Q: Faraday’s law in words: What occurs when expanding magnetic lines of force sweep across (cut) a conductor?

Introduction / Context: Electromagnetic induction explains the fundamental operation of generators, transformers, and many sensors. The qualitative description ”moving/expanding lines of force cut a conductor” is a classic way of stating the condition for inducing an electromotive force (EMF). Given Data / Assumptions: There is relative motion between magnetic flux and a conductor, or the flux changes with time. We focus on the immediate electrical effect in the conductor. Resistive load or closed path determines whether current subsequently flows. Concept / Approach: Faraday’s law states that the induced EMF around a loop equals the negative time rate of change of the magnetic flux linking that loop: e = −dΦ/dt. When expanding lines of force cross a conductor, flux linkage changes, so an EMF (voltage) is induced. Whether current flows depends on circuit closure and impedance; the primary, universal effect is induced voltage. Step-by-Step Solution: Identify the condition: changing flux linkage (expanding or moving field intersecting the conductor).Apply Faraday’s law: change in flux → induced EMF.Conclude: the first effect is an induced voltage; current follows only if a path exists. Verification / Alternative check: Generators rotate coils in magnetic fields, creating continuous change in Φ and thus a sinusoidal EMF. Transformers induce voltage in a stationary secondary via time-varying flux produced by the primary. Why Other Options Are Wrong: Current increases: Not necessarily; it depends on circuit impedance and whether the circuit is closed. Resistance reduced / Power increased: Neither is a direct consequence of changing flux; they depend on materials and loads. Common Pitfalls: Equating induced voltage with guaranteed current flow; an open-circuited loop will have voltage but no current. Final Answer: Voltage is induced.

Question 1

Magnetic materials — In general systems and specifically in magnetics, hysteresis refers to which behavior?

Accepted Answer

Introduction: Hysteresis appears in magnetic cores, ferroelectrics, and other systems where the output depends on current and past inputs. In transformers and inductors, magnetic hysteresis directly affects core losses and B-H loop behavior. Given Data / Assumptions: Considering magnetization M vs. applied magnetic field H. Core material exhibits a B-H loop (non-linear, path-dependent response). Sinusoidal steady-state or cycling conditions. Concept / Approach: Hysteresis is a path-dependent lag: the material’s response (magnetization/flux density) does not follow the instant value of excitation alone; it lags and depends on history. The loop area represents hysteresis loss per cycle, contributing to heating at a rate proportional to frequency and core volume. Step-by-Step Solution: Apply increasing H → B increases along an ascending curve.Reduce H → B does not return the same path; remanence remains at H = 0.Reverse H → coercive field is required to drive B to zero.The enclosed loop area is the hysteresis loss; lag behavior defines hysteresis. Verification / Alternative check: Core datasheets specify hysteresis loops and coercivity. Loss modeling (Steinmetz parameters) includes a hysteresis term proportional to frequency and peak flux, confirming the lagging, path-dependent nature. Why Other Options Are Wrong: lead between cause and effect: Opposite of hysteresis behavior. lead/lag between voltage and current: That describes reactive phase shift, not hysteresis per se. instantaneous proportional response: Would be a linear, memoryless system, not hysteretic. Common Pitfalls: Confusing hysteresis (memory effect) with phase shift from reactance. Ignoring that hysteresis persists even at low frequency, though losses scale with frequency. Final Answer: lag between cause and effect

Question 2

Ideal transformer current scaling — A 100 Vac primary draws 500 mA. For a 300 Vac secondary on the same core (ideal assumptions), what secondary current is expected?

Accepted Answer

Introduction: In an ideal transformer, power is conserved (neglecting losses). Therefore, voltage and current scale inversely with the turns ratio. This question practices applying the fundamental V–I relationships to predict secondary current from known primary conditions. Given Data / Assumptions: Primary voltage Vp = 100 Vac, primary current Ip = 0.5 A. Secondary voltage Vs = 300 Vac. Ideal transformer: Vp * Ip ≈ Vs * Is (ignoring losses), same frequency on both sides. Concept / Approach: For an ideal transformer, power in equals power out. Hence Is = (Vp * Ip) / Vs. Voltage is stepped up by a factor of 3 (300/100), so current is stepped down by the same factor to conserve power. Step-by-Step Solution: Compute input power: Pin = Vp * Ip = 100 * 0.5 = 50 W.Equate to output power: Pout ≈ 50 W (ideal).Solve for secondary current: Is = Pout / Vs = 50 / 300 = 0.1667 A.Convert to milliamps: 0.1667 A ≈ 167 mA. Verification / Alternative check: Using ratios: Vs/Vp = 3, so Is/Ip = Vp/Vs = 1/3 → Is = 0.5/3 = 0.1667 A, confirming 167 mA. Why Other Options Are Wrong: 1500 mA: Would imply an impossible power increase. 200 mA / 150 mA / 50 mA: Do not match the exact inverse scaling by 3. Common Pitfalls: Confusing which side is stepped up; higher voltage side carries proportionally lower current. Ignoring real transformer losses; they slightly reduce output current for a given load but not to this magnitude. Final Answer: 167 mA

Question 3

Fixed inductors — Which option is NOT among the three major fixed-inductor core types commonly encountered in electronics?

