FAD1022: Minimal Drill — Just the Easy Picks
Objective: Master ONLY the 8 question groups you'll actually answer in the exam.
Total problems: ~30
Estimated time: 60–90 min
Cheat Sheet (Memorize First)
Physics Constants
| Constant | Symbol | Value |
|---|---|---|
| Coulomb constant | $k$ | $8.99 \times 10^{9}\ \text{N·m}^{2}\text{·C}^{-2}$ |
| Permittivity of free space | $\varepsilon_0$ | $8.85 \times 10^{-12}\ \text{F·m}^{-1}$ |
| Permeability of free space | $\mu_0$ | $4\pi \times 10^{-7}\ \text{T·m·A}^{-1}$ |
| Elementary charge | $e$ | $1.60 \times 10^{-19}\ \text{C}$ |
| Electron mass | $m_e$ | $9.11 \times 10^{-31}\ \text{kg}$ |
| Planck's constant | $h$ | $6.63 \times 10^{-34}\ \text{J·s}$ |
| $hc$ (useful) | $hc$ | $1240\ \text{eV·nm}$ |
| Bohr radius | $a_0$ | $0.529\ \text{Å} = 5.29 \times 10^{-11}\ \text{m}$ |
All Formulas Needed (18 total)
| # | Topic | Formula | Notes |
|---|---|---|---|
| 1 | Capacitance | $C = \kappa\varepsilon_0 A/d$ | Parallel plate |
| 2 | Capacitor energy | $U = \frac12 CV^2 = Q^2/(2C)$ | Pick convenient form |
| 3 | RC charging | $q(t) = Q(1 - e^{-t/\tau})$ | $\tau = RC$ |
| 4 | RC discharging | $q(t) = Q_0 e^{-t/\tau}$ | |
| 5 | Voltage divider | $V_{out} = V_{in} R_2/(R_1+R_2)$ | Unloaded |
| 6 | Voltage divider (loaded) | $V_{out} = V_{in} R_{eq}/(R_1+R_{eq})$ | $R_{eq} = R_2 \parallel R_L$ |
| 7 | Coulomb's law | $\vec{F} = k q_1 q_2 / r^2\ \hat{r}$ | |
| 8 | E-field of point charge | $\vec{E} = k Q / r^2\ \hat{r}$ | Superposition: $\vec{E}_{net} = \sum \vec{E}_i$ |
| 9 | E-field projectile | $a_y = qE/m$, $y = \frac12 a_y (L/v_0)^2$ | |
| 10 | Amplifier (inverting) | $V_{out} = -(R_f/R_{in})V_{in}$ | 180$^\circ$ phase shift |
| 11 | Amplifier (non-inv) | $V_{out} = (1 + R_f/R_{in})V_{in}$ | No phase shift |
| 12 | Ampere inside wire | $B = \mu_0 I r / (2\pi R^2)$ | $r < R$ |
| 13 | AC power triangle | $S^2 = P^2 + (Q_L - Q_C)^2$ | $\cos\phi = P/S$ |
| 14 | Motor torque | $\tau = NIAB\sin\theta$ | Max at $\theta = 90^\circ$ |
| 15 | Self-inductance | $L = \mu_0 N^2 A / \ell$ | Solenoid |
| 16 | Transformer | $V_s/V_p = N_s/N_p$ | Also $I_s/I_p = N_p/N_s$ |
| 17 | Bohr radius & energy | $r_n = n^2 a_0$, $E_n = -13.6/n^2$ eV | $a_0 = 0.529$ Å |
| 18 | Photoelectric | $K_{max} = hf - \phi$, $eV_s = K_{max}$ | $f_0 = \phi/h$, $\lambda_c = hc/\phi$ |
CIVIL Mnemonic
In a Capacitor, I leads V; In an Inductor, V leads I (opposite in L).
Dielectric Effects
| Condition | Insert dielectric | Remove dielectric |
|---|---|---|
| Battery connected ($V$ constant) | $C \uparrow$, $Q \uparrow$, $U \uparrow$ | $C \downarrow$, $Q \downarrow$, $U \downarrow$ |
| Battery disconnected ($Q$ constant) | $C \uparrow$, $V \downarrow$, $U \downarrow$ | $C \downarrow$, $V \uparrow$, $U \uparrow$ |
Part A — Structured Practice (15 problems)
Target: 2 min per question. All compulsory.
A1–A3: Capacitor Charging & Discharging
-
A $220\ \mu\text{F}$ capacitor charges through a $4.7\ \text{k}\Omega$ resistor from a $15\ \text{V}$ battery. What is the time constant?
