Resistor–Capacitor (RC) Circuits - AP Physics C Electricity and Magnetism

A capacitor with an initial charge $Q_0$ is connected in a simple loop with a resistor $R$ at $t=0$ . Let $Q$ be the charge on the capacitor's positive plate at time $t$ . Which of the following differential equations, derived from Kirchhoff's loop rule, describes this discharging circuit?
Answer: $\frac{Q}{C} + \frac{dQ}{dt}R = 0$ Applying Kirchhoff's loop rule for a simple discharging RC circuit, the sum of potential differences around the loop is zero. Traversing the loop in the direction of current flow, the potential drops across the resistor by $V_R = IR$ . The potential increases across the capacitor by $V_C = Q/C$ (going from the negative to the positive plate). So, $V_C - V_R = 0$ , or $Q/C - IR = 0$ . The current $I$ is the rate at which charge leaves the positive plate, so $I = -dQ/dt$ . Substituting this into the loop equation gives $Q/C - (-dQ/dt)R = 0$ , which simplifies to $\frac{Q}{C} + \frac{dQ}{dt}R = 0$ .
A $10 \, \mu\text{F}$ capacitor is charged to a potential difference of $20 \text{ V}$ . It is then disconnected from the charging source and connected at $t=0$ to a network of resistors. The network consists of a $2 \, \Omega$ resistor in series with a parallel combination of a $3 \, \Omega$ resistor and a $6 \, \Omega$ resistor. What is the initial current flowing from the capacitor?
Answer: $5.0 \text{ A}$ First, find the equivalent resistance of the resistor network. The parallel combination of the $3 \, \Omega$ and $6 \, \Omega$ resistors has a resistance of $R_p = (\frac{1}{3} + \frac{1}{6})^{-1} = (\frac{3}{6})^{-1} = 2 \, \Omega$ . This is in series with the $2 \, \Omega$ resistor, so the total equivalent resistance is $R_{eq} = 2 \, \Omega + R_p = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega$ . The initial potential difference across this network is the capacitor's initial voltage, $V_0 = 20 \text{ V}$ . The initial current is given by Ohm's Law: $I_0 = V_0 / R_{eq} = 20 \text{ V} / 4 \, \Omega = 5.0 \text{ A}$ .
A capacitor of capacitance $C$ holds an initial charge $Q_0$ . At time $t=0$ , a switch is closed to connect the capacitor in series with a resistor of resistance $R$ . Which expression represents the current $I(t)$ in the resistor as a function of time, where positive current is defined as flowing away from the initially positive plate?
Answer: $I(t) = \frac{Q_0}{RC} e^{-t/RC}$ During discharge, the charge on the capacitor decreases exponentially: $Q(t) = Q_0 e^{-t/RC}$ . The current $I(t)$ 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 $I(t) = -dQ/dt$ . Differentiating $Q(t)$ with respect to time gives $dQ/dt = Q_0(-\frac{1}{RC})e^{-t/RC}$ . Therefore, $I(t) = -(-\frac{Q_0}{RC}e^{-t/RC}) = \frac{Q_0}{RC}e^{-t/RC}$ .
Consider the RC circuit shown. A $6.0\,\text{V}$ capacitor initially at $6.0\,\text{V}$ discharges through $R=3.0\,\text{k}\Omega$ , $C=100\,\mu\text{F}$ ; what is $V_C$ after $0.60\,\text{s}$ ?
Answer: $0.81\,\text{V}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically analyzing capacitor discharge in RC circuits. During discharge, voltage follows V(t) = V₀ $e^{-t/RC}$ , where V₀ is the initial voltage across the capacitor. With R = 3.0 kΩ and C = 100 μF, the time constant is τ = RC = (3.0×10³)(100×10⁻⁶) = 0.30 s, and the capacitor starts at V₀ = 6.0 V. Choice A is correct because after t = 0.60 s (two time constants), $V_C$ = 6.0× $e^{-0.60/0.30}$ = 6.0× $e^{-2}$ = 6.0×0.135 = 0.81 V. Choice B might result from using only one time constant (6.0× $e^{-1}$ ≈ 2.2 V) or calculation error, while choice D might incorrectly use the charging formula. To help students: remember that after two time constants, voltage drops to about 13.5% of initial value ( $e^{-2}$ ≈ 0.135), and practice recognizing multiples of the time constant. Watch for confusion between charging and discharging scenarios and ensure proper exponential calculations.
