FAD1022: Rapid-Fire Drill Pack — CQ 2 Capacitor & DC

Objective: Achieve mechanical fluency in capacitor calculations, RC transient analysis, and DC circuit fundamentals.
Target: 60–90 seconds per problem. If you stall >3 minutes, skip and mark it.
Total problems: 66
Estimated time: 70–100 minutes

Cheat Sheet (Memorize First)

Capacitor Formulas

Formula Description
$C = \dfrac{Q}{\Delta V}$ Definition of capacitance
$C = \varepsilon_0 \dfrac{A}{d}$ Parallel-plate (vacuum/air)
$C = \kappa \varepsilon_0 \dfrac{A}{d}$ With dielectric
$\kappa = \dfrac{\varepsilon}{\varepsilon_0} = \dfrac{C}{C_0}$ Dielectric constant
$E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0}$ Field between plates (vacuum)
$E = \dfrac{\Delta V}{d}$ Uniform field
$\vec{E} = \dfrac{\vec{E}_0}{\kappa}$ Net field with dielectric
$\Delta V = \dfrac{\Delta V_0}{\kappa}$ Voltage on isolated capacitor with dielectric

Combinations & Energy

Formula Description
$\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots$ Series
$C_{\text{eq}} = C_1 + C_2 + \dots$ Parallel
$U = \dfrac{1}{2} C (\Delta V)^2 = \dfrac{1}{2} Q \Delta V = \dfrac{Q^2}{2C}$ Stored energy
$u = \dfrac{1}{2} \varepsilon_0 E^2$ Energy density (vacuum)
$u = \dfrac{1}{2} \kappa \varepsilon_0 E^2$ Energy density (dielectric)

RC Transients

Formula Description
$\tau = RC$ Time constant
$q(t) = Q_{\max}(1 - e^{-t/\tau})$ Charging
$q(t) = Q_0 e^{-t/\tau}$ Discharging
$I(t) = I_0 e^{-t/\tau}$ Current during discharge
$I(t) = \dfrac{\mathcal{E}}{R} e^{-t/\tau}$ Current during charge
At $t = \tau$: $q \approx 63%$ of max; at $t = 5\tau$: $>99%$

DC Circuits

Formula Description
$V = IR$ Ohm's law
$P = IV = I^2 R = \dfrac{V^2}{R}$ Power
$R_{\text{eq}} = R_1 + R_2 + \dots$ Resistors in series
$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ Resistors in parallel
$V_{\text{terminal}} = \mathcal{E} - Ir$ Terminal voltage
$\sum I_{\text{in}} = \sum I_{\text{out}}$ Kirchhoff's Current Law (KCL)
$\sum \Delta V = 0$ (loop) Kirchhoff's Voltage Law (KVL)

Mnemonic — SERIES vs PARALLEL for Capacitors:

  • Series → Same charge $Q$; Sum of reciprocals; total $C$ Smaller than smallest.
  • Parallel → Pame voltage $V$; Plain sum; total $C$ Pigger than largest.

Part A: Basic Capacitor Calculations

Target: 45–60 seconds per problem.

Set A1 — Definition $C = Q/V$ (6 problems)

Apply $C = Q/\Delta V$ and its rearrangements directly.

  1. A $47 , \mu\text{F}$ capacitor is connected to a $9.0 , \text{V}$ battery. How much charge is stored on each plate?
  2. A capacitor stores $3.6 \times 10^{-4} , \text{C}$ of charge at $120 , \text{V}$. What is its capacitance in $\mu\text{F}$?
  3. A $220 , \mu\text{F}$ capacitor holds $2.64 \times 10^{-3} , \text{C}$. What is the potential difference across it?
  4. A capacitor is marked $50 , \text{V}$, $1000 , \mu\text{F}$. What is the maximum charge it can safely hold?
  5. If the voltage across a $10 , \mu\text{F}$ capacitor is tripled from $4 , \text{V}$ to $12 , \text{V}$, by what factor does the stored charge change?
  6. Two identical capacitors each store charge $Q$ at voltage $V$. They are connected in parallel. What is the total charge and the voltage across the combination?

