Question 1

AC generators and waveform constancy: will a practical AC generator always produce a perfectly constant, unchanging waveform regardless of speed and loading?

Accepted Answer

Introduction / Context: AC generators (alternators) convert mechanical rotation into electrical waveforms. Idealized models produce a pure sine wave of fixed amplitude and frequency. Real machines deviate due to speed variations, winding distribution, magnetic saturation, and load dynamics. The statement that a generator always produces a “constant waveform” is therefore misleading. Given Data / Assumptions: Conventional rotating AC generator with field and armature windings. Grid-independent operation (stand-alone) for illustration. Non-idealities: finite regulation, mechanical speed ripple, load-dependent voltage drop. Concept / Approach: Frequency f = P * n / 120 (for a synchronous machine with P poles and speed n in rpm). Any change in mechanical speed alters frequency. Terminal voltage is affected by excitation and internal impedance; load changes alter voltage due to synchronous reactance and resistance, distorting the waveform under nonlinear loads. Even grid-tied machines rely on the grid to enforce frequency/shape; the generator itself is not inherently “constant.” Step-by-Step Solution: 1) Recognize that waveform quality depends on magnetic symmetry and sinusoidal excitation.2) Note that speed droop, shaft torque ripple, and load changes perturb amplitude and frequency.3) Nonlinear loads (rectifiers, SMPS) draw harmonic currents, producing voltage distortion through source impedance.4) Conclude the statement of guaranteed constancy is incorrect. Verification / Alternative check: Compare no-load vs full-load waveforms on an oscilloscope; observe changes in amplitude and possible distortion. Standards limit, but do not eliminate, variation. Why Other Options Are Wrong: Correct: ignores practical dependencies. “Only for three-phase” and “True if 50 Hz”: phase count or nominal frequency does not guarantee perfection. Common Pitfalls: Confusing grid stability with generator intrinsic behavior; assuming mechanical systems provide perfectly constant speed without control. Final Answer: Incorrect

Question 2

Electromagnetism basics: does any steady current flowing through a conductor produce a magnetic field in the surrounding space?

Accepted Answer

Introduction / Context: Electromagnetism links electric currents and magnetic fields. Recognizing that even a simple straight wire carrying current generates a magnetic field is foundational for understanding inductors, transformers, motors, and EMC behavior on circuit boards and cables. Given Data / Assumptions: Long, straight conductor carrying current I. Quasi-static conditions; displacement current effects negligible for DC. Uniform surrounding medium (e.g., air). Concept / Approach: The magnetic field magnitude around a long straight conductor is given by B(r) = μ * I / (2 * π * r), encircling the wire per the right-hand rule. This holds for DC and AC instantaneous currents alike; AC simply varies the field with time. No ferromagnetic material is required to create the field; such material only concentrates or guides it. Step-by-Step Solution: 1) State the qualitative rule: a current produces a circumferential magnetic field.2) For DC, the field is steady; for AC, the field is time-varying at the same frequency.3) Apply the right-hand rule to determine field direction around the wire.4) Conclude that any nonzero current results in a real magnetic field in space. Verification / Alternative check: Place a compass near a current-carrying wire: deflection confirms the field. Clamp-on ammeters exploit this principle to measure current non-invasively. Why Other Options Are Wrong: Incorrect: contradicts the basic current-field relationship. “True only for AC” or “only with ferromagnetics”: the phenomenon exists for DC and without special cores; materials only modify field intensity and distribution. Common Pitfalls: Assuming magnetism requires magnets or iron; overlooking that PCB traces and cables radiate/receive due to these very fields. Final Answer: Correct

Question 3

AC inductive reactance and current: in a sinusoidal circuit with fixed voltage and frequency, will increasing the inductance reduce the current drawn?

Accepted Answer

Introduction / Context: Inductive reactance determines how an inductor impedes AC. Designers often adjust inductance to control current in chokes, ballasts, and filters. This question asks whether increasing inductance decreases current in an AC circuit with otherwise fixed conditions. Given Data / Assumptions: Sinusoidal steady-state at fixed frequency f and RMS voltage V. Series circuit whose impedance includes inductive reactance XL = 2 * π * f * L. Linear inductor (no saturation), small series resistance for first-order reasoning. Concept / Approach: Inductive reactance is proportional to inductance: XL ∝ L. The magnitude of current in a series circuit is I = V / |Z|. As L increases at fixed f, XL rises, increasing |Z| and therefore reducing I. This is the basis for inductive ballasts and chokes that limit current without resistive losses equivalent to brute-force series resistors. Step-by-Step Solution: 1) Compute XL = 2 * π * f * L.2) Form impedance magnitude, e.g., |Z| = sqrt(R^2 + XL^2) for a simple series RL.3) Write current: I = V / |Z|.4) Observe that increasing L increases XL and |Z|, thus decreasing I. Verification / Alternative check: Hold f constant and double L; XL doubles, so I approximately halves if R is small compared to XL—consistent with practical choke behavior. Why Other Options Are Wrong: Incorrect: contradicts the direct proportionality between XL and L. “Only true at DC”: at DC, XL = 0, so inductance does not limit steady current; the effect is specifically AC. Common Pitfalls: Confusing transient DC behavior (inductor initially resists change) with steady-state DC; overlooking frequency′s role via XL = 2 * π * f * L. Final Answer: Correct

Question 4

Electromagnet definition in basic electricity: A magnet formed by a coil of wire wound around a soft iron or steel core and energized by current is called an electromagnet. Decide whether this statement is accurate.