- A) $0.33\ \text{s}$ B) $1.03\ \text{s}$ C) $2.15\ \text{s}$ D) $4.70\ \text{s}$
-
After how many time constants does a discharging capacitor retain approximately $5%$ of its initial charge?
- A) $1\tau$ B) $2\tau$ C) $3\tau$ D) $5\tau$
-
A $33\ \mu\text{F}$ capacitor is fully charged to $25\ \text{V}$. How much energy is stored?
- A) $5.2\ \text{mJ}$ B) $10.3\ \text{mJ}$ C) $20.6\ \text{mJ}$ D) $41.3\ \text{mJ}$
A4–A6: Kirchhoff's Current Law
-
At a junction, three wires meet. Currents of $3\ \text{A}$ and $2\ \text{A}$ flow into the junction, and $I$ flows out. What is $I$?
- A) $1\ \text{A}$ B) $5\ \text{A}$ C) $-5\ \text{A}$ D) $6\ \text{A}$
-
A node has four branches: $I_1 = 10\ \text{A}$ (entering), $I_2 = 4\ \text{A}$ (leaving), $I_3 = 3\ \text{A}$ (leaving), $I_4$ (entering). Find $I_4$.
- A) $3\ \text{A}$ entering B) $3\ \text{A}$ leaving C) $7\ \text{A}$ entering D) $7\ \text{A}$ leaving
-
KCL states $\sum I_{in} = \sum I_{out}$ at a junction. This is a direct consequence of:
- A) Conservation of energy B) Conservation of charge C) Ohm's law D) Faraday's law
A7–A9: Amplifier Formulas
-
An inverting op-amp has $R_f = 220\ \text{k}\Omega$ and $R_{in} = 22\ \text{k}\Omega$. If $V_{in} = 0.3\ \text{V}$, what is $V_{out}$?
- A) $+3.0\ \text{V}$ B) $-3.0\ \text{V}$ C) $+6.6\ \text{V}$ D) $-6.6\ \text{V}$
-
A non-inverting op-amp has $R_f = 100\ \text{k}\Omega$ and $R_{in} = 25\ \text{k}\Omega$. The voltage gain is:
- A) $4$ B) $5$ C) $-4$ D) $-5$
-
An inverting amplifier has $V_{out} = -8.0\ \text{V}$ when $V_{in} = 0.4\ \text{V}$ and $R_{in} = 12\ \text{k}\Omega$. The feedback resistor $R_f$ is:
- A) $60\ \text{k}\Omega$ B) $120\ \text{k}\Omega$ C) $240\ \text{k}\Omega$ D) $360\ \text{k}\Omega$
A10–A12: Ampere's Law Inside Wire ($r < R$)
-
A cylindrical wire of radius $R = 3.0\ \text{mm}$ carries $I = 15\ \text{A}$. What is $B$ at $r = 1.5\ \text{mm}$?
- A) $2.0 \times 10^{-4}\ \text{T}$ B) $3.0 \times 10^{-4}\ \text{T}$ C) $5.0 \times 10^{-4}\ \text{T}$ D) $1.0 \times 10^{-3}\ \text{T}$
-
For the same wire, what is $B$ at $r = R$ (the surface)?
- A) $2.0 \times 10^{-4}\ \text{T}$ B) $5.0 \times 10^{-4}\ \text{T}$ C) $1.0 \times 10^{-3}\ \text{T}$ D) $2.0 \times 10^{-3}\ \text{T}$
-
Inside a current-carrying wire, the magnetic field $B(r)$ varies with $r$ as:
- A) $B \propto 1/r$ B) $B \propto r$ C) $B \propto r^2$ D) $B$ is constant
A13–A15: Mixed Topics
-
Amplifier: A non-inverting op-amp has $V_{in} = 0.2\ \text{V}$, $R_f = 50\ \text{k}\Omega$, $R_{in} = 10\ \text{k}\Omega$. What is $V_{out}$?
- A) $-1.0\ \text{V}$ B) $+1.0\ \text{V}$ C) $+1.2\ \text{V}$ D) $-1.2\ \text{V}$
-
KCL + Capacitor: A $50\ \mu\text{F}$ capacitor charges to $63%$ of its final voltage after $0.5\ \text{s}$. If $R$ is in series, the current at $t = 0$ is:
- A) $I = V/R$ B) $I = 0$ C) $I = 0.37\ V/R$ D) $I = 0.63\ V/R$
-
Ampere: A wire of radius $R = 2.0\ \text{mm}$ carries current $I$. At $r = 0.5\ \text{mm}$, $B = 1.25 \times 10^{-4}\ \text{T}$. The total current $I$ is:
- A) $5.0\ \text{A}$ B) $10.0\ \text{A}$ C) $15.0\ \text{A}$ D) $20.0\ \text{A}$
Score: ___/15
Part B1 — Electrostatic & Projectile (12 marks)
(a) E-field from Point Charges (4 marks)
Two point charges are on the $x$-axis: $Q_1 = +6\ \mu\text{C}$ at $x = 0$ and $Q_2 = -4\ \mu\text{C}$ at $x = 8\ \text{cm}$. Calculate the net electric field $\vec{E}_{net}$ (magnitude and direction) at $x = 3\ \text{cm}$.