Score: ___/6

Set A2 — Parallel-Plate Geometry & Dielectrics (6 problems)

Use $C = \varepsilon_0 A/d$, $C = \kappa \varepsilon_0 A/d$, and the isolated-capacitor dielectric rules. Take $\varepsilon_0 = 8.85 \times 10^{-12} , \text{F m}^{-1}$.

  1. A parallel-plate capacitor has plates of area $25 , \text{cm}^2$ separated by $0.50 , \text{mm}$ in air. Calculate its capacitance in pF.
  2. The capacitor in Q7 is connected to a $12 , \text{V}$ battery. Calculate the charge on each plate in pC.
  3. A parallel-plate capacitor with air gap $d$ has capacitance $C_0$. The plate separation is doubled while the capacitor remains connected to the battery. What is the new capacitance and what happens to the charge?
  4. A $200 , \text{pF}$ air-filled capacitor is charged to $100 , \text{V}$ and then isolated. A dielectric with $\kappa = 4.0$ is inserted. Find the new capacitance, voltage, and charge.
  5. A Teflon-filled ($\kappa = 2.1$) parallel-plate capacitor has plate area $0.30 , \text{m}^2$ and plate separation $0.10 , \text{mm}$. Calculate its capacitance in nF.
  6. For the capacitor in Q11 connected to a $15 , \text{V}$ battery, find the electric field strength between the plates.

Score: ___/6

Part B: Energy Storage & Dielectric Effects

Target: 60–90 seconds per problem.

Set B1 — Stored Energy (5 problems)

Use $U = \tfrac{1}{2} C V^2 = \tfrac{1}{2} Q V = Q^2/(2C)$.

  1. A $4.7 , \mu\text{F}$ capacitor is connected to a $12 , \text{V}$ battery. How much energy is stored?
  2. A capacitor stores $5.0 \times 10^{-4} , \text{J}$ at $50 , \text{V}$. Find its capacitance in $\mu\text{F}$.
  3. A $100 , \mu\text{F}$ capacitor holds $2.0 \times 10^{-3} , \text{C}$. What is the stored energy?
  4. Compare the energy stored in a $10 , \mu\text{F}$ capacitor at $6 , \text{V}$ versus the same capacitor at $12 , \text{V}$. By what factor does the energy increase?
  5. Two identical $20 , \mu\text{F}$ capacitors are each charged to $10 , \text{V}$. They are then connected in parallel. What is the total energy stored in the combination?

Score: ___/5

Set B2 — Dielectric Energy & Isolated vs Connected (3 problems)

  1. An isolated charged capacitor stores energy $U_0$. A dielectric with constant $\kappa$ is inserted. By what factor does the stored energy change? Explain physically why it changes.
  2. A capacitor remains connected to a battery while a dielectric is inserted. Does the stored energy increase, decrease, or stay the same? Calculate the factor of change in terms of $\kappa$.
  3. A paper-filled ($\kappa = 3.7$) capacitor is connected to a $20 , \text{V}$ battery. The vacuum-filled version would store $1.0 \times 10^{-5} , \text{J}$. How much energy does the paper-filled version store?

Score: ___/3

Part C: Series & Parallel Combinations

Target: 60–90 seconds per problem.

Set C1 — Series Capacitors (4 problems)

  1. Three capacitors ($2 , \mu\text{F}$, $3 , \mu\text{F}$, $6 , \mu\text{F}$) are connected in series across a $12 , \text{V}$ battery. Find the equivalent capacitance.
  2. For the combination in Q21, what is the charge on each capacitor?
  3. For the combination in Q21, find the voltage across the $3 , \mu\text{F}$ capacitor.
  4. Two capacitors in series have an equivalent capacitance of $2.4 , \mu\text{F}$. One of them is $4 , \mu\text{F}$. What is the other?

