AP Physics C Electricity and Magnetism Quiz: Resistor Capacitor Rc Circuits
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Resistor Capacitor Rc CircuitsQuestion 1 of 20

The charge on a capacitor in a charging RC circuit is given by the function Q(t)=(10 μC)(1−e−t/(2.0 s))Q(t) = (10 \, \mu\text{C})(1 - e^{-t/(2.0\text{ s})}). What is the current I(t)I(t) flowing into the capacitor at t=0t=0?

2.5 μA2.5 \, \mu\text{A}
5.0 μA5.0 \, \mu\text{A}
10 μA10 \, \mu\text{A}
20 μA20 \, \mu\text{A}
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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Resistor Capacitor Rc Circuits

Practice Resistor Capacitor Rc Circuits in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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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 C Electricity and Magnetism.

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Question 1

The charge on a capacitor in a charging RC circuit is given by the function Q(t)=(10 μC)(1−e−t/(2.0 s))Q(t) = (10 \, \mu\text{C})(1 - e^{-t/(2.0\text{ s})}). What is the current I(t)I(t) flowing into the capacitor at t=0t=0?

  1. 2.5 μA2.5 \, \mu\text{A}
  2. 5.0 μA5.0 \, \mu\text{A} (correct answer)
  3. 10 μA10 \, \mu\text{A}
  4. 20 μA20 \, \mu\text{A}
Explanation: Current is the time derivative of charge, I(t)=dQ/dtI(t) = dQ/dt. Differentiating the given function: I(t)=ddt[(10 μC)(1−e−t/(2.0 s))]=(10 μC)(−(−12.0 s)e−t/(2.0 s))=(5.0 μA)e−t/(2.0 s)I(t) = \frac{d}{dt} [(10 \, \mu\text{C})(1 - e^{-t/(2.0\text{ s})})] = (10 \, \mu\text{C}) (-(-\frac{1}{2.0\text{ s}})e^{-t/(2.0\text{ s})}) = (5.0 \, \mu\text{A}) e^{-t/(2.0\text{ s})}. To find the current at t=0t=0, we evaluate this expression at t=0t=0: I(0)=(5.0 μA)e0=5.0 μAI(0) = (5.0 \, \mu\text{A}) e^{0} = 5.0 \, \mu\text{A}.

Question 2

For a series RC circuit connected to a DC voltage source at t=0t=0, which of the following best describes the graph of the potential difference across the resistor, VRV_R, versus time?

  1. A straight line with a negative slope, starting from the source emf EE and decreasing to 0.
  2. An exponential decay curve, starting from the source emf EE and asymptotically approaching 0. (correct answer)
  3. An exponential growth curve, starting from 0 and asymptotically approaching the source emf EE.
  4. A curve that increases from 0 to a maximum value and then exponentially decreases back to 0.
Explanation: At t=0t=0, the capacitor is uncharged and the current is maximum, I0=E/RI_0 = E/R. Thus, the initial potential difference across the resistor is VR(0)=I0R=EV_R(0) = I_0 R = E. As the capacitor charges, the current decreases exponentially as I(t)=(E/R)e−t/RCI(t) = (E/R) e^{-t/RC}. Consequently, the voltage across the resistor also decreases exponentially: VR(t)=I(t)R=Ee−t/RCV_R(t) = I(t)R = E e^{-t/RC}. This function represents an exponential decay from an initial value of EE to a final value of 0.

Question 3

A circuit contains an ideal 12 V12 \text{ V} battery, a 2 Ω2 \, \Omega resistor (resistor 1), a 4 Ω4 \, \Omega resistor (resistor 2), and an uncharged 10 μF10 \, \mu\text{F} capacitor. Resistor 1 is in series with the battery. This combination is connected in series with the parallel combination of resistor 2 and the capacitor. A switch is closed at t=0t=0.

What is the current through resistor 1 at time t=0t=0, and what is the current through resistor 1 after a very long time (t→∞t \to \infty)?

  1. At t=0t=0, I1=2.0 AI_1=2.0 \text{ A}; at t→∞t \to \infty, I1=4.0 AI_1=4.0 \text{ A}
  2. At t=0t=0, I1=4.0 AI_1=4.0 \text{ A}; at t→∞t \to \infty, I1=2.0 AI_1=2.0 \text{ A}
  3. At t=0t=0, I1=6.0 AI_1=6.0 \text{ A}; at t→∞t \to \infty, I1=2.0 AI_1=2.0 \text{ A} (correct answer)
  4. At t=0t=0, I1=6.0 AI_1=6.0 \text{ A}; at t→∞t \to \infty, I1=0 AI_1=0 \text{ A}
Explanation: At time t=0t=0, the uncharged capacitor acts as a short circuit (a wire with zero resistance). Resistor 2 is in parallel with this short circuit, so the equivalent resistance of the parallel branch is 0. The total resistance of the circuit is just R1=2 ΩR_1 = 2 \, \Omega. The initial current through resistor 1 is I1(0)=E/R1=12 V/2 Ω=6.0 AI_1(0) = E/R_1 = 12 \text{ V} / 2 \, \Omega = 6.0 \text{ A}. After a very long time (t→∞t \to \infty), the capacitor is fully charged and acts as an open circuit. No current flows through the capacitor's branch. The circuit then behaves as a simple series circuit with resistor 1 and resistor 2. The total resistance is Rtotal=R1+R2=2 Ω+4 Ω=6 ΩR_{total} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega. The final current through resistor 1 is I1(∞)=E/Rtotal=12 V/6 Ω=2.0 AI_1(\infty) = E/R_{total} = 12 \text{ V} / 6 \, \Omega = 2.0 \text{ A}.

Question 4

A capacitor of capacitance CC is charged to a potential difference EE by a battery through a resistor RR. What is the total energy dissipated as heat by the resistor during the entire charging process?

  1. 14CE2\frac{1}{4}CE^2
  2. 12CE2\frac{1}{2}CE^2 (correct answer)
  3. CE2CE^2
  4. 2CE22CE^2
Explanation: The total energy supplied by the battery is the total charge that flows, Qfinal=CEQ_{final} = CE, multiplied by the battery emf EE, which is Wbatt=QfinalE=(CE)E=CE2W_{batt} = Q_{final}E = (CE)E = CE^2. The energy stored in the fully charged capacitor is UC=12CV2=12CE2U_C = \frac{1}{2}CV^2 = \frac{1}{2}CE^2. By conservation of energy, the energy dissipated in the resistor is the difference between the energy supplied by the battery and the energy stored in the capacitor: Edissipated=Wbatt−UC=CE2−12CE2=12CE2E_{dissipated} = W_{batt} - U_C = CE^2 - \frac{1}{2}CE^2 = \frac{1}{2}CE^2.

Question 5

A 12 V12 \text{ V} battery is connected in series with a 2.0 Ω2.0 \, \Omega resistor and a combination of two capacitors. A 3.0 μF3.0 \, \mu\text{F} capacitor is connected in parallel with a 6.0 μF6.0 \, \mu\text{F} capacitor. What is the time constant for the charging of this circuit?

  1. 4.0 μs4.0 \, \mu\text{s}
  2. 12 μs12 \, \mu\text{s}
  3. 18 μs18 \, \mu\text{s} (correct answer)
  4. 24 μs24 \, \mu\text{s}
Explanation: For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances: Ceq=C1+C2=3.0 μF+6.0 μF=9.0 μFC_{eq} = C_1 + C_2 = 3.0 \, \mu\text{F} + 6.0 \, \mu\text{F} = 9.0 \, \mu\text{F}. The time constant of an RC circuit is given by the product of the resistance and the equivalent capacitance: τ=RCeq=(2.0 Ω)(9.0 μF)=18 μsτ = RC_{eq} = (2.0 \, \Omega)(9.0 \, \mu\text{F}) = 18 \, \mu\text{s}.

Question 6

A capacitor with capacitance CC is initially charged to a potential difference V0V_0. At t=0t=0, it is connected in a simple loop with a resistor of resistance RR. Which of the following expressions describes the magnitude of the potential difference across the resistor, VR(t)V_R(t), as a function of time tt?

