FAD1022 CQ2: Rapid-Fire Drill Pack — Capacitor & DC

Objective: Master all Capacitor & DC Circuit concepts for CQ2.
Target: 3-4 min per problem (CQ2 style: 4 questions in 15 min).
Total problems: 28
Estimated time: ~100 min

Cheat Sheet (Memorize First)

Constants for CQ2

Constant Symbol Value
Coulomb constant $k$ $9.0 \times 10^{9}\ \text{N·m}^{2}\text{·C}^{-2}$
Permittivity of free space $\varepsilon_{0}$ $8.85 \times 10^{-12}\ \text{F·m}^{-1}$

Capacitor Formulas

Formula Description
$C = \frac{Q}{\Delta V}$ Definition of capacitance
$C = \frac{\varepsilon_0 A}{d}$ Parallel-plate capacitor (air)
$C = \frac{\kappa\varepsilon_0 A}{d}$ With dielectric ($\kappa$ = dielectric constant)
$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$ Energy stored
$C_{\text{eq}} = C_1 + C_2 + ...$ Parallel combination
$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + ...$ Series combination

RC Circuit Formulas

Formula Description
$\tau = RC$ Time constant
$q(t) = Q_0(1 - e^{-t/\tau})$ Charging: charge on capacitor
$i(t) = \frac{V}{R}e^{-t/\tau}$ Charging: current
$V_C(t) = V_0(1 - e^{-t/\tau})$ Charging: capacitor voltage
$q(t) = Q_0 e^{-t/\tau}$ Discharging: charge
$V_C(t) = V_0 e^{-t/\tau}$ Discharging: voltage

DC Circuit Formulas

Formula Description
$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$ Voltage divider (unloaded)
$\sum I_{\text{in}} = \sum I_{\text{out}}$ Kirchhoff's Current Law
$\sum V_{\text{loop}} = 0$ Kirchhoff's Voltage Law

Key Values to Remember

Condition Value
After $t = \tau$ (charging) $V_C \approx 0.632V_0$, $63.2%$ charged
After $t = 2\tau$ (charging) $V_C \approx 0.865V_0$
After $t = 5\tau$ (charging) $V_C \approx 0.993V_0$ (~fully charged)
After $t = \tau$ (discharging) $V_C \approx 0.368V_0$, $36.8%$ remaining
After $t = 5\tau$ (discharging) $V_C \approx 0.007V_0$ (~fully discharged)

Part A: Capacitance & Geometry (6 problems)

Target: 3 min per problem.

Set A1 — Basic Capacitance (3 problems)

  1. A parallel-plate capacitor has plates of area $A = 25\ \text{cm}^2$ separated by $d = 1.0\ \text{mm}$. Calculate its capacitance.

  2. A capacitor stores $Q = 50\ \mu\text{C}$ of charge when connected to $V = 10\ \text{V}$. Find: (a) The capacitance (b) The energy stored

  3. How much charge is stored on a $100\ \mu\text{F}$ capacitor connected to a $12\ \text{V}$ battery? What is the energy stored?

Score: ___/3

Set A2 — Capacitance with Dielectric (3 problems)

  1. A parallel-plate capacitor has $A = 50\ \text{cm}^2$, $d = 2.0\ \text{mm}$, and is filled with a dielectric of $\kappa = 5.0$. Calculate the capacitance.

  2. An air-filled parallel-plate capacitor has $C_0 = 20\ \text{pF}$. If the space between plates is filled with mica ($\kappa = 7.0$), what is the new capacitance?

  3. A capacitor with dielectric ($\kappa = 4.0$) is connected to a $24\ \text{V}$ battery and stores $U = 1.44\ \text{mJ}$. Calculate: (a) The capacitance (b) The charge stored (c) The electric field if $d = 0.5\ \text{mm}$

Score: ___/3

Part B: Capacitor Combinations (6 problems)

Target: 3-4 min per problem.

Set B1 — Series & Parallel (3 problems)

  1. Two capacitors $C_1 = 6.0\ \mu\text{F}$ and $C_2 = 12\ \mu\text{F}$ are connected in series. Find the equivalent capacitance.