Score: ___/4

Set C2 — Parallel Capacitors (4 problems)

  1. Three capacitors ($5 , \mu\text{F}$, $10 , \mu\text{F}$, $15 , \mu\text{F}$) are connected in parallel across a $9.0 , \text{V}$ battery. Find the equivalent capacitance.
  2. For the combination in Q25, what is the total charge supplied by the battery?
  3. For the combination in Q25, how much charge is on the $10 , \mu\text{F}$ capacitor?
  4. Two capacitors in parallel have an equivalent capacitance of $50 , \mu\text{F}$. One of them is $30 , \mu\text{F}$. What is the other?

Score: ___/4

Set C3 — Mixed Networks (4 problems)

  1. A $6 , \mu\text{F}$ and a $12 , \mu\text{F}$ capacitor are connected in series. This combination is connected in parallel with a $4 , \mu\text{F}$ capacitor across a $24 , \text{V}$ battery. Find the equivalent capacitance of the network.
  2. For the network in Q29, find the total charge stored.
  3. In the network below, $C_1 = 2 , \mu\text{F}$, $C_2 = 4 , \mu\text{F}$, and $C_3 = 4 , \mu\text{F}$. $C_2$ and $C_3$ are in parallel, and this pair is in series with $C_1$. The network is connected to $12 , \text{V}$. Find the voltage across $C_1$.
  4. For the network in Q31, find the charge on $C_2$.

Score: ___/4

Part D: RC Transient Analysis

Target: 60–120 seconds per problem.

Set D1 — Charging RC Circuits (5 problems)

Use $q(t) = Q_{\max}(1 - e^{-t/\tau})$ and $\tau = RC$.

  1. A $50 , \mu\text{F}$ capacitor is in series with a $20 , \text{k}\Omega$ resistor and a $10 , \text{V}$ battery. Calculate the time constant $\tau$.
  2. For the circuit in Q33, how long does it take for the capacitor to reach $80%$ of its maximum charge?
  3. An RC circuit has $\tau = 4.0 , \text{s}$. At what time does the capacitor reach half its maximum charge?
  4. A capacitor charges through a resistor. After $3\tau$, approximately what percentage of the maximum charge has accumulated?
  5. A $100 , \mu\text{F}$ capacitor is charged through a $50 , \text{k}\Omega$ resistor from a $20 , \text{V}$ source. What is the voltage across the capacitor after one time constant?

Score: ___/5

Set D2 — Discharging RC Circuits (5 problems)

Use $q(t) = Q_0 e^{-t/\tau}$ and $I(t) = I_0 e^{-t/\tau}$.

  1. A fully charged $200 , \mu\text{F}$ capacitor at $50 , \text{V}$ is discharged through a $10 , \text{k}\Omega$ resistor. Find the time constant.
  2. For the circuit in Q38, what is the charge remaining on the capacitor after $2\tau$?
  3. A capacitor discharges through a resistor. After how many time constants has the charge fallen to less than $1%$ of its initial value?
  4. A $12 , \mu\text{F}$ capacitor charged to $30 , \text{V}$ is discharged through a $500 , \text{k}\Omega$ resistor. Find the initial current.
  5. For the circuit in Q41, what is the current after $6.0 , \text{s}$?

Score: ___/5

Part E: DC Circuit Basics

Target: 60–90 seconds per problem.

Set E1 — Ohm's Law, Power, & Resistor Networks (5 problems)

  1. A $12 , \text{V}$ battery is connected to a $48 , \Omega$ resistor. Find the current and the power dissipated.
  2. Three resistors ($2 , \Omega$, $3 , \Omega$, $6 , \Omega$) are connected in series to a $12 , \text{V}$ battery. Find the equivalent resistance and the current through each resistor.
  3. The same three resistors are connected in parallel to the same $12 , \text{V}$ battery. Find the equivalent resistance and the total current drawn from the battery.
  4. A $60 , \text{W}$ light bulb operates on $120 , \text{V}$. What is its resistance and the current it draws?
  5. A resistor network has $R_1 = 6 , \Omega$ and $R_2 = 12 , \Omega$ in parallel. This combination is in series with $R_3 = 4 , \Omega$. The network is connected to $12 , \text{V}$. Find the total current.