Accepted Answer

Introduction / Context: An electromagnet is a foundational concept in electromagnetism and electric machines. It is created when current flows through a coil (solenoid), generating a magnetic field that is intensified by a ferromagnetic core such as soft iron or certain steels. This question checks recognition of the standard definition used in electronics and physics courses. Given Data / Assumptions: A coil of wire is wound around a soft iron or suitable steel core. Current is supplied to the coil from a source (DC or AC). No permanent magnet is required; magnetism is current-induced. Concept / Approach: When current I passes through a coil of N turns, a magnetomotive force is established: MMF = N * I. A ferromagnetic core provides a low-reluctance path, concentrating flux and increasing field strength. Because the magnetic field exists only while current flows (and scales with it), this assembly is termed an electromagnet, distinct from a permanent magnet that retains magnetization without supply current. Step-by-Step Solution: Identify the structure: a current-carrying coil plus a soft iron/steel core.Recall the definition: a magnet produced by electric current is an electromagnet.Note that the core amplifies flux via high permeability.Conclude the statement matches the standard definition. Verification / Alternative check: Turn current on: the assembly strongly attracts ferrous objects. Turn current off: magnetic effect largely disappears (except minor residual magnetism), confirming electromagnet behavior. Why Other Options Are Wrong: Incorrect: Contradicts the established definition. Applies only to AC: DC also creates steady flux; both AC and DC can produce an electromagnet. True only with permanent magnets: Not required; the core is ferromagnetic but not permanently magnetized. Valid only above 1 A: There is no magic current threshold; effect scales with N * I. Common Pitfalls: Confusing permanent magnets with current-driven magnets; assuming steel must be permanently magnetized to be useful as a core; forgetting that soft iron is preferred for easy magnetization and demagnetization. Final Answer: Correct

Question 5

Ideal inductor energy dissipation concept: An ideal inductor is a lossless element that stores energy in its magnetic field but does not dissipate energy as heat. Evaluate whether this statement is correct.

Accepted Answer

Introduction / Context: Component idealizations are used in circuit theory to separate storage from loss. The ideal inductor is defined to have purely inductive reactance with no series resistance and no core losses. Understanding this ideal is essential before adding real-world parasitics in power electronics, filters, and RF design. Given Data / Assumptions: Inductor is ideal: R_series = 0, core losses = 0. It stores energy as E = 0.5 * L * I^2. Any heat generation is ignored in the ideal model. Concept / Approach: A lossless reactive element exchanges energy with the source each cycle without net dissipation. For an ideal inductor, voltage and current are 90 degrees out of phase in steady-state sinusoidal excitation; average power over a cycle is zero because instantaneous power alternates positive and negative symmetrically. In time domain, the inductor’s energy increases or decreases with I^2 but is not converted to heat in the ideal case. Step-by-Step Solution: Recall energy relation: E = 0.5 * L * I^2.Ideal inductor has no resistive component; thus P_average = 0 for sinusoidal steady state.Any power delivered to the inductor is later returned to the source.Therefore, no net energy is dissipated as heat in the ideal model. Verification / Alternative check: Compare with a resistor where P = I^2 * R > 0 always. Setting R = 0 (ideal inductor) eliminates the mechanism for heat dissipation, leaving only storage and release of magnetic energy. Why Other Options Are Wrong: Incorrect: Conflicts with the definition of an ideal inductor. True only at 50 Hz / superconducting wire: Idealization is frequency-independent; superconductivity is not required in theory. Depends on core saturation: Saturation affects inductance value and waveforms but does not introduce idealized loss. Common Pitfalls: Mixing the ideal concept with real inductors that have copper and core losses; assuming “no dissipation” means “no energy” (it stores, but does not dissipate). Final Answer: Correct

Question 6

Schematic literacy — variable inductor symbols: In standard electronics notation, not every inductor symbol shown in reference figures indicates a variable inductor. Assess the claim that “all the illustrated inductor symbols are variable inductors.”

Accepted Answer

Introduction / Context: Reading schematics accurately requires recognizing graphical cues. Inductors can be fixed or variable. Variable types are shown with clear modifiers (for example, an arrow through the coil or an adjustable core indicator). This question addresses whether all commonly shown coil symbols necessarily indicate variability. Given Data / Assumptions: Standard IEC/ANSI symbol conventions. Variable components are marked with adjustment indicators (arrow, diagonal slash, or movable core symbol). Plain coil symbols without modifiers denote fixed inductors. Concept / Approach: Symbol semantics: a base coil symbol indicates an inductor. Added annotation conveys special properties: iron core (lines/slug), tapped coil (tap symbol), or adjustability (arrow). Therefore, unless an adjust feature is explicitly drawn, one must assume the inductor is fixed. Stating that “all shown symbols are variable” is overgeneralization and incorrect. Step-by-Step Solution: Identify adjustment indicator: presence of an arrow or movable slug denotes variability.Check each depicted symbol (conceptually): fixed coils lack the adjust indicator.Conclude that only the subset with adjustment marks are variable; others are fixed.Therefore the blanket claim is false. Verification / Alternative check: Cross-reference with any electronics symbol chart: you will find separate entries for “inductor,” “variable inductor,” “tapped inductor,” and “iron-core inductor,” confirming the need for explicit modifiers to signal adjustability. Why Other Options Are Wrong: Every inductor symbol implies adjustability: Incorrect; adjustability must be shown. Valid only in RF schematics / depends on core / many turns: RF usage, core material, or drawing style does not change the meaning of the symbol modifiers. Common Pitfalls: Assuming an iron-core or slugged symbol is always adjustable; confusing tapped inductors with variable inductors; misreading an arrow that actually denotes a variable resistor nearby. Final Answer: Incorrect — only symbols with an adjust arrow/slug are variable

Question 7

Transformer basics — step-up ratio: A step-up transformer has a turns ratio greater than 1 (secondary turