Resistor–Capacitor (RC) Circuits - AP Physics 2

A capacitor is discharging through a resistor. Compared with using resistance $R$ , the circuit is rebuilt with resistance $2R$ (same initial capacitor voltage). Which statement best describes the discharge current at later times?
Answer: It decreases more slowly and is smaller at any given time. This question tests understanding of resistor-capacitor (RC) circuits. During discharge, the current at any time is I = ( $V_C$ /R) $e^{-t/RC}$ , where the time constant τ = RC determines the decay rate. With doubled resistance (2R), the initial current is halved ( $V_C$ /2R versus $V_C$ /R) and the time constant doubles (2RC versus RC), making the exponential decay twice as slow. This means at any given time t > 0, the current with 2R is less than half the current with R because it started smaller and decays more slowly. Choice B incorrectly suggests faster decay with larger current, representing the misconception that higher resistance somehow increases current flow—Ohm's law clearly shows resistance opposes current. To analyze modified RC circuits, consider how changes affect both the initial value and the time constant of the exponential behavior.
A $12\,\text{V}$ battery is connected in series with a resistor and an initially uncharged capacitor; the switch closes at $t=0$ so the capacitor is charging. Which statement best describes the current as time increases?
Answer: It decreases from a maximum value toward zero. This question tests understanding of resistor-capacitor (RC) circuits. When a switch closes to begin charging an initially uncharged capacitor, the capacitor acts like a short circuit (zero resistance) at t=0, allowing maximum current I₀ = V/R to flow. As time increases, charge accumulates on the capacitor plates, creating a voltage that opposes the battery voltage and reduces the current. Eventually, the capacitor voltage equals the battery voltage, no more charge can flow, and the current approaches zero. Choice B incorrectly assumes constant current, failing to recognize that the capacitor's increasing voltage reduces the potential difference across the resistor. To analyze RC circuits, always consider the initial state (capacitor acts like short/open) and final state (no current flows when fully charged/discharged).
A circuit has a battery, resistor $R$ , and capacitor $C$ in series. The capacitor is charging from $0\,\text{V}$ when the switch closes at $t=0$ . Which statement best describes the voltage across the resistor $V_R$ as time increases?
Answer: It decreases from the battery voltage toward $0$ . This question tests understanding of resistor-capacitor (RC) circuits. In a series RC circuit with a battery, Kirchhoff's voltage law requires V_battery = $V_R$ + $V_C$ at all times. Initially, when the uncharged capacitor acts like a short circuit ( $V_C$ = 0), all the battery voltage appears across the resistor ( $V_R$ = V_battery), creating maximum current. As the capacitor charges and $V_C$ increases toward the battery voltage, $V_R$ must decrease to maintain the sum, and since $V_R$ = IR, the current also decreases. Choice B incorrectly assumes $V_R$ stays constant, ignoring that the capacitor's increasing voltage reduces the voltage available across the resistor. To solve RC problems, use Kirchhoff's laws: the sum of voltages around any loop must equal zero.
A capacitor is charging through a resistor from an ideal $9.0\,\text{V}$ battery after a switch closes at $t=0$ . At long times, which statement best describes $V_C$ ?
Answer: It approaches $9.0\,\text{V}$ and the current approaches zero. This question tests understanding of resistor-capacitor (RC) circuits. When charging a capacitor from a battery, the capacitor voltage increases exponentially according to $V_C$ = V_battery(1 - $e^{-t/RC}$ ), starting from 0V and asymptotically approaching the battery voltage. As $V_C$ approaches 9.0V, the voltage across the resistor ( $V_R$ = V_battery - $V_C$ ) approaches zero, which means the current (I = $V_R$ /R) also approaches zero. Once fully charged, the capacitor acts like an open circuit with no current flow. Choice C incorrectly suggests $V_C$ can exceed the battery voltage, violating energy conservation—a capacitor cannot store more energy than the source provides. Remember that in steady state (t → ∞), capacitors act like open circuits with $V_C$ = V_battery and I = 0.
Two different RC circuits each use the same battery and start with uncharged capacitors, so both are charging. Circuit 1 has $R$ and $C$ ; Circuit 2 has $2R$ and $C$ . Which statement best describes the initial current at $t=0^+$ ?
Answer: Circuit 2 has half the initial current of Circuit 1. This question tests understanding of resistor-capacitor (RC) circuits. At t = 0⁺ when charging begins, uncharged capacitors act like short circuits with zero voltage across them, so the initial current is determined solely by the battery voltage and resistance: I₀ = V/R. Circuit 1 has initial current V/R while Circuit 2 has initial current V/(2R) = (1/2)(V/R), making Circuit 2's initial current half that of Circuit 1. The capacitance value doesn't affect the initial current because the capacitor voltage is initially zero regardless of its capacitance. Choice C incorrectly assumes capacitance determines initial current, representing the misconception that larger capacitors somehow draw more initial current—in reality, only resistance limits initial current in RC circuits. When comparing RC circuits, remember that initial conditions depend only on resistance while time constants depend on both R and C.
