FAD1022: Max Score Strategy — Minimal Study

[!tip] The Goal Score ~129/129 by picking the RIGHT questions and studying ONLY the topics you'll actually answer.

Total formulas to memorise: ~18 Total study time: 2–3 hours

The Maths

Section	Format	Marks	Your Pick
A	15 structured × 3 (all compulsory)	45	All 15
B	4 Q → pick 3 × 12	36	B2 + B1 + B3 ✅
C	6 Q → pick 4 × 12	48	C2 + C3 + C5 + C6 ✅
Total attempted		129

Section A — Structured Questions (all compulsory, 45 marks)

A1: Capacitor Charging & Discharging

Charging:
- $$q(t) = Q(1 - e^{-t/\tau})$$
- $$i(t) = \frac{V}{R}e^{-t/\tau}$$
- $$V_C(t) = V(1 - e^{-t/\tau})$$
Discharging:
- $$q(t) = Q_0 e^{-t/\tau}$$
- $$i(t) = -I_0 e^{-t/\tau}$$
- $$V_C(t) = V_0 e^{-t/\tau}$$
Time constant: $$\tau = RC$$
Key graph shapes:
- Charging: exponential rise to Q (asymptotic)
- Discharging: exponential decay to 0
At $t = \tau$: charging reaches ~63% of final value; discharging falls to ~37% of initial

A2: Kirchhoff's Current Law

$$\sum I_{in} = \sum I_{out}$$ at any junction
Equivalent: $$\sum I = 0$$ (currents entering positive, leaving negative)
Simple conservation of charge — currents entering a node must equal currents leaving it
Typical Section A: given two currents entering and one leaving, find the missing current

A3: Amplifier Formulas

Inverting:
- $$V_{out} = -\frac{R_f}{R_{in}} V_{in}$$
- Negative sign means 180° phase shift
Non-inverting:
- $$V_{out} = \left(1 + \frac{R_f}{R_{in}}\right) V_{in}$$
- Positive gain, no phase shift
Voltage follower (non-inverting special case, no feedback resistors): $$V_{out} = V_{in}$$
Just memorise these two formulas — plug and chug

A4: Ampere's Law — Inside Wire (r < R)

Formula:
- $$B(r) = \frac{\mu_0 I r}{2\pi R^2}$$
Derivation for Section A:
- $$\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = \mu_0 I_{enc}$$
- $$I_{enc} = I \cdot \frac{\pi r^2}{\pi R^2} = I\frac{r^2}{R^2}$$
- ∴ $$B(2\pi r) = \mu_0 I \frac{r^2}{R^2} \implies B = \frac{\mu_0 I r}{2\pi R^2}$$
Check: at r = R, $$B = \frac{\mu_0 I}{2\pi R}$$ which matches the outside-wire formula

A5: Lenz Law

🚫 NOT TESTED — skip entirely

Section B — Pick 3 of 4 (36 marks)

✅ PICK B2: Capacitor + Dielectric + DC

Parts a & b: Capacitance & Energy

Parallel plate capacitance:
- $$C = \kappa\varepsilon_0 \frac{A}{d}$$
- $\kappa$ = dielectric constant (1 for vacuum/air)
Energy stored:
- $$U = \frac12 CV^2 = \frac12 QV = \frac{Q^2}{2C}$$
Dielectric effects:
- Battery connected (V constant): inserting dielectric → C↑ (×κ), Q↑ (×κ), U↑ (×κ)
- Battery disconnected (Q constant): inserting dielectric → C↑ (×κ), V↓ (÷κ), U↓ (÷κ)
- Removing dielectric reverses the effect in each case

Part c: Voltage Divider (Loaded & Unloaded)

Unloaded:
- $$V_{out} = V_{in} \frac{R_2}{R_1 + R_2}$$
Loaded (with $R_L$ across $R_2$):
- $$R_{parallel} = \frac{R_2 R_L}{R_2 + R_L}$$
- $$V_{out} = V_{in} \frac{R_{parallel}}{R_1 + R_{parallel}}$$
Loading effect: Adding a load reduces $V_{out}$ compared to the unloaded value — the smaller $R_L$, the more severe the loading
Rule of thumb: To minimise loading, make $R_L \gg R_2$

