Grade 13 — Term 1
14 study notes · 202 practice questions · Aligned with NIE syllabus
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This term in the Idasara Academy app has 14 lessons (NIE Units 7, 8). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Electric Current & Charge Flow
What you must know
- Conventional Current Direction: By historical convention, direction of electric current is defined as the direction in which positive charges would flow; conduction electrons (negative) physically drift in the OPPOSITE direction to conventional current.
- Charge Carriers across Media: In metallic conductors: Conduction electrons; In electrolytes: Positive and negative ions (cations and anions); In semiconductor crystal lattices: Electrons and holes; In ionized gases/plasmas: Positive ions and free electrons.
- Vector nature of current: Although current has an associated direction, it is a SCALAR quantity because it obeys ordinary scalar addition laws (Kirchhoff's Current Law), not vector addition rules.
Try one question
A copper wire of cross-sectional area A = 1.0 × 10⁻⁶ m² carries a steady current I = 3.2 A. If the free electron density in copper is n = 8.0 × 10²⁸ m⁻³ and electron charge is e = 1.6 × 10⁻¹⁹ C, what is the drift velocity v_d of conduction electrons in the wire?
- A. 2.5 × 10⁻⁴ m s⁻¹
- B. 2.5 × 10⁻² m s⁻¹
- C. 5.0 × 10⁻⁴ m s⁻¹
- D. 1.25 × 10⁻³ m s⁻¹
Show the answer
Answer: A. 2.5 × 10⁻⁴ m s⁻¹. The relationship between electric current and electron drift velocity is I = n A e v_d. Rearranging: v_d = I / (n A e) = 3.2 A / [(8.0 × 10²⁸ m⁻³) × (1.0 × 10⁻⁶ m²) × (1.6 × 10⁻¹⁹ C)] = 3.2 / (1.28 × 10⁴) = 2.5 × 10⁻⁴ m s⁻¹.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Ohm’s Law, Resistance & Resistivity
What you must know
- Ohmic vs Non-Ohmic Devices: Ohmic conductors exhibit linear I-V characteristic passing through origin (constant R); Non-Ohmic conductors exhibit non-linear I-V curve (e.g., filament lamp curves as filament heats up, semiconductor diode conducts only in forward bias, thermistor resistance drops with temperature).
- Negative Temperature Coefficient (NTC): Semiconductors and electrolytes have NEGATIVE temperature coefficients of resistance (α < 0); resistance DECREASES with heating because thermal generation of charge carriers dramatically increases carrier density n.
- Superconductivity: Phenomenon wherein certain metals and alloys lose ALL electrical resistance (R = 0 identically) below a characteristic critical transition temperature T_c (e.g., Mercury at 4.2 K).
Try one question
A uniform metallic wire of resistance R is drawn out mechanically so that its length is doubled while its volume remains constant. What is the new resistance of the wire?
- A. R / 2
- B. 4R
- C. 2R
- D. R / 4
Show the answer
Answer: B. 4R. Resistance is R = ρ L / A. When volume V = A · L remains constant and length L' = 2L, the cross-sectional area must halve: A' = A / 2. Therefore, the new resistance is R' = ρ L' / A' = ρ (2L) / (A / 2) = 4 (ρ L / A) = 4R.
A model question written by Idasara on the exam pattern. It is not from a past paper.
EMF, Internal Resistance & Cells
What you must know
- Maximum Power Transfer Theorem: A real power source of emf E and internal resistance r transfers MAXIMUM power to an external variable load R when load resistance equals internal resistance: R = r; Maximum Power P_max = E² / (4r); efficiency at maximum power transfer is exactly 50%.
- Measuring EMF vs Terminal Voltage: A standard voltmeter draws current, thus always measuring V = E - Ir < E; to measure true EMF E, use a Potentiometer (null-deflection, draws zero current at balance point).
- Graph of V vs I: Linear plot of V = -r I + E yields vertical intercept E and negative slope -r.
Try one question
A cell of electromotive force (EMF) E = 6.0 V and internal resistance r = 1.0 Ω is connected across an external resistor R. According to the Maximum Power Transfer Theorem, what is the maximum power that can be delivered to the external resistor?