Score: ___/5

Set E2 — Kirchhoff's Laws & Terminal Voltage (5 problems)

  1. A non-ideal cell has emf $1.5 , \text{V}$ and internal resistance $0.50 , \Omega$. What is its terminal voltage when delivering $0.40 , \text{A}$?
  2. Two identical cells (emf $1.5 , \text{V}$, internal resistance $r$) are connected in series to an external resistor $R = 5.0 , \Omega$. The current is $0.50 , \text{A}$. Find $r$.
  3. In a single-loop circuit, a $9.0 , \text{V}$ battery drives current through resistors of $2.0 , \Omega$ and $4.0 , \Omega$ in series. Write the KVL equation and solve for the current.
  4. At a circuit junction, currents of $2.0 , \text{A}$ and $3.0 , \text{A}$ flow inward, and one current $I$ flows outward. Write the KCL equation and solve for $I$.
  5. A cell has terminal voltage $1.48 , \text{V}$ when delivering $0.20 , \text{A}$ and $1.32 , \text{V}$ when delivering $0.80 , \text{A}$. Find its emf $\mathcal{E}$ and internal resistance $r$.

Score: ___/5

Part F: Mixed & Reverse Engineering

Target: 90–150 seconds per problem.

Set F1 — Reverse & Multi-Concept Problems (4 problems)

  1. Reverse: A capacitor network has $C_{\text{eq}} = 6 , \mu\text{F}$ when two identical capacitors are connected in parallel. What is the capacitance of each capacitor?
  2. Reverse: An RC charging circuit reaches $75%$ of maximum charge in $8.0 , \text{s}$. Find the time constant $\tau$.
  3. Multi-concept: A $10 , \mu\text{F}$ capacitor is charged to $50 , \text{V}$, then disconnected and connected across an uncharged $15 , \mu\text{F}$ capacitor. Find the final common voltage and the final energy stored. Compare with the initial energy. Where did the "missing" energy go?
  4. Multi-concept: A parallel-plate capacitor has $C_0 = 50 , \text{pF}$, plate area $A = 40 , \text{cm}^2$, and is charged to $V_0 = 80 , \text{V}$. It is isolated, and a dielectric with $\kappa = 5.0$ is inserted. Find the new voltage, new energy, and the energy density before and after insertion (assume uniform field).

Score: ___/4

Set F2 — Capacitor + DC Circuit Hybrids (4 problems)

  1. A DC circuit contains a $12 , \text{V}$ battery, a $2.0 , \Omega$ resistor, and a $4.0 , \Omega$ resistor in series. A voltmeter is connected across the $4.0 , \Omega$ resistor. What does it read?
  2. In the circuit of Q57, a $1000 , \mu\text{F}$ capacitor is placed in parallel with the $4.0 , \Omega$ resistor (in steady state). What charge does it hold?
  3. A $6.0 , \text{V}$ battery with negligible internal resistance is connected to a network of $R = 2.0 , \text{k}\Omega$ and $C = 100 , \mu\text{F}$ in series. At $t = 0$, the switch is closed. How long after closing does the capacitor voltage reach $3.0 , \text{V}$?
  4. In a two-loop circuit, the outer loop contains a $12 , \text{V}$ battery, a $4 , \Omega$ resistor, and a $2 , \Omega$ resistor. The inner loop contains a $6 , \text{V}$ battery (opposing) and the $2 , \Omega$ resistor. Apply KVL to both loops and solve for the current through the $2 , \Omega$ resistor.

Score: ___/4

Part G: Traps & Common Mistakes

Target: 60–90 seconds per problem. These problems target the most common exam slip-ups.