Accepted Answer

Introduction: Fixed inductors are built on a variety of cores optimized for frequency range, loss, and size. Recognizing standard core families helps with selection and expectations for Q factor and saturation characteristics. Given Data / Assumptions: Focus on widely used fixed inductor types in RF and power electronics. Common categories include air core, ferrite core, and iron/iron-powder core. Exotic or nonmagnetic metals like magnesium do not serve as magnetic cores. Concept / Approach: Magnetic behavior hinges on permeability. Ferrite and iron-based cores provide high permeability; air cores have μ close to μ0 and are used at high frequencies to avoid core losses. Magnesium is non-ferromagnetic and is not used as a magnetic core in standard fixed inductors. Step-by-Step Solution: List standard families: air, ferrite, iron/iron-powder.Evaluate “magnesium core”: lacks the required magnetic properties.Therefore, “magnesium core” is not one of the major fixed inductor types.Powdered iron is a common iron-based variant (included as a valid type). Verification / Alternative check: Component catalogs categorize inductors by core as air, ferrite, and iron-powder among others. None list magnesium cores for inductors. Why Other Options Are Wrong: air core / ferrite core / iron core / powdered iron: All standard and widely used. Common Pitfalls: Confusing “iron core” with laminated transformer cores vs. iron-powder toroids; both are legitimate iron-based cores. Assuming any metal can be a magnetic core; only materials with suitable permeability and loss characteristics are used. Final Answer: magnesium core

Question 4

Energy and opposition in AC — The opposition to current flow that does not dissipate energy (i.e., stores and releases it cyclically) is called:

Accepted Answer

Introduction: In AC circuits, not all opposition to current results in heat. Reactive elements (inductors and capacitors) store energy in fields and return it to the source within each cycle. Understanding the terminology distinguishes lossless opposition from dissipative resistance. Given Data / Assumptions: Sine-wave steady state AC. Considering ideal components (no winding resistance, no dielectric loss). Focus on the inductor’s frequency-dependent opposition. Concept / Approach: Resistance R dissipates energy as heat (P = I^2 * R). Reactance X stores energy and returns it each cycle, causing phase shift without net energy loss. Specifically, an inductor contributes inductive reactance X_L = 2 * pi * f * L, which increases with frequency and inductance. Step-by-Step Solution: Define reactance: magnitude of opposition due to energy storage, not dissipation.For inductors: X_L = 2 * pi * f * L (ohms), voltage leads current by 90 degrees in the ideal case.Energy oscillates in the magnetic field, returning to the source each half-cycle.Thus, the lossless opposition is inductive reactance (for an inductor). Verification / Alternative check: Power over a full cycle for a pure inductor averages to zero; instantaneous power alternates positive and negative, confirming storage without net dissipation. Why Other Options Are Wrong: resistance: Dissipative, converts energy to heat. counter emf: The induced voltage opposing change, not a measure of steady AC opposition in ohms. impedance: Generalized opposition combining R and X; includes dissipation if R ≠ 0. conductance: The reciprocal of resistance, not opposition. Common Pitfalls: Equating impedance entirely with loss; impedance may be purely reactive. Forgetting frequency dependence of X_L; it is zero at DC. Final Answer: inductive reactance

Question 5

Magnetic materials: What is the typical relative permeability (μ_r) for common transformer iron (silicon steel used in power transformers)?