(b) Force on Test Charge (3 marks)
A test charge $q = +5\ \text{nC}$ is placed at $x = 3\ \text{cm}$. Calculate the electrostatic force $\vec{F}$ acting on it.
(c) Projectile Between Parallel Plates (5 marks)
A proton ($q = +e = 1.60 \times 10^{-19}\ \text{C}$, $m = 1.67 \times 10^{-27}\ \text{kg}$) enters horizontally at $v_0 = 2.0 \times 10^6\ \text{m/s}$ into a uniform electric field $E = 8000\ \text{N/C}$ directed upward between parallel plates of length $L = 8.0\ \text{cm}$.
(i) Find the vertical acceleration $a_y$ of the proton.
(ii) How long does the proton take to traverse the plates?
(iii) Determine the vertical deflection $y$ at exit.
(iv) Find the exit velocity components $v_x$ and $v_y$, and the exit angle $\theta$.
Score: ___/12
Part B2 — Capacitor + Dielectric + Voltage Divider (12 marks)
(a) Parallel Plate Capacitance with Dielectric (3 marks)
A parallel-plate capacitor has plates of area $A = 80\ \text{cm}^2$ separated by $d = 0.50\ \text{mm}$ in air. A dielectric slab with $\kappa = 4.2$ is inserted filling the gap. Calculate the capacitance $C$.
(b) Energy and Dielectric Effects (5 marks)
The capacitor from part (a) is connected to a $30\ \text{V}$ battery.
(i) Calculate the charge $Q$ on each plate and the stored energy $U$.
(ii) While the battery remains connected, the dielectric is removed. Explain what happens to $Q$ and $U$. Calculate the new charge $Q'$ and new energy $U'$.
(iii) If instead the battery had been disconnected before removing the dielectric, explain how the result would differ.
(c) Voltage Divider — Loaded vs Unloaded (4 marks)
A voltage divider consists of $R_1 = 12\ \text{k}\Omega$ and $R_2 = 18\ \text{k}\Omega$ across a $24\ \text{V}$ supply.
(i) Calculate the unloaded $V_{out}$ across $R_2$.
(ii) A load $R_L = 9\ \text{k}\Omega$ is connected across $R_2$. Find the new $V_{out}$.
(iii) Calculate the percentage drop in $V_{out}$ due to loading.
Score: ___/12
Part B3 — AC Power & Phasor (12 marks)
(a) Phasor Diagram (4 marks)
An RLC series circuit has $R = 60\ \Omega$, $X_L = 100\ \Omega$, $X_C = 40\ \Omega$, and $I_{rms} = 1.5\ \text{A}$.
(i) Calculate $V_R$, $V_L$, and $V_C$.
(ii) Draw a phasor diagram with $V_R$ along $0^\circ$, $V_L$ at $+90^\circ$, and $V_C$ at $-90^\circ$.
(iii) Calculate the magnitude of the total voltage $V_T$ and the phase angle $\phi$.
(b) Power Calculations (4 marks)
For the same circuit:
(i) Calculate real power $P$, inductive reactive power $Q_L$, and capacitive reactive power $Q_C$.
(ii) Calculate apparent power $S$ and power factor $\cos\phi$.
(iii) Is the circuit inductive or capacitive? Explain using $\phi$ and CIVIL.
(c) Pure Capacitive Circuit (4 marks)
A $100\ \mu\text{F}$ capacitor is connected across a $120\ \text{V}_{rms}$, $60\ \text{Hz}$ supply.
(i) Calculate $X_C$ and the rms current.
(ii) Find $P$, $Q_C$, and $S$.
(iii) State the phase angle $\phi$. Use CIVIL to justify the sign.
Score: ___/12
Part C2 — AC Motor Torque (12 marks)
(a) Torque Calculation (7 marks)
A rectangular coil of $N = 200$ turns, width $w = 5.0\ \text{cm}$, and height $h = 8.0\ \text{cm}$ carries current $I = 3.0\ \text{A}$. It is placed in a uniform magnetic field $B = 0.50\ \text{T}$.
(i) Define each symbol in $\tau = NIAB\sin\theta$.
(ii) Calculate the area $A$ of the coil and the magnetic dipole moment $\mu$.