  2. Three capacitors $C_1 = 4.0\ \mu\text{F}$, $C_2 = 6.0\ \mu\text{F}$, and $C_3 = 12\ \mu\text{F}$ are connected in parallel across a $9.0\ \text{V}$ battery. Find: (a) The equivalent capacitance (b) The total charge stored (c) The charge on each capacitor

  3. For the capacitors in problem 7, if connected across $V = 12\ \text{V}$, find: (a) The total charge stored (b) The voltage across each capacitor (c) The energy stored in each capacitor

Score: ___/3

Set B2 — Mixed Combinations (3 problems)

  1. In the circuit shown, $C_1 = 3.0\ \mu\text{F}$ and $C_2 = 6.0\ \mu\text{F}$ are in parallel, and this combination is in series with $C_3 = 2.0\ \mu\text{F}$. Find the equivalent capacitance.

  2. Capacitors $C_1 = 2.0\ \mu\text{F}$ and $C_2 = 4.0\ \mu\text{F}$ are in series. This combination is in parallel with $C_3 = 6.0\ \mu\text{F}$. Find the equivalent capacitance.

  3. Four capacitors are arranged: $C_1 = 2.0\ \mu\text{F}$ and $C_2 = 4.0\ \mu\text{F}$ in series, connected in parallel with $C_3 = 3.0\ \mu\text{F}$ and $C_4 = 6.0\ \mu\text{F}$ in series. The combination is connected to $V = 12\ \text{V}$. Find the total energy stored.

Score: ___/3

Part C: RC Circuits — Time Constant & Transients (8 problems)

Target: 3-4 min per problem.

Set C1 — Time Constant Calculations (4 problems)

  1. A $50\ \mu\text{F}$ capacitor is connected in series with a $20\ \text{k}\Omega$ resistor. Calculate the time constant $\tau$.

  2. An RC circuit has $R = 100\ \text{k}\Omega$ and $\tau = 2.5\ \text{s}$. Find the capacitance.

  3. A capacitor discharges through a resistor. After $t = 3\ \text{s}$, the voltage drops to $40%$ of its initial value. If $C = 100\ \mu\text{F}$, find $R$.

  4. Design an RC circuit with $\tau = 1.0\ \text{s}$ using a capacitor $C = 47\ \mu\text{F}$. What resistor value is needed?

Score: ___/4

Set C2 — Charging & Discharging (4 problems)

  1. A $100\ \mu\text{F}$ capacitor is charged through a $50\ \text{k}\Omega$ resistor from a $20\ \text{V}$ source. Calculate: (a) The time constant (b) The voltage across the capacitor after $t = 2.5\ \text{s}$ (c) The current after $t = 2.5\ \text{s}$

  2. A fully charged $50\ \mu\text{F}$ capacitor at $V_0 = 30\ \text{V}$ discharges through a $200\ \text{k}\Omega$ resistor. Calculate: (a) The time constant (b) The voltage after $t = 5\ \text{s}$ (c) The time for the voltage to drop to $10\ \text{V}$

  3. How long does it take for a charging RC circuit to reach $90%$ of its final voltage? Express in terms of $\tau$.

  4. An RC circuit charges from zero. At what time $t$ is the rate of change of voltage maximum? What is that maximum rate?

Score: ___/4

Part D: DC Circuits — Voltage Divider & KCL (8 problems)

Target: 3 min per problem.

Set D1 — Voltage Divider (4 problems)

  1. A voltage divider consists of $R_1 = 4.0\ \text{k}\Omega$ and $R_2 = 6.0\ \text{k}\Omega$ connected across a $15\ \text{V}$ supply. Calculate the output voltage across $R_2$.

  2. Design a voltage divider to provide $V_{\text{out}} = 5.0\ \text{V}$ from a $12\ \text{V}$ supply. The total resistance should be approximately $R_1 + R_2 = 20\ \text{k}\Omega$.

  3. A voltage divider has $R_1 = 10\ \text{k}\Omega$, $R_2 = 15\ \text{k}\Omega$, and $V_{\text{in}} = 9.0\ \text{V}$. A load resistor $R_L = 30\ \text{k}\Omega$ is connected across $R_2$. Find the loaded output voltage.

  4. For the voltage divider in problem 23, calculate the percent change in output voltage when the load is connected.

Score: ___/4

Set D2 — Kirchhoff's Current Law (4 problems)

  1. At a junction, currents $I_1 = 3.0\ \text{A}$ (entering), $I_2 = 5.0\ \text{A}$ (leaving), and $I_3$ (entering) meet. Find $I_3$.

  2. Three wires meet at a node. $I_1 = 2.5\ \text{A}$ flows into the node, $I_2 = -1.5\ \text{A}$ (negative means opposite to defined direction). If $I_3$ is defined as leaving the node, find $I_3$.