Accepted Answer

Introduction / Context: Transformer cores are made from ferromagnetic materials (often silicon steel) chosen for their high magnetic permeability. Knowing a realistic order of magnitude for relative permeability helps students estimate magnetizing current, core size, and inductance, and avoids confusing absolute permeability with the dimensionless relative permeability μ_r used in most transformer discussions. Given Data / Assumptions: The question refers to typical transformer iron used in power-frequency transformers (not exotic alloys). Permeability is the relative permeability μ_r (dimensionless), not absolute μ in H/m. We assume unsaturated operation near the linear region of the B–H curve at 50/60 Hz. Concept / Approach: Relative permeability μ_r = μ / μ0 indicates how many times more a material supports magnetic flux compared with free space. Transformer-grade steels generally have μ_r from a few hundred to a few thousand depending on lamination, silicon content, stress, and flux density. Values around 1,000–5,000 are common, while specialized alloys (e.g., permalloy) can be much higher. A value of about 1,500 is a widely cited nominal for plain transformer iron under typical conditions. Step-by-Step Solution: List plausible μ_r range for transformer steels: hundreds to a few thousands.Compare options: 50 (too low), 450 (low end but possible for poor conditions), 1,500 (solid mid-range nominal), 5,500 (too high for ordinary transformer iron).Select the most representative typical value: 1,500. Verification / Alternative check: Engineering handbooks and manufacturer datasheets for silicon-steel laminations show μ_r varying with B and frequency; nominal catalog points often lie near 1,000–2,000 for practical, linear operation, supporting the selection of 1,500 as a typical figure. Why Other Options Are Wrong: 50: Characteristic of weakly magnetic materials, not transformer iron. 450: Possible but on the low side; not representative as a general “typical” value. 5,500: More representative of specialty high-permeability alloys, not standard transformer lamination steel. Common Pitfalls: Confusing relative and absolute permeability or quoting values measured at very low flux density that do not reflect typical operating points. Final Answer: 1,500

Question 6

AC filters (RL): A series RL filter has the output taken across the inductor. Which type of frequency-selective behavior will this network exhibit?

Accepted Answer

Introduction / Context: First-order RL networks are foundational building blocks in analog electronics. Whether the output is taken across the resistor or the inductor changes the filter's behavior. Understanding this helps in predicting Bode plots, time-domain response, and in designing simple tone controls or sensor front ends. Given Data / Assumptions: Series RL circuit driven by an AC source. Output is measured across the inductor L. Ideal components; no parasitic effects beyond R and L. Concept / Approach: The inductor's reactance is XL = 2 * π * f * L, which increases with frequency. In a series path, the voltage division is frequency dependent: V_L / V_in = XL / sqrt(R^2 + XL^2). At low f, XL is small, so most input appears across R (little across L). At high f, XL dominates, so most input appears across L. Therefore, with output taken across L, higher frequencies are passed better than lower ones—this is a high-pass characteristic (first order). Step-by-Step Solution: Write inductor reactance: XL = 2 * π * f * L.Form the divider: V_L = V_in * XL / sqrt(R^2 + XL^2).Consider limits: as f → 0, XL → 0 → V_L ≈ 0; as f → ∞, XL ≫ R → V_L ≈ V_in.Conclude that higher frequencies see larger output across L. Verification / Alternative check: Plot the magnitude response: the -3 dB cutoff occurs near f_c = R / (2 * π * L). Frequency sweep measurements will show increasing output with frequency when probing across the inductor. Why Other Options Are Wrong: Passing lower frequencies / blocking higher frequencies: Describe a low-pass response (which would be seen if the output were across R). Blocking a certain range: That suggests a band-stop (not first-order RL). Common Pitfalls: Mixing up where the output is taken; swapping L and R or moving the probe changes the apparent filter type. Final Answer: passing of the higher frequencies

Question 7

Real inductor modeling: Which option best represents a practical equivalent circuit for an inductor including its dominant parasitics?

Accepted Answer

Introduction / Context: No physical inductor is ideal. Winding resistance and interwinding capacitance alter behavior at both low and high frequencies. A suitable equivalent circuit helps predict resonances, Q factor, and losses—critical in filters, RF chokes, and switch-mode power supplies. Given Data / Assumptions: Winding resistance (R_s) is in series with the inductance (L). Parasitic self-capacitance (C_p) appears effectively in parallel with the winding. We focus on the simplest widely used small-signal model. Concept / Approach: A common small-signal model is L in series with R_s, representing copper loss, with a shunt (parallel) capacitance C_p modeling self-capacitance between turns or layers. This topology exhibits a self-resonant frequency where the inductive reactance and capacitive reactance cancel, beyond which the inductor can appear capacitive. Such a model matches measured impedance-versus-frequency curves for practical components. Step-by-Step Solution: Identify series elements: L + R_s capture inductance and DC resistance.Add parallel C_p to represent turn-to-turn capacitance.Compare with options and select the one describing C in parallel with (R + L in series). Verification / Alternative check: Impedance analyzers reveal a peak and phase crossover at the self-resonant frequency, predicted by the R_s–L series branch shunted by C_p. Why Other Options Are Wrong: Series L–R–C: Places capacitance in series, not representative of interwinding capacitance. L in parallel with (R–C series): Not the standard dominant-parasitic model. Only inductance: Ignores real-world losses and capacitance. Common Pitfalls: Neglecting C_p at high frequency, leading to incorrect assumptions that inductors stay inductive across all frequencies. Final Answer: A capacitance in parallel with the series combination of a resistance and an inductance

Question 8

Magnetism fundamentals: Magnetic flux is defined as the total amount of magnetic field passing through a surface. In SI units, magnetic flux is measured in what unit?

