Question 1

Resistor color code interpretation – brown–black–red–gold A carbon film resistor shows color bands: brown (1), black (0), red (multiplier 10^2), gold (tolerance). What is the resistance value including tolerance?

Accepted Answer

Introduction / Context: Color bands on resistors encode their nominal value and tolerance. Mastering the code enables quick identification in labs and field work without measuring each part. The standard four-band system uses two significant digits, a multiplier, and a tolerance band. Given Data / Assumptions: Band 1 (brown) = 1, Band 2 (black) = 0. Multiplier (red) = 10^2. Tolerance (gold) = ±5%. Assume standard E24 series conventions for four-band resistors. Concept / Approach: The first two bands form the significant figures; the third band scales them by the power of ten multiplier; the fourth band gives tolerance. Combining these yields the nominal resistance and permissible variation range. Step-by-Step Solution: Significant digits: 1 (brown), 0 (black) → '10'.Multiplier: red → × 10^2 = × 100.Nominal resistance: 10 × 100 = 1000 Ω.Tolerance: gold → ±5%. Verification / Alternative check: Cross-check with standard tables: brown-black-red equals 1.0 kΩ; gold tolerance equals ±5% → range 950 Ω to 1050 Ω. Why Other Options Are Wrong: 100 Ω ignores the multiplier; 1000 Ω without tolerance omits key spec; ±10% corresponds to silver, not gold; 10 kΩ ± 5% would require orange or a different pattern. Common Pitfalls: Reading the bands in reverse (especially with five-band parts), confusing gold and silver meanings, or misplacing the multiplier band. Final Answer: 1000 Ω ± 5%

Question 2

Magnetic units – what physical quantity is measured in Tesla (T)? Identify the electromagnetic quantity for which the SI unit Tesla (T) is used.

Accepted Answer

Introduction / Context: Correct units link physical intuition to formulas. In magnetics, we distinguish between magnetic flux Φ (weber), magnetic flux density B (tesla), magnetic field strength H (ampere per metre), and magnetomotive force (ampere-turns). This clarity is essential for interpreting datasheets and designing magnetic circuits. Given Data / Assumptions: SI base and derived units are used. Magnetic flux density B relates to flux per unit area. Flux Φ and field strength H have different units from B. Concept / Approach: The Tesla (T) is defined as 1 weber per square metre. It measures magnetic flux density B, i.e., how much magnetic flux passes through a unit area. This is distinct from H, which has units A/m and describes the magnetizing field, and from Φ (Wb), the total flux threading a surface. Step-by-Step Solution: Recall: B = Φ / A.Units: Φ in Wb; A in m^2; thus B in Wb/m^2 = T.Therefore, Tesla corresponds to magnetic flux density. Verification / Alternative check: Permanent magnets are often specified in tesla for surface flux density (e.g., 0.3–1.4 T), and MRI systems in tesla for field intensity of the B field experienced by spins. Why Other Options Are Wrong: Flux uses weber; field strength (H) uses A/m; MMF uses ampere-turns; reluctance uses inverse henry. Hence only 'flux density' is correct. Common Pitfalls: Interchanging B and H or assuming tesla measures 'field strength' generically; forgetting that B incorporates material response via B = μ0 μr H. Final Answer: flux density

Question 3

Atomic number of germanium – count of protons in the nucleus How many protons are present in one neutral atom of germanium (chemical symbol Ge)?

Accepted Answer

Introduction / Context: The atomic number Z equals the number of protons in an element’s nucleus and determines its position in the periodic table and chemical identity. In semiconductor physics, recognizing Z for common materials like silicon (14) and germanium (32) helps relate to electron configuration and bonding behavior in crystals. Given Data / Assumptions: Germanium’s chemical symbol is Ge. Atomic number Z defines proton count and, in a neutral atom, equals electron count. Periodic table conventions apply. Concept / Approach: Locate germanium in group 14 (IVA) of the periodic table, period 4. Its atomic number is 32, meaning its nucleus contains 32 protons. In a neutral Ge atom, there are also 32 electrons arranged to yield 4 valence electrons, which explains Ge’s tetravalent bonding and its use as a semiconductor. Step-by-Step Solution: Identify element: Ge (germanium).Read atomic number: Z = 32.Deduce proton count = Z = 32.Neutral atom electron count = 32 (contextual note). Verification / Alternative check: Any standard periodic table lists germanium with Z = 32, atomic mass about 72.63 u, and electron configuration [Ar] 3d^10 4s^2 4p^2 (four valence electrons consistent with group 14). Why Other Options Are Wrong: 4 and 16 correspond to beryllium and sulfur; 28 corresponds to nickel; 34 corresponds to selenium. None match germanium. Common Pitfalls: Confusing atomic number with mass number; mixing up neighboring group 14 elements such as silicon (14), tin (50), and lead (82). Final Answer: 32