  1. VR(t)=V0(1−e−t/RC)V_R(t) = V_0(1 - e^{-t/RC})
  2. VR(t)=V0e−t/RCV_R(t) = V_0 e^{-t/RC} (correct answer)
  3. VR(t)=V0(1−e−RC/t)V_R(t) = V_0(1 - e^{-RC/t})
  4. VR(t)=V0e−RC/tV_R(t) = V_0 e^{-RC/t}
Explanation: During the discharge of a capacitor through a resistor in a simple loop, the potential difference across the capacitor decreases exponentially according to VC(t)=V0e−t/RCV_C(t) = V_0 e^{-t/RC}. By Kirchhoff's loop rule for this circuit, the magnitude of the potential difference across the resistor must be equal to the potential difference across the capacitor at all times. Thus, VR(t)=VC(t)=V0e−t/RCV_R(t) = V_C(t) = V_0 e^{-t/RC}.

Question 7

Consider the RC circuit shown: a capacitor initially at 10 V discharges through R=4.7×103 ΩR=4.7\times10^{3}\,\Omega with C=4.7×10−6 FC=4.7\times10^{-6}\,\text{F}. What is the remaining voltage across the capacitor after discharging for 0.022 s?

  1. VC≈3.7 VV_C\approx 3.7\,\text{V} (correct answer)
  2. VC≈6.3 VV_C\approx 6.3\,\text{V}
  3. VC≈0.47 VV_C\approx 0.47\,\text{V}
  4. VC≈9.5 VV_C\approx 9.5\,\text{V}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically analyzing RC circuits during capacitor discharge. For a discharging capacitor, the voltage decreases exponentially according to V(t) = V₀e^(-t/RC), where V₀ is the initial voltage. In this circuit, with R = 4.7×10³ Ω and C = 4.7×10⁻⁶ F, the time constant is τ = RC = 0.0221 s, and we need the voltage after t = 0.022 s. Choice A is correct because V(0.022) = 10×e^(-0.022/0.0221) = 10×e^(-0.995) = 10×0.370 ≈ 3.7 V. Choice B is incorrect because it appears to use the charging formula V = V₀(1-e^(-t/RC)) instead of the discharging formula, giving approximately 6.3 V. To help students: clearly distinguish between charging and discharging formulas, emphasize that discharge starts from initial voltage and decays to zero, and practice identifying initial conditions. Watch for: mixing up charging and discharging equations, which is the most common error in RC circuit problems.

Question 8

A series RC circuit with a time constant ττ is connected to an ideal battery at time t=0t=0. The time constant represents the time required for the charge on the initially uncharged capacitor to reach approximately what percentage of its maximum possible value?

  1. 37% of its maximum value.
  2. 50% of its maximum value.
  3. 63% of its maximum value. (correct answer)
  4. 100% of its maximum value.
Explanation: In a charging RC circuit, the charge on the capacitor as a function of time is given by Q(t)=Qmax(1−e−t/τ)Q(t) = Q_{max}(1 - e^{-t/τ}). At time t=τt = τ, the charge is Q(τ)=Qmax(1−e−1)Q(τ) = Q_{max}(1 - e^{-1}). Since e−1≈0.37e^{-1} \approx 0.37, the charge is Qmax(1−0.37)=0.63QmaxQ_{max}(1 - 0.37) = 0.63 Q_{max}, which is 63% of the maximum charge.

Question 9

A circuit consists of an uncharged capacitor with capacitance CC, a resistor with resistance RR, and an ideal battery with emf EE, all connected in series. At time t=0t=0, a switch is closed. What is the current in the circuit immediately after the switch is closed?

  1. 00
  2. E/RE/R (correct answer)
  3. E/(2R)E/(2R)
  4. E/CE/C
Explanation: At time t=0t=0, an uncharged capacitor offers no opposition to current flow, so its effective resistance is zero. The potential difference across the capacitor is zero. Therefore, the circuit behaves as if it only contains the battery and the resistor. According to Ohm's law, the initial current is I=E/RI = E/R.

