Quiz: RC Circuit Charging and Discharging — 17 questions

1. Which component of a series RC circuit supplies energy to the circuit rather than storing charge and energy?

A. The previously discharged capacitor

B. The ideal voltage generator

C. The ohmic resistor

D. The switching device

Answer: (B)

Explanation

The ideal voltage generator provides the electromotive force and supplies energy to the circuit. The capacitor stores charge and energy, while the resistor dissipates energy and the switch controls the circuit.

2. What happens when switch K is closed at $t=0$ in the described RC circuit?

A. The capacitor immediately reaches its permanent charge

B. The capacitor discharges through the resistor

C. The generator stops supplying energy to the circuit

D. The capacitor begins charging until the switch is opened

Answer: (D)

Explanation

Closing switch K at $t=0$ starts the capacitor-charging process, which continues until the switch is opened at $t=40\,\text{ms}$. Opening the switch interrupts charging rather than initiating it.

3. For a series RC circuit governed by $RC\frac{dq(t)}{dt}+q(t)=CE$, which identification of the differential-equation constants is correct?

A. $\alpha=RC$ and $\beta=CE$

B. $\alpha=CE$ and $\beta=RC$

C. $\alpha=R+C$ and $\beta=E$

D. $\alpha=E/R$ and $\beta=RC$

Answer: (A)

Explanation

Matching the equation $\alpha\frac{dq}{dt}+q=\beta$ gives $\alpha=RC$ and $\beta=CE$. The quantity $RC$ is also the time constant, but the constant term on the right side is $CE$.

4. What physical property does the RC time constant $\tau=RC$ characterize?

A. The voltage drop across the resistor at equilibrium

B. The final charge stored by the capacitor

C. The speed of capacitor charging

D. The initial current supplied by the generator

Answer: (C)

Explanation

The time constant $\tau=RC$ describes how rapidly the capacitor charges. The final charge is instead determined by $Q_0=CE$, so it represents the charging result rather than the charging speed.

5. For an initially uncharged capacitor, which expression gives the charge at time $t$?

A. $q(t)=RC\left(1-e^{-t/(CE)}\right)$

B. $q(t)=CEe^{-t/(RC)}$

C. $q(t)=\frac{E}{R}\left(1-e^{-t/(RC)}\right)$

D. $q(t)=CE\left(1-e^{-t/(RC)}\right)$

Answer: (D)

Explanation

An initially uncharged capacitor follows $q(t)=CE\left(1-e^{-t/(RC)}\right)$ and approaches the permanent charge $Q_0=CE$. The exponential-only form describes a decaying quantity such as current, not the accumulated charge.

6. How does the charging current behave in an RC circuit?

A. It starts at $E$ and decreases toward $R$

B. It remains constant at $E/R$ throughout charging

C. It starts at zero and increases toward $E/R$

D. It starts at $E/R$ and decreases toward zero

Answer: (D)

Explanation

The current is $i(t)=\frac{E}{R}e^{-t/(RC)}$, so its initial value is $I_0=E/R$ and it decays toward zero. The capacitor's increasing charge causes the current to diminish as charging progresses.

7. During capacitor charging, which description correctly compares the resistor and capacitor voltages?

A. $u_R(t)=Ee^{-t/\tau}$ decreases while $u_C(t)=E(1-e^{-t/\tau})$ increases

B. Both voltages decrease from $E$ toward zero

C. $u_R(t)=E(1-e^{-t/\tau})$ increases while $u_C(t)=Ee^{-t/\tau}$ decreases

D. Both voltages increase from zero toward $E$

Answer: (A)

Explanation

The resistor voltage is $u_R(t)=Ee^{-t/\tau}$, so it falls from $E$ toward zero, while the capacitor voltage is $u_C(t)=E(1-e^{-t/\tau})$, so it rises from zero toward $E$. The two voltages therefore evolve in opposite directions during charging.

8. A charging experiment is observed at $t=40\,\text{ms}$, but the current and resistor-voltage curve are still above zero. Has the circuit reached the permanent regime?

A. Yes, because any finite observation time defines the final state

B. No, because the current has not yet fallen to zero

C. No, because the charging time must always exceed $40\,\text{ms}$

D. Yes, because the capacitor voltage has begun approaching its final value

Answer: (B)

Explanation

The permanent regime requires the current and resistor voltage to have reached zero while the capacitor voltage has reached its final value. A finite-time measurement is not sufficient when the current remains nonzero.

9. How is the time constant $\tau$ determined from a graph of the capacitor voltage during charging?

A. By extending the final asymptote until it intersects the time axis

B. By extending the tangent at $t=0$ until it meets the final asymptote

C. By measuring the time at which the current first becomes nonzero

D. By reading the voltage value where the final asymptote begins

Answer: (B)

Explanation

The tangent drawn at $t=0$ intersects the final asymptote at the time corresponding to $\tau$. The asymptote itself identifies the permanent value rather than the time constant.

10. A capacitor is charged through a circuit until it reaches the permanent regime. Which expression gives the stored electrostatic energy?