Question 4

Magnetic attraction and material type – which metal is not attracted by a magnet? Identify the material that is not appreciably attracted by a typical permanent magnet.

Accepted Answer

Introduction / Context: Materials respond to magnetic fields in different ways: ferromagnetic (strong attraction and possible permanent magnetization), paramagnetic (weak attraction), and diamagnetic (weak repulsion). Recognizing which everyday metals are ferromagnetic is important for separation, sensing, and design of magnetic devices. Given Data / Assumptions: Permanent magnet producing a moderate static magnetic field. Common engineering metals: iron, nickel, cobalt (ferromagnetic); copper (diamagnetic). No special alloys or heat treatments that alter magnetic phases are considered. Concept / Approach: Iron, nickel, and cobalt are ferromagnetic at room temperature and are strongly attracted by magnets. Copper is diamagnetic with a very small negative susceptibility, meaning it is not attracted and can even exhibit weak repulsion in nonuniform fields. Therefore, among the listed options, copper is the material a magnet does not attract appreciably. Step-by-Step Solution: List material classes: Fe, Ni, Co → ferromagnetic; Cu → diamagnetic.Ferromagnetic materials show strong attraction to magnets; diamagnetic show negligible attraction (slight repulsion).Therefore, choose 'copper' as not attracted by a magnet. Verification / Alternative check: Simple experiments show paper clips (steel/iron) cling to magnets, nickel coins may stick depending on composition, while copper wire does not stick. Eddy currents in moving magnets can exert forces on copper, but that is a dynamic effect unrelated to static attraction. Why Other Options Are Wrong: Nickel and iron are categorically ferromagnetic and strongly attracted; 'iron and copper' wrongly includes iron; cobalt (not listed as an option originally but provided here for contrast) is also ferromagnetic. Common Pitfalls: Confusing dynamic eddy-current braking effects in copper with static magnetic attraction; assuming all metals are attracted by magnets equally. Final Answer: copper

Question 5

Intrinsic semiconductors at absolute zero (0 K) At 0 K, silicon and germanium have their valence bands completely filled and conduction bands empty. Do they behave like monoatomic crystals in terms of crystal basis and bonding at this temperature?

Accepted Answer

Introduction / Context: This conceptual question checks your understanding of crystal structure versus electrical behavior at absolute zero. Silicon and germanium are classic covalent semiconductors with a diamond-cubic lattice. The phrase “monoatomic crystal” relates to the basis of the lattice (how many atoms per primitive cell), not to whether carriers exist at 0 K. Given Data / Assumptions: Temperature is 0 K (absolute zero). Materials: silicon (Si) and germanium (Ge). Consider crystallography (basis) and bonding, not just carrier concentration. Concept / Approach: Si and Ge crystallize in the diamond cubic structure, which can be viewed as an fcc Bravais lattice with a two-atom basis. A “monoatomic” crystal has one atom per basis (e.g., simple fcc of a monatomic metal). Electrically, at 0 K, the valence band is full and the conduction band is empty, so they act as perfect insulators; however, that does not convert their lattice into a monoatomic basis. Structure and electronic occupation are distinct ideas. Step-by-Step Solution: Identify lattice: diamond cubic for both Si and Ge (two-atom basis).Interpret “monoatomic crystal”: one atom per primitive cell basis (not true for diamond structure).At 0 K: carriers freeze out, but the structural basis remains unchanged.Therefore, the statement that they “behave like monoatomic crystals” is incorrect. Verification / Alternative check: The well-known transition of iron between bcc and fcc shows structure is independent of electronic carrier freeze-out; similarly, Si and Ge remain diamond-structured irrespective of temperature-induced carrier changes. Why Other Options Are Wrong: True: wrongly equates absence of carriers with monoatomic basis. “Only silicon/germanium” variants: both share the same diamond-cubic two-atom basis. Common Pitfalls: Confusing electronic properties (insulating at 0 K) with crystallographic basis; assuming “no free electrons” implies a monoatomic lattice. Final Answer: False

