Physics 2 Quiz: Rc Circuits Voltage Current Vs Time
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Rc Circuits Voltage Current Vs TimeQuestion 1 of 8

A fully charged capacitor (capacitance CC, initial voltage V0V_0) discharges through two resistors: R1R_1 and R2R_2 connected in series with the capacitor. A voltmeter with very high (but finite) internal resistance RV≫R1,R2R_V \gg R_1, R_2 is connected directly across R2R_2. Which expression best represents the voltage the voltmeter reads at t=0+t = 0^+ (immediately after discharge begins)?

Vmeter=V0V_{\text{meter}} = V_0, because the capacitor is fully charged at t=0+t = 0^+ and the voltmeter reads the full capacitor voltage.
Vmeter=V0⋅R2R1+R2V_{\text{meter}} = V_0 \cdot \frac{R_2}{R_1 + R_2}, because the voltmeter resistance is very large so the parallel combination R2∥RV≈R2R_2 \parallel R_V \approx R_2, and the voltage divides between R1R_1 and R2R_2.
Vmeter=V0⋅RVR1+RVV_{\text{meter}} = V_0 \cdot \frac{R_V}{R_1 + R_V}, because the voltmeter is in parallel with R2R_2 and so effectively removes R2R_2 from the circuit for the purpose of calculating the voltage at t=0+t=0^+.
Vmeter=0V_{\text{meter}} = 0, because at t=0+t = 0^+ no current has yet flowed through the circuit and the resistors have not developed any voltage drop.
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Physics 2 Quiz: Rc Circuits Voltage Current Vs Time

Practice Rc Circuits Voltage Current Vs Time in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A fully charged capacitor (capacitance CC, initial voltage V0V_0) discharges through two resistors: R1R_1 and R2R_2 connected in series with the capacitor. A voltmeter with very high (but finite) internal resistance RV≫R1,R2R_V \gg R_1, R_2 is connected directly across R2R_2. Which expression best represents the voltage the voltmeter reads at t=0+t = 0^+ (immediately after discharge begins)?

  1. Vmeter=V0V_{\text{meter}} = V_0, because the capacitor is fully charged at t=0+t = 0^+ and the voltmeter reads the full capacitor voltage.
  2. Vmeter=V0⋅R2R1+R2V_{\text{meter}} = V_0 \cdot \frac{R_2}{R_1 + R_2}, because the voltmeter resistance is very large so the parallel combination R2∥RV≈R2R_2 \parallel R_V \approx R_2, and the voltage divides between R1R_1 and R2R_2. (correct answer)
  3. Vmeter=V0⋅RVR1+RVV_{\text{meter}} = V_0 \cdot \frac{R_V}{R_1 + R_V}, because the voltmeter is in parallel with R2R_2 and so effectively removes R2R_2 from the circuit for the purpose of calculating the voltage at t=0+t=0^+.
  4. Vmeter=0V_{\text{meter}} = 0, because at t=0+t = 0^+ no current has yet flowed through the circuit and the resistors have not developed any voltage drop.
Explanation: When a capacitor discharges through resistors, treat the capacitor as a voltage source equal to its current charge. At t=0+t = 0^+, the capacitor still holds V0V_0, so current immediately begins flowing — this is the key insight that eliminates wrong answers right away. The circuit at t=0+t = 0^+ looks like: a voltage source V0V_0 driving current through R1R_1 in series with the parallel combination of R2R_2 and RVR_V. Since RV≫R2R_V \gg R_2, the parallel combination R2∥RV≈R2R_2 \parallel R_V \approx R_2. The voltage across R2R_2 (which is what the voltmeter reads) is simply the voltage divider result: Vmeter=V0⋅R2R1+R2V_{\text{meter}} = V_0 \cdot \frac{R_2}{R_1 + R_2}, confirming answer B. Answer A is wrong because the voltmeter cannot read V0V_0 unless R1=0R_1 = 0. With R1R_1 in series, some voltage drops across it the moment current flows — and current does flow at t=0+t = 0^+. Answer C contains a subtle trap: it imagines R2R_2 is "removed" because the voltmeter is in parallel with it. In reality, the parallel combination still approximates R2R_2 (not infinity), so R2R_2 very much participates in the voltage divider. The formula in C would only apply if R2R_2 were truly negligible compared to RVR_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+t = 0^+. Study tip: Whenever a voltmeter with high RVR_V appears in a circuit problem, your first move should be to replace R2∥RV≈R2R_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.

Question 2

A capacitor with capacitance CC is initially charged to voltage V0V_0 and then connected at t=0t = 0 to a resistor RR in series with an uncharged capacitor of capacitance 2C2C. There is no battery in the circuit.

Which of the following correctly describes the final voltage across the originally charged capacitor (capacitance CC) as t→∞t \to \infty?

  1. Vf=V03V_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. (correct answer)
  2. Vf=V02V_f = \frac{V_0}{2}, because the two capacitors share the initial voltage equally once the transient current decays to zero.
  3. Vf=0V_f = 0, because the capacitor fully discharges through the resistor just as it would discharge into a wire, leaving no stored energy.
  4. Vf=2V03V_f = \frac{2V_0}{3}, because the larger capacitor stores proportionally more charge, so the smaller capacitor retains a larger fraction of the original voltage.
Explanation: 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 CC holds charge Q0=CV0Q_0 = CV_0, and the uncharged capacitor 2C2C holds zero charge. Once connected, charge redistributes until both capacitors sit at the same final voltage VfV_f. Setting up the conservation equation: the total charge is preserved, so Q0=CVf+2C⋅Vf=3CVfQ_0 = CV_f + 2C \cdot V_f = 3CV_f. Solving gives Vf=Q03C=V03V_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 2C2C 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 Qtotal=(C1+C2)VfQ_{total} = (C_1 + C_2)V_f to find the final voltage — conservation of charge is always your starting point.

Question 3

A resistor R1=10 kΩR_1 = 10\,\text{k}\Omega and a capacitor C=1 μFC = 1\,\mu\text{F} are connected in series with an ideal battery of EMF E\mathcal{E}. A second resistor R2=10 kΩR_2 = 10\,\text{k}\Omega is connected in parallel with the capacitor only (not with R1R_1). The circuit is assembled at t=0t = 0 with the capacitor initially uncharged. What is the time constant for the charging transient?

  1. τ=R2C=10 ms\tau = R_2 C = 10\,\text{ms}, because the capacitor charges through R2R_2 alone once the battery drives current through the network.
  2. τ=(R1+R2)C=20 ms\tau = (R_1 + R_2)C = 20\,\text{ms}, because from the capacitor's perspective both resistors appear in series when the battery is treated as a short circuit for the Thévenin equivalent.
  3. τ=R1R2R1+R2C=5 ms\tau = \frac{R_1 R_2}{R_1 + R_2}C = 5\,\text{ms}, because the Thévenin resistance seen by the capacitor is R1∥R2R_1 \parallel R_2 when the battery is replaced by a short circuit. (correct answer)
  4. τ=R1C=10 ms\tau = R_1 C = 10\,\text{ms}, because R2R_2 is in parallel with the capacitor and therefore carries no net current during the transient, leaving only R1R_1 to set the time constant.
Explanation: 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 RthR_{th}, and the time constant is simply τ=RthC\tau = R_{th} C. With the battery shorted, trace what the capacitor "sees": R1R_1 is connected from the positive terminal node to the junction, and R2R_2 is connected across the capacitor's terminals. From the capacitor's perspective, R1R_1 and R2R_2 share both terminal nodes, placing them in parallel. So Rth=R1∥R2=(10)(10)10+10=5 kΩR_{th} = R_1 \parallel R_2 = \frac{(10)(10)}{10+10} = 5\,\text{k}\Omega, giving τ=5 kΩ×1 μF=5 ms\tau = 5\,\text{k}\Omega \times 1\,\mu\text{F} = 5\,\text{ms}. That confirms C is correct. Choice A wrongly ignores R1R_1 entirely — even though R2R_2 is directly across the capacitor, current must still flow through R1R_1 to reach the circuit, so R1R_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, R1R_1 and R2R_2 are clearly in parallel, not series. Choice D makes the opposite error from A — it keeps R1R_1 but discards R2R_2, reasoning that a parallel element "carries no net current." In reality, R2R_2 provides an additional discharge/charge path that actively reduces the effective resistance. Study tip: Always find τ\tau by computing Thévenin resistance at the capacitor's terminals with sources zeroed — never just pick whichever resistor "looks closest" to the capacitor.

Question 4

Two identical RC circuits (each with resistance RR and capacitance CC) are connected to the same ideal battery of EMF E\mathcal{E}. In Circuit 1, the resistor and capacitor are in series with the battery. In Circuit 2, the resistor and capacitor are both in parallel with the battery (the resistor directly across the battery, the capacitor directly across the battery).

Which of the following correctly compares how the current through the resistor varies with time after the switch is closed at t=0t = 0 in each circuit, assuming capacitors are initially uncharged?

  1. In Circuit 1 the resistor current decays exponentially from E/R\mathcal{E}/R toward zero; in Circuit 2 the resistor current is constant at E/R\mathcal{E}/R at all times t≥0t \geq 0. (correct answer)
  2. In Circuit 1 the resistor current decays exponentially from E/R\mathcal{E}/R toward zero; in Circuit 2 the resistor current grows from zero toward E/R\mathcal{E}/R with time constant RCRC.
  3. In both circuits the resistor current decays from E/R\mathcal{E}/R toward zero with the same time constant RCRC, because both circuits contain the same RR and CC values.
  4. In Circuit 1 the resistor current decays from E/R\mathcal{E}/R toward zero; in Circuit 2 no current flows through the resistor at any time because the capacitor, once charged, blocks all DC current in the parallel branch.
Explanation: 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=0t = 0, the uncharged capacitor acts like a short circuit, so all the voltage drops across R, giving an initial current of E/R\mathcal{E}/R. As the capacitor charges, it opposes the battery, and the current decays exponentially: i(t)=ERe−t/RCi(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 E\mathcal{E} across its terminals, so by Ohm's law it always carries E/R\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≥0t \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 E\mathcal{E} — full stop.

Question 5

An RC circuit with R=1 MΩR = 1\,\text{M}\Omega and C=1 μFC = 1\,\mu\text{F} (so τ=1 s\tau = 1\,\text{s}) is driven by a square-wave voltage source that alternates between +V0+V_0 and 00 with a half-period of 0.1 s0.1\,\text{s} (much less than τ\tau). The capacitor is in series with the resistor.

In steady-state operation (after many cycles), which of the following best describes the voltage waveform across the capacitor?

  1. The capacitor voltage follows the source almost exactly, charging and discharging nearly to +V0+V_0 and 00 respectively during each half-cycle, because RCRC sets a short charging time.
  2. The capacitor voltage is exactly zero at all times in steady state, because the average value of a square wave alternating between +V0+V_0 and 00 causes equal charging and discharging that cancel perfectly.
  3. The capacitor voltage grows without bound over successive cycles because the square wave continuously adds charge faster than the RC circuit can dissipate it through the resistor.
  4. The capacitor voltage is nearly constant, fluctuating only slightly above and below an approximately constant DC level near V0/2V_0/2, because the half-period is much shorter than τ\tau and the capacitor barely charges or discharges during each half-cycle. (correct answer)
Explanation: When an RC circuit is driven by a periodic signal, the key question is how the half-period compares to the time constant τ=RC\tau = RC. Here, τ=1 s\tau = 1\,\text{s} but the half-period is only 0.1 s0.1\,\text{s} — just one-tenth of τ\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≈9.5%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 +V0+V_0 phase exactly balance the tiny downward discharges during the 00 phase. The equilibrium settles near the average value of the square wave, which is V0/2V_0/2, with only small ripples around it. That makes D correct. A has the logic backwards — a short RCRC relative to the period would allow nearly complete charging/discharging. Here RCRC 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 V0/2V_0/2, not zero, so the capacitor settles near V0/2V_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 τ\tau. If period ≪τ\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.

Question 6

A capacitor CC is fully charged to voltage V0V_0 and then at t=0t = 0 is connected to a network consisting of two resistors: R1R_1 in series with the parallel combination of R2R_2 and R3R_3. The capacitor, R1R_1, and the parallel pair form a single loop.

Which of the following expressions correctly gives the initial rate of change of the capacitor voltage, dVCdt∣t=0\left.\frac{dV_C}{dt}\right|_{t=0}?

  1. dVCdt∣t=0=−V0C(R1+R2+R3)\left.\frac{dV_C}{dt}\right|_{t=0} = -\frac{V_0}{C(R_1 + R_2 + R_3)}, because all three resistors are effectively in series for the initial current calculation.
  2. dVCdt∣t=0=−V0C(R1+R2R3R2+R3)\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 V0V_0 divided by the total series-parallel resistance, and dVC/dt=−I/CdV_C/dt = -I/C. (correct answer)
  3. dVCdt∣t=0=−V0CR1\left.\frac{dV_C}{dt}\right|_{t=0} = -\frac{V_0}{C R_1}, because R2R_2 and R3R_3 in parallel approach a short circuit for the initial transient, so only R1R_1 limits the current.
  4. dVCdt∣t=0=−V0C(R2+R3)\left.\frac{dV_C}{dt}\right|_{t=0} = -\frac{V_0}{C(R_2 + R_3)}, because R1R_1 is in series with the capacitor and acts as an internal resistance that does not appear in the voltage-divider calculation for dVC/dtdV_C/dt.
Explanation: 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=0t = 0, the capacitor acts like a voltage source of V0V_0. The current it drives must flow through R1R_1 in series with the parallel combination of R2R_2 and R3R_3. The total resistance is therefore Req=R1+R2R3R2+R3R_{eq} = R_1 + \frac{R_2 R_3}{R_2 + R_3}, giving an initial current I0=V0ReqI_0 = \frac{V_0}{R_{eq}}. Since I=−CdVCdtI = -C\frac{dV_C}{dt}, you get dVCdt∣t=0=−V0C ⁣(R1+R2R3R2+R3)\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 (R1+R2+R3R_1 + R_2 + R_3), ignoring that R2R_2 and R3R_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 R2R_2 or R3R_3 were literally zero; finite resistors in parallel always contribute finite resistance, never zero. Answer D drops R1R_1 entirely from the denominator, incorrectly treating it as an "internal" resistance that doesn't limit current — but since R1R_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 ReqR_{eq}, and then apply dVC/dt=−I/CdV_C/dt = -I/C. Getting the topology right is the whole battle.

Question 7

A series circuit contains a battery (EMF E\mathcal{E}, internal resistance rr), an external resistor RR, and a capacitor CC. The capacitor is initially uncharged. The switch is closed at t=0t = 0.

A student claims: 'The final voltage across the capacitor equals E⋅RR+r\mathcal{E} \cdot \frac{R}{R+r} because the internal resistance rr permanently reduces the voltage available to the capacitor, just as in a resistive voltage divider.' Which of the following correctly evaluates this claim?