(iii) Calculate the torque when $\theta = 30^\circ$ and when $\theta = 60^\circ$.
(iv) What is the maximum torque? At what orientation does it occur?
(b) Max/Min Torque Conditions (5 marks)
(i) State the condition for maximum torque on a current loop in a magnetic field. What is $\tau_{max}$ in terms of $\mu$ and $B$?
(ii) State the condition for minimum (zero) torque.
(iii) Explain why the torque in an AC motor, though it varies sinusoidally, produces continuous rotation in one direction.
Score: ___/12
Part C3 — Transformer (12 marks)
(a) Self-Inductance of Solenoid (4 marks)
A solenoid of length $\ell = 0.40\ \text{m}$, cross-sectional area $A = 5.0\ \text{cm}^2$, and $N = 600$ turns carries a current $I = 2.0\ \text{A}$.
(i) Derive $L = \mu_0 N^2 A / \ell$ starting from $B = \mu_0 n I$.
(ii) Calculate the numerical value of $L$ in mH.
(iii) Calculate the energy stored in the solenoid's magnetic field.
(b) Mutual Inductance (4 marks)
Two coils are placed near each other. Coil 1 has $L_1 = 8.0\ \text{mH}$, coil 2 has $L_2 = 18\ \text{mH}$. The coupling coefficient is $k = 0.75$.
(i) Calculate the mutual inductance $M$.
(ii) If $dI_1/dt = 150\ \text{A/s}$, what is the induced EMF in coil 2?
(iii) State the reciprocity principle for mutual inductance.
(c) Ideal Transformer (4 marks)
An ideal transformer has $N_p = 150$ turns and $N_s = 450$ turns. The primary is connected to $120\ \text{V}_{rms}$, $60\ \text{Hz}$ and draws $I_p = 1.2\ \text{A}$.
(i) Is this step-up or step-down?
(ii) Calculate $V_s$ and $I_s$.
(iii) A resistive load is connected to the secondary. Calculate the load resistance $R_L$.
Score: ___/12
Part C5 — Bohr Atom (12 marks)
(a) Derive Bohr Radius (5 marks)
(i) State Bohr's angular momentum quantization condition.
(ii) Set Coulomb force equal to centripetal force for an electron in a hydrogen atom.
(iii) Derive $r_n = n^2 a_0$, where $a_0 = 0.529\ \text{Å}$. Show all algebraic steps.
(iv) Calculate $r_2$ and $r_3$ in terms of $a_0$ and in Angstroms.
(b) Energy Levels & Transitions (4 marks)
(i) Calculate $E_n$ for $n = 1, 2, 3$ in eV.
(ii) A hydrogen atom transitions from $n = 4$ to $n = 2$. Find $\Delta E$ and the wavelength $\lambda$ of the emitted photon (use $hc = 1240\ \text{eV·nm}$).
(iii) Identify the spectral series and state whether the light is visible.
(c) Series Identification (3 marks)
Identify the spectral series for each transition and state the region of the EM spectrum:
- (i) $n = 5 \to n = 1$
- (ii) $n = 4 \to n = 3$
- (iii) $n = 6 \to n = 2$
Score: ___/12
Part C6 — Photoelectric Effect (12 marks)
(a) Thresholds (4 marks)
A metal surface has a work function $\phi = 2.80\ \text{eV}$. Light of wavelength $\lambda = 380\ \text{nm}$ is incident.
(i) Calculate the photon energy in eV (use $hc = 1240\ \text{eV·nm}$).
(ii) Calculate the threshold frequency $f_0$ and cutoff wavelength $\lambda_c$.
(iii) Will photoelectrons be emitted? Justify.
(b) Kinetic Energy, Stopping Potential, and Speed (5 marks)
(i) Calculate $K_{max}$ of the emitted electrons in eV and Joules.
(ii) Find the stopping potential $V_s$.
(iii) Calculate the maximum speed $v_{max}$ of ejected electrons. ($m_e = 9.11 \times 10^{-31}\ \text{kg}$)
(c) Graph and Parameter Effects (3 marks)
(i) Sketch $K_{max}$ vs $f$. Label slope, $y$-intercept, and $x$-intercept with physical meanings.
(ii) If the light intensity is doubled at constant frequency, how does the photocurrent change? How does $K_{max}$ change?
(iii) If a different metal with $\phi = 3.50\ \text{eV}$ is used with the same light source ($380\ \text{nm}$), will photoelectrons be emitted?
