Differential Equations Quiz: Rc Circuit Models

What this quiz covers

This quiz focuses on Rc Circuit Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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. $Q_{total} = V_0 C$
2. $Q_{total} = 2V_0 C$ (correct answer)
3. $Q_{total} = \frac{V_0 C}{2}$
4. $Q_{total} = V_0 C\sqrt{2}$

Explanation: In steady state, no current flows through the resistors, so the voltage across each capacitor equals the applied voltage $V_0$. Since the circuits are identical and connected in parallel, each capacitor independently charges to $q = CV_0$. The total charge is the sum: $Q_{total} = q_1 + q_2 = CV_0 + CV_0 = 2CV_0$. This is equivalent to having an effective capacitance of $2C$ charged to voltage $V_0$. Choice A treats the combination as a single capacitor $C$. Choice C incorrectly assumes voltage division. Choice D applies an incorrect factor related to impedance calculations.

Question 2

1. $RC \ln(2)$ (correct answer)
2. $RC$
3. $\frac{RC}{\ln(2)}$
4. $2RC$

Explanation: The charge on a charging capacitor is given by $Q(t) = Q_{max}(1 - e^{-t/RC})$, where $Q_{max} = CE_0$. We want to find the time $t$ when $Q(t) = \frac{1}{2}Q_{max}$. Setting up the equation: $\frac{1}{2}Q_{max} = Q_{max}(1 - e^{-t/RC})$. This simplifies to $\frac{1}{2} = 1 - e^{-t/RC}$, which gives $e^{-t/RC} = \frac{1}{2}$. Taking the natural logarithm of both sides, we get $-t/RC = \ln(1/2) = -\ln(2)$. Therefore, $t = RC \ln(2)$.

Question 3

1. $R=500 \, \Omega$, $C=2 \, \mathrm{F}$, $E(t)=20 \cos(10t) \, \mathrm{V}$
2. $R=2 \, \Omega$, $C=500 \, \mathrm{F}$, $E(t)=20 \cos(10t) \, \mathrm{V}$
3. $R=1 \, \Omega$, $C=4 \, \mathrm{mF}$, $E(t)=10 \cos(10t) \, \mathrm{V}$
4. $R=2 \, \Omega$, $C=2 \, \mathrm{mF}$, $E(t)=20 \cos(10t) \, \mathrm{V}$ (correct answer)

Explanation: RC circuit problems require you to match the given differential equation with the standard form. For a series RC circuit, Kirchhoff's voltage law gives us $R\frac{dQ}{dt} + \frac{Q}{C} = E(t)$, where $R$ is resistance, $C$ is capacitance, and $E(t)$ is the voltage source. To find the circuit parameters, rewrite your given equation $2\frac{dQ}{dt} + 500Q = 20\cos(10t)$ in standard form by comparing coefficients. The coefficient of $\frac{dQ}{dt}$ tells us that $R = 2\,\Omega$. The coefficient of $Q$ equals $\frac{1}{C}$, so $\frac{1}{C} = 500$, which means $C = \frac{1}{500} = 0.002\,\mathrm{F} = 2\,\mathrm{mF}$. The right side directly gives us $E(t) = 20\cos(10t)\,\mathrm{V}$. Choice A incorrectly swaps the resistance and capacitance relationships—it treats the coefficient 500 as resistance rather than $\frac{1}{C}$, and treats 2 as capacitance rather than resistance. Choice B makes the same coefficient confusion as A, additionally giving an impossibly large capacitance of 500 F. Choice C gets the resistance calculation wrong ($R$ should be 2, not 1), miscalculates the capacitance as 4 mF instead of 2 mF, and incorrectly halves the voltage amplitude to 10 V instead of 20 V. Remember this pattern: in the standard RC equation $R\frac{dQ}{dt} + \frac{Q}{C} = E(t)$, the coefficient of the derivative term is always the resistance, and the coefficient of $Q$ is always the reciprocal of capacitance. This systematic approach prevents coefficient mix-ups.

Question 4

1. $2.5 \ln(3) \, \mathrm{s}$
2. $5 \ln(2) \, \mathrm{s}$
3. $5 \ln(3) \, \mathrm{s}$
4. $5 \ln(2+\sqrt{3}) \, \mathrm{s}$ (correct answer)

Explanation: The energy stored in a capacitor is $U(t) = \frac{Q(t)^2}{2C}$. The maximum energy is $U_{max} = \frac{(CE_0)^2}{2C}$. We want the time $t$ when $U(t) = 0.75 U_{max}$, which means $Q(t) = \sqrt{0.75} CE_0 = \frac{\sqrt{3}}{2} CE_0$. The charge during charging is $Q(t) = CE_0(1-e^{-t/RC})$. Setting $\frac{\sqrt{3}}{2} = 1-e^{-t/RC}$ gives $e^{-t/RC} = 1 - \frac{\sqrt{3}}{2} = \frac{2-\sqrt{3}}{2}$. Solving for $t$: $t = RC \ln\left(\frac{2}{2-\sqrt{3}}\right) = RC \ln(2+\sqrt{3})$. With $\tau=RC = 5 \, \mathrm{s}$, we get $t=5\ln(2+\sqrt{3}) \, \mathrm{s}$.