- A. 36.0 W
- B. 4.5 W
- C. 9.0 W
- D. 18.0 W
Show the answer
Answer: C. 9.0 W. By the Maximum Power Transfer Theorem, power delivered to external load R is maximized when R = r = 1.0 Ω. At this condition, circuit current is I = E / (R + r) = 6.0 V / (1.0 + 1.0) Ω = 3.0 A. Maximum power delivered is P_max = I² R = (3.0 A)² × 1.0 Ω = 9.0 W (or P_max = E² / (4r) = 36 / 4 = 9.0 W).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Series & Parallel Circuits
What you must know
- Ammeter & Voltmeter Design: Moving Coil Galvanometer (resistance G, full-scale deflection current Ig): Converted to Ammeter by connecting low-resistance Shunt S in PARALLEL: S = (Ig G) / (I - Ig); Converted to Voltmeter by connecting high-resistance Multiplier R in SERIES: R = (V / Ig) - G.
- Ideal Meter Characteristics: Ideal Ammeter has ZERO resistance (zero circuit disturbance); Ideal Voltmeter has INFINITE resistance (draws zero current).
- Equivalent resistance bounding rules: Series R_eq is always GREATER than the largest individual resistor; Parallel R_eq is always SMALLER than the smallest individual resistor.
Try one question
Three identical resistors each of resistance R = 12 Ω are connected in a network such that two resistors are in parallel, and this combination is connected in series with the third resistor. What is the equivalent resistance of the entire network between terminals?
- A. 18 Ω
- B. 36 Ω
- C. 24 Ω
- D. 8 Ω
Show the answer
Answer: A. 18 Ω. The two parallel resistors have an equivalent resistance of R_p = (12 × 12) / (12 + 12) = 6 Ω. This combination is in series with the third 12 Ω resistor: R_eq = R_p + R₃ = 6 Ω + 12 Ω = 18 Ω.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Kirchhoff’s Laws
What you must know
- Linear independence of loops: When applying KVL to multiple loops, ensure each new loop includes at least ONE new branch not contained in previous loops to maintain mathematical independence.
- Handling internal resistances in KVL: Always treat internal resistance r as a series resistor physically attached to the cell: loop equation includes I * r.
- Negative current result: If solving circuit equations yields a negative value for current (I < 0), the magnitude is physically correct, but the true current flows in the OPPOSITE direction to your initial assumed arrow.
Try one question
Kirchhoff's Junction Rule (Current Law, ΣI = 0) and Kirchhoff's Loop Rule (Voltage Law, ΣV = 0) are direct consequences of which fundamental physical conservation laws, respectively?
- A. Conservation of energy and Conservation of linear momentum
- B. Conservation of mass and Conservation of magnetic flux
- C. Conservation of electric charge and Conservation of angular momentum
- D. Conservation of electric charge and Conservation of energy
Show the answer
Answer: D. Conservation of electric charge and Conservation of energy. Kirchhoff's first law (Junction Rule) asserts that charge cannot accumulate at an electrical node, embodying Conservation of Electric Charge. Kirchhoff's second law (Loop Rule) states that the sum of potential differences around any closed circuit loop is zero, embodying Conservation of Energy.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Wheatstone Bridge & Meter Bridge
What you must know
- Maximum Sensitivity of Wheatstone Bridge: Bridge sensitivity is maximized when all four arms have roughly equal resistance values (P ≈ Q ≈ R ≈ S); balance point on Meter Bridge is most accurate near center (L1 ≈ 50 cm).
- Zero current independence: At balance (I_g = 0), the balance condition P/Q = R/S is completely independent of the galvanometer's internal resistance and the battery's emf/internal resistance.
- Interchanging Battery and Galvanometer: Interchanging the positions of the battery and galvanometer does not affect the balance condition.
Try one question
In a meter bridge experiment, an unknown resistor X is placed in the left gap and a standard 15 Ω resistor is placed in the right gap. The balance point is obtained at a distance of 40 cm from the zero end (left side). What is the value of the unknown resistance X?