For a series RC circuit connected to a DC voltage source at $t=0$ , which of the following best describes the graph of the potential difference across the resistor, $V_R$ , versus time?
Answer: An exponential decay curve, starting from the source emf $E$ and asymptotically approaching 0. At $t=0$ , the capacitor is uncharged and the current is maximum, $I_0 = E/R$ . Thus, the initial potential difference across the resistor is $V_R(0) = I_0 R = E$ . As the capacitor charges, the current decreases exponentially as $I(t) = (E/R) e^{-t/RC}$ . Consequently, the voltage across the resistor also decreases exponentially: $V_R(t) = I(t)R = E e^{-t/RC}$ . This function represents an exponential decay from an initial value of $E$ to a final value of 0.
Circuit X consists of a resistor $R$ and a capacitor $C$ in series with a battery. Circuit Y consists of a resistor $2R$ and a capacitor $C/2$ in series with an identical battery. How does the time constant $τ_X$ of Circuit X compare to the time constant $τ_Y$ of Circuit Y?
Answer: $τ_X = τ_Y$ The time constant of an RC circuit is given by the formula $τ = RC$ . For Circuit X, the time constant is $τ_X = RC$ . For Circuit Y, the time constant is $τ_Y = (2R)(C/2) = RC$ . Therefore, the time constants of the two circuits are equal, $τ_X = τ_Y$ .
Consider the RC circuit shown. A precharged capacitor $C=2.0\times10^{-4}\,\text{F}$ at $9.0\,\text{V}$ discharges through $R=1.5\,\text{k}\Omega$ ; what is $V_C$ after $0.30\,\text{s}$ ?
Answer: $3.3\,\text{V}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically analyzing RC circuits during capacitor discharge. For a discharging capacitor, the voltage follows V(t) = V₀ $e^{-t/RC}$ , where V₀ is the initial voltage and τ = RC is the time constant. In this circuit with R = 1.5 kΩ and C = 2.0×10⁻⁴ F, the time constant is τ = (1.5×10³)(2.0×10⁻⁴) = 0.30 s, with initial voltage V₀ = 9.0 V. Choice A is correct because after t = 0.30 s (one time constant), $V_C$ = 9.0× $e^{-0.30/0.30}$ = 9.0× $e^{-1}$ = 9.0×0.368 = 3.3 V. Choice B might result from incorrectly using the charging formula (9.0×(1- $e^{-1}$ ) ≈ 5.7 V) instead of the discharging formula. To help students: emphasize the difference between charging and discharging equations, and remember that after one time constant of discharge, voltage drops to about 36.8% of initial value. Watch for sign errors and confusion about which formula applies to which scenario.
What is the current through resistor 1 at time $t=0$ , and what is the current through resistor 1 after a very long time ( $t \to \infty$ )?
Answer: At $t=0$ , $I_1=6.0 \text{ A}$ ; at $t \to \infty$ , $I_1=2.0 \text{ A}$ At time $t=0$ , 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 $R_1 = 2 \, \Omega$ . The initial current through resistor 1 is $I_1(0) = E/R_1 = 12 \text{ V} / 2 \, \Omega = 6.0 \text{ A}$ . After a very long time ( $t \to \infty$ ), 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 $R_{total} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega$ . The final current through resistor 1 is $I_1(\infty) = E/R_{total} = 12 \text{ V} / 6 \, \Omega = 2.0 \text{ A}$ .
A capacitor of capacitance $C$ is charged to a potential difference $E$ by a battery through a resistor $R$ . What is the total energy dissipated as heat by the resistor during the entire charging process?