✅ PICK B1: Electrostatic — E-field & Projectile Motion

E-field Vector

Point charge:
- $$\vec{E} = \frac{kQ}{r^2} \hat{r}$$
- $k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\ \text{N·m}^2/\text{C}^2$
Force on test charge:
- $$\vec{F} = q\vec{E}$$
Superposition principle:
- $$\vec{E}_{net} = \sum \vec{E}_i$$
- Add fields as vectors (resolve x and y components)

Projectile Motion in E-field

Setup: Charge q enters uniform E-field between parallel plates with initial horizontal velocity v₀
Vertical acceleration:
- $$a_y = \frac{F}{m} = \frac{qE}{m}$$
Time in field:
- $$t = \frac{L}{v_0}$$ where L = plate length
Vertical deflection at exit:
- $$y = \frac12 a_y t^2 = \frac12 \cdot \frac{qE}{m} \cdot \left(\frac{L}{v_0}\right)^2$$
Exit velocity components:
- $v_x = v_0$ (constant, no horizontal force)
- $v_y = a_y t = \frac{qE}{m} \cdot \frac{L}{v_0}$
Exit angle:
- $$\tan\theta = \frac{v_y}{v_x} = \frac{qEL}{mv_0^2}$$
Analogy: treat like gravity projectile but with $a_y = qE/m$ instead of g

✅ PICK B3: AC (PRC + PLC + PCC)

Phasor Diagrams

Resistive (PRC): V and I in phase (0° phase difference)
Inductive (PLC): Voltage leads current by 90° (V ahead of I)
Capacitive (PCC): Current leads voltage by 90° (I ahead of V)
CIVIL mnemonic:
- C — I leads V
- V leads I — L

Power Types

Power	Symbol	Formula	Unit	Description
Real/Active (PRC)	P	$V_{rms}I_{rms}\cos\phi$	W (watts)	Dissipated by resistor
Inductive Reactive (PLC)	Q_L	$V_{rms}I_{rms}\sin\phi$	VAR	Stored/released by inductor
Capacitive Reactive (PCC)	Q_C	$V_{rms}I_{rms}\sin\phi$	VAR	Stored/released by capacitor
Apparent	S	$V_{rms}I_{rms}$	VA	Total power supplied

Power Triangle:
- $$S^2 = P^2 + (Q_L - Q_C)^2$$
Power Factor:
- $$\cos\phi = \frac{P}{S} = \frac{R}{Z}$$
- Power factor = 1 → purely resistive (maximum real power)
- Power factor = 0 → purely reactive (no real power dissipation)

RLC Series Circuit

Impedance:
- $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
- $X_L = \omega L$ (inductive reactance)
- $X_C = \frac{1}{\omega C}$ (capacitive reactance)
Phase angle:
- $$\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$
Interpretation:
- $X_L > X_C$ → positive φ → inductive → V leads I
- $X_C > X_L$ → negative φ → capacitive → I leads V
- $X_L = X_C$ → resonance → φ = 0 → purely resistive

❌ SKIP B4: AC Claims & RLC

Conceptual questions about identifying true/false statements. Requires precise judgement on edge cases. Trickier mark allocation and more ambiguous than the formulaic B1-B3. Not worth the study time.

Section C — Pick 4 of 6 (48 marks)

✅ PICK C2: EM — AC Motor Torque

Single formula, extended across parts:

$$\tau = NIAB\sin\theta$$
- N = number of turns in the coil
- I = current through the coil
- A = area of the loop ($A = l \times w$)
- B = uniform magnetic field strength
- $\theta$ = angle between B-field and normal to the loop's plane
Maximum torque:
- Occurs at $\theta = 90^\circ$ (plane of loop parallel to B-field)
- $$\tau_{max} = NIAB$$
Minimum torque:
- Occurs at $\theta = 0^\circ$ (plane of loop perpendicular to B-field)
- $$\tau = 0$$ — forces are colinear, no turning effect
Magnetic dipole moment:
- $$\vec{\mu} = NIA\ \hat{n}$$ where $\hat{n}$ is normal to the loop
- Alternative: $$\tau = \mu B \sin\theta$$
AC motor behaviour:
- Current alternates → torque varies sinusoidally with time
- Average torque is non-zero → continuous rotation
Direction: Use Fleming's Left-Hand Rule (thumb = motion, index = field, middle = current)