Set G1 — Exam Traps (6 problems)

  1. Trap — Area units: A parallel-plate capacitor has plates $5.0 , \text{cm} \times 4.0 , \text{cm}$ separated by $2.0 , \text{mm}$. A student calculates $C = 8.85 \times 10^{-12} \times 20 / 0.002$ and gets $8.85 \times 10^{-8} , \text{F}$. What did they forget, and what is the correct answer?
  2. Trap — Series charge sharing: Three capacitors in series ($1 , \mu\text{F}$, $2 , \mu\text{F}$, $3 , \mu\text{F}$) are connected to $11 , \text{V}$. A student claims the voltage across the $1 , \mu\text{F}$ capacitor is $11/3 , \text{V}$ because voltages "split equally." What is the correct voltage?
  3. Trap — Isolated vs connected: A charged capacitor (battery connected) has a dielectric inserted while the battery remains connected. A student says $V$ drops to $V_0/\kappa$. Is this correct? What actually happens to $V$ and $Q$?
  4. Trap — Discharge time: A student says "after $5\tau$, the capacitor is completely empty." Explain why this is technically incorrect and state the actual remaining charge as a percentage.
  5. Trap — Internal resistance: A $9.0 , \text{V}$ battery with internal resistance $1.0 , \Omega$ is short-circuited. A student calculates the short-circuit current as $I = 9.0 / 0 = \infty$. What is the actual short-circuit current?
  6. Trap — Power sign: In a circuit with two opposing batteries, a student calculates the power dissipated in a resistor as $P = I^2R = (-0.5)^2 \times 4 = 1.0 , \text{W}$ and concludes the negative current means negative power. What is the correct power dissipated, and why does the sign of $I$ not matter for power?

Score: ___/6

Final Scorecard

Part Sets Problems Raw Score
A — Basic Capacitor Calculations A1, A2 12 ___/12
B — Energy & Dielectrics B1, B2 8 ___/8
C — Series & Parallel Combinations C1, C2, C3 12 ___/12
D — RC Transient Analysis D1, D2 10 ___/10
E — DC Circuit Basics E1, E2 10 ___/10
F — Mixed & Reverse Engineering F1, F2 8 ___/8
G — Traps & Common Mistakes G1 6 ___/6
TOTAL 66 ___/66

Proficiency Benchmarks

  • 47/66 (71%) — Proficient. You can handle standard exam problems.
  • 56/66 (85%) — Solid. Fast and accurate on most patterns.
  • 62/66 (94%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <70 min: Excellent mechanical fluency.
  • 70–95 min: Good. Review missed patterns.
  • >95 min: Drill the specific sets you scored lowest on again tomorrow.

Error Log Template

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

Problem Reason
e.g. 7 unit conversion

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

Answer Key

Set A1

  1. $Q = CV = 47 \times 10^{-6} \times 9.0 = 4.23 \times 10^{-4} , \text{C} = 423 , \mu\text{C}$
  2. $C = Q/V = 3.6 \times 10^{-4} / 120 = 3.0 \times 10^{-6} , \text{F} = 3.0 , \mu\text{F}$
  3. $V = Q/C = 2.64 \times 10^{-3} / (220 \times 10^{-6}) = 12 , \text{V}$
  4. $Q_{\max} = 1000 \times 10^{-6} \times 50 = 0.050 , \text{C} = 50 , \text{mC}$
  5. Charge triples (factor of 3), since $Q \propto V$ for fixed $C$
  6. Total charge $= 2Q$, voltage $= V$ (parallel preserves voltage)