Two circuits are identical except for resistance. In each, an initially uncharged capacitor is charging from the same ideal battery. Circuit 1 has resistance $R$ ; circuit 2 has resistance $2R$ . Which statement best compares the initial current at $t=0^+$ ?
Answer: Circuit 2 has half the initial current of circuit 1. This question tests understanding of resistor-capacitor (RC) circuits. At t=0⁺ when charging begins, an uncharged capacitor acts like a short circuit (zero voltage drop), so the entire battery voltage appears across the resistor, making the initial current I₀ = V_battery/R by Ohm's law. In circuit 1 with resistance R, the initial current is V/R, while in circuit 2 with resistance 2R, the initial current is V/(2R) = (1/2)(V/R), which is half the initial current of circuit 1. The capacitor value doesn't affect the initial current because at t=0⁺ the capacitor voltage is zero regardless of capacitance. Choice C incorrectly claims both circuits have the same initial current, representing the misconception that initial conditions are independent of circuit resistance. To find initial currents in RC circuits, use the fact that uncharged capacitors initially act like short circuits, so I₀ = V_battery/R_total.
An uncharged capacitor is charging in series with a resistor from a battery. Which statement best describes the charge $Q$ on the capacitor as time increases?
Answer: $Q$ increases and approaches a maximum value. This question tests understanding of resistor-capacitor (RC) circuits. During charging, current flows onto the capacitor plates, accumulating charge according to Q = CV_C, where $V_C$ increases exponentially toward the battery voltage. The charge starts at zero and increases following Q = CV_battery(1 - $e^{-t/RC}$ ), asymptotically approaching the maximum value Q_max = CV_battery when the capacitor is fully charged. The charging process is gradual, not instantaneous, because the resistor limits the current flow rate. Choice C incorrectly suggests unlimited linear growth, representing the misconception that capacitors can store infinite charge—in reality, charge accumulation stops when the capacitor voltage equals the source voltage. When analyzing capacitor charging, remember that charge, voltage, and stored energy all follow the same exponential approach to their maximum values.
A $10\,\text{V}$ battery charges a capacitor through a resistor; the capacitor is charging. Which statement best describes the capacitor voltage $V_C$ as time increases?
Answer: It rises toward $10\,\text{V}$ but does not exceed it. This question tests understanding of resistor-capacitor (RC) circuits. During charging, the capacitor voltage rises exponentially from zero toward the battery voltage, asymptotically approaching but never exceeding it due to the decreasing current as VC approaches Vbattery. The voltage across the capacitor increases as charge accumulates on its plates, following VC = Vbattery(1 - $e^{-t/RC}$ ). Choice C incorrectly suggests the capacitor voltage can exceed the source voltage, which violates energy conservation in a passive circuit. To predict final values in RC circuits, use the principle that capacitors charge to match the applied voltage in steady state.
A $100\ \mu\text{F}$ capacitor is initially uncharged. At $t=0$ , a switch connects it in series with a $2.0\ \text{k}\Omega$ resistor to an ideal $9.0\ \text{V}$ battery, so the capacitor is charging. Which statement best describes the current in the circuit as time increases?
Answer: It decreases toward zero as the capacitor's voltage increases. This question tests understanding of resistor-capacitor (RC) circuits. When a capacitor begins charging through a resistor, the initial current is at its maximum value (I₀ = V/R) because the capacitor acts like a short circuit with zero voltage across it. As time progresses, charge accumulates on the capacitor plates, creating a voltage that opposes the battery voltage, which reduces the voltage across the resistor and therefore decreases the current exponentially toward zero. At long times, the capacitor becomes fully charged to the battery voltage, no more charge flows, and the current becomes zero. Choice C incorrectly suggests current increases, which violates the fundamental behavior that current must decrease as the capacitor's opposing voltage builds up—this represents the misconception that charging means increasing current. To analyze RC circuits correctly, always consider the initial state (capacitor acts like short/open for charging/discharging) and the final steady state (no current flows when voltages equilibrate).
A capacitor initially at $V_C=10\,\text{V}$ is connected to a resistor and allowed to discharge starting at $t=0$ . Which statement best describes the current direction and magnitude as time increases?
