply the voltage divider result: $V_{\text{meter}} = V_0 \cdot \frac{R_2}{R_1 + R_2}$, confirming answer B. Answer A is wrong because the voltmeter cannot read $V_0$ unless $R_1 = 0$. With $R_1$ in series, some voltage drops across it the moment current flows — and current does flow at $t = 0^+$. Answer C contains a subtle trap: it imagines $R_2$ is "removed" because the voltmeter is in parallel with it. In reality, the parallel combination still approximates $R_2$ (not infinity), so $R_2$ very much participates in the voltage divider. The formula in C would only apply if $R_2$ were truly negligible compared to $R_V$. Answer D is a classic misconception: confusing capacitor behavior with inductor behavior. Inductors resist sudden current changes; capacitors do not. A charged capacitor drives current instantly at $t = 0^+$. Study tip: Whenever a voltmeter with high $R_V$ appears in a circuit problem, your first move should be to replace $R_2 \parallel R_V \approx R_2$ and proceed with standard circuit analysis — don't let the voltmeter complicate what is essentially a voltage divider.
8. Which of the following expressions correctly gives the initial rate of change of the capacitor voltage, $\left.\frac{dV_C}{dt}\right|_{t=0}$?
   Answer: $\left.\frac{dV_C}{dt}\right|_{t=0} = -\frac{V_0}{C\left(R_1 + \frac{R_2 R_3}{R_2+R_3}\right)}$, because the initial current is $V_0$ divided by the total series-parallel resistance, and $dV_C/dt = -I/C$. When an RC circuit discharges through a resistor network, your first job is to find the effective resistance seen by the capacitor — because that determines how quickly charge flows off the plates. At $t = 0$, the capacitor acts like a voltage source of $V_0$. The current it drives must flow through $R_1$ in series with the parallel combination of $R_2$ and $R_3$. The total resistance is therefore $R_{eq} = R_1 + \frac{R_2 R_3}{R_2 + R_3}$, giving an initial current $I_0 = \frac{V_0}{R_{eq}}$. Since $I = -C\frac{dV_C}{dt}$, you get $\left.\frac{dV_C}{dt}\right|_{t=0} = -\frac{V_0}{C\!\left(R_1 + \frac{R_2 R_3}{R_2+R_3}\right)}$, confirming answer B. Answer A treats all three resistors as if they're in series ($R_1 + R_2 + R_3$), ignoring that $R_2$ and $R_3$ share the same two nodes — that's a topology error. Answer C claims the parallel pair acts like a short circuit, which would only be true if $R_2$ or $R_3$ were literally zero; finite resistors in parallel always contribute finite resistance, never zero. Answer D drops $R_1$ entirely from the denominator, incorrectly treating it as an "internal" resistance that doesn't limit current — but since $R_1$ is in series with the capacitor, it absolutely restricts current flow and belongs in the denominator. Study tip: Any time a capacitor discharges into a mixed resistor network, sketch the circuit, identify which resistors are truly in series versus parallel, compute $R_{eq}$, and then apply $dV_C/dt = -I/C$. Getting the topology right is the whole battle.

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