Answer: $\frac{1}{2}CE^2$ The total energy supplied by the battery is the total charge that flows, $Q_{final} = CE$ , multiplied by the battery emf $E$ , which is $W_{batt} = Q_{final}E = (CE)E = CE^2$ . The energy stored in the fully charged capacitor is $U_C = \frac{1}{2}CV^2 = \frac{1}{2}CE^2$ . 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: $E_{dissipated} = W_{batt} - U_C = CE^2 - \frac{1}{2}CE^2 = \frac{1}{2}CE^2$ .
A series RC circuit with a time constant $τ$ is connected to an ideal battery at time $t=0$ . 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?
Answer: 63% of its maximum value. In a charging RC circuit, the charge on the capacitor as a function of time is given by $Q(t) = Q_{max}(1 - e^{-t/τ})$ . At time $t = τ$ , the charge is $Q(τ) = Q_{max}(1 - e^{-1})$ . Since $e^{-1} \approx 0.37$ , the charge is $Q_{max}(1 - 0.37) = 0.63 Q_{max}$ , which is 63% of the maximum charge.
A series circuit contains an ideal battery with emf $E$ , a resistor $R$ , an uncharged capacitor $C$ , and a switch. The switch is closed at $t=0$ . Which differential equation correctly describes the charge $Q$ on the capacitor as a function of time $t$ during the charging process?
Answer: $E - \frac{Q}{C} - \frac{dQ}{dt}R = 0$ 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 $E$ . There are potential drops across the capacitor ( $V_C = Q/C$ ) and the resistor ( $V_R = IR$ ). Since current is the rate of change of charge, $I = dQ/dt$ . Thus, the loop equation is $E - V_C - V_R = 0$ , which becomes $E - \frac{Q}{C} - \frac{dQ}{dt}R = 0$ .
Consider the RC circuit shown. A $12\,\text{V}$ capacitor initially at $12\,\text{V}$ discharges through $R=10\,\text{k}\Omega$ , $C=10\,\mu\text{F}$ ; what is $V_C$ after $0.050\,\text{s}$ ?
Answer: $7.3\,\text{V}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating voltage during capacitor discharge. For a discharging capacitor, V(t) = V₀ $e^{-t/RC}$ , where V₀ is the initial voltage. With R = 10 kΩ and C = 10 μF, the time constant is τ = RC = (10×10³)(10×10⁻⁶) = 0.10 s, and the capacitor starts at V₀ = 12 V. Choice A is correct because after t = 0.050 s (half a time constant), $V_C$ = 12× $e^{-0.050/0.10}$ = 12× $e^{-0.5}$ = 12×0.606 = 7.3 V. Choice B might result from using the wrong exponential value or calculation error, while choice C might incorrectly apply the charging formula. To help students: memorize that after half a time constant of discharge, voltage drops to about 60.6% of initial value, and practice quick mental calculations with common exponential values. Watch for confusion between charging and discharging formulas and ensure accurate exponential calculations.
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?
Answer: The time is doubled; the final charge is unchanged. The time constant of the circuit is $τ = RC$ . Doubling the resistance $R$ 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, $Q_{final} = CE$ , depends only on the capacitance and the emf of the source, neither of which has changed. Thus, the final charge is unchanged.
In a series RC circuit with time constant $τ$ , an initially uncharged capacitor is being charged by a battery with emf $E$ . At what time $t$ will the potential difference across the capacitor be equal to half of the battery's emf?
Answer: $t = τ \ln(2)$ The potential difference across a charging capacitor is given by $V_C(t) = E(1 - e^{-t/τ})$ . We need to find the time $t$ when $V_C(t) = E/2$ . Setting the expressions equal: $E/2 = E(1 - e^{-t/τ})$ . This simplifies to $1/2 = 1 - e^{-t/τ}$ , which gives $e^{-t/τ} = 1/2$ . Taking the natural logarithm of both sides yields $-t/τ = \ln(1/2) = -\ln(2)$ . Therefore, $t = τ \ln(2)$ .