Set A2

  1. $C = \varepsilon_0 A/d = 8.85 \times 10^{-12} \times 25 \times 10^{-4} / (0.50 \times 10^{-3}) = 44.25 \times 10^{-12} , \text{F} = 44.25 , \text{pF}$
  2. $Q = CV = 44.25 \times 10^{-12} \times 12 = 531 , \text{pC}$
  3. New capacitance $= C_0/2$; charge halves because battery keeps $V$ fixed and $Q = CV$
  4. $C = 800 , \text{pF}$; $V = 25 , \text{V}$; $Q = 2.0 \times 10^{-8} , \text{C}$\ 3(unchanged)
  5. $C = 2.1 \times 8.85 \times 10^{-12} \times 0.30 / (0.10 \times 10^{-3}) = 55.8 \times 10^{-9} , \text{F} = 55.8 , \text{nF}$
  6. $E = V/d = 15 / (0.10 \times 10^{-3}) = 1.5 \times 10^{5} , \text{V m}^{-1}$

Set B1

  1. $U = \tfrac{1}{2} \times 4.7 \times 10^{-6} \times 12^2 = 3.38 \times 10^{-4} , \text{J} \approx 0.34 , \text{mJ}$
  2. $C = 2U/V^2 = 2(5.0 \times 10^{-4})/50^2 = 4.0 \times 10^{-7} , \text{F} = 0.40 , \mu\text{F}$
  3. $U = Q^2/(2C) = (2.0 \times 10^{-3})^2 / (2 \times 100 \times 10^{-6}) = 0.020 , \text{J}$
  4. Factor of 4, since $U \propto V^2$
  5. $U_{\text{total}} = 2 \times \tfrac{1}{2} (20 \times 10^{-6})(10)^2 = 2.0 \times 10^{-3} , \text{J} = 2.0 , \text{mJ}$

Set B2

  1. Energy decreases by factor $1/\kappa$; work is done by the field on the dielectric as it is drawn in
  2. Increases by factor $\kappa$; battery does work to maintain voltage while $C$ increases
  3. $U = \kappa U_0 = 3.7 \times 1.0 \times 10^{-5} = 3.7 \times 10^{-5} , \text{J}$

Set C1

  1. $1/C_{\text{eq}} = 1/2 + 1/3 + 1/6 = 1 \Rightarrow C_{\text{eq}} = 1 , \mu\text{F}$
  2. $Q = C_{\text{eq}} V = 1 \times 12 = 12 , \mu\text{C}$ (same on each)
  3. $V_3 = Q/C_3 = 12/3 = 4 , \text{V}$
  4. $1/C_2 = 1/2.4 - 1/4 = 1/6 \Rightarrow C_2 = 6 , \mu\text{F}$

Set C2

  1. $C_{\text{eq}} = 5 + 10 + 15 = 30 , \mu\text{F}$
  2. $Q_{\text{total}} = 30 \times 10^{-6} \times 9.0 = 2.7 \times 10^{-4} , \text{C} = 270 , \mu\text{C}$
  3. $Q_{10} = 10 \times 10^{-6} \times 9.0 = 90 , \mu\text{C}$
  4. $C_2 = 50 - 30 = 20 , \mu\text{F}$

Set C3

  1. Series pair: $1/C_s = 1/6 + 1/12 = 1/4 \Rightarrow C_s = 4 , \mu\text{F}$. Total: $C_{\text{eq}} = 4 + 4 = 8 , \mu\text{F}$
  2. $Q_{\text{total}} = 8 \times 10^{-6} \times 24 = 1.92 \times 10^{-4} , \text{C} = 192 , \mu\text{C}$
  3. Parallel pair: $C_p = 4 + 4 = 8 , \mu\text{F}$. Series with $C_1$: $1/C_{\text{eq}} = 1/2 + 1/8 = 5/8 \Rightarrow C_{\text{eq}} = 1.6 , \mu\text{F}$. $Q_{\text{total}} = 1.6 \times 12 = 19.2 , \mu\text{C}$. $V_1 = Q/C_1 = 19.2/2 = 9.6 , \text{V}$
  4. $V_p = 12 - 9.6 = 2.4 , \text{V}$. $Q_2 = C_2 V_p = 4 \times 2.4 = 9.6 , \mu\text{C}$