Accepted Answer

Introduction / Context: Magnetic quantities often confuse beginners because multiple units and legacy terms exist. Distinguishing magnetic flux (Φ), flux density (B), and field strength (H) is vital for analyzing transformers, inductors, and machines, and for converting between device specifications and formulas. Given Data / Assumptions: Magnetic flux Φ is a scalar quantity representing the total field passing through an area. SI unit for flux is the weber (Wb). Flux density B is measured in tesla (T), where 1 T = 1 Wb/m^2. Concept / Approach: Flux quantifies the ”amount” of magnetic field, analogous to electric flux in electromagnetism. In the SI system, Φ is measured in webers. Older CGS literature uses the maxwell, where 1 Wb = 10^8 maxwells. Tesla measures flux density, not flux. Therefore, when a question asks “what is magnetic flux,” the correct unit is the weber and descriptions referring to “lines of force” are colloquial placeholders mapping to webers. Step-by-Step Solution: Recall definitions: Φ in webers; B in teslas; H in A/m.Relate flux and flux density: B = Φ / A (for uniform fields over area A).Select the description identifying webers as the correct SI unit of flux. Verification / Alternative check: Textbook relationships for inductors (Φ = L * I / N in simplified cases) and transformer EMF (E = N * dΦ/dt) consistently use webers for flux to yield volts for induced EMF. Why Other Options Are Wrong: Maxwell: A CGS unit, not SI; included here as a distractor. Tesla: Unit of flux density, not total flux. “Flux density”: A different physical quantity; the option is self-contradictory. Common Pitfalls: Mixing up flux and flux density when reading datasheets; overlooking unit conversions between CGS and SI. Final Answer: the number of lines of force in webers

Question 9

Inductance units and definitions: When a current changes at a rate of 1 ampere per second and induces 1 volt across a coil, what is the unit of inductance defined by this relationship?

Accepted Answer

Introduction / Context: Inductance quantifies how effectively a coil resists changes in current by developing an induced voltage. Recognizing the definition tied to basic units allows quick sanity checks in transient circuit analysis and in sizing inductors for filters or power converters. Given Data / Assumptions: The induced voltage v across an inductor is v = L * di/dt. Given di/dt = 1 A/s results in v = 1 V. SI base units are used throughout. Concept / Approach: By definition, an inductance of 1 henry produces 1 volt of induced EMF when the current through it changes at 1 ampere per second. The equation v = L * di/dt directly encodes this. The henry (H) is the SI unit of inductance; alternative options list unrelated units (ohm for resistance, farad for capacitance) or a nonstandard term that does not exist in SI (lenz). Step-by-Step Solution: Start with v = L * di/dt.Set di/dt = 1 A/s and v = 1 V → L must be 1 H.Therefore, the unit matching this definition is the henry. Verification / Alternative check: Dimensional analysis: [L] = V / (A/s) = V * s / A, which is the dimension of the henry. Why Other Options Are Wrong: Ohm: Unit of resistance (V/A). Farad: Unit of capacitance (C/V or s/Ω). “Lenz”: Not an SI unit; Lenz's law describes polarity of induced EMF, not a unit. Common Pitfalls: Confusing the roles of L, R, and C in differential equations; always tie the unit back to v = L * di/dt. Final Answer: a henry

Question 10

Magnetic field properties: Which statement best characterizes a magnetic field in classical electromagnetism?

Accepted Answer

Introduction / Context: The visualization of magnetic fields using ”lines of force” (field lines) is a standard pedagogical tool. While field lines are not physical objects, they help reason about direction, relative strength, and interaction with materials—especially in problems involving flux, flux density, and induced EMF. Given Data / Assumptions: Classical, quasi-static fields (no relativistic or quantum effects). Field lines illustrate both direction (tangent gives direction of B) and density (number per area suggests magnitude). We distinguish the field itself from sources and material properties. Concept / Approach: A magnetic field B is commonly depicted via lines of force that originate conceptually at the north pole of a magnet and terminate at the south pole (externally), forming closed loops. Magnetomotive force (MMF) is a source quantity measured in ampere-turns; reluctance is an opposition to flux, not a ”line.” Thus, when describing what a magnetic field ”has,” the most appropriate choice is ”lines of force.” Step-by-Step Solution: Recall: Field lines are a representation of B and help visualize interactions.Differentiate: MMF drives flux; reluctance impedes flux; neither are features of the field itself.Choose the statement that correctly describes the field's representation: lines of force. Verification / Alternative check: Iron filings and magnetic viewing films reveal patterns corresponding to field lines, confirming their utility as a representation of magnetic fields. Why Other Options Are Wrong: Polar fields: Vague phrasing; magnetic fields are vector fields not described as ”polar fields.” Magnetomotive force: A source quantity (ampere-turns), not a property the field ”has.” Lines of reluctance: Reluctance is a scalar property of a path/material, not a line-based representation. Common Pitfalls: Confusing the drivers of flux (MMF) with the field itself, or treating field lines as physical objects. Final Answer: lines of force