✅ PICK C3: Transformer — Self & Mutual Induction

Self-Inductance

Definition:
- $$L = \frac{N\Phi}{I}$$
- Units: Henry (H)
Solenoid:
- $$L = \frac{\mu_0 N^2 A}{\ell}$$
- $\ell$ = length of solenoid, A = cross-sectional area
Back EMF:
- $$\mathcal{E} = -L\frac{dI}{dt}$$
- Negative sign from Lenz Law (opposes change)
Energy stored in inductor:
- $$U = \frac12 LI^2$$

Mutual Inductance

Definition:
- $$M = \frac{N_2\Phi_{21}}{I_1} = \frac{N_1\Phi_{12}}{I_2}$$
With coupling coefficient:
- $$M = k\sqrt{L_1 L_2}$$
- k = 1 for perfect coupling (all flux links both coils)
- k = 0 for no coupling
Reciprocity:
- $$M_{12} = M_{21} = M$$

Transformer

Turns ratio (voltage):
- $$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$
Current ratio:
- $$\frac{I_s}{I_p} = \frac{N_p}{N_s}$$
Step-up: $N_s > N_p$ → $V_s > V_p$, $I_s < I_p$
Step-down: $N_s < N_p$ → $V_s < V_p$, $I_s > I_p$
Ideal transformer (power conserved):
- $$V_p I_p = V_s I_s$$
- $$P_{in} = P_{out}$$ (negligible losses)
Transformer equation:
- $$\frac{V_p}{N_p} = \frac{V_s}{N_s}$$ (volts per turn is constant)

✅ PICK C5: Atomic — Bohr Radius Derivation

Step-by-step derivation (memorise the flow):

Step 1: Coulomb attraction = centripetal force $$\frac{ke^2}{r^2} = \frac{mv^2}{r}$$

Step 2: Angular momentum is quantised $$mvr = n\hbar \quad \text{where } \hbar = \frac{h}{2\pi}$$

Step 3: From Step 1: $$mv^2 = \frac{ke^2}{r}$$

Step 4: From Step 2: $$v = \frac{n\hbar}{mr}$$

Step 5: Substitute v into Step 3: $$m\left(\frac{n\hbar}{mr}\right)^2 = \frac{ke^2}{r} \quad\implies\quad \frac{n^2\hbar^2}{mr^2} = \frac{ke^2}{r}$$

Step 6: Solve for r: $$r_n = \frac{n^2\hbar^2}{mke^2} = n^2 a_0$$ where $$a_0 = \frac{\hbar^2}{mke^2} = 0.529\ \text{Å}$$ (Bohr radius)

Energy Levels

Total energy: $$E_n = \frac12 mv^2 - \frac{ke^2}{r} = -\frac{ke^2}{2r}$$
Substitute $r_n$: $$E_n = -\frac{ke^2}{2a_0} \cdot \frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}$$
Ground state (n = 1): E₁ = -13.6 eV
Ionisation energy: 13.6 eV (energy to remove electron from ground state)
Transition energy: $$\Delta E = E_f - E_i = 13.6\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) \text{eV}$$
Wavelength: $$\lambda = \frac{hc}{\Delta E} = \frac{1240\ \text{eV·nm}}{\Delta E\ (\text{in eV})}$$

Spectral Series

Series	n_f	Region	First line (n_i→n_f)
Lyman	1	UV	2 → 1 (121.6 nm)
Balmer	2	Visible	3 → 2 (656.3 nm)
Paschen	3	Infrared	4 → 3 (1875 nm)

✅ PICK C6: Photoelectric Effect

Key Equations

Photon energy: $$E = hf = \frac{hc}{\lambda}$$
- h = 6.626 × 10⁻³⁴ J·s (Planck's constant)
- c = 3.0 × 10⁸ m/s
- hc = 1240 eV·nm (useful shortcut)
Einstein's photoelectric equation: $$K_{max} = hf - \phi$$
- $K_{max}$ = maximum kinetic energy of ejected electrons
- $\phi$ = work function of the metal (minimum energy to eject)
Stopping potential: $$eV_s = K_{max} \quad\implies\quad V_s = \frac{hf - \phi}{e}$$
- $V_s$ = potential needed to stop the most energetic electrons
Threshold frequency: $$f_0 = \frac{\phi}{h} \quad (\text{when } K_{max} = 0)$$
Cutoff wavelength: $$\lambda_c = \frac{hc}{\phi}$$
Maximum electron speed: $$v_{max} = \sqrt{\frac{2K_{max}}{m_e}}$$
- $m_e = 9.11 \times 10^{-31}$ kg