Question 6

Capacitance per unit length of a coaxial cable Two coaxial metallic cylinders have inner radius R1 and outer radius R2. The space between them is uniformly filled with a dielectric of relative permittivity εr. What is the capacitance per metre lengt…

Accepted Answer

Introduction / Context: The capacitance per unit length of a coaxial line is a staple result in electromagnetics and transmission line theory. It follows from Gauss’s law under cylindrical symmetry with a homogeneous dielectric between the conductors. Given Data / Assumptions: Inner conductor radius R1, outer conductor inner radius R2 (R2 > R1). Dielectric is linear, homogeneous, isotropic with permittivity ε = ε0 * εr. Length considered is 1 metre (per-unit-length capacitance C′). Concept / Approach: Use Gauss’s law in integral form over a cylindrical surface to find the electric field E(r). Integrate the potential difference V between R1 and R2. The per-unit-length capacitance is C′ = Q / V for a 1 m length. The result contains a logarithm due to the 1/r dependence of E in cylindrical coordinates. Step-by-Step Solution: Choose Gaussian surface of radius r, length 1 m: ∮ D · dA = Q_free ⇒ D_r * (2 * π * r * 1) = Q ⇒ D_r = Q / (2 * π * r).Relate fields: D = εE ⇒ E(r) = D_r / ε = Q / (2 * π * ε * r).Potential: V = ∫_{R1}^{R2} E(r) dr = Q / (2 * π * ε) * ln(R2 / R1).Capacitance per metre: C′ = Q / V = 2 * π * ε / ln(R2 / R1) = 2 * π * ε0 * εr / ln(R2 / R1). Verification / Alternative check: Dimensional analysis: ε has units F/m; logarithm is dimensionless; result has units F/m as required. For εr = 1, this reduces to the vacuum formula. Why Other Options Are Wrong: Halving the numerator (π instead of 2π) underestimates C′ by 2. Multiplying by ln instead of dividing reverses the dependence. Using (R2 − R1) ignores cylindrical field variation (not a parallel-plate capacitor). Common Pitfalls: Confusing natural log with log base 10; forgetting ε = ε0 * εr; using diameter rather than radius. Final Answer: C′ = 2 * π * ε0 * εr / ln(R2 / R1)

Question 7

Magnetic hysteresis and domain wall motion Is the hysteresis loop of a ferromagnet closely associated with the motion and pinning of magnetic domain walls?

Accepted Answer

Introduction / Context: Ferromagnetic materials consist of domains—regions where spins align. When an external magnetic field changes, magnetization changes occur via two main processes: movement of domain walls and rotation of domain magnetization. Hysteresis emerges because real materials have imperfections that impede these processes. Given Data / Assumptions: Ferromagnetic specimen with domains and defects. Quasi-static magnetization cycling to trace the B–H loop. Pinning sites exist (dislocations, grain boundaries, inclusions). Concept / Approach: The hysteresis loop records the history-dependent response. Domain walls must overcome pinning barriers to move; this requires finite field increments, leading to irreversible jumps (Barkhausen effect) and area in the B–H loop corresponding to energy loss per cycle. Thus, hysteresis is intimately tied to domain wall motion and rotation impeded by microstructural obstacles. Step-by-Step Solution: Apply field H: domains favorably oriented grow by wall motion.Encounter pinning: extra field is needed to unpin walls, creating path dependence.Reverse H: walls do not retrace exactly; the loop exhibits coercivity and remanence. Verification / Alternative check: Barkhausen noise recordings and microscopy of magnetic domains confirm discrete wall jumps correlated with hysteresis behavior. Why Other Options Are Wrong: “False” ignores well-established domain physics. Temperature or paramagnet caveats are irrelevant; paramagnets do not show ferromagnetic hysteresis. Common Pitfalls: Assuming hysteresis is due only to rotation; in many soft magnets, wall motion dominates low-field behavior. Final Answer: True

Question 8

Direction of induced magnetic dipole moment For atoms or molecules without permanent magnetic moments, the induced magnetic dipole moment produced by an applied magnetic field is in which direction relative to the field?