Question 11

Ideal transformer calculation: A transformer has a 300-turn secondary producing 360 Vac. If the primary has 150 turns, what input (primary) voltage is required (assume an ideal transformer)?

Accepted Answer

Introduction / Context: Transformer voltage ratios follow directly from the turns ratio. Quick mental calculations using V ∝ N allow you to size windings, anticipate secondary voltages, and verify lab measurements, assuming ideal behavior (no losses, perfect coupling). Given Data / Assumptions: Secondary turns N_s = 300 with V_s = 360 Vac. Primary turns N_p = 150; find V_p. Ideal transformer: V_p / V_s = N_p / N_s. Concept / Approach: For ideal transformers, the voltage ratio equals the turns ratio: V_p / V_s = N_p / N_s. If the primary has fewer turns than the secondary (N_p < N_s), the primary voltage must be proportionally lower than the secondary voltage for the same flux level. Step-by-Step Solution: Write ratio: V_p / 360 = 150 / 300.Compute turns ratio: 150 / 300 = 0.5.Solve for V_p: V_p = 360 * 0.5 = 180 Vac. Verification / Alternative check: Cross-check direction: Secondary has more turns (step-up 2:1 from primary), so for a given secondary of 360 Vac the primary should be half of that—180 Vac. Units and proportions are consistent. Why Other Options Are Wrong: 2.4 Vac: Off by a factor of 75; not consistent with 2:1 ratio. 150 Vac: Does not match the exact ratio; would imply N_p/N_s = 150/300 = 0.4167, which is incorrect arithmetic. 720 Vac: Would correspond to reversing ratio; not applicable here. Common Pitfalls: Inverting the ratio or mixing up which side is primary vs secondary; always align V with its corresponding N. Final Answer: 180 Vac

Question 12

Faraday’s law in words: What occurs when expanding magnetic lines of force sweep across (cut) a conductor?

Accepted Answer

Introduction / Context: Electromagnetic induction explains the fundamental operation of generators, transformers, and many sensors. The qualitative description ”moving/expanding lines of force cut a conductor” is a classic way of stating the condition for inducing an electromotive force (EMF). Given Data / Assumptions: There is relative motion between magnetic flux and a conductor, or the flux changes with time. We focus on the immediate electrical effect in the conductor. Resistive load or closed path determines whether current subsequently flows. Concept / Approach: Faraday’s law states that the induced EMF around a loop equals the negative time rate of change of the magnetic flux linking that loop: e = −dΦ/dt. When expanding lines of force cross a conductor, flux linkage changes, so an EMF (voltage) is induced. Whether current flows depends on circuit closure and impedance; the primary, universal effect is induced voltage. Step-by-Step Solution: Identify the condition: changing flux linkage (expanding or moving field intersecting the conductor).Apply Faraday’s law: change in flux → induced EMF.Conclude: the first effect is an induced voltage; current follows only if a path exists. Verification / Alternative check: Generators rotate coils in magnetic fields, creating continuous change in Φ and thus a sinusoidal EMF. Transformers induce voltage in a stationary secondary via time-varying flux produced by the primary. Why Other Options Are Wrong: Current increases: Not necessarily; it depends on circuit impedance and whether the circuit is closed. Resistance reduced / Power increased: Neither is a direct consequence of changing flux; they depend on materials and loads. Common Pitfalls: Equating induced voltage with guaranteed current flow; an open-circuited loop will have voltage but no current. Final Answer: Voltage is induced.

Question 13

Magnetization (B–H) characteristic: On a standard B–H curve used in electrical engineering materials, which two physical quantities are plotted on the horizontal and vertical axes, respectively?

