AP Physics 2 Quiz: Resistor Capacitor Rc Circuits

What this quiz covers

This quiz focuses on Resistor Capacitor Rc Circuits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP 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. It stays at $12\ \text{V}$ because the resistor fixes the voltage.
2. It rises from $0\ \text{V}$ toward $12\ \text{V}$ over time.
3. It exceeds $12\ \text{V}$ briefly due to stored charge release.
4. It drops from $12\ \text{V}$ toward $0\ \text{V}$ over time. (correct answer)

Explanation: 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.

Question 2

1. It reaches $9.0\,\text{V}$ instantly at $t=0$.
2. It approaches $9.0\,\text{V}$ and the current approaches zero. (correct answer)
3. It exceeds $9.0\,\text{V}$ because charge continues accumulating.
4. It approaches $0\,\text{V}$ and the current stays constant.

Explanation: 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.

Question 3

1. It reverses direction repeatedly while decreasing in magnitude.
2. It stays in one direction and decreases toward zero. (correct answer)
3. It stays constant in one direction because $R$ is constant.
4. It is zero at first and then increases as the capacitor empties.

Explanation: 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 = I₀e^(-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.

Question 4

1. It is largest at early times and nearly zero at long times. (correct answer)
2. It is nearly zero at early times and largest at long times.
3. It is constant at all times because the battery is ideal.
4. It becomes negative at long times because the capacitor overcharges.

Explanation: 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.

Question 5

1. It remains constant because the capacitor sets the current.
2. It becomes zero immediately because larger resistance stops discharge.
3. It decreases more slowly and is smaller at any given time. (correct answer)
4. It decreases more quickly and is larger at any given time.

Explanation: 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.

Question 6

1. It remains equal to its initial value.
2. It approaches the battery voltage and stays there.
3. It increases above its initial value.
4. It approaches $0\,\text{V}$. (correct answer)

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, it initially has some voltage V₀ and drives current through the resistor. As charge flows off the capacitor plates, the voltage across the capacitor decreases exponentially according to V_C = V₀e^(-t/RC). Since there's no battery to maintain voltage, the capacitor continues losing charge until it's completely discharged and V_C approaches 0V. Choice A incorrectly suggests the capacitor voltage approaches battery voltage, but there's no battery in a discharging circuit—only the resistor provides a path for current. When analyzing RC circuits, remember that discharging means the capacitor is the only energy source, and it will eventually deplete completely.

Question 7

1. There is no current because a capacitor cannot be a source of potential difference.
2. It is from the positively charged plate toward the negatively charged plate. (correct answer)
3. It alternates direction repeatedly because capacitors reverse polarity as they discharge.
4. It is from the negatively charged plate toward the positively charged plate.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. During discharge, the charged capacitor acts as a voltage source with the positive plate at higher potential than the negative plate. Conventional current always flows from higher to lower potential through the external circuit (the resistor), so current flows from the positive plate, through the resistor, to the negative plate, gradually neutralizing the charge separation. This current direction is the same as when the capacitor was charging, just with the battery removed—positive charges effectively move from the positive plate to neutralize negative charges on the other plate. The current magnitude decreases exponentially but maintains the same direction throughout the discharge. Choice A incorrectly reverses the current direction, representing the misconception that current flows from negative to positive or that discharge current opposes charging current. To determine current direction in RC circuits, identify which terminal is at higher potential and remember that conventional current flows from high to low potential through resistors.

Question 8

1. Circuit 1 has twice the initial current of Circuit 2. (correct answer)
2. Both initial currents are zero because the capacitors are uncharged.
3. Both circuits have the same initial current.
4. Circuit 2 has twice the initial current of Circuit 1.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. At t = 0+, an uncharged capacitor acts like a short circuit (zero voltage), so the full battery voltage appears across the resistor, giving initial current I₀ = V/R. Circuit 1 with resistance R has initial current V/R, while Circuit 2 with resistance 2R has initial current V/(2R), which is half as large. Choice D incorrectly assumes zero initial current, confusing the final steady state (when capacitors are fully charged) with the initial transient state. To find initial conditions in RC circuits, treat uncharged capacitors as short circuits and fully charged capacitors as open circuits.

Question 9

1. It decreases toward $0\,\text{A}$ as the capacitor charges. (correct answer)
2. It increases toward a constant nonzero value.
3. It stays constant because $R$ is constant.
4. It becomes negative after the capacitor reaches $12\,\text{V}$.

Explanation: 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).

Question 10

1. It increases from zero toward a maximum value.
2. It remains constant because the battery voltage is constant.
3. It decreases from a maximum value toward zero. (correct answer)
4. It drops instantly to zero as soon as the switch closes.

Explanation: 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).