Accepted Answer

Introduction / Context: This question targets basic diamagnetism. In many materials with no permanent dipoles (closed electron shells), an applied magnetic field induces orbital currents that oppose the change in magnetic flux, creating an induced dipole moment. Given Data / Assumptions: No permanent magnetic moment (diamagnetic response). Weak applied field, linear regime. Atomic/electronic origin of induced magnetism. Concept / Approach: Lenz’s law and the classical Langevin viewpoint show that induced currents generate magnetic moments opposing the applied field. Quantum mechanically, Larmor precession leads to an induced magnetic moment antiparallel to the applied field. This universal, weak effect is diamagnetism, present in all materials but usually masked when paramagnetic or ferromagnetic effects exist. Step-by-Step Solution: Apply magnetic field B to a closed-shell atom.Induced currents are such that the magnetic moment opposes the field (minimizing energy).Therefore, the induced magnetic dipole is opposite to the applied field direction. Verification / Alternative check: Magnetic susceptibility of diamagnets is small and negative, confirming antiparallel induced magnetization (M opposite to H). Why Other Options Are Wrong: Perpendicular or random orientations contradict the linear diamagnetic response. Along the field corresponds to paramagnetic alignment of permanent moments, not induced diamagnetism. Common Pitfalls: Confusing induced diamagnetism with alignment of permanent moments; overlooking that most materials exhibit at least weak diamagnetism. Final Answer: Opposite to the applied field

Question 9

Minority-carrier density in an n-type semiconductor (use mass-action law) An n-type semiconductor at 300 K has electron concentration n = 6.25 × 10^18 cm^-3. The intrinsic carrier concentration is n_i = 2.5 × 10^13 cm^-3. What is the hole concentrat…

Accepted Answer

Introduction / Context: Carrier concentrations in semiconductors at equilibrium obey the mass-action law. Given majority carrier density in an n-type sample and intrinsic carrier concentration, you can compute the minority carrier density (holes) directly. This is fundamental for diode and transistor analysis. Given Data / Assumptions: Electron concentration n = 6.25 × 10^18 cm^-3. Intrinsic concentration n_i = 2.5 × 10^13 cm^-3 at 300 K. Thermal equilibrium (no significant generation-recombination transients). Concept / Approach: The mass-action law states n * p = n_i^2 for a homogeneous semiconductor at equilibrium. Solve for p = n_i^2 / n. Keep careful track of powers of ten to avoid arithmetic mistakes. Step-by-Step Solution: Compute n_i^2 = (2.5 × 10^13)^2 = 6.25 × 10^26 cm^-6.Use p = n_i^2 / n = (6.25 × 10^26) / (6.25 × 10^18) cm^-3.Cancel 6.25: p = 10^8 cm^-3.Thus, the minority-hole density is 1 × 10^8 cm^-3. Verification / Alternative check: Order-of-magnitude check: heavy n-type doping (10^18) should force holes down to a very low value (10^8), consistent with n * p = 10^26. Why Other Options Are Wrong: 10^6 or 10^10 or 10^12 cm^-3 do not satisfy n * p = n_i^2 with the given n and n_i. Common Pitfalls: Squaring 2.5 incorrectly; misplacing exponents; forgetting that units of n and p are per cm^3. Final Answer: 10^8 cm^-3

Question 10

Lorentz force on an electron An electron of charge −e moves with velocity v in simultaneous electric field E and magnetic flux density B. Is the force given by F = −e (E + v × B)?

Accepted Answer

Introduction / Context: The Lorentz force law is the foundation of electromagnetics and beam physics. For a particle with charge q, the total electromagnetic force is q (E + v × B). The sign of q matters, so for an electron (q = −e), the force reverses direction relative to a positive charge. Given Data / Assumptions: Particle is an electron with charge −e. Fields are E (electric) and B (magnetic flux density). Classical, non-relativistic expression is adequate. Concept / Approach: Start with F = q (E + v × B). Substitute q = −e to get F = −e (E + v × B). This holds regardless of the relative orientation of v and B and in vacuum or material media (as a local equation of motion for the particle). Step-by-Step Solution: Write Lorentz law: F = qE + q (v × B).Substitute q = −e: F = −e E − e (v × B) = −e (E + v × B).Conclude the statement is correct. Verification / Alternative check: Right-hand-rule cross products correctly predict curvature direction in magnetic fields for positive charges; electrons curve oppositely because q is negative, consistent with the formula. Why Other Options Are Wrong: Restrictions like v ⟂ B or “in vacuum only” are unnecessary; the law is general. Common Pitfalls: Forgetting the negative sign for electrons; mixing H and B (the Lorentz force uses B). Final Answer: True

Question 11

Crystal structure of iron at room temperature At ordinary (room) temperature, does iron exhibit a body-centred cubic (bcc) crystal structure?