- A. 22.5 Ω
- B. 15 Ω
- C. 10 Ω
- D. 6.0 Ω
Show the answer
Answer: C. 10 Ω. Using the balanced meter bridge relation: X / R_standard = l₁ / (100 - l₁). With l₁ = 40 cm, the remaining length is 100 - 40 = 60 cm. Thus: X / 15 Ω = 40 / 60 = 2 / 3 => X = (2/3) × 15 Ω = 10 Ω.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Potentiometer
What you must know
- Primary Circuit Mandatory Conditions: 1) Driver battery EMF E0 MUST be greater than test EMF E (E0 > E); 2) Positive terminal of driver cell and positive terminal of test cell MUST be connected to the SAME zero-index terminal (node A).
- Increasing Potentiometer Sensitivity: To detect smaller potential differences, make potential gradient k smaller (flatter) by increasing wire length L or adding series rheostat in primary driver circuit.
- Calibration of Potentiometer: Determining exact numerical value of potential gradient k using a standard cell of certified emf (e.g., Weston cadmium cell E_std = 1.0186 V): k = E_std / l_std.
Try one question
A potentiometer wire of length 100 cm and resistance 5.0 Ω is connected in series with a 4.0 V driver battery of negligible internal resistance and a series limiting resistor R_s = 15.0 Ω. What is the potential gradient along the potentiometer wire?
- A. 0.5 V m⁻¹ (0.005 V cm⁻¹)
- B. 2.0 V m⁻¹ (0.02 V cm⁻¹)
- C. 4.0 V m⁻¹ (0.04 V cm⁻¹)
- D. 1.0 V m⁻¹ (0.01 V cm⁻¹)
Show the answer
Answer: D. 1.0 V m⁻¹ (0.01 V cm⁻¹). Total circuit resistance R_total = R_wire + R_s = 5.0 Ω + 15.0 Ω = 20.0 Ω. Current in potentiometer wire I = V_driver / R_total = 4.0 V / 20.0 Ω = 0.20 A. Potential drop across the wire V_wire = I × R_wire = 0.20 A × 5.0 Ω = 1.0 V. Potential gradient k = V_wire / L = 1.0 V / 1.0 m = 1.0 V m⁻¹ (0.01 V cm⁻¹).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Magnetic Fields & Magnetic Force
What you must know
- Charged particle trajectory in uniform magnetic field: Injected perpendicular (θ = 90°): executes Uniform Circular Motion with radius r = (m v) / (q B) and cyclotron frequency f = (q B) / (2π m); Injected at angle θ: executes Helical path with pitch p = v_parallel * T = (2π m v cos θ) / (q B).
- Velocity Selector: Crossed uniform electric field E and magnetic field B (E ⊥ B); particles passing undeflected satisfy q E = q v B => v = E / B, regardless of charge or mass.
- Torque on Current Loop: Rectangular coil of N turns and area A carrying current I in magnetic field B: Torque τ = N I A B sin θ = m × B, where magnetic dipole moment m = N I A n̂.
Try one question
A proton of charge q = 1.6 × 10⁻¹⁹ C and mass m = 1.67 × 10⁻²⁷ kg enters perpendicularly into a uniform magnetic field B = 0.50 T with velocity v = 2.0 × 10⁶ m s⁻¹. What is the radius of the circular path traversed by the proton?
- A. 1.05 cm (0.0105 m)
- B. 8.35 cm (0.0835 m)
- C. 2.09 cm (0.0209 m)
- D. 4.18 cm (0.0418 m)
Show the answer
Answer: D. 4.18 cm (0.0418 m). The magnetic Lorentz force provides the centripetal force: q v B = m v² / r => r = m v / (q B). Substituting values: r = (1.67 × 10⁻²⁷ kg × 2.0 × 10⁶ m s⁻¹) / (1.6 × 10⁻¹⁹ C × 0.50 T) = (3.34 × 10⁻²¹) / (0.80 × 10⁻¹⁹) ≈ 0.04175 m ≈ 4.18 cm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Magnetic Fields of Conductors & Solenoids
What you must know
- Force between two long parallel conductors: Magnetic force per unit length between two infinite wires separated by distance d carrying currents I1 and I2: F / L = (μ0 I1 I2) / (2π d); currents in SAME direction ATTRACT; currents in OPPOSITE directions REPEL.
- Formal SI Definition of Ampere: The Ampere is that constant current which, maintained in two straight parallel conductors of infinite length and negligible cross-section placed 1 meter apart in vacuum, produces between them a force equal to 2 × 10⁻⁷ N per meter of length.