Accepted Answer

Introduction / Context: The B–H curve is a foundational graph in electromagnetism and magnetic materials. It captures how a material’s magnetic flux density responds to an applied magnetizing force. This relationship underpins transformer core design, inductor sizing, saturation analysis, and hysteresis loss estimation, making it essential knowledge for electrical and electronics engineers. Given Data / Assumptions: B represents magnetic flux density, typically measured in tesla (T) or weber per square meter (Wb/m^2). H represents magnetizing force (magnetic field intensity), typically in ampere-turns per meter (A·turn/m or A/m). The standard graph places H on the horizontal axis and B on the vertical axis. We consider generic ferromagnetic material behavior, including possible hysteresis when cycled. Concept / Approach: Flux density B quantifies how much magnetic flux per unit area is established within the material. Magnetizing force H quantifies the effort applied by currents (ampere-turns) to magnetize the material. Plotting B versus H shows key phenomena: initial magnetization, knee of the curve (onset of saturation), and the closed hysteresis loop under alternating excitation which indicates core loss. Step-by-Step Solution: Identify the axes: B on the vertical axis and H on the horizontal axis.Interpretation: as H increases, B increases nonlinearly, approaching saturation.Under AC cycling, the path forms a loop (hysteresis), whose enclosed area equals hysteresis energy loss per cycle.Therefore, the two plotted quantities are flux density (B) and magnetizing force (H). Verification / Alternative check: Standard transformer and inductor textbooks show B–H plots with H along x-axis and B along y-axis. Laboratory B–H tracers also output B vs H directly, confirming the convention. Why Other Options Are Wrong: Reluctance/permeability are material parameters, not the direct plotted axes.Magnetizing force and permeability: permeability is a slope (dB/dH) or ratio B/H, not an axis quantity. Common Pitfalls: Confusing B–H with the B versus time or H versus time plots. Also, note that permeability varies with H; it is not constant across the whole curve. Final Answer: flux density and magnetizing force

Question 14

Residential distribution transformer (utility pole): A single utility pole transformer feeding a typical home provides two 120 V “legs.” Which construction detail on the secondary winding is used to deliver these two phases (split-phase supply)?

Accepted Answer

Introduction / Context: Many residential services (e.g., North American split-phase) use a single-phase transformer on the pole to provide two 120 V legs that are 180 degrees apart. Understanding how the transformer produces both legs helps with panel wiring, balancing loads, and troubleshooting voltage complaints. Given Data / Assumptions: The primary is single-phase and connected to the distribution line. The secondary provides two 120 V legs and a neutral. Loads may be 120 V (leg to neutral) or 240 V (leg to leg). Concept / Approach: Split-phase service is obtained by center-tapping the secondary of a single-phase transformer. The two ends of the secondary provide opposite polarities with respect to the center tap (neutral). Thus, each end to neutral is about 120 V, while end-to-end is about 240 V. This arrangement does not require a two-phase primary; the “two phases” seen at the home are simply two halves of one secondary winding referenced to its center tap. Step-by-Step Solution: Use a single secondary winding with a center tap taken as neutral.Measure from either end to neutral → approximately 120 V.Measure end-to-end → approximately 240 V (legs are 180° apart).Therefore, the hardware feature enabling this is the center tap on the secondary. Verification / Alternative check: Service entrance diagrams show a three-wire service: L1, N (center tap), and L2. Meter and panelboards accept this configuration for both 120 V and 240 V circuits. Why Other Options Are Wrong: Two phases on the primary: utilities use single-phase primaries for such transformers.Separate ground for each phase: grounding does not create two legs.Small coefficient of coupling: unrelated to split-phase creation. Common Pitfalls: Calling split-phase “two-phase”; it is a center-tapped single-phase system. Final Answer: a center tap on the secondary

Question 15

Terminology of wound components: A length of insulated wire wound into turns may be referred to by various names in electronics. Which description(s) apply?

Accepted Answer

Introduction / Context: Wound components appear in power supplies, filters, RF circuits, and energy storage applications. The same physical object (a wire wound into turns) can be called a coil, an inductor, or a choke depending on its function and frequency range. Recognizing these interchangeable terms prevents confusion when reading datasheets or schematics. Given Data / Assumptions: We consider a generic wound component: a wire formed into multiple turns, possibly on a core. Terminology may vary by application domain (power vs RF). Magnetic core or air core may be used. Concept / Approach: The component stores energy in a magnetic field when current flows (L * I^2 / 2). In general terms it is a “coil” or “inductor.” The word “choke” often implies use as a series element intended to impede AC (especially high-frequency ripple) while passing DC—common in filters and line EMI suppression. Despite nuanced meanings, all labels refer to a wound inductor. Step-by-Step Solution: Identify the physical form: wound turns of wire.Function: energy storage and impedance to changing current.Use cases: RF coils (tuned circuits), power inductors (buck/boost), chokes (filtering).Conclusion: all listed names correctly describe a wire winding used as an inductor. Verification / Alternative check: Component catalogs label similar parts as “power inductors,” “RF coils,” and “common-mode chokes,” confirming overlapping terminology. Why Other Options Are Wrong: Each individual term is accurate—but the most comprehensive, correct choice is the inclusive option. Common Pitfalls: Assuming “choke” is a different device; it is an inductor tailored to choking (impeding) AC components. Final Answer: All of the above

Question 16

Counter-emf law for inductors: The magnitude of induced voltage across an inductor depends on what parameter(s), holding all else the same?