Accepted Answer

Introduction / Context: Iron exhibits allotropic transformations: its crystal structure changes with temperature. Understanding these phases is crucial in metallurgy and heat treatment of steels, where phase transformations determine mechanical properties. Given Data / Assumptions: Consider pure iron (no alloying elements). Room temperature conditions (~20–30 °C). Ambient pressure. Concept / Approach: Pure iron has the following phases: α-Fe (ferrite) is bcc from room temperature up to about 912 °C; γ-Fe (austenite) is fcc from ~912 °C to ~1394 °C; δ-Fe is bcc again from ~1394 °C to the melting point. Thus, at room temperature, iron is indeed bcc (α-Fe). Step-by-Step Solution: Identify phase ranges: α-Fe bcc (RT to 912 °C), γ-Fe fcc (912–1394 °C), δ-Fe bcc (1394 °C to melt).Room temperature lies in the α-Fe range.Therefore, iron is bcc at room temperature. Verification / Alternative check: X-ray diffraction data and iron–carbon phase diagrams confirm the bcc structure of ferrite at room temperature. Why Other Options Are Wrong: “False” contradicts the α-Fe phase field. “Only above 912 °C” describes austenite (fcc), not bcc. Curie temperature concerns magnetism, not lattice type. Common Pitfalls: Confusing magnetic transition (Curie point ~770 °C) with structural change at 912 °C. Final Answer: True

Question 12

Application of sulphur hexafluoride (SF₆) in power systems SF₆ gas is widely used in high-voltage equipment because of its excellent dielectric strength and arc-quenching properties. Which equipment is the most typical application?

Accepted Answer

Introduction / Context: Sulphur hexafluoride (SF₆) is an inert, non-flammable, high-electron-affinity gas with outstanding dielectric properties. Power engineers employ it where strong insulation and effective arc extinction are required under high voltage and current interruptions. Given Data / Assumptions: Consider common high-voltage apparatus. Focus on roles requiring gaseous insulation and arc quenching. Environmental aspects (GWP) aside for the physics question. Concept / Approach: SF₆ supports high dielectric withstand and quickly quenches arcs by electron attachment and thermal properties. Hence, it is the standard medium in gas-insulated switchgear (GIS) and high-voltage circuit breakers. Transformers typically use mineral oil or ester fluids; generators use solid and air/hydrogen cooling; electromechanical relays do not rely on a special insulating gas for contacts. Step-by-Step Solution: Identify need for strong gaseous insulation and arc extinction → circuit breakers, GIS.Assess other devices: transformers (oil dielectric), generators (insulation varnish/air/H₂), relays (air/solid).Thus, SF₆ is primarily used in circuit breakers and GIS. Verification / Alternative check: Manufacturer datasheets and substation layouts routinely specify SF₆ breakers and GIS bays for compact, high-voltage installations. Why Other Options Are Wrong: Transformers: usually oil-paper systems. Generators: air or hydrogen cooling; solid insulation. Relays: do not require SF₆ environments. Common Pitfalls: Assuming any HV equipment uses SF₆; in practice it is specific to switchgear and breakers for insulation and arc control. Final Answer: High-voltage circuit breakers and gas-insulated switchgear

Question 13

Electronic polarizability and relative permittivity of a dilute rare gas A rare gas has N atoms per m^3. If the electronic polarizability of a single atom is α_e (SI units), which relation correctly connects ε_r and α_e in the dilute-gas limit?