- Circular Loop Center Field: At the center of a circular coil of N turns and radius R: B = (μ0 N I) / (2 R).
Try one question
A long air-core solenoid of length 0.50 m has 1000 turns of closely wound insulated wire. If a steady current of 2.0 A flows through the winding, what is the magnetic flux density B at the center of the solenoid? (μ₀ = 4π × 10⁻⁷ T m A⁻¹)
- A. 2.51 × 10⁻³ T
- B. 1.26 × 10⁻³ T
- C. 5.03 × 10⁻³ T (1.6π × 10⁻³ T)
- D. 1.01 × 10⁻² T
Show the answer
Answer: C. 5.03 × 10⁻³ T (1.6π × 10⁻³ T). The magnetic field inside a long solenoid is B = μ₀ n I, where n = N / L is the turn density. Here n = 1000 / 0.50 m = 2000 turns m⁻¹. Thus: B = (4π × 10⁻⁷ T m A⁻¹) × (2000 m⁻¹) × (2.0 A) = 1.6π × 10⁻³ T ≈ 5.03 × 10⁻³ T.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Electromagnetic Induction
What you must know
- Eddy Currents: Closed circulating loops of induced current generated inside solid conducting masses subjected to varying magnetic fields; generate unwanted Joule heating (I²R loss); minimized by laminating cores into insulated sheets (transformers) or utilized constructively in induction cookers and electromagnetic braking.
- Self-Inductance (L) and Mutual Inductance (M): Self-induced emf E = - L (dI/dt); Mutual induced emf E2 = - M (dI1/dt); Inductance of solenoid L = μ0 n² A L; Energy stored in inductor U = (1/2) L I².
- Rotating Coil in Magnetic Field (AC Generator): Coil of N turns and area A rotating at angular velocity ω: Instantaneous flux Φ = B A cos(ωt); Induced emf E = - dΦ/dt = E0 sin(ωt) where peak emf E0 = N B A ω.
Try one question
A horizontal straight conducting rod of length 0.40 m moves horizontally at right angles to its length with a velocity of 5.0 m s⁻¹ through a vertical downward magnetic field B = 0.30 T. What motional electromotive force (EMF) is induced across the ends of the rod?
- A. 1.20 V
- B. 0.60 V
- C. 0.30 V
- D. 0.15 V
Show the answer
Answer: B. 0.60 V. The motional electromotive force induced across a conductor moving perpendicular to a magnetic field is given by Faraday's / Lorentz relation: ε = B l v. Substituting: ε = 0.30 T × 0.40 m × 5.0 m s⁻¹ = 0.60 V.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Alternating Current & RMS
What you must know
- Hot-Wire / Moving-Iron Meters: Conventional moving-coil DC meters read ZERO on AC because average torque is zero; AC measuring instruments (hot-wire or moving-iron) utilize I² heating or magnetic repulsion to read TRUE RMS values directly.
- Form Factor and Peak Factor: Form Factor = V_rms / V_avg = (V0/√2) / (2V0/π) = π / (2√2) ≈ 1.11 (for pure sine wave); Peak Factor = V0 / V_rms = √2 ≈ 1.414.
- Peak insulation stress: A 230 V RMS AC line has a maximum peak voltage of V0 = 230 * √2 ≈ 325 V; electrical insulation must withstand peak voltage, not RMS.
Try one question
A sinusoidal alternating voltage is given by v(t) = 311 sin(100π t) V. What are the root-mean-square (RMS) voltage V_rms and the frequency f of this alternating supply?
- A. V_rms = 220 V, f = 50 Hz
- B. V_rms = 311 V, f = 100 Hz
- C. V_rms = 220 V, f = 100 Hz
- D. V_rms = 155 V, f = 50 Hz
Show the answer
Answer: A. V_rms = 220 V, f = 50 Hz. Standard equation: v(t) = V₀ sin(2π f t). Peak voltage V₀ = 311 V. V_rms = V₀ / √2 = 311 / 1.414 ≈ 220 V. Angular frequency ω = 2π f = 100π => f = 50 Hz.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Inductive & Capacitive Reactance
What you must know
- Series Resonance Condition: Occurs when inductive reactance equals capacitive reactance: X_L = X_C => ω0 L = 1 / (ω0 C); Resonant frequency f0 = 1 / [ 2π √(L C) ].