Set D1

  1. $\tau = RC = 20 \times 10^3 \times 50 \times 10^{-6} = 1.0 , \text{s}$
  2. $0.80 = 1 - e^{-t/\tau} \Rightarrow t = -\tau \ln(0.20) = 1.0 \times 1.61 = 1.61 , \text{s}$
  3. $0.50 = 1 - e^{-t/\tau} \Rightarrow t = \tau \ln 2 = 4.0 \times 0.693 = 2.77 , \text{s} \approx 2.8 , \text{s}$
  4. $q/Q_{\max} = 1 - e^{-3} \approx 1 - 0.0498 = 0.950 = 95.0%$
  5. $V_C = \mathcal{E}(1 - e^{-1}) = 20 \times 0.632 = 12.6 , \text{V}$

Set D2

  1. $\tau = 10 \times 10^3 \times 200 \times 10^{-6} = 2.0 , \text{s}$
  2. $Q_0 = CV_0 = 200 \times 10^{-6} \times 50 = 0.010 , \text{C} = 10 , \text{mC}$. After $2\tau$: $Q = Q_0 e^{-2} = 10 \times 0.135 = 1.35 , \text{mC}$
  3. $e^{-t/\tau} < 0.01 \Rightarrow t/\tau > \ln(100) \approx 4.6$, so after 5 time constants
  4. $I_0 = V_0/R = 30/(500 \times 10^3) = 6.0 \times 10^{-5} , \text{A} = 60 , \mu\text{A}$
  5. $\tau = 500 \times 10^3 \times 12 \times 10^{-6} = 6.0 , \text{s}$. $I = I_0 e^{-6/6} = 60 \times e^{-1} = 22.1 , \mu\text{A}$

Set E1

  1. $I = 12/48 = 0.25 , \text{A}$; $P = V^2/R = 144/48 = 3.0 , \text{W}$
  2. $R_{\text{eq}} = 2 + 3 + 6 = 11 , \Omega$; $I = 12/11 = 1.09 , \text{A}$ (same through all)
  3. $1/R_{\text{eq}} = 1/2 + 1/3 + 1/6 = 1 \Rightarrow R_{\text{eq}} = 1 , \Omega$; $I_{\text{total}} = 12/1 = 12 , \text{A}$
  4. $R = V^2/P = 120^2/60 = 240 , \Omega$; $I = P/V = 60/120 = 0.50 , \text{A}$
  5. Parallel: $R_p = (6 \times 12)/(6+12) = 4 , \Omega$. Total: $R_{\text{eq}} = 4 + 4 = 8 , \Omega$. $I = 12/8 = 1.5 , \text{A}$

Set E2

  1. $V_{\text{terminal}} = 1.5 - (0.40)(0.50) = 1.30 , \text{V}$
  2. Total emf $= 3.0 , \text{V}$; total resistance $= 2r + 5.0$. $I = 3.0/(2r + 5.0) = 0.50 \Rightarrow 2r + 5.0 = 6.0 \Rightarrow r = 0.50 , \Omega$
  3. KVL: $9.0 - I(2.0) - I(4.0) = 0 \Rightarrow 9.0 = 6.0I \Rightarrow I = 1.5 , \text{A}$
  4. KCL: $2.0 + 3.0 = I \Rightarrow I = 5.0 , \text{A}$ (outward)
  5. $1.48 = \mathcal{E} - 0.20r$ and $1.32 = \mathcal{E} - 0.80r$. Subtract: $0.16 = 0.60r \Rightarrow r = 0.267 , \Omega$. Then $\mathcal{E} = 1.48 + 0.20(0.267) = 1.53 , \text{V}$