Accepted Answer

Introduction / Context: In gases at low density, the macroscopic polarization is simply the sum of independent atomic dipoles induced by the field. Connecting single-atom polarizability to bulk permittivity reveals how ε_r approaches 1 and increases linearly with number density N. Given Data / Assumptions: Dilute, non-interacting gas of atoms (rare gas). Electronic polarizability per atom: α_e (SI). Linear response, low fields. Concept / Approach: The polarization P equals dipole moment per unit volume: P = N * α_e * E. Also, D = ε0 E + P = ε E = ε0 ε_r E. Therefore, ε0 (ε_r − 1) E = P = N α_e E, giving ε0 (ε_r − 1) = N α_e. This is the dilute-gas limit of the more general Clausius–Mossotti relation, which introduces local-field corrections at higher densities. Step-by-Step Solution: Write P = N α_e E.Use D = ε0 E + P = ε0 ε_r E.Equate: ε0 (ε_r − 1) E = N α_e E ⇒ ε0 (ε_r − 1) = N α_e. Verification / Alternative check: As N → 0, ε_r → 1, consistent with vacuum; the linear dependence on N confirms the low-density assumption. Why Other Options Are Wrong: Options a and b miss the vacuum term; d uses “−2”, which is unphysical in this context. Common Pitfalls: Mixing cgs and SI forms; forgetting that polarization measures excess over vacuum response. Final Answer: ε0 * (ε_r − 1) = N * α_e

Question 14

Magnetization–permeability relationship In linear magnetic materials (SI units), is the magnetization given by M = (μ_r − 1) * H?

Accepted Answer

Introduction / Context: Relating magnetization M, magnetic field strength H, and relative permeability μ_r is fundamental in electromagnetics. In linear, isotropic media, a simple proportionality connects these quantities through magnetic susceptibility χ_m. Given Data / Assumptions: Linear, isotropic magnetic response. Definitions: B = μ0 (H + M) and μ = μ0 μ_r. Magnetic susceptibility χ_m = M / H. Concept / Approach: From B = μ0 (H + M) and B = μ H = μ0 μ_r H, equate to get μ0 μ_r H = μ0 H + μ0 M ⇒ μ_r H = H + M ⇒ M = (μ_r − 1) H. Also, since χ_m = μ_r − 1, we have the standard relation M = χ_m H. This holds for linear media (small-signal regime). Step-by-Step Solution: Start: B = μ0 (H + M) = μ0 μ_r H.Rearrange: μ_r H − H = M.Conclude: M = (μ_r − 1) H = χ_m H. Verification / Alternative check: In vacuum, μ_r = 1 ⇒ M = 0, consistent with no magnetization; for paramagnets/diamagnetics, |χ_m| ≪ 1, so M is small compared with H. Why Other Options Are Wrong: “False” contradicts standard linear-material relations. Limiting to ferromagnets or vacuum is unnecessary; the relation is general for linear regimes. Common Pitfalls: Confusing H and B; forgetting that μ_r = 1 + χ_m in SI units. Final Answer: True

Question 15

Temperature alarm circuits – which basic sensor is typically used? In practical electronic temperature alarm circuits (e.g., overheat warning), which component is most commonly employed as the temperature-sensing element?

Accepted Answer

Introduction / Context: Temperature alarm circuits require a sensor whose electrical property varies predictably with temperature and is easy to interface. The classic, low-cost choice in many appliances and industrial panels is the thermistor, especially the negative-temperature-coefficient (NTC) type. Given Data / Assumptions: Moderate temperature range near ambient to a few hundred °C. Low-cost analog circuits using comparators or microcontrollers. Need for large, monotonic resistance change with temperature. Concept / Approach: NTC thermistors exhibit a steep decrease in resistance with increasing temperature, modeled by R(T) ≈ R0 exp(β(1/T − 1/T0)). This yields strong sensitivity for threshold detection. They can be placed in a voltage divider feeding a comparator to trigger alarms. Other components listed do not primarily sense temperature: photoconductors respond to light; varistors respond to voltage surges; piezo materials sense stress; a plain transistor can be used as a temperature sensor but is not the common, standalone choice in simple alarm circuits compared to a dedicated thermistor. Step-by-Step Solution: Select a component with R = f(T) strongly monotonic → thermistor.Implement divider + reference threshold → comparator output drives alarm.Add hysteresis to avoid chatter; calibrate via resistor selection. Verification / Alternative check: Household thermostats, battery packs, and HVAC safety circuits frequently employ NTC thermistors for both monitoring and alarm due to their sensitivity and low cost. Why Other Options Are Wrong: Photoconductor senses light; varistor is for surge protection; transistor usually needs linearization and biasing and is less common in off-the-shelf alarm modules. Common Pitfalls: Ignoring self-heating of thermistors at high currents; use small sense currents and proper thermal placement. Final Answer: Thermistor

Question 16

Identify the device from its nonlinear V–I curve (voltage-dependent resistance) The shown V–I characteristic (with pronounced nonlinearity and a knee, resistance decreasing as voltage increases) corresponds to which component?