  3. In a circuit node: $I_1 = 4.0\ \text{A}$ entering, $I_2 = 2.5\ \text{A}$ leaving, $I_3 = 1.5\ \text{A}$ entering, and $I_4$ leaving. Find $I_4$.

  4. Four branches meet at a node with currents: $I_1 = 6.0\ \text{A}$ (in), $I_2 = 3.0\ \text{A}$ (out), $I_3 = 2.0\ \text{A}$ (in), and $I_4$ unknown. Find $I_4$ (magnitude and direction).

Score: ___/4

Final Scorecard

Part Topic Problems Raw Score
A — Capacitance Sets A1-A2 6 ___/6
B — Combinations Sets B1-B2 6 ___/6
C — RC Circuits Sets C1-C2 8 ___/8
D — DC Circuits Sets D1-D2 8 ___/8
TOTAL 28 ___/28

Proficiency Benchmarks for CQ2

Level Score Meaning
Exam-Ready $\ge 25/28$ (89%+) Ready for CQ2. Focus on speed.
Solid $22-24/28$ (79-86%) Good grasp. Review missed concepts.
Developing $17-21/28$ (61-75%) Drill weak areas again.
Needs Work $< 17/28$ (<61%) Re-study cheat sheet, retry tutorial.

Speed Benchmarks

  • < 80 min: Excellent speed for CQ2 conditions.
  • 80–100 min: Good. Practice under time pressure.
  • > 100 min: Need more pattern recognition drills.

Error Log Template

After grading, list every wrong problem with a one-word reason:

Problem Reason

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A — Capacitance & Geometry

  1. $C = \frac{\varepsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(25 \times 10^{-4})}{1.0 \times 10^{-3}} = 2.21 \times 10^{-11}\ \text{F} = 22.1\ \text{pF}$

  2. (a) $C = Q/V = (50 \times 10^{-6})/10 = 5.0\ \mu\text{F}$ (b) $U = \frac{1}{2}CV^2 = 0.5(5 \times 10^{-6})(100) = 2.5 \times 10^{-4}\ \text{J} = 0.25\ \text{mJ}$

  3. $Q = CV = (100 \times 10^{-6})(12) = 1.2 \times 10^{-3}\ \text{C} = 1.2\ \text{mC}$ $$U = \frac{1}{2}CV^2 = 0.5(100 \times 10^{-6})(144) = 7.2 \times 10^{-3}\ \text{J} = 7.2\ \text{mJ}$$

  4. $C = \frac{\kappa\varepsilon_0 A}{d} = \frac{(5.0)(8.85 \times 10^{-12})(50 \times 10^{-4})}{2.0 \times 10^{-3}} = 1.11 \times 10^{-10}\ \text{F} = 111\ \text{pF}$

  5. $C = \kappa C_0 = 7.0 \times 20 = 140\ \text{pF}$

  6. (a) $U = \frac{1}{2}CV^2$ → $C = 2U/V^2 = 2(1.44 \times 10^{-3})/(24)^2 = 5.0\ \mu\text{F}$ (b) $Q = CV = (5 \times 10^{-6})(24) = 1.2 \times 10^{-4}\ \text{C} = 0.12\ \text{mC}$ (c) $E = V/d = 24/(0.5 \times 10^{-3}) = 4.8 \times 10^{4}\ \text{V/m}$

Part B — Capacitor Combinations

  1. $\frac{1}{C_{\text{eq}}} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4}$ → $C_{\text{eq}} = 4.0\ \mu\text{F}$

  2. (a) $C_{\text{eq}} = 4 + 6 + 12 = 22\ \mu\text{F}$ (b) $Q_{\text{total}} = C_{\text{eq}}V = (22 \times 10^{-6})(9) = 1.98 \times 10^{-4}\ \text{C} = 0.198\ \text{mC}$ (c) $Q_1 = 36\ \mu\text{C}$, $Q_2 = 54\ \mu\text{C}$, $Q_3 = 108\ \mu\text{C}$

  3. (a) $Q = C_{\text{eq}}V = (4 \times 10^{-6})(12) = 48\ \mu\text{C}$ (b) $V_1 = Q/C_1 = 48/6 = 8.0\ \text{V}$, $V_2 = 48/12 = 4.0\ \text{V}$ (c) $U_1 = \frac{1}{2}C_1V_1^2 = 0.5(6)(64) = 192\ \mu\text{J}$, $U_2 = 0.5(12)(16) = 96\ \mu\text{J}$