Accepted Answer

Introduction / Context: Inductors generate a counter electromotive force (emf) whenever the current through them changes. This phenomenon is central to understanding switching transients, snubbers, and the behavior of RL circuits under step inputs. The quantitative relationship is captured by the inductor voltage equation. Given Data / Assumptions: Ideal inductor model is used. Current varies with time, producing an induced voltage. We compare proportionalities while other variables are held constant. Concept / Approach: The governing relation is v_L = L * di/dt. The induced voltage increases with larger inductance L and with faster current changes di/dt. For a fixed current change occurring over a shorter time, di/dt increases, producing a larger voltage spike. Conversely, a small inductance or slow current change produces a smaller induced voltage. Step-by-Step Solution: Write the inductor law: v_L = L * di/dt.Hold di/dt constant → v_L ∝ L (directly proportional to inductance).Hold L constant → v_L ∝ di/dt (larger change in current per unit time gives larger voltage).Therefore, the correct proportionality with respect to L alone is “directly proportional to the inductance in henrys.” Verification / Alternative check: Observe switching in a lab: doubling inductance in a given RL step raises the induced voltage during rapid transients proportionally, matching v_L = L * di/dt. Why Other Options Are Wrong: Inversely proportional to the change in current: conflicts with the di/dt term.Directly proportional to change in time: for a fixed current change, longer time lowers di/dt and thus reduces v_L.Inversely proportional to inductance: contradicts v_L ∝ L. Common Pitfalls: Confusing di/dt with Δi or Δt independently; always consider the ratio. Also, note the sign of v_L follows Lenz’s law, opposing the change in current. Final Answer: directly proportional to the inductance in henrys

Question 17

Series RL phase relationship: A 100 mH inductor has inductive reactance of 6 kΩ at the operating frequency. It is in series with a 1 kΩ resistor and driven by a source. What is the phase angle between source voltage and current?

Accepted Answer

Introduction / Context: In AC analysis, series RL circuits exhibit a phase shift between the source voltage and the circuit current due to the inductor’s reactance. Estimating this angle quickly helps in phasor calculations, power factor estimation, and filter design. Given Data / Assumptions: Inductive reactance XL = 6 kΩ. Resistance R = 1 kΩ in series. Sinusoidal steady-state operation. Concept / Approach: The phase angle φ of a series RL circuit (voltage referenced to current) is given by φ = arctan(XL / R). A larger ratio XL/R yields a larger lagging angle, since current lags voltage in inductive circuits. We compute the arctangent using the given values. Step-by-Step Solution: Compute ratio: XL / R = 6000 / 1000 = 6.Find angle: φ = arctan(6) ≈ 80.5 degrees (rounded).Compare to options: closest available is 81.0 degrees.Therefore, the phase angle is approximately 81.0 degrees (current lags voltage). Verification / Alternative check: Compute impedance magnitude: |Z| = sqrt(R^2 + XL^2) = sqrt(1^2 + 6^2) kΩ = sqrt(37) kΩ ≈ 6.083 kΩ. Then cos φ = R / |Z| = 1 / 6.083 ≈ 0.164, so φ ≈ arccos(0.164) ≈ 80.6°, consistent with arctan calculation. Why Other Options Are Wrong: 0.1° and 9.0°: far too small; would require XL much less than R.61.0°: corresponds to a smaller XL/R ratio (about 1.8), not 6. Common Pitfalls: Forgetting that angle is referenced as the voltage leading the current in inductive circuits; current lags by φ in an RL series network. Final Answer: 81.0 degrees

Question 18

Purely inductive AC branch: In an ideal purely inductive circuit driven by a sinusoidal source, what is the phase relationship between voltage and current?