Question 10

A circuit consists of an uncharged capacitor with capacitance CC, a resistor with resistance RR, and an ideal battery with emf EE, all connected in series. A switch is closed at t=0t=0. After a very long time (t→∞t \to \infty), what is the potential difference across the capacitor?

  1. 00
  2. E/2E/2
  3. EE (correct answer)
  4. ER/CE R/C
Explanation: After a very long time, the capacitor becomes fully charged, and the current in the circuit drops to zero. With no current flowing (I=0I=0), the potential drop across the resistor (VR=IRV_R = IR) is zero. According to Kirchhoff's loop rule (E−VC−VR=0E - V_C - V_R = 0), the potential difference across the capacitor must equal the emf of the battery, VC=EV_C = E.

Question 11

A series circuit contains an ideal battery with emf EE, a resistor RR, an uncharged capacitor CC, and a switch. The switch is closed at t=0t=0. Which differential equation correctly describes the charge QQ on the capacitor as a function of time tt during the charging process?

  1. E−QC−dQdtR=0E - \frac{Q}{C} - \frac{dQ}{dt}R = 0 (correct answer)
  2. E+QC+dQdtR=0E + \frac{Q}{C} + \frac{dQ}{dt}R = 0
  3. E−CQ−dQdtR=0E - \frac{C}{Q} - \frac{dQ}{dt}R = 0
  4. E−QC+dQdtR=0E - \frac{Q}{C} + \frac{dQ}{dt}R = 0
Explanation: Applying Kirchhoff's loop rule to the charging RC circuit, the sum of the potential changes around the closed loop must be zero. The battery provides a potential increase of EE. There are potential drops across the capacitor (VC=Q/CV_C = Q/C) and the resistor (VR=IRV_R = IR). Since current is the rate of change of charge, I=dQ/dtI = dQ/dt. Thus, the loop equation is E−VC−VR=0E - V_C - V_R = 0, which becomes E−QC−dQdtR=0E - \frac{Q}{C} - \frac{dQ}{dt}R = 0.

Question 12

Circuit X consists of a resistor RR and a capacitor CC in series with a battery. Circuit Y consists of a resistor 2R2R and a capacitor C/2C/2 in series with an identical battery. How does the time constant τXτ_X of Circuit X compare to the time constant τYτ_Y of Circuit Y?

  1. τX=τYτ_X = τ_Y (correct answer)
  2. τX=2τYτ_X = 2τ_Y
  3. τX=12τYτ_X = \frac{1}{2}τ_Y
  4. τX=4τYτ_X = 4τ_Y
Explanation: The time constant of an RC circuit is given by the formula τ=RCτ = RC. For Circuit X, the time constant is τX=RCτ_X = RC. For Circuit Y, the time constant is τY=(2R)(C/2)=RCτ_Y = (2R)(C/2) = RC. Therefore, the time constants of the two circuits are equal, τX=τYτ_X = τ_Y.

Question 13

In a series RC circuit with time constant ττ, an initially uncharged capacitor is being charged by a battery with emf EE. At what time tt will the potential difference across the capacitor be equal to half of the battery's emf?

  1. t=τln⁡(2)t = τ \ln(2) (correct answer)
  2. t=τ/2t = τ / 2
  3. t=2τt = 2τ
  4. t=τln⁡(1/2)t = τ \ln(1/2)
Explanation: The potential difference across a charging capacitor is given by VC(t)=E(1−e−t/τ)V_C(t) = E(1 - e^{-t/τ}). We need to find the time tt when VC(t)=E/2V_C(t) = E/2. Setting the expressions equal: E/2=E(1−e−t/τ)E/2 = E(1 - e^{-t/τ}). This simplifies to 1/2=1−e−t/τ1/2 = 1 - e^{-t/τ}, which gives e−t/τ=1/2e^{-t/τ} = 1/2. Taking the natural logarithm of both sides yields −t/τ=ln⁡(1/2)=−ln⁡(2)-t/τ = \ln(1/2) = -\ln(2). Therefore, t=τln⁡(2)t = τ \ln(2).