Score: ___/12
Scorecard
| Part | Problems | Raw Score |
|---|---|---|
| A — Structured | 15 | ___/15 |
| B1 — Electrostatic & Projectile | 1 (3 parts) | ___/12 |
| B2 — Capacitor + Dielectric + DC | 1 (3 parts) | ___/12 |
| B3 — AC Power & Phasor | 1 (3 parts) | ___/12 |
| C2 — Motor Torque | 1 (2 parts) | ___/12 |
| C3 — Transformer | 1 (3 parts) | ___/12 |
| C5 — Bohr Atom | 1 (3 parts) | ___/12 |
| C6 — Photoelectric | 1 (3 parts) | ___/12 |
| TOTAL | ~30 | ___/129 |
Proficiency Benchmarks
- 90/129 (70%) — Pass. You know the basics.
- 110/129 (85%) — Solid. Most formulas memorised.
- 120/129 (93%) — Exam-ready. Any mistake is careless.
- 129/129 (100%) — Perfect paper.
Error Log Template
After grading, list every wrong problem number with a one-word reason:
| Problem | Reason |
|---|---|
| e.g. A1 | time constant formula |
Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.
Answer Key
Section A — Structured
- B — $\tau = RC = 4.7 \times 10^3 \times 220 \times 10^{-6} = 1.034\ \text{s} \approx 1.03\ \text{s}$
- C — $e^{-3} \approx 0.0498 \approx 5%$, so $3\tau$
- B — $U = \frac12 CV^2 = \frac12 (33 \times 10^{-6})(25)^2 = \frac12 (33 \times 10^{-6})(625) = 1.031 \times 10^{-2}\ \text{J} = 10.3\ \text{mJ}$
- B — $3 + 2 = I$ so $I = 5\ \text{A}$ out
- B — $10 = 4 + 3 + I_4 \Rightarrow I_4 = 3\ \text{A}$ (if defined as leaving). If $I_4$ enters: $10 + I_4 = 4 + 3 \Rightarrow I_4 = -3$, meaning $3\ \text{A}$ leaving
- B — Conservation of charge (charge cannot accumulate at a junction)
- B — $V_{out} = -(220/22)(0.3) = -10 \times 0.3 = -3.0\ \text{V}$
- B — Gain $= 1 + 100/25 = 1 + 4 = 5$
- C — $-8.0 = -(R_f/12)(0.4) \Rightarrow R_f = 8.0 \times 12 / 0.4 = 240\ \text{k}\Omega$
- C — $B = \mu_0 I r / (2\pi R^2) = (4\pi \times 10^{-7})(15)(0.0015) / (2\pi \times 0.003^2) = (4\pi \times 10^{-7} \times 0.0225) / (2\pi \times 9 \times 10^{-6}) = (2.827 \times 10^{-8}) / (5.655 \times 10^{-5}) = 5.0 \times 10^{-4}\ \text{T}$
- C — At $r = R$: $B = \mu_0 I / (2\pi R) = (4\pi \times 10^{-7})(15) / (2\pi \times 0.003) = (6\pi \times 10^{-6}) / (6\pi \times 10^{-3}) = 1.0 \times 10^{-3}\ \text{T}$
- B — $B(r) \propto r$ inside the wire
- C — $V_{out} = (1 + 50/10)(0.2) = (6)(0.2) = +1.2\ \text{V}$
- A — At $t = 0$, capacitor is uncharged (acts as short circuit), so $I = V/R$
- B — $B = \mu_0 I r / (2\pi R^2) \Rightarrow I = (2\pi R^2 B) / (\mu_0 r) = (2\pi \times 0.002^2 \times 1.25 \times 10^{-4}) / (4\pi \times 10^{-7} \times 0.0005) = (2\pi \times 4 \times 10^{-6} \times 1.25 \times 10^{-4}) / (4\pi \times 10^{-7} \times 5 \times 10^{-4}) = (3.142 \times 10^{-9}) / (6.283 \times 10^{-10}) = 10.0\ \text{A}$
Section B1 — Electrostatic & Projectile
(a) $r_1 = 0.03\ \text{m}$, $r_2 = 0.05\ \text{m}$.
$E_1 = kQ_1/r_1^2 = (9 \times 10^9)(6 \times 10^{-6})/(0.03)^2 = 54 \times 10^3 / (9 \times 10^{-4}) = 6.0 \times 10^7\ \text{N/C}$ (away from $+$, so $+x$ direction).
$E_2 = k|Q_2|/r_2^2 = (9 \times 10^9)(4 \times 10^{-6})/(0.05)^2 = 36 \times 10^3 / (2.5 \times 10^{-3}) = 1.44 \times 10^7\ \text{N/C}$ (toward the negative charge at $x = 8$, so $-x$ direction).
$E_{net} = E_1 - E_2 = (6.0 - 1.44) \times 10^7 = 4.56 \times 10^7\ \text{N/C}$, direction $+x$ (rightward).