- Characteristics at Series Resonance: 1) Minimum impedance Z_min = R (purely resistive); 2) Maximum current I_max = V / R; 3) Phase angle φ = 0 (power factor cos φ = 1, in-phase); 4) Voltage magnification across L and C (Q-factor times source voltage).
- Quality Factor (Q-Factor): Measure of circuit sharpness/selectivity at resonance: Q = (ω0 L) / R = (1 / (ω0 C R)) = (1 / R) √(L / C) = f0 / Bandwidth.
Try one question
In a series LCR alternating current circuit, an inductor of inductance L = 0.10 H and a capacitor of capacitance C = 10 μF are connected with a resistor. What is the electrical resonant frequency f₀ of the circuit?
- A. 50 Hz
- B. 1000 Hz
- C. 318 Hz
- D. 159 Hz
Show the answer
Answer: D. 159 Hz. Resonance occurs when inductive reactance equals capacitive reactance: X_L = X_C => ω₀ L = 1 / (ω₀ C) => f₀ = 1 / (2π √(LC)). Substituting: f₀ = 1 / [2π √(0.10 × 10 × 10⁻⁶)] = 1 / [2π √(10⁻⁶)] = 1 / [2π × 10⁻³] = 1000 / (2π) ≈ 159.15 Hz.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Transformers
What you must know
- High Voltage AC Power Transmission: Grid power transmitted at ultra-high voltages (e.g., 132 kV - 400 kV) because high V drastically minimizes transmission current I (I = P/V); Joulean heat dissipation in transmission cables (P_loss = I² R_line = P² R_line / V²) decreases inversely with square of transmission voltage (1/V²).
- DC Incompatibility: Transformers operate strictly on alternating current (AC) with time-varying magnetic flux; connecting a DC voltage source produces constant flux (dΦ/dt = 0), inducing ZERO secondary voltage, and burning primary coil due to low DC resistance.
- Impedance Matching: A transformer of turns ratio Np/Ns matches load resistance RL to primary source with equivalent input resistance R_in = (Np / Ns)² * RL.
Try one question
Why is the soft iron core of an electrical power transformer constructed using thin, mutually insulated laminations rather than a solid iron block?
- A. To increase the secondary turns ratio
- B. To minimize energy losses caused by induced eddy currents
- C. To eliminate magnetic hysteresis loss entirely
- D. To shield the transformer from external electrostatic charge
Show the answer
Answer: B. To minimize energy losses caused by induced eddy currents. Alternating magnetic flux in the core induces circulating eddy currents in the iron, dissipating electrical energy as heat (I²R loss). Laminating the core breaks the closed conductive loops, drastically increasing electrical resistance to eddy currents and minimizing energy loss.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Generators & Motors
What you must know
- Motor Startup Surge & Starters: At startup (motor stationary, ω = 0), Back EMF Eb = 0; initial starting current Ia_start = V / Ra is dangerously high (can burn windings); starting rheostats add temporary series resistance until motor accelerates and builds protective back emf.
- Motor Mechanical Power Output: Total electrical power supplied V * Ia = Ia² Ra (Joule heat loss) + Eb * Ia (converted to mechanical power P_mech); Mechanical Power is maximized when Eb = V / 2.
- Generator vs Motor Rule Distinction: Generator uses Fleming's RIGHT-Hand Rule (Mechanical + Field => Induced Current); Motor uses Fleming's LEFT-Hand Rule (Current + Field => Mechanical Force).
Try one question
A direct-current (DC) electric motor with an armature resistance of 1.0 Ω is operated from a 200 V DC supply. When running at full operating speed, the armature draws a current of 10 A. What back electromotive force (back EMF, E_b) is developed in the spinning armature?
- A. 200 V
- B. 100 V
- C. 210 V
- D. 190 V
Show the answer
Answer: D. 190 V. In a running DC motor, the applied voltage V overcomes the back EMF E_b and supplies the ohmic drop across armature resistance: V = E_b + I_a R_a. Rearranging: E_b = V - I_a R_a = 200 V - (10 A × 1.0 Ω) = 200 V - 10 V = 190 V.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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