  1. The student's formula is self-contradictory: they correctly note that steady-state current is zero, but then apply a voltage-divider formula that requires nonzero current. The formula ER/(R+r)\mathcal{E}R/(R+r) gives the wrong answer.
  2. The student is correct. The internal resistance rr acts like the lower leg of a voltage divider, permanently reducing the final capacitor voltage to E⋅R/(R+r)\mathcal{E} \cdot R/(R+r), just as it would reduce the terminal voltage under a resistive load.
  3. The student's conclusion is wrong. At steady state the capacitor is an open circuit, so current is zero, voltage drops across both RR and rr are zero, and the full EMF E\mathcal{E} appears across the capacitor. The internal resistance rr only affects the time constant (R+r)C(R+r)C, not the final voltage. (correct answer)
  4. The student's formula has the fraction inverted. The correct final voltage is E⋅r/(R+r)\mathcal{E} \cdot r/(R+r), because the internal resistance rr, not RR, is the element that charges the capacitor while RR dissipates the remaining voltage.
Explanation: 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 RR is VR=IR=0V_R = IR = 0, and the voltage drop across the internal resistance is Vr=Ir=0V_r = Ir = 0. By Kirchhoff's voltage law, all of the EMF must appear across the capacitor: VC=EV_C = \mathcal{E}. The internal resistance rr is irrelevant to the final voltage — it only affects how quickly the capacitor charges, giving a time constant τ=(R+r)C\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=∞t = \infty.

Question 8

An ideal switch, battery (EMF E\mathcal{E}), resistor RR, and capacitor CC are arranged so that when the switch is in position A (t<0t < 0), the capacitor is fully charged through RR to voltage E\mathcal{E}. At t=0t = 0, the switch moves instantaneously to position B, disconnecting the battery and connecting a second resistor 2R2R in series with the first resistor RR and the capacitor (forming a closed loop with no battery).

Which expression correctly gives the current through the capacitor as a function of time for t>0t > 0?

  1. I(t)=ER e−t/(RC)I(t) = \frac{\mathcal{E}}{R}\,e^{-t/(RC)}, because the capacitor discharges through only RR (the resistor that was in the original charging circuit) with the original time constant.
  2. I(t)=E3R e−t/(RC)I(t) = \frac{\mathcal{E}}{3R}\,e^{-t/(RC)}, because the initial current is limited by the total resistance 3R3R but the time constant is still set by only RR and CC since 2R2R was not present during charging.
  3. I(t)=E2R e−t/(2RC)I(t) = \frac{\mathcal{E}}{2R}\,e^{-t/(2RC)}, because the larger resistor 2R2R dominates the discharge and sets both the initial amplitude and the time constant.
  4. I(t)=E3R e−t/(3RC)I(t) = \frac{\mathcal{E}}{3R}\,e^{-t/(3RC)}, because both RR and 2R2R are now in series in the discharge loop, giving total resistance 3R3R and time constant 3RC3RC. (correct answer)
Explanation: 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+t = 0^+, the capacitor holds voltage E\mathcal{E} (fully charged). The moment the switch moves to position B, the capacitor drives current through a closed loop containing both RR and 2R2R in series. The total resistance is R+2R=3RR + 2R = 3R. Applying Ohm's law at the initial moment gives I(0)=E/3RI(0) = \mathcal{E}/3R. The time constant for any RC discharge is τ=Rtotal⋅C=3RC\tau = R_{\text{total}} \cdot C = 3RC. Putting it together: I(t)=E3R e−t/(3RC)I(t) = \frac{\mathcal{E}}{3R}\,e^{-t/(3RC)}, confirming D. Choice A ignores 2R2R 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 E/3R\mathcal{E}/3R but then incorrectly uses RCRC as the time constant, as if 2R2R only affects amplitude but not the rate of decay. Both resistors in series slow the discharge — 2R2R absolutely affects τ\tau. Choice C arbitrarily elevates 2R2R 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 τ=RC\tau = RC uses the total resistance seen by the capacitor in the current loop — not the resistance from a previous phase.