(b) $F = qE = (5 \times 10^{-9})(4.56 \times 10^7) = 0.228\ \text{N}$, direction $+x$.
(c) (i) $F_y = qE = (+1.60 \times 10^{-19})(8000) = 1.28 \times 10^{-15}\ \text{N}$ upward. $a_y = F_y/m = 1.28 \times 10^{-15} / 1.67 \times 10^{-27} = 7.66 \times 10^{11}\ \text{m/s}^2$ upward.
(ii) $t = L/v_0 = 0.08 / (2.0 \times 10^6) = 4.0 \times 10^{-8}\ \text{s} = 40\ \text{ns}$.
(iii) $y = \frac12 a_y t^2 = \frac12 (7.66 \times 10^{11})(4.0 \times 10^{-8})^2 = \frac12 (7.66 \times 10^{11})(1.6 \times 10^{-15}) = \frac12 (1.226 \times 10^{-3}) = 6.13 \times 10^{-4}\ \text{m} \approx 0.613\ \text{mm}$ (upward).
(iv) $v_x = v_0 = 2.0 \times 10^6\ \text{m/s}$ (constant). $v_y = a_y t = (7.66 \times 10^{11})(4.0 \times 10^{-8}) = 3.06 \times 10^4\ \text{m/s}$. $\theta = \tan^{-1}(v_y/v_x) = \tan^{-1}(3.06 \times 10^4 / 2.0 \times 10^6) = \tan^{-1}(0.0153) \approx 0.877^\circ$ above horizontal.
Section B2 — Capacitor + Dielectric + Voltage Divider
(a) $C = \kappa\varepsilon_0 A/d = 4.2(8.85 \times 10^{-12})(80 \times 10^{-4}) / (0.50 \times 10^{-3}) = 4.2(8.85 \times 10^{-12})(0.0080) / (5.0 \times 10^{-4}) = 4.2 \times 7.08 \times 10^{-14} / (5.0 \times 10^{-4}) = 2.97 \times 10^{-13} / (5.0 \times 10^{-4}) = 5.95 \times 10^{-10}\ \text{F} \approx 595\ \text{pF}$.
(b) (i) $Q = CV = (5.95 \times 10^{-10})(30) = 1.785 \times 10^{-8}\ \text{C} \approx 17.9\ \text{nC}$. $U = \frac12 CV^2 = \frac12 (5.95 \times 10^{-10})(900) = 2.68 \times 10^{-7}\ \text{J} \approx 268\ \text{nJ}$.
(ii) Battery connected ($V$ constant). Removing dielectric: $C' = C/\kappa = 5.95 \times 10^{-10}/4.2 = 1.417 \times 10^{-10}\ \text{F}$. $Q' = C'V = 1.417 \times 10^{-10} \times 30 = 4.25 \times 10^{-9}\ \text{C}$ (decreases). $U' = \frac12 C'V^2 = \frac12 (1.417 \times 10^{-10})(900) = 6.38 \times 10^{-8}\ \text{J}$ (decreases). The battery draws charge back.
(iii) If battery disconnected ($Q$ constant): removing dielectric makes $C \downarrow$ so $V = Q/C \uparrow$ (increases). $U = Q^2/(2C) \uparrow$ (increases because you do work pulling the dielectric out).
(c) (i) $V_{out} = 24 \times 18/(12 + 18) = 24 \times 18/30 = 14.4\ \text{V}$.
(ii) $R_{eq} = R_2 \parallel R_L = (18 \times 9)/(18 + 9) = 162/27 = 6.0\ \text{k}\Omega$. $V_{out}' = 24 \times 6.0/(12 + 6.0) = 24 \times 6/18 = 8.0\ \text{V}$.
(iii) Percentage drop $= (14.4 - 8.0)/14.4 \times 100% = 44.4%$.
Section B3 — AC Power & Phasor
(a) (i) $V_R = IR = 1.5 \times 60 = 90\ \text{V}$. $V_L = IX_L = 1.5 \times 100 = 150\ \text{V}$. $V_C = IX_C = 1.5 \times 40 = 60\ \text{V}$.
(ii) Phasor diagram: $V_R$ right ($0^\circ$), $V_L$ up ($+90^\circ$), $V_C$ down ($-90^\circ$). Net vertical: $150 - 60 = 90\ \text{V}$ up.
(iii) $V_T = \sqrt{90^2 + 90^2} = 90\sqrt{2} \approx 127.3\ \text{V}$. $\phi = \tan^{-1}(90/90) = 45^\circ$ (voltage leads current).
(b) (i) $P = I^2 R = (1.5)^2(60) = 2.25 \times 60 = 135\ \text{W}$. $Q_L = I^2 X_L = 2.25 \times 100 = 225\ \text{VAR}$. $Q_C = I^2 X_C = 2.25 \times 40 = 90\ \text{VAR}$.