Question 14

A capacitor is being charged through a resistor by a DC source. If the resistance of the resistor is doubled, what is the effect on the time required to charge the capacitor to 95% of its final charge, and what is the effect on the final charge stored?

  1. The time is doubled; the final charge is doubled.
  2. The time is doubled; the final charge is unchanged. (correct answer)
  3. The time is halved; the final charge is halved.
  4. The time is halved; the final charge is unchanged.
Explanation: The time constant of the circuit is τ=RCτ = RC. Doubling the resistance RR will double the time constant. The time to reach any specific percentage of the final charge is directly proportional to the time constant, so this time will also be doubled. The final charge stored on the capacitor, Qfinal=CEQ_{final} = CE, depends only on the capacitance and the emf of the source, neither of which has changed. Thus, the final charge is unchanged.

Question 15

A capacitor of capacitance CC holds an initial charge Q0Q_0. At time t=0t=0, a switch is closed to connect the capacitor in series with a resistor of resistance RR. Which expression represents the current I(t)I(t) in the resistor as a function of time, where positive current is defined as flowing away from the initially positive plate?

  1. I(t)=Q0RCet/RCI(t) = \frac{Q_0}{RC} e^{t/RC}
  2. I(t)=−Q0RC(1−e−t/RC)I(t) = -\frac{Q_0}{RC} (1 - e^{-t/RC})
  3. I(t)=Q0Ce−t/RCI(t) = \frac{Q_0}{C} e^{-t/RC}
  4. I(t)=Q0RCe−t/RCI(t) = \frac{Q_0}{RC} e^{-t/RC} (correct answer)
Explanation: During discharge, the charge on the capacitor decreases exponentially: Q(t)=Q0e−t/RCQ(t) = Q_0 e^{-t/RC}. The current I(t)I(t) is the rate at which charge flows through the resistor. Since the charge on the capacitor is decreasing, the current flowing out of the positive plate is I(t)=−dQ/dtI(t) = -dQ/dt. Differentiating Q(t)Q(t) with respect to time gives dQ/dt=Q0(−1RC)e−t/RCdQ/dt = Q_0(-\frac{1}{RC})e^{-t/RC}. Therefore, I(t)=−(−Q0RCe−t/RC)=Q0RCe−t/RCI(t) = -(-\frac{Q_0}{RC}e^{-t/RC}) = \frac{Q_0}{RC}e^{-t/RC}.

Question 16

The potential difference VCV_C across a capacitor in a discharging RC circuit is measured as a function of time tt. A graph of ln⁡(VC)\ln(V_C) versus tt is created, and it is found to be a straight line with a slope of −2.5 s−1-2.5 \text{ s}^{-1}.

What is the time constant ττ of the circuit?

  1. −2.5 s-2.5 \text{ s}
  2. 0.40 s0.40 \text{ s} (correct answer)
  3. 2.5 s2.5 \text{ s}
  4. ln⁡(2.5) s\ln(2.5) \text{ s}
Explanation: For a discharging capacitor, the voltage is given by VC(t)=V0e−t/τV_C(t) = V_0 e^{-t/τ}. Taking the natural logarithm of both sides gives ln⁡(VC)=ln⁡(V0)−t/τ\ln(V_C) = \ln(V_0) - t/τ. This equation is in the form of a line, y=b+mxy = b + mx, where y=ln⁡(VC)y = \ln(V_C), x=tx = t, the y-intercept b=ln⁡(V0)b = \ln(V_0), and the slope m=−1/τm = -1/τ. Given that the slope is −2.5 s−1-2.5 \text{ s}^{-1}, we have −1/τ=−2.5 s−1-1/τ = -2.5 \text{ s}^{-1}. Solving for ττ gives τ=1/2.5=0.40 sτ = 1/2.5 = 0.40 \text{ s}.

Question 17

A circuit contains a battery with emf EE, a capacitor CC, and two resistors R1R_1 and R2R_2. A switch S can connect the battery and resistor R1R_1 to the capacitor (position A), or connect the charged capacitor to resistor R2R_2 (position B). Initially, the switch S has been in position A for a very long time.