Question 5

1. $I_A(0)=2I_B(0)$ and $Q_{A,final}=Q_{B,final}$
2. $I_A(0)=I_B(0)/2$ and $Q_{A,final}=Q_{B,final}/2$
3. $I_A(0)=2I_B(0)$ and $Q_{A,final}=2Q_{B,final}$ (correct answer)
4. $I_A(0)=I_B(0)$ and $Q_{A,final}=2Q_{B,final}$

Explanation: When analyzing RC circuits, you need to understand two key behaviors: initial current depends on resistance alone (since uncharged capacitors act like short circuits), while final charge depends on capacitance and applied voltage. For initial current, apply Ohm's law at $t=0$. Since both capacitors start uncharged, they initially act as short circuits, so $I(0) = E_0/R$. Circuit A gives $I_A(0) = E_0/R_A$, while Circuit B gives $I_B(0) = E_0/R_B = E_0/(2R_A)$. Therefore, $I_A(0) = 2I_B(0)$. For final charge, use $Q_{final} = CE_0$ (when the capacitor is fully charged, current drops to zero). Circuit A reaches $Q_{A,final} = C_A E_0$, while Circuit B reaches $Q_{B,final} = C_B E_0 = (C_A/2)E_0$. Therefore, $Q_{A,final} = 2Q_{B,final}$. Looking at the choices: Choice A incorrectly states the final charges are equal—this ignores that Circuit B has half the capacitance. Choice B gets both relationships backwards, suggesting Circuit A has lower initial current and final charge. Choice D correctly identifies equal initial currents, but this contradicts Ohm's law since the resistances differ. Choice C correctly captures both relationships: Circuit A has twice the initial current (due to half the resistance) and twice the final charge (due to twice the capacitance). Study tip: Remember that at $t=0$, capacitors are shorts (focus on resistance), while at $t=\infty$, capacitors are open circuits (focus on capacitance and voltage).

Question 6

1. The initial current is doubled, and the circuit charges faster.
2. The initial current is unchanged, and the steady-state charge is halved.
3. The initial current is halved, and the steady-state charge is halved.
4. The initial current is halved, and the steady-state charge is unchanged. (correct answer)

Explanation: When analyzing RC circuits, you need to understand how resistance affects both the initial current and the steady-state behavior. The key is recognizing that initial current depends on Ohm's law, while steady-state charge depends on the capacitor's voltage. For an RC charging circuit, the initial current occurs when the capacitor acts like a short circuit (no voltage drop across it yet). Using Ohm's law: $I_0 = \frac{E_0}{R}$. When resistance doubles from $R$ to $2R$, the initial current becomes $I_{new} = \frac{E_0}{2R} = \frac{I_0}{2}$, so it's halved. At steady state, the capacitor is fully charged and no current flows. The voltage across the capacitor equals the source voltage $E_0$, regardless of resistance (since there's no current through the resistor). The steady-state charge is $Q = CE_0$, which depends only on capacitance and source voltage, not resistance. Answer A incorrectly suggests doubling the resistance doubles the initial current, but Ohm's law shows current decreases when resistance increases. Answer B wrongly claims the steady-state charge is halved, but this charge depends on $CE_0$, not resistance. Answer C makes both errors: claiming halved steady-state charge and misunderstanding that increased resistance actually slows charging (larger time constant $\tau = RC$). Answer D correctly identifies that initial current is halved (Ohm's law) while steady-state charge remains unchanged (depends only on $C$ and $E_0$). Study tip: In RC circuits, resistance affects the rate of charging and initial current, but never the final steady-state charge or voltage across the capacitor.

Question 7

1. $120 e^{-0.5} \, \mu\mathrm{C}$ (correct answer)
2. $120 e^{-1} \, \mu\mathrm{C}$
3. $120 e^{-1/3} \, \mu\mathrm{C}$
4. $120(1 - e^{-0.5}) \, \mu\mathrm{C}$

Explanation: First, find the initial condition for the discharging phase. After being connected to the source for a long time, the capacitor is fully charged to $Q(0) = CE_0 = (10 \times 10^{-6} \, \mathrm{F})(12 \, \mathrm{V}) = 120 \, \mu\mathrm{C}$. At $t=0$, the switch moves, and the capacitor discharges through $R_2$. The discharging equation is $Q(t) = Q(0)e^{-t/\tau_2}$, where the time constant is $\tau_2 = R_2 C = (2 \times 10^3 \, \Omega)(10 \times 10^{-6} \, \mathrm{F}) = 0.02 \, \mathrm{s}$. We need to find the charge at $t=0.01 \, \mathrm{s}$: $Q(0.01) = 120 \, \mu\mathrm{C} \times e^{-0.01/0.02} = 120 e^{-0.5} \, \mu\mathrm{C}$.