(ii) $S = IV_T = 1.5 \times 127.3 \approx 191\ \text{VA}$. $\cos\phi = P/S = 135/191 \approx 0.707$.
(iii) Inductive — $X_L > X_C$, so $\phi > 0$ (voltage leads current).
(c) (i) $X_C = 1/(2\pi f C) = 1/(2\pi \times 60 \times 100 \times 10^{-6}) = 1/(0.0377) \approx 26.5\ \Omega$. $I_{rms} = V/X_C = 120/26.5 \approx 4.53\ \text{A}$.
(ii) $P = 0$ (pure capacitor, no resistance). $Q_C = V_{rms} I_{rms} = 120 \times 4.53 \approx 544\ \text{VAR}$. $S = 544\ \text{VA}$.
(iii) $\phi = -90^\circ$. By CIVIL: In a Capacitor, I leads V, so current leads voltage by $90^\circ$, meaning $\phi = -90^\circ$.
Section C2 — AC Motor Torque
(a) (i) $\tau = NIAB\sin\theta$: $N$ = number of turns, $I$ = current, $A$ = area of loop, $B$ = magnetic field, $\theta$ = angle between $\vec{B}$ and normal to loop plane.
(ii) $A = w \times h = 0.05 \times 0.08 = 0.004\ \text{m}^2$. $\mu = NIA = 200 \times 3.0 \times 0.004 = 2.4\ \text{A·m}^2$.
(iii) At $\theta = 30^\circ$: $\tau = NIAB\sin 30^\circ = 200 \times 3.0 \times 0.004 \times 0.50 \times 0.5 = 0.60\ \text{N·m}$.
At $\theta = 60^\circ$: $\tau = NIAB\sin 60^\circ = 0.60\ \text{N·m}$ (wait — let me recalculate properly). $\tau = 200 \times 3.0 \times 0.004 \times 0.50 = 1.2$ (that's $NIAB$). So $\tau = 1.2\sin\theta$. At $\theta = 30^\circ$: $\tau = 1.2 \times 0.5 = 0.60\ \text{N·m}$. At $\theta = 60^\circ$: $\tau = 1.2 \times 0.866 = 1.04\ \text{N·m}$.
(iv) $\tau_{max} = NIAB = 1.2\ \text{N·m}$ at $\theta = 90^\circ$ (plane of coil parallel to $\vec{B}$, normal $\perp \vec{B}$).
(b) (i) Maximum torque when $\theta = 90^\circ$: $\tau_{max} = NIAB = \mu B$. The plane of the loop is parallel to $\vec{B}$.
(ii) Minimum (zero) torque when $\theta = 0^\circ$ (or $180^\circ$): the normal to the loop is parallel/antiparallel to $\vec{B}$. The plane is perpendicular to $\vec{B}$.
(iii) In an AC motor, the current reverses direction each half-cycle. The reversal occurs when the coil passes through $\theta = 0^\circ$ (neutral plane), so the torque direction remains consistent — always driving rotation in the same direction. The average torque is $\tau_{avg} = \frac{2}{\pi} NIAB$, which is non-zero, producing continuous rotation.
Section C3 — Transformer
(a) (i) $B = \mu_0 n I = \mu_0 (N/\ell) I$. Flux through one turn: $\Phi_B = BA = \mu_0 N I A / \ell$. Flux linkage: $N\Phi_B = \mu_0 N^2 I A / \ell$. Then $L = N\Phi_B / I = \mu_0 N^2 A / \ell$.
(ii) $L = (4\pi \times 10^{-7})(600)^2(5.0 \times 10^{-4})/(0.40) = (4\pi \times 10^{-7})(3.6 \times 10^5)(5.0 \times 10^{-4})/0.40 = (4\pi \times 10^{-7})(180)/0.40 = (2.262 \times 10^{-4})/0.40 = 5.655 \times 10^{-4}\ \text{H} \approx 0.566\ \text{mH}$.
(iii) $U = \frac12 L I^2 = \frac12 (5.655 \times 10^{-4})(2.0)^2 = \frac12 (5.655 \times 10^{-4})(4) = 1.131 \times 10^{-3}\ \text{J} \approx 1.13\ \text{mJ}$.
(b) (i) $M = k\sqrt{L_1 L_2} = 0.75\sqrt{8.0 \times 10^{-3} \times 18 \times 10^{-3}} = 0.75\sqrt{1.44 \times 10^{-4}} = 0.75 \times 0.012 = 9.0 \times 10^{-3}\ \text{H} = 9.0\ \text{mH}$.