At time t=0t=0, the switch is moved to position B. What is the current through resistor R2R_2 immediately after the switch is moved?

  1. ER1\frac{E}{R_1}
  2. ER2\frac{E}{R_2} (correct answer)
  3. ER1+R2\frac{E}{R_1+R_2}
  4. 00
Explanation: When the switch is in position A for a long time, the capacitor becomes fully charged. In this steady state, no current flows, so there is no voltage drop across R1R_1. The potential difference across the capacitor becomes equal to the battery's emf, VC=EV_C = E. Immediately after the switch is moved to position B at t=0t=0, the capacitor begins to discharge through resistor R2R_2. The initial voltage across R2R_2 is the voltage on the capacitor, EE. By Ohm's law, the initial current through R2R_2 is I2(0)=VC/R2=E/R2I_2(0) = V_C/R_2 = E/R_2.

Question 18

A 10 μF10 \, \mu\text{F} capacitor is charged to a potential difference of 20 V20 \text{ V}. It is then disconnected from the charging source and connected at t=0t=0 to a network of resistors. The network consists of a 2 Ω2 \, \Omega resistor in series with a parallel combination of a 3 Ω3 \, \Omega resistor and a 6 Ω6 \, \Omega resistor. What is the initial current flowing from the capacitor?

  1. 2.0 A2.0 \text{ A}
  2. 4.0 A4.0 \text{ A}
  3. 5.0 A5.0 \text{ A} (correct answer)
  4. 10.0 A10.0 \text{ A}
Explanation: First, find the equivalent resistance of the resistor network. The parallel combination of the 3 Ω3 \, \Omega and 6 Ω6 \, \Omega resistors has a resistance of Rp=(13+16)−1=(36)−1=2 ΩR_p = (\frac{1}{3} + \frac{1}{6})^{-1} = (\frac{3}{6})^{-1} = 2 \, \Omega. This is in series with the 2 Ω2 \, \Omega resistor, so the total equivalent resistance is Req=2 Ω+Rp=2 Ω+2 Ω=4 ΩR_{eq} = 2 \, \Omega + R_p = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega. The initial potential difference across this network is the capacitor's initial voltage, V0=20 VV_0 = 20 \text{ V}. The initial current is given by Ohm's Law: I0=V0/Req=20 V/4 Ω=5.0 AI_0 = V_0 / R_{eq} = 20 \text{ V} / 4 \, \Omega = 5.0 \text{ A}.

Question 19

A fully charged capacitor is discharged through a resistor. The time constant ττ of the circuit represents the time required for the charge on the capacitor to decrease to approximately what percentage of its initial value?

  1. 0% of its initial value.
  2. 37% of its initial value. (correct answer)
  3. 50% of its initial value.
  4. 63% of its initial value.
Explanation: During the discharge of a capacitor, the charge remaining on it as a function of time is given by Q(t)=Q0e−t/τQ(t) = Q_0 e^{-t/τ}, where Q0Q_0 is the initial charge. At time t=τt = τ, the charge is Q(τ)=Q0e−1Q(τ) = Q_0 e^{-1}. Since e−1≈0.37e^{-1} \approx 0.37, the charge has decreased to approximately 37% of its initial value.

Question 20

A capacitor with capacitance CC is connected in series with a resistor RR and a battery with emf EE. The circuit is allowed to reach a steady state, and the energy stored is UU. If the capacitance is then changed to 2C2C while the battery and resistor remain the same, what is the new energy stored in the capacitor after a new steady state is reached?

  1. U/2U/2
  2. UU
  3. 2U2U (correct answer)
  4. 4U4U
Explanation: The energy stored in a capacitor is given by the formula U=12CV2U = \frac{1}{2}CV^2. In a steady state after being charged by a battery with emf EE, the potential difference across the capacitor is V=EV = E. The initial energy stored is U=12CE2U = \frac{1}{2}CE^2. When the capacitance is changed to 2C2C, the new steady-state energy will be Unew=12(2C)E2=2(12CE2)=2UU_{new} = \frac{1}{2}(2C)E^2 = 2(\frac{1}{2}CE^2) = 2U. The stored energy is doubled.