  4. Parallel: $C_{12} = 3 + 6 = 9\ \mu\text{F}$. Series: $\frac{1}{C_{\text{eq}}} = \frac{1}{9} + \frac{1}{2} = \frac{11}{18}$ → $C_{\text{eq}} = 1.64\ \mu\text{F}$

  5. Series: $\frac{1}{C_{12}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ → $C_{12} = 1.33\ \mu\text{F}$. Parallel: $C_{\text{eq}} = 1.33 + 6 = 7.33\ \mu\text{F}$

  6. First series: $C_{12} = \frac{2 \times 4}{2+4} = 1.33\ \mu\text{F}$. Second series: $C_{34} = \frac{3 \times 6}{3+6} = 2.0\ \mu\text{F}$. Parallel: $C_{\text{eq}} = 1.33 + 2.0 = 3.33\ \mu\text{F}$ $$U = \frac{1}{2}CV^2 = 0.5(3.33 \times 10^{-6})(144) = 2.4 \times 10^{-4}\ \text{J} = 0.24\ \text{mJ}$$

Part C — RC Circuits

  1. $\tau = RC = (20 \times 10^3)(50 \times 10^{-6}) = 1.0\ \text{s}$

  2. $C = \tau/R = 2.5/(100 \times 10^3) = 2.5 \times 10^{-5}\ \text{F} = 25\ \mu\text{F}$

  3. $V = V_0 e^{-t/\tau}$ → $0.40 = e^{-3/\tau}$ → $\tau = 3/\ln(2.5) = 3.27\ \text{s}$ $$R = \tau/C = 3.27/(100 \times 10^{-6}) = 32.7\ \text{k}\Omega$$

  4. $R = \tau/C = 1.0/(47 \times 10^{-6}) = 21.3\ \text{k}\Omega$

  5. (a) $\tau = RC = (50 \times 10^3)(100 \times 10^{-6}) = 5.0\ \text{s}$ (b) $V_C = 20(1 - e^{-2.5/5}) = 20(1 - e^{-0.5}) = 20(0.393) = 7.86\ \text{V}$ (c) $i = (20/50000)e^{-0.5} = (0.4 \times 10^{-3})(0.607) = 0.243\ \text{mA}$

  6. (a) $\tau = RC = (200 \times 10^3)(50 \times 10^{-6}) = 10\ \text{s}$ (b) $V = 30e^{-5/10} = 30e^{-0.5} = 30(0.607) = 18.2\ \text{V}$ (c) $10 = 30e^{-t/10}$ → $t = -10\ln(1/3) = 10\ln(3) = 10.99\ \text{s}$

  7. $0.90 = 1 - e^{-t/\tau}$ → $e^{-t/\tau} = 0.10$ → $t = \tau\ln(10) = 2.30\tau$

  8. $\frac{dV}{dt} = \frac{V_0}{\tau}e^{-t/\tau}$. Maximum at $t = 0$: $\left.\frac{dV}{dt}\right|_{\text{max}} = \frac{V_0}{\tau} = \frac{V_0}{RC}$

Part D — DC Circuits

  1. $V_{\text{out}} = 15 \times \frac{6.0}{4.0 + 6.0} = 15 \times 0.60 = 9.0\ \text{V}$

  2. $\frac{R_2}{R_1 + R_2} = \frac{5}{12}$. With $R_1 + R_2 = 20\ \text{k}\Omega$: $R_2 = 20 \times (5/12) = 8.33\ \text{k}\Omega$, $R_1 = 11.67\ \text{k}\Omega$

  3. $R_{2L} = \frac{15 \times 30}{15 + 30} = 10\ \text{k}\Omega$. $V_{\text{out}} = 9.0 \times \frac{10}{10+10} = 4.5\ \text{V}$

  4. Unloaded: $V_{\text{out}} = 9.0 \times (15/25) = 5.4\ \text{V}$. Loaded: $4.5\ \text{V}$. Change: $\frac{5.4-4.5}{5.4} \times 100% = 16.7%$

  5. KCL: $3.0 + I_3 = 5.0$ → $I_3 = 2.0\ \text{A}$ entering

  6. Net entering: $2.5 + (-1.5) = 1.0\ \text{A}$. So $I_3 = 1.0\ \text{A}$ leaving

  7. Net entering: $4.0 - 2.5 + 1.5 = 3.0\ \text{A}$. So $I_4 = 3.0\ \text{A}$ leaving

  8. Net entering: $6.0 - 3.0 + 2.0 = 5.0\ \text{A}$. So $I_4 = 5.0\ \text{A}$ leaving

Related Resources