(ii) $\mathcal{E}_2 = -M\ dI_1/dt = -(9.0 \times 10^{-3})(150) = -1.35\ \text{V}$. Magnitude: $1.35\ \text{V}$.
(iii) Reciprocity: $M_{12} = M_{21} = M$ — mutual inductance from coil 1 to coil 2 equals that from coil 2 to coil 1.
(c) (i) Step-up ($N_s/N_p = 450/150 = 3 > 1$).
(ii) $V_s = V_p \times N_s/N_p = 120 \times 3 = 360\ \text{V}$. $I_s = I_p \times N_p/N_s = 1.2 \times 150/450 = 1.2 \times 1/3 = 0.4\ \text{A}$.
(iii) $R_L = V_s/I_s = 360/0.4 = 900\ \Omega$.
Section C5 — Bohr Atom
(a) (i) Bohr's quantization: $mvr = n\hbar$ where $\hbar = h/(2\pi)$ and $n = 1, 2, 3, \dots$
(ii) Coulomb force = centripetal force: $k e^2 / r^2 = m v^2 / r$. So $k e^2 / r = m v^2$.
(iii) From (ii): $v^2 = k e^2 / (m r)$. From (i): $v = n\hbar / (m r)$, so $v^2 = n^2 \hbar^2 / (m^2 r^2)$.
Equate: $n^2 \hbar^2 / (m^2 r^2) = k e^2 / (m r)$.
Multiply both sides by $m r^2$: $n^2 \hbar^2 / m = k e^2 r$.
Solve: $r_n = n^2 \hbar^2 / (m k e^2) = n^2 a_0$ where $a_0 = \hbar^2 / (m k e^2) = 0.529\ \text{Å}$.
(iv) $r_2 = 4 a_0 = 4 \times 0.529 = 2.116\ \text{Å}$. $r_3 = 9 a_0 = 9 \times 0.529 = 4.761\ \text{Å}$.
(b) (i) $E_1 = -13.6\ \text{eV}$, $E_2 = -13.6/4 = -3.40\ \text{eV}$, $E_3 = -13.6/9 \approx -1.51\ \text{eV}$.
(ii) $\Delta E = E_4 - E_2 = (-13.6/16) - (-3.40) = -0.85 + 3.40 = 2.55\ \text{eV}$.
$\lambda = hc / \Delta E = 1240 / 2.55 \approx 486\ \text{nm}$.
(iii) Balmer series ($n \to 2$). Yes, $486\ \text{nm}$ is visible blue-green light (visible range ~400–700 nm).
(c) (i) $n = 5 \to n = 1$: Lyman series, UV.
(ii) $n = 4 \to n = 3$: Paschen series, infrared.
(iii) $n = 6 \to n = 2$: Balmer series, visible.
Section C6 — Photoelectric Effect
(a) (i) $E = hc/\lambda = 1240/380 \approx 3.26\ \text{eV}$.
(ii) $f_0 = \phi/h = (2.80 \times 1.60 \times 10^{-19}) / (6.63 \times 10^{-34}) = 4.48 \times 10^{-19} / 6.63 \times 10^{-34} \approx 6.76 \times 10^{14}\ \text{Hz}$.
$\lambda_c = hc/\phi = 1240/2.80 \approx 443\ \text{nm}$.
(iii) Yes. Photon energy $3.26\ \text{eV} > \phi = 2.80\ \text{eV}$ (equivalently, $\lambda = 380\ \text{nm} < \lambda_c = 443\ \text{nm}$).
(b) (i) $K_{max} = hf - \phi = 3.26 - 2.80 = 0.46\ \text{eV}$. In Joules: $0.46 \times 1.60 \times 10^{-19} = 7.36 \times 10^{-20}\ \text{J}$.
(ii) $V_s = K_{max}/e = 0.46\ \text{V}$.
(iii) $v_{max} = \sqrt{2K_{max}/m_e} = \sqrt{2(7.36 \times 10^{-20}) / (9.11 \times 10^{-31})} = \sqrt{1.616 \times 10^{11}} \approx 4.02 \times 10^5\ \text{m/s}$.
(c) (i) Graph: straight line $K_{max} = hf - \phi$. Slope $= h$ (Planck's constant). $y$-intercept $= -\phi$ (negative work function). $x$-intercept $= f_0 = \phi/h$ (threshold frequency).
(ii) Doubling intensity at same frequency: photocurrent doubles (more photons $\to$ more electrons ejected per second). $K_{max}$ unchanged (depends only on $f$ and $\phi$, not intensity).
(iii) For $\phi = 3.50\ \text{eV}$: $E = 3.26\ \text{eV} < \phi = 3.50\ \text{eV}$. No photoelectrons emitted. The photon energy is below the work function threshold.