Answer: It stays in one direction and decreases toward zero. This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, current flows from the positive plate through the resistor to the negative plate, maintaining one consistent direction throughout the discharge. The current magnitude starts at I₀ = $V_C$ /R and decreases exponentially as I = I₀ $e^{-t/RC}$ because the capacitor voltage driving the current also decreases exponentially. The current never reverses direction because the capacitor polarity remains the same—it just loses charge magnitude. Choice A incorrectly suggests oscillating current, confusing RC circuits with LC circuits which can oscillate. In RC circuits, all quantities ( $V_C$ , I, and Q) decay exponentially without oscillation—think of it as energy dissipating in the resistor, not bouncing back and forth.
An initially uncharged capacitor is charging from an ideal battery through a resistor. Which statement best describes the electric field magnitude between the capacitor plates as time increases?
Answer: It increases from near zero toward a steady nonzero value. This question tests understanding of resistor-capacitor (RC) circuits. The electric field between capacitor plates is directly proportional to the charge stored: E = Q/(ε₀A), where A is the plate area. Initially, an uncharged capacitor has zero charge and thus zero electric field between its plates. As the capacitor charges through the resistor, charge accumulates on the plates (+Q on one plate, -Q on the other), creating an increasing electric field that points from the positive to the negative plate. The charge approaches its maximum value Q_max = CV_battery asymptotically, so the electric field increases from zero toward its maximum steady value E_max = V_battery/d (where d is the plate separation). Choice B incorrectly assumes the field stays constant, representing the misconception that electric field depends only on applied voltage, not on the actual charge present. To understand capacitor behavior in RC circuits, remember that electric field builds up gradually as charge accumulates during the charging process.
A series RC circuit is connected to an ideal battery; the capacitor is charging starting at $t=0$ . Which statement best describes the sum $V_R+V_C$ as time increases?
Answer: It remains equal to the battery voltage at all times. This question tests understanding of resistor-capacitor (RC) circuits. By Kirchhoff's voltage law, the sum of voltage drops around a closed loop must equal the source voltage: V_battery = $V_R$ + $V_C$ at all times during the charging process. Initially, $V_C$ = 0 and $V_R$ = V_battery, but as the capacitor charges, $V_C$ increases while $V_R$ decreases such that their sum always equals V_battery. This is a fundamental constraint that cannot be violated regardless of the charging state. Choice A incorrectly suggests the sum increases, while choice C suggests it decreases, both violating Kirchhoff's law which requires the sum to equal the constant battery voltage. When solving RC circuits, always apply Kirchhoff's laws—they provide constraints that must be satisfied at every instant.
A capacitor is discharging through a resistor. Which statement best describes the magnitude of the current as time increases?
Answer: It decreases toward zero because the capacitor's voltage decreases. This question tests understanding of resistor-capacitor (RC) circuits. During discharge, a charged capacitor acts as a temporary voltage source, driving current through the resistor in a direction that removes charge from the capacitor plates. Initially, the full capacitor voltage appears across the resistor, creating maximum current determined by Ohm's law (I = $V_C$ /R). As charge flows off the plates, the capacitor voltage decreases exponentially, which reduces the voltage across the resistor and thus the current, following the same exponential decay. Eventually, when all stored charge has been dissipated as heat in the resistor, both the capacitor voltage and current approach zero. Choice B incorrectly suggests constant current, reflecting the misconception that current depends only on resistance, not on the changing driving voltage. To analyze discharge circuits, remember that both voltage and current decay exponentially together as the capacitor's stored energy is converted to heat in the resistor.
A series $RC$ circuit is connected to a constant battery at $t=0$ , so the capacitor is charging. Which statement best describes the rate at which charge accumulates on the capacitor plates as time increases?
Answer: It decreases toward zero as the current decreases. This question tests understanding of resistor-capacitor (RC) circuits. The rate at which charge accumulates on the capacitor is simply the current (I = dQ/dt), which starts at maximum when the capacitor is uncharged and decreases exponentially toward zero as the capacitor charges. Initially, the full battery voltage drives current through the resistor, but as the capacitor voltage increases, less voltage appears across the resistor, reducing the current and thus the charging rate. Choice C incorrectly suggests the battery voltage is changing, when actually it's the voltage division between R and C that changes. To understand charging rates, remember that current represents charge flow rate, and current decreases as the capacitor fills.
A charged capacitor is connected to a resistor only, so it is discharging. Which statement best describes the current direction and magnitude as time increases?
Answer: It keeps the same direction and decreases toward zero. This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, the initial current is determined by the capacitor's voltage and the resistance (I₀ = $V_C$ /R), flowing in the direction that depletes the capacitor's charge. As charge leaves the capacitor, its voltage decreases, which proportionally decreases the current through the resistor, creating an exponential decay toward zero current. The current maintains the same direction throughout the discharge process because charge always flows from higher to lower potential. Choice A incorrectly suggests alternating current, representing the misconception that capacitors inherently create AC behavior—this only occurs in AC circuits, not in DC discharge scenarios. To analyze discharge circuits, remember that current direction is set by initial charge distribution and magnitude always decreases monotonically.