A fully charged capacitor is discharged through a resistor. The time constant $τ$ of the circuit represents the time required for the charge on the capacitor to decrease to approximately what percentage of its initial value?
Answer: 37% of its initial value. During the discharge of a capacitor, the charge remaining on it as a function of time is given by $Q(t) = Q_0 e^{-t/τ}$ , where $Q_0$ is the initial charge. At time $t = τ$ , the charge is $Q(τ) = Q_0 e^{-1}$ . Since $e^{-1} \approx 0.37$ , the charge has decreased to approximately 37% of its initial value.
What is the time constant $τ$ of the circuit?
Answer: $0.40 \text{ s}$ For a discharging capacitor, the voltage is given by $V_C(t) = V_0 e^{-t/τ}$ . Taking the natural logarithm of both sides gives $\ln(V_C) = \ln(V_0) - t/τ$ . This equation is in the form of a line, $y = b + mx$ , where $y = \ln(V_C)$ , $x = t$ , the y-intercept $b = \ln(V_0)$ , and the slope $m = -1/τ$ . Given that the slope is $-2.5 \text{ s}^{-1}$ , we have $-1/τ = -2.5 \text{ s}^{-1}$ . Solving for $τ$ gives $τ = 1/2.5 = 0.40 \text{ s}$ .
A $12 \text{ V}$ battery is connected in series with a $2.0 \, \Omega$ resistor and a combination of two capacitors. A $3.0 \, \mu\text{F}$ capacitor is connected in parallel with a $6.0 \, \mu\text{F}$ capacitor. What is the time constant for the charging of this circuit?
Answer: $18 \, \mu\text{s}$ For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances: $C_{eq} = C_1 + C_2 = 3.0 \, \mu\text{F} + 6.0 \, \mu\text{F} = 9.0 \, \mu\text{F}$ . The time constant of an RC circuit is given by the product of the resistance and the equivalent capacitance: $τ = RC_{eq} = (2.0 \, \Omega)(9.0 \, \mu\text{F}) = 18 \, \mu\text{s}$ .
Consider the RC circuit shown: a capacitor initially at 10 V discharges through $R=4.7\times10^{3}\,\Omega$ with $C=4.7\times10^{-6}\,\text{F}$ . What is the remaining voltage across the capacitor after discharging for 0.022 s?
Answer: $V_C\approx 3.7\,\text{V}$ 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) = V₀ $e^{-t/RC}$ , 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.
A capacitor with capacitance $C$ is connected in series with a resistor $R$ and a battery with emf $E$ . The circuit is allowed to reach a steady state, and the energy stored is $U$ . If the capacitance is then changed to $2C$ while the battery and resistor remain the same, what is the new energy stored in the capacitor after a new steady state is reached?
Answer: $2U$ The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$ . In a steady state after being charged by a battery with emf $E$ , the potential difference across the capacitor is $V = E$ . The initial energy stored is $U = \frac{1}{2}CE^2$ . When the capacitance is changed to $2C$ , the new steady-state energy will be $U_{new} = \frac{1}{2}(2C)E^2 = 2(\frac{1}{2}CE^2) = 2U$ . The stored energy is doubled.
The charge on a capacitor in a charging RC circuit is given by the function $Q(t) = (10 \, \mu\text{C})(1 - e^{-t/(2.0\text{ s})})$ . What is the current $I(t)$ flowing into the capacitor at $t=0$ ?
Answer: $5.0 \, \mu\text{A}$ Current is the time derivative of charge, $I(t) = dQ/dt$ . Differentiating the given function: $I(t) = \frac{d}{dt} [(10 \, \mu\text{C})(1 - e^{-t/(2.0\text{ s})})] = (10 \, \mu\text{C}) (-(-\frac{1}{2.0\text{ s}})e^{-t/(2.0\text{ s})}) = (5.0 \, \mu\text{A}) e^{-t/(2.0\text{ s})}$ . To find the current at $t=0$ , we evaluate this expression at $t=0$ : $I(0) = (5.0 \, \mu\text{A}) e^{0} = 5.0 \, \mu\text{A}$ .

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