Question 11

1. It remains constant at a nonzero value because the battery voltage is constant.
2. It immediately becomes zero because the capacitor charges instantly to $6.0\,\text{V}$.
3. It decreases from a maximum value toward zero as the capacitor's voltage rises. (correct answer)
4. It increases from zero toward a maximum value as charge builds on the capacitor.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. When an uncharged capacitor begins charging through a resistor, the initial current is at its maximum value (I₀ = V/R = 6.0V/2000Ω = 3.0 mA) because the capacitor initially 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 thus 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 from zero, which represents the common misconception that capacitors initially block all current like an open circuit. To analyze RC circuits correctly, always consider the extreme time limits: at t=0⁺, an uncharged capacitor acts like a short circuit (maximum current), and at t→∞, it acts like an open circuit (zero current).

Question 12

1. It is zero because the capacitor initially has no charge.
2. It is half as large because the initial current is $V/R$. (correct answer)
3. It is twice as large because the capacitor initially acts like a short.
4. It is the same because the capacitor initially acts like an open circuit.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. At t=0 when charging begins, an uncharged capacitor acts like a short circuit (zero resistance), so the initial current is determined solely by Ohm's law: I₀ = V/R. Circuit 1 has initial current I₁ = V/R, while Circuit 2 with twice the resistance has initial current I₂ = V/(2R) = (1/2)(V/R), which is half as large. The capacitance C affects the charging time constant τ = RC but not the initial current. Choice C incorrectly doubles the current instead of halving it, confusing the effect of increased resistance. To find initial conditions in RC circuits, remember that uncharged capacitors act like shorts (V_C = 0) and fully charged capacitors act like opens (I = 0).

Question 13

1. Both circuits have constant nonzero long-time current set by $V/R$.
2. Both circuits have zero long-time current because the capacitor eventually empties. (correct answer)
3. Circuit 1 has a larger long-time current because it discharges faster.
4. Circuit 2 has a larger long-time current because it stores more charge.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. During capacitor discharge, the current decreases exponentially as I(t) = I₀e^(-t/RC), where the time constant τ = RC determines how quickly the discharge occurs. While circuit 2 with capacitance 2C has a longer time constant (2RC) and thus discharges more slowly than circuit 1, both circuits eventually reach the same final state: zero current. This is because discharge continues until all stored charge has been removed from the capacitor plates and converted to heat in the resistor, regardless of the initial charge amount or discharge rate. At long times (t >> RC), the exponential term approaches zero for both circuits, making the current effectively zero. Choice A incorrectly suggests circuit 2 maintains a larger current, reflecting the misconception that larger capacitors can sustain current indefinitely. To analyze long-time behavior in RC circuits, remember that capacitors cannot maintain current once their stored charge is depleted.

Question 14

1. Both have the same initial current because $C$ is the same.
2. Circuit 2 has twice the initial current of Circuit 1.
3. Circuit 2 has half the initial current of Circuit 1. (correct answer)
4. Both have zero initial current because the capacitor blocks current.

Explanation: 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.

Question 15

1. It instantly becomes zero because the capacitor charges immediately.
2. It remains constant because the battery voltage is constant.
3. It decreases toward zero as the capacitor's voltage increases. (correct answer)
4. It increases toward a maximum as charge builds on the capacitor.

Explanation: 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).

Question 16

1. It remains constant because the plate separation is constant.
2. It decreases from a maximum value toward zero.
3. It increases from near zero toward a steady nonzero value. (correct answer)
4. It increases without bound and can exceed any value set by the battery.

Explanation: 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.

Question 17

1. It remains constant because no battery is present.
2. It increases above its initial value due to induced emf.
3. It instantly drops to $0\,\text{V}$ at $t=0$.
4. It decreases toward $0\,\text{V}$. (correct answer)

Explanation: 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.

Question 18

1. It rises above $10\,\text{V}$ due to charge buildup.
2. It instantly becomes $10\,\text{V}$ at $t=0$.
3. It stays at $0\,\text{V}$ because the resistor blocks voltage change.
4. It rises toward $10\,\text{V}$ but does not exceed it. (correct answer)

Explanation: 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.

Question 19

1. It increases because the capacitor voltage increases.
2. It decreases toward zero as the current decreases. (correct answer)
3. It becomes negative after the capacitor voltage equals the battery voltage.
4. It stays constant because the battery voltage is constant.

Explanation: 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.

Question 20

1. It stays constant because plate separation is constant.
2. It increases toward a constant value. (correct answer)
3. It decreases toward zero because charge leaves the plates.
4. It instantly reaches its final value at $t=0$.

Explanation: This question tests understanding of resistor-capacitor (RC) circuits. The electric field between capacitor plates is proportional to the charge stored (E = Q/ε₀A), and during charging, charge accumulates on the plates, increasing the electric field from zero toward its maximum value. As current flows onto the plates, positive charge builds on one plate and negative on the other, creating an increasing electric field that eventually reaches a constant value when the capacitor is fully charged. Choice C incorrectly describes discharge rather than charging, confusing the two processes. When analyzing capacitor behavior, remember that electric field strength is directly proportional to the amount of stored charge.