A capacitor is charging through a resistor from a battery. Which statement best describes how the circuit current compares at early times versus long times?
Answer: It is largest at early times and nearly zero at long times. This question tests understanding of resistor-capacitor (RC) circuits. During charging, the current follows I = (V_battery/R) $e^{-t/RC}$ , starting at its maximum value I₀ = V_battery/R when the capacitor has zero voltage and acts like a short circuit. As time progresses and the capacitor voltage builds up to oppose the battery, the current decreases exponentially, approaching zero at long times when the capacitor is fully charged and acts like an open circuit. The current is always positive (same direction) but decreases monotonically from maximum to zero. Choice B incorrectly reverses the time behavior, representing the misconception that capacitors initially block current like they do in steady-state DC circuits—in reality, uncharged capacitors offer no initial opposition to current flow. To remember RC behavior, think of capacitors as initially transparent to current when uncharged but eventually blocking DC current when fully charged.
A capacitor has been fully charged to $12\ \text{V}$ . At $t=0$ , the battery is removed and the capacitor is connected across a single $4.0\ \text{k}\Omega$ resistor, so it is discharging. Which statement best describes the voltage across the resistor as time increases?
Answer: It drops from $12\ \text{V}$ toward $0\ \text{V}$ over time. This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, the capacitor initially maintains its full voltage (12V in this case), which appears entirely across the resistor since they are in a simple loop. As the capacitor discharges, its stored charge decreases, causing both the capacitor voltage and the resistor voltage to decrease exponentially toward zero. The current flows in one direction only, depleting the capacitor's charge until no voltage remains across either component. Choice A incorrectly assumes the resistor somehow maintains a fixed voltage, which represents the misconception that resistors generate or maintain voltages independently rather than having voltage drops proportional to current (V = IR). To solve RC problems, remember that in a discharging circuit, both the capacitor and resistor voltages decay together following the same exponential time constant.
A series $RC$ circuit with battery $V_b$ is switched on at $t=0$ ; the capacitor is charging. At long times, which statement best describes the voltages across $R$ and $C$ ?
Answer: $V_R\to 0$ and $V_C\to V_b$ . This question tests understanding of resistor-capacitor (RC) circuits. During charging, current flows through both the resistor and onto the capacitor plates, but as the capacitor voltage increases, the voltage across the resistor must decrease (since VR + VC = Vb by Kirchhoff's voltage law). Eventually, the capacitor becomes fully charged to the battery voltage, no more current flows, and the resistor voltage drops to zero. Choice C incorrectly assumes the voltage divides equally, but this only occurs instantaneously in circuits with two resistors, not in RC circuits where the division changes with time. When analyzing RC circuits at long times, remember that capacitors act like open circuits in DC steady state.
A $12\,\text{V}$ battery, resistor $R$ , and initially uncharged capacitor $C$ are connected in series; the switch is closed at $t=0$ , so the capacitor is charging. Which statement best describes the current as time increases?
Answer: It decreases toward $0\,\text{A}$ as the capacitor charges. This question tests understanding of resistor-capacitor (RC) circuits. When a capacitor charges through a resistor, the current starts at its maximum value (V/R) when the capacitor has no voltage across it, then decreases exponentially toward zero as the capacitor voltage approaches the battery voltage. Initially, the full battery voltage appears across the resistor, driving maximum current, but as charge accumulates on the capacitor, the voltage across the resistor decreases, reducing the current. Choice A incorrectly assumes current depends only on resistance, ignoring that the voltage across R changes as the capacitor charges. To analyze RC circuits, always consider the initial state (capacitor acts like a short circuit) and the final steady state (capacitor acts like an open circuit, so current approaches zero).
A charged capacitor is connected across a resistor $R$ at $t=0$ by closing a switch, so the capacitor is discharging. Which statement best describes the capacitor's voltage $V_C$ as time increases?
Answer: It decreases toward $0\,\text{V}$ . This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, the capacitor voltage decreases exponentially from its initial value toward zero as charge flows off the plates through the resistor. The current through the circuit is driven by the capacitor's stored energy, and as charge leaves the plates, both the voltage and current decrease together. Choice D incorrectly suggests an instantaneous drop, but RC circuits always involve exponential time dependence due to the relationship between current (I = V/R) and charge loss rate (I = -dQ/dt). To solve RC problems, remember that all quantities change exponentially with time constant τ = RC.

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