Question 8

1. $\tau/2$
2. $\tau$
3. $2\tau$ (correct answer)
4. $\tau \ln(2)$

Explanation: For a discharging capacitor, the charge is $Q(t) = Q_0 e^{-t/\tau}$. The current is $I(t) = \frac{dQ}{dt} = -\frac{Q_0}{\tau}e^{-t/\tau}$. The initial current at $t=0$ is $I(0) = -Q_0/\tau$, and its magnitude is $|I(0)| = Q_0/\tau$. The magnitude of the current at time $t$ is $|I(t)| = \frac{Q_0}{\tau}e^{-t/\tau}$. We want to find $t$ such that $|I(t)| = \frac{1}{e^2}|I(0)|$. This gives $\frac{Q_0}{\tau}e^{-t/\tau} = \frac{1}{e^2} \frac{Q_0}{\tau}$, which simplifies to $e^{-t/\tau} = e^{-2}$. Therefore, $t/\tau = 2$, or $t=2\tau$.

Question 9

1. $0.1(1-e^{-3.5}) \, \mathrm{C}$
2. $0.1(1-e^{-2.5})e^{-1} \, \mathrm{C}$ (correct answer)
3. $0.1(e^{-1} - e^{-3.5}) \, \mathrm{C}$
4. $0.1(1-e^{-2.5}) \, \mathrm{C}$

Explanation: This is a two-stage problem. First, the capacitor charges for 5 seconds. The time constant is $\tau = RC = (200)(10 \times 10^{-3}) = 2 \, \mathrm{s}$. The charge at time $t$ for $0 \le t < 5$ is $Q(t) = CE_0(1-e^{-t/\tau}) = (0.01)(10)(1-e^{-t/2}) = 0.1(1-e^{-t/2})$. At $t=5$, the charge is $Q(5) = 0.1(1-e^{-5/2}) = 0.1(1-e^{-2.5}) \, \mathrm{C}$. For $t \ge 5$, the voltage source is turned off, and the capacitor discharges from this charge. The equation for discharging is $Q_d(t') = Q_{initial}e^{-t'/\tau}$, where $t'$ is the time since discharging began ($t'=t-5$). The initial charge for this phase is $Q(5)$. So, at $t=7$, $t'=2$, and the charge is $Q(7) = Q(5) e^{-2/\tau} = 0.1(1-e^{-2.5})e^{-2/2} = 0.1(1-e^{-2.5})e^{-1} \, \mathrm{C}$.

Question 10

1. $50 \, \mathrm{k}\Omega$
2. $100 \, \mathrm{k}\Omega$ (correct answer)
3. $200 \, \mathrm{k}\Omega$
4. $400 \, \mathrm{k}\Omega$

Explanation: The voltage across a discharging capacitor is given by $V(t) = V_0 e^{-t/RC}$, where $V_0$ is the initial voltage. We are given $V_0 = 50 \, \mathrm{V}$, $C = 20 \, \mu\mathrm{F} = 20 \times 10^{-6} \, \mathrm{F}$, and at $t=4 \, \mathrm{s}$, $V(4) = 50/e^2 \, \mathrm{V}$. Substituting these values into the equation: $50/e^2 = 50 e^{-4/(R \cdot 20 \times 10^{-6})}$. Dividing by 50 gives $e^{-2} = e^{-4/(20 \times 10^{-6} R)}$. Equating the exponents, we have $2 = \frac{4}{20 \times 10^{-6} R}$. Solving for R: $R = \frac{4}{2 \cdot 20 \times 10^{-6}} = \frac{4}{40 \times 10^{-6}} = \frac{1}{10 \times 10^{-6}} = 10^5 \, \Omega$, which is $100 \, \mathrm{k}\Omega$.