Set F1

  1. $2C = 6 \Rightarrow C = 3 , \mu\text{F}$
  2. $0.75 = 1 - e^{-8/\tau} \Rightarrow e^{-8/\tau} = 0.25 \Rightarrow -8/\tau = \ln(0.25) = -1.386 \Rightarrow \tau = 5.77 , \text{s} \approx 5.8 , \text{s}$
  3. Initial charge $Q_0 = 10 \times 50 = 500 , \mu\text{C}$. Final $V = Q_0/(C_1+C_2) = 500/(10+15) = 20 , \text{V}$. Final energy $U_f = \tfrac{1}{2}(25)(20)^2 = 5.0 \times 10^{-3} , \text{J} = 5.0 , \text{mJ}$. Initial $U_i = \tfrac{1}{2}(10)(50)^2 = 12.5 , \text{mJ}$. "Missing" energy dissipated as heat/sparks during charge redistribution.
  4. New $V = V_0/\kappa = 80/5 = 16 , \text{V}$. New $C = 250 , \text{pF}$. New $U = \tfrac{1}{2}(250 \times 10^{-12})(16)^2 = 3.2 \times 10^{-8} , \text{J} = 32 , \text{nJ}$. Initial $U_0 = \tfrac{1}{2}(50 \times 10^{-12})(80)^2 = 1.6 \times 10^{-7} , \text{J} = 160 , \text{nJ}$. $u_0 = U_0/(A d)$; need $d = \varepsilon_0 A/C_0 = 8.85 \times 10^{-12} \times 40 \times 10^{-4} / (50 \times 10^{-12}) = 7.08 \times 10^{-4} , \text{m}$. $u_0 = 1.6 \times 10^{-7} / (40 \times 10^{-4} \times 7.08 \times 10^{-4}) = 0.0565 , \text{J m}^{-3}$. After: $u = u_0/\kappa = 0.0113 , \text{J m}^{-3}$ (or use $u = \tfrac{1}{2}\kappa\varepsilon_0 E^2$).

Set F2

  1. Total resistance $= 6.0 , \Omega$; $I = 12/6 = 2.0 , \text{A}$. $V_4 = IR = 2.0 \times 4.0 = 8.0 , \text{V}$
  2. In steady state, capacitor voltage $= 8.0 , \text{V}$. $Q = CV = 1000 \times 10^{-6} \times 8.0 = 8.0 \times 10^{-3} , \text{C} = 8.0 , \text{mC}$
  3. $\tau = 2000 \times 100 \times 10^{-6} = 0.20 , \text{s}$. $V_C = 6.0(1 - e^{-t/0.20}) = 3.0 \Rightarrow 1 - e^{-t/0.20} = 0.5 \Rightarrow t = 0.20 \ln 2 = 0.139 , \text{s} \approx 0.14 , \text{s}$
  4. KVL outer: $12 - 4I_1 - 2I_2 = 0$. KVL inner: $-6 + 2I_2 = 0 \Rightarrow I_2 = 3.0 , \text{A}$ (through $2 , \Omega$ resistor).

Set G1

  1. Forgot to convert area to $\text{m}^2$; correct $A = 20 \times 10^{-4} , \text{m}^2$, so $C = 8.85 \times 10^{-12} \times 20 \times 10^{-4} / 0.002 = 8.85 \times 10^{-12} , \text{F} = 8.85 , \text{pF}$
  2. Correct voltage is $6 , \text{V}$. Voltages split in inverse ratio to capacitance: $V_1 : V_2 : V_3 = 1/1 : 1/2 : 1/3 = 6:3:2$. $V_1 = (6/11) \times 11 = 6 , \text{V}$
  3. No. For connected capacitor, $V$ stays at $V_0$ (battery fixes it), and $Q$ increases by factor $\kappa$
  4. After $5\tau$, $q = Q_0 e^{-5} \approx 0.0067 Q_0$, so about $0.67%$ remains — not zero
  5. $I_{\text{sc}} = \mathcal{E}/r = 9.0/1.0 = 9.0 , \text{A}$
  6. Correct power $= I^2R = (0.5)^2 \times 4 = 1.0 , \text{W}$. The sign of current only indicates direction; power dissipated is always $I^2R$ (positive).

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