Accepted Answer

Introduction / Context: Many passive components are nonlinear. A varistor (VDR), including metal-oxide varistors (MOVs), exhibits a strongly voltage-dependent resistance: at low voltages it has very high resistance; as voltage rises toward a knee, its resistance collapses and significant current flows, clamping the voltage. Recognizing this V–I signature is essential in surge protection design. Given Data / Assumptions: Curve shows symmetrical nonlinearity around the origin (bidirectional behavior). No clear threshold like a diode forward turn-on; instead a soft knee and power-law I–V. Two-quadrant conduction typical of MOVs under AC. Concept / Approach: Varistors obey I ≈ k V^α (α > 1). On a V–I plot this produces a characteristic knee; differential resistance decreases with increasing V. Thermistors primarily show R = f(T), not voltage; LDRs show R = f(light); Zeners exhibit sharp breakdown only in one polarity and are active semiconductor diodes, not symmetric passive two-terminal ceramics. Hence the depicted curve maps to a varistor. Step-by-Step Solution: Observe symmetry about origin → suggests bidirectional passive element.Note decreasing slope resistance with increasing voltage → varistor behavior.Exclude diode-like one-sided breakdown and temperature/light sensitive parts. Verification / Alternative check: Datasheets for MOVs show log–log I–V straight lines with exponents α between about 10 and 50 in the operating region, matching the soft-knee appearance. Why Other Options Are Wrong: Thermistors and LDRs are resistance-vs-environment devices; a Zener conducts in reverse beyond VZ only in one polarity; the curve described is symmetric. Common Pitfalls: Confusing MOV curves with gas arresters (which have sharper breakdown and hysteresis) or with diodes. Final Answer: Varistor (VDR/MOV)

Question 17

Ohm’s law for metallic conduction (current density vs. electric field) Which relation correctly expresses Ohm’s law in differential form for a homogeneous metal?

Accepted Answer

Introduction / Context: Ohm’s law at the continuum level relates current density J to electric field E via the material’s conductivity σ. This microscopic (point-form) statement underpins resistor design, field analysis, and interpretation of bulk measurements where R = ρ L/A is derived from σ = 1/ρ. Given Data / Assumptions: Homogeneous, isotropic metal. Steady-state conditions; linear response (no significant heating nonlinearity). Definitions: σ is conductivity, ρ = 1/σ is resistivity. Concept / Approach: In differential form: J = σ E. Integrating over cross-sectional area and length recovers the familiar V = I R with R = ρ L/A. The law is empirical but arises from drift of conduction electrons with average drift velocity v_d and scattering time τ, yielding σ = n e^2 τ / m in the Drude model. Step-by-Step Solution: Start from Drude: J = n e v_d and v_d = (e τ / m) E.Thus J = n e (e τ / m) E = (n e^2 τ / m) E.Identify σ = n e^2 τ / m → J = σ E. Verification / Alternative check: Dimensional check: [σ] = S/m so [J] = S/m * V/m = A/m^2, consistent. Why Other Options Are Wrong: (b) uses resistivity ρ; the correct relation with ρ is J = E/ρ. (c) and (d) are vague proportionalities, not equations. (e) is dimensionally wrong. Common Pitfalls: Mixing up σ and ρ or forgetting vector nature (J and E are generally colinear in isotropic media). Final Answer: J = σ E

Question 18

Soft iron for temporary magnets – statement evaluation Is the following statement correct? “Soft iron is suitable for making temporary magnets.”