Question 11

1. $C = 10 \, \mathrm{mF}$, $E_0 = 0.4 \, \mathrm{V}$
2. $C = 10 \, \mathrm{mF}$, $E_0 = 400 \, \mathrm{V}$
3. $C = 1 \, \mathrm{mF}$, $E_0 = 4000 \, \mathrm{V}$ (correct answer)
4. $C = 1 \, \mathrm{mF}$, $E_0 = 4 \, \mathrm{V}$

Explanation: When analyzing RC circuits, you need to understand how the standard form of the charge equation relates to circuit parameters. The general solution for charge in a series RC circuit with constant voltage is $Q(t) = CE_0(1 - e^{-t/(RC)})$, where $CE_0$ represents the steady-state charge and $1/(RC)$ is the time constant. Given $Q(t) = 4(1 - e^{-5t})$, you can directly compare this to the standard form. The coefficient 4 tells you that $CE_0 = 4$ coulombs, and the exponent $-5t$ means $\frac{1}{RC} = 5$, so $RC = 0.2$ seconds. Since $R = 200\,\Omega$ and $RC = 0.2$, you get $C = \frac{0.2}{200} = 0.001$ farads = $1\,\mathrm{mF}$. From $CE_0 = 4$ with $C = 0.001\,\mathrm{F}$, you find $E_0 = \frac{4}{0.001} = 4000\,\mathrm{V}$. Answer A incorrectly uses $C = 10\,\mathrm{mF}$, which would require $RC = 2$ seconds, not 0.2. Answer B makes the same capacitance error and compounds it by using the wrong relationship between charge and voltage. Answer D correctly identifies the capacitance but drastically underestimates the voltage—this represents a units error where someone might have used capacitance in millifarads instead of farads when calculating $E_0$. The correct answer is C: $C = 1\,\mathrm{mF}$, $E_0 = 4000\,\mathrm{V}$. Study tip: Always match your given equation to the standard form first, then extract the time constant and steady-state values. Watch your units carefully—capacitance problems often involve millifarads, which can lead to power-of-10 errors in voltage calculations.

Question 12

1. $\tau_{new} = \tau_{old}$ and $Q_{new} = \frac{1}{2} Q_{old}$ (correct answer)
2. $\tau_{new} = \tau_{old}$ and $Q_{new} = Q_{old}$
3. $\tau_{new} = 4 \tau_{old}$ and $Q_{new} = \frac{1}{2} Q_{old}$
4. $\tau_{new} = \frac{1}{4} \tau_{old}$ and $Q_{new} = Q_{old}$

Explanation: The time constant of an RC circuit is given by $\tau = RC$. The new parameters are $R_{new} = 2R$ and $C_{new} = C/2$. Thus, the new time constant is $\tau_{new} = R_{new}C_{new} = (2R)(C/2) = RC = \tau_{old}$. The steady-state (maximum) charge on the capacitor is given by $Q_{max} = CE_0$. The new steady-state charge is $Q_{new} = C_{new}E_0 = (C/2)E_0 = \frac{1}{2}CE_0 = \frac{1}{2}Q_{old}$.

Question 13

1. $I_{ss} = \frac{k}{R} = \frac{100}{R}\,\text{A}$
2. $I_{ss} = kC = 100C\,\text{A}$ (correct answer)
3. $I_{ss} = \frac{k}{R + \frac{1}{C}} = \frac{100R}{R + \frac{1}{C}}\,\text{A}$
4. $I_{ss} = k\sqrt{RC} = 100\sqrt{RC}\,\text{A}$

Explanation: The circuit equation is $V_R + V_C = kt$, where $V_R = IR$ and $V_C = \frac{q}{C}$. Since $I = \frac{dq}{dt}$, we have $IR + \frac{q}{C} = kt$. Differentiating: $R\frac{dI}{dt} + \frac{I}{C} = k$. In steady state, $\frac{dI}{dt} = 0$, so $\frac{I_{ss}}{C} = k$, giving $I_{ss} = kC = 100C\,\text{A}$. Physically, this makes sense: with a linearly increasing voltage, the capacitor must charge at a constant rate to maintain a constant voltage difference across the resistor, requiring constant current $I = C\frac{dV_C}{dt} = Ck$. Choice A treats it as a purely resistive circuit. Choice C incorrectly combines resistance and capacitive reactance. Choice D has no physical basis for this relationship.

Question 14

1. $t = 2\,\text{ms}$
2. $t = 4\,\text{ms}$ (correct answer)
3. $t = 1\,\text{ms}$
4. $t = \sqrt{2}\,\text{ms}$

Explanation: In the original circuit, $q(t) = q_0 e^{-t/RC}$. When $q = q_0/e$, we have $\frac{q_0}{e} = q_0 e^{-t/RC}$, so $t = RC = 2\,\text{ms}$. When two identical resistors of resistance $R$ each are connected in series, the total resistance becomes $2R$. The new time constant is $\tau_{new} = (2R)C = 2RC = 4\,\text{ms}$. Therefore, it takes $4\,\text{ms}$ for the charge to decrease to $q_0/e$. Choice A assumes the time constant doesn't change. Choice C assumes the resistors are in parallel instead of series. Choice D incorrectly applies a square root relationship.