Accepted Answer

Introduction / Context: Understanding phase relationships in ideal reactive elements is essential for AC circuit analysis and power calculations. Inductors and capacitors shift current and voltage by 90°, but in opposite directions. Remembering which quantity leads in which element is a key conceptual skill. Given Data / Assumptions: Purely inductive branch (no resistance). Sinusoidal steady-state excitation. Ideal inductor behavior (no core losses or winding resistance). Concept / Approach: For an inductor, v_L = L * di/dt. If current is sinusoidal, its derivative (and therefore voltage) leads the current by 90°. Equivalently, in phasor form, the inductor’s impedance is jXL, which has a phase of +90°, indicating that voltage across the inductor leads the current through it by 90°. Step-by-Step Solution: Assume i(t) = I_peak * sin(ωt).Then v(t) = L * di/dt = L * I_peak * ω * cos(ωt) = L * I_peak * ω * sin(ωt + 90°).Thus, the voltage waveform is shifted +90° relative to current.Therefore, in a purely inductive circuit, voltage leads current by 90°. Verification / Alternative check: Phasor impedance Z_L = jXL; multiplying current phasor by j rotates it +90°, producing the voltage phasor—graphically confirming the lead. Why Other Options Are Wrong: Current leads by 90° or in-phase: describe capacitive or resistive cases, not inductive.Voltage lags current by 90°: opposite of inductor behavior; that is for capacitors (current leads in capacitors). Common Pitfalls: Mixing up inductor and capacitor phase rules. A popular mnemonic: “ELI the ICE man” (E leads I in L; I leads E in C). Final Answer: voltage leads current by 90°

Question 19

Transformer loading and ampere-turn balance: In a transformer, if the secondary current (and thus the secondary ampere-turns and magnetic field) increases due to increased load, what happens to the primary current Ip?

Accepted Answer

Introduction / Context: Transformers operate by mutual induction and the balance of ampere-turns. The primary and secondary currents interact via the core flux. Understanding how primary current responds to secondary loading is vital for interpreting input current measurements and selecting protective devices. Given Data / Assumptions: Ideal transformer approximation (tight coupling, negligible leakage and copper loss). Primary is fed by a constant-voltage AC source. Secondary load current increases. Concept / Approach: Core flux is set primarily by applied primary voltage and frequency. When secondary load current increases, its ampere-turns produce a demagnetizing component of flux that would reduce net core flux. The transformer responds by drawing additional primary current to supply the load’s power and to maintain the original core flux (ampere-turns balance). Therefore, increased secondary current results in increased primary current according to the turns ratio and power balance. Step-by-Step Solution: Primary voltage sets a near-constant magnetizing flux in the core.Secondary load current produces opposing ampere-turns (demagnetizing).To keep net flux near constant, primary draws extra current so Np * Ip approximately cancels Ns * Is (plus magnetizing current).Hence, Ip increases as secondary load increases. Verification / Alternative check: Power balance: P_primary ≈ P_secondary (neglecting losses). If secondary power rises, primary apparent power and hence current must rise. Why Other Options Are Wrong: Ip canceled or remains the same: contradict power and ampere-turn balance.Ip decreases: opposite of observed behavior under increased load. Common Pitfalls: Assuming core flux varies widely with load; in practice, with constant applied voltage, flux stays nearly constant while currents adjust. Final Answer: Ip will increase.

Question 20

RL step response sizing: A DC source Vs = 12 V is applied to a series RL circuit with R = 60 Ω and L = 24 mH. After exactly 2 time constants following the step, what is the current in the inductor (current rise value)?

Accepted Answer

Introduction / Context: The RL step response describes how current builds up when a DC voltage is applied. Designers use the time constant τ to predict rise time and choose component values that meet transient requirements in drivers, relays, and filters. Given Data / Assumptions: Supply Vs = 12 V DC. Series elements: R = 60 Ω, L = 24 mH. We evaluate current at t = 2τ after the step is applied. Concept / Approach: For a series RL driven by a DC step, the current is i(t) = I_final * (1 − e^(−t/τ)), where I_final = Vs / R and τ = L / R. Substituting the given values yields the numerical current at t = 2τ. Step-by-Step Solution: Compute time constant: τ = L / R = 0.024 H / 60 Ω = 0.0004 s = 0.4 ms.Final current: I_final = Vs / R = 12 V / 60 Ω = 0.2 A = 200 mA.Evaluate exponent at t = 2τ: e^(−2) ≈ 0.135335.Current at t = 2τ: i(2τ) = 0.2 A * (1 − 0.135335) = 0.2 * 0.864665 ≈ 0.172933 A.Rounded to three significant figures: 173.0 mA. Verification / Alternative check: Rule of thumb: After 1τ current ≈ 63.2% of final; after 2τ ≈ 86.5%; after 3τ ≈ 95.0%. With I_final = 200 mA, 0.865 * 200 mA ≈ 173 mA, confirming the computation. Why Other Options Are Wrong: 79.9 mA: near 40% of final, much earlier than 2τ.126.4 mA: around 63% of final, corresponds to 1τ.198.6 mA: almost full-scale; closer to 3–4τ or steady state, not 2τ. Common Pitfalls: Mixing time constant definitions or forgetting to convert millihenries to henries. Always compute τ = L / R with consistent SI units. Final Answer: 173.0 mA

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