Graphs

$K_{max}$ vs $f$:
- Straight line: slope = h (Planck's constant)
- y-intercept = -φ (negative work function)
- x-intercept = f₀ (threshold frequency)
$V_s$ vs $f$:
- Straight line: slope = h/e
- y-intercept = -φ/e
Effect of increasing intensity:
- Doubles photocurrent (more photons → more electrons)
- Does NOT change $K_{max}$ or $V_s$
Effect of higher work function:
- Graph shifts right (higher f₀)
- Graph shifts down (more negative intercept)

Past Year Pattern (from leak)

Part	Typical Question	Marks
(a)	Define terms / state Einstein's equation	2–3
(b)	Calculate $K_{max}$, $V_s$, $v_{max}$, or $\lambda$	5–6
(c)	Sketch graph, label axes, explain effect of changing parameters	3–4

❌ SKIP — These Questions (Don't Study)

Question	Why Skip
A5 (Lenz Law)	Explicitly NOT TESTED per leak
B4 (AC Claims)	Tricky conceptual true/false; ambiguous marking
C1 (Magnetism — Gauss & Ampere, 4-wire)	Heaviest derivation in the paper; too much time
C4 (Semi & Biasing)	Many sub-parts; clipper circuits need specialised practice

Quick Reference: All Formulas You Need

#	Topic	Formula	Notes
1	Capacitance	$$C = \kappa\varepsilon_0 \frac{A}{d}$$	Parallel plate
2	Capacitor energy	$$U = \frac12 CV^2 = \frac{Q^2}{2C}$$	Pick the 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	$$V_{out} = V_{in}\frac{R_2}{R_1+R_2}$$	Unloaded
6	Voltage divider (loaded)	$$V_{out} = V_{in}\frac{R_{eq}}{R_1+R_{eq}}$$	$R_{eq} = R_2 \parallel R_L$
7	E-field	$$\vec{E} = \frac{kQ}{r^2} \hat{r}$$	$k = 1/4\pi\varepsilon_0$
8	Acceleration in E-field	$$a_y = \frac{qE}{m}$$	Analogy: gravity but with qE/m
9	Deflection in E-field	$$y = \frac12 \frac{qE}{m} \left(\frac{L}{v_0}\right)^2$$
10	Amplifier (inverting)	$$V_{out} = -\frac{R_f}{R_{in}}V_{in}$$	180° phase shift
11	Amplifier (non-inv)	$$V_{out} = \left(1+\frac{R_f}{R_{in}}\right)V_{in}$$	No phase shift
12	Ampere inside wire	$$B = \frac{\mu_0 I r}{2\pi R^2}$$	r < R
13	Power triangle	$$S^2 = P^2 + (Q_L - Q_C)^2$$	$\cos\phi = P/S$
14	Motor torque	$$\tau = NIAB\sin\theta$$	Max at 90°
15	Self-inductance	$$L = \frac{\mu_0 N^2 A}{\ell}$$	Solenoid
16	Transformer	$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$	Also $I_s/I_p = N_p/N_s$
17	Bohr radius	$$r_n = n^2 a_0$$	$a_0 = 0.529\ \text{Å}$
18	Bohr energy	$$E_n = -\frac{13.6}{n^2}\ \text{eV}$$	Ionisation = 13.6 eV
19	Photoelectric	$$K_{max} = hf - \phi$$
20	Stopping potential	$$eV_s = K_{max}$$

Study Order (highest yield first)

Step	Topic	Time	Rationale
1	B2 formulas — capacitance + voltage divider	15 min	High marks, few formulas, formulaic
2	C6 formulas — photoelectric equation + graphs	15 min	3-part pattern repeats yearly
3	C2 formula — motor torque	5 min	Single formula → 12 marks
4	C3 formulas — transformer ratios + inductance	15 min	Turns ratio is plug-and-chug
5	C5 derivation — Bohr radius + energy levels	20 min	Step-by-step → full derivation marks
6	B1 formulas — E-field + projectile kinematics	15 min	Standard mechanics with E-field
7	B3 formulas — power triangle + phasors	15 min	CIVIL mnemonic + 3 equations
8	A formulas — quick-fire revision	20 min	All compulsory, covers all topics
9	Targeted drill pack — apply everything	60 min	FAD1022 Rapid-Fire Drill — Leak Topics
	Total	~3 hours

FAD1022 Max Score Strategy — Minimal Study