A. $E_e=\frac{1}{2}C^2E=\frac{Q_0^2}{2C^2}$

B. $E_e=\frac{1}{2}CE^2=\frac{Q_0^2}{2C}$

C. $E_e=\frac{1}{2}CE=\frac{Q_0}{2C}$

D. $E_e=CE^2=\frac{Q_0^2}{C}$

Answer: (B)

Explanation

At the permanent regime, the capacitor energy is $E_e=\frac{1}{2}CE^2$, equivalently $\frac{Q_0^2}{2C}$. Omitting the factor of one-half or changing the powers of capacitance gives an incorrect energy expression.

11. Two resistors $R_0$ and $R$ are connected in series with a capacitor $C$. What determines the circuit time constant?

A. $\tau=(R_0+R)C$

B. $\tau=R_0C$

C. $\tau=RC$

D. $\tau=\frac{C}{R_0+R}$

Answer: (A)

Explanation

The capacitor experiences the total series resistance, so the time constant is $\tau=(R_0+R)C$. Using either resistor alone neglects part of the resistance that controls the transient.

12. In a series circuit containing $R_0$, $R$, and a capacitor, how are the resistor voltages and capacitor voltage related?

A. $u_R(t)=\frac{R}{R_0}u_{R_0}(t)$ and $u_C(t)=E-u_R(t)-u_{R_0}(t)$

B. $u_R(t)=u_{R_0}(t)$ and $u_C(t)=E-u_R(t)+u_{R_0}(t)$

C. $u_R(t)=\frac{R}{R_0}u_{R_0}(t)$ and $u_C(t)=E+u_R(t)-u_{R_0}(t)$

D. $u_R(t)=\frac{R_0}{R}u_{R_0}(t)$ and $u_C(t)=E+u_R(t)+u_{R_0}(t)$

Answer: (A)

Explanation

Series resistors carry the same current, so their voltage ratio equals their resistance ratio, and Kirchhoff’s voltage law gives $u_C(t)=E-u_R(t)-u_{R_0}(t)$. Reversing the resistance ratio or changing the voltage signs violates these relationships.

13. A capacitor reaches $99\%$ of its maximum voltage after a duration $\theta$. What is the corresponding relation between $\theta$ and the time constant?

A. $\theta=\tau\ln(100)\approx4.6\tau$

B. $\theta=\tau\ln(10)\approx2.3\tau$

C. $\theta=\tau\ln(20)\approx3\tau$

D. $\theta=\frac{\tau}{\ln(100)}\approx0.22\tau$

Answer: (A)

Explanation

Reaching $99\%$ means the remaining voltage difference is $1\%$, leading to $\theta=\tau\ln(100)\approx4.6\tau$. About $3\tau$ corresponds to roughly $95\%$ charging, not the 99% criterion.

14. A capacitor is initially charged and then connected through resistors $R_0$ and $R_1$. Which pair gives the initial current and time constant?

A. $I_0=\frac{E}{R_0+R_1}$ and $\tau=(R_0+R_1)C$

B. $I_0=\frac{E}{R_1}$ and $\tau=R_0C$

C. $I_0=\frac{E}{R_0+R_1}$ and $\tau=\frac{C}{R_0+R_1}$

D. $I_0=\frac{E}{R_0R_1}$ and $\tau=(R_0+R_1)C$

Answer: (A)

Explanation

At the instant the circuit is closed, the series resistance limits the current, giving $I_0=\frac{E}{R_0+R_1}$, while the same total resistance gives $\tau=(R_0+R_1)C$. The product of resistance and capacitance, rather than their quotient, determines the time constant.

15. What is the current through an initially uncharged capacitor at the instant charging begins?

A. $I_0=\frac{E}{C}$

B. $I_0=\frac{C}{R}$

C. $I_0=\frac{R}{E}$

D. $I_0=\frac{E}{R}$

Answer: (D)

Explanation

At the start, the capacitor voltage is zero, so the full emf drives the current through the resistance, giving $I_0=\frac{E}{R}$. The expression $\frac{E}{C}$ does not represent the initial current and confuses capacitance with resistance in the circuit relation.

16. Why does a capacitor take a finite time to reach its final charge during charging?

A. Its resistance rises until the current becomes permanently constant.

B. Its final voltage is reached before any charge accumulates.

C. Its charge increases progressively toward the final value.

D. Its capacitance decreases as the voltage across it rises.

Answer: (C)

Explanation

The charging curve rises progressively rather than making an immediate jump, showing that charge accumulates over a finite time. A changing capacitance is not the stated cause, and the other choices do not describe the observed charging behavior.

17. A circuit must store the same final capacitor energy while charging more slowly. If the emf and capacitance remain fixed, what change is required?

A. Decrease the resistance so the charging current becomes smaller.

B. Increase the emf so the charging process takes more time.

C. Increase the resistance so the time constant becomes larger.

D. Increase the capacitance so the final energy remains unchanged.

Answer: (C)

Explanation

The final energy $E_e=\frac12CE^2$ remains unchanged when $C$ and $E$ are fixed, while the time constant $\tau=RC$ increases when resistance increases. Decreasing resistance would speed up charging, and changing capacitance or emf would alter the final energy under the stated conditions.