Accepted Answer

Introduction / Context: Magnetic materials are broadly chosen as either soft (easily magnetized and demagnetized) or hard (retain magnetization). Understanding this distinction is vital for cores, relays, transformers, and temporary electromagnets. Given Data / Assumptions: Soft iron has high permeability, low coercivity, and low hysteresis loss. Temporary magnet means it should follow the applied field and lose magnetization when the field is removed. Contrast with hard magnetic materials (e.g., AlNiCo, ferrites) used for permanent magnets. Concept / Approach: Soft iron’s low coercive force means small fields can magnetize it strongly and removing the field collapses magnetization. This is ideal for temporary magnets and AC magnetic circuits. Hard magnets require large coercive fields to demagnetize and are not suitable for temporary applications. Step-by-Step Solution: Identify requirement: easy magnetization/demagnetization → low coercivity.Soft iron exhibits μ_r ≫ 1 and narrow hysteresis loop → small losses.Conclusion: soft iron is the correct choice for temporary magnets. Verification / Alternative check: Transformer laminations, relay cores, and electromagnet yokes are made from soft magnetic steels or irons, confirming the practical preference. Why Other Options Are Wrong: (b) contradicts magnetic design practice; (d) confuses permanent vs. temporary needs; (e) alloying can tune properties but is not a prerequisite for the truth of the statement. Common Pitfalls: Assuming “strong magnet” always equals “good magnet”; application determines whether soft or hard magnetic behavior is desirable. Final Answer: True

Question 19

Current density for electron drift in a metal A specimen of metal contains n conduction (valence) electrons per m^3. If v_x is the average drift velocity of these electrons along x and each electron carries charge e, what is the current density J (x…

Accepted Answer

Introduction / Context: In conductors, conventional current density J points in the direction of positive-charge flow. Electrons carry negative charge, so their drift direction is opposite to J. Getting the sign right is important in microscopic transport equations and when relating J to drift velocity in the Drude model. Given Data / Assumptions: Electron number density n (per m^3). Average electron drift velocity component v_x (along x). Electron charge magnitude e > 0; electron charge is −e. Concept / Approach: By definition, J = charge density in motion × velocity. For electrons: charge density ρ_e = (−e) n. Hence, J_x = ρ_e v_x = (−e n) v_x. If v_x is positive (electrons moving +x), J_x is negative (conventional current in −x). Many circuit texts quote the magnitude J = n e |v_d| and treat direction separately, but the vector equation includes the minus sign. Step-by-Step Solution: Electron charge density: ρ_e = − e n.Current density: J = ρ_e v → J_x = − n e v_x.Therefore, J = − n e v_x (x-component). Verification / Alternative check: Combine with Drude result v_d = (− e τ / m) E to obtain J = n e^2 τ / m E = σ E, consistent with Ohm’s law and sign conventions. Why Other Options Are Wrong: (a) omits the negative sign; (c), (d), (e) are dimensionally incorrect or unrelated. Common Pitfalls: Forgetting that v_x refers to electron motion, not conventional current direction; always include the sign associated with the carrier charge. Final Answer: J = − n e v_x

Question 20

Equation of motion for an electron in a uniform electric field in a metal A metal has a large number of free electrons. Let e be the elementary charge and m the electron mass. When a uniform electric field E is applied along the x-direction, the ele…

Accepted Answer

Introduction / Context: The Drude model starts from Newton’s second law applied to electrons in an applied electric field. Because electrons carry negative charge (−e), the force and acceleration directions are opposite to the electric field. Establishing the correct sign is crucial to derive drift velocity, mobility, and conductivity expressions. Given Data / Assumptions: Electron charge q_e = − e with e > 0 by convention. Uniform, static electric field E along +x. Instantaneous motion before averaging over scattering time τ. Concept / Approach: Lorentz force (electric part) on an electron is F = q_e E = (− e) E. Newton’s law gives m dv/dt = F, hence m dv_x/dt = − e E along x. After including collisions, the average steady-state drift velocity is v_d = − (e τ / m) E, leading to current density J = n e v_d (with the sign handled conventionally) and conductivity σ = n e^2 τ / m. Step-by-Step Solution: Write force: F_x = q_e E = − e E.Apply Newton’s law: m dv_x/dt = F_x → m dv_x/dt = − e E.Average with scattering → v_d = − (e τ/m) E (not required for the instantaneous relation). Verification / Alternative check: Direction check: E along +x pushes electrons toward −x (negative acceleration), consistent with the negative sign. Why Other Options Are Wrong: (b) and (c) have wrong sign; (d) gives the magnitude but wrong sign for negative carriers; (e) ignores the negative charge in the momentum growth. Common Pitfalls: Dropping the sign of the electron charge or confusing electron drift with conventional current direction. Final Answer: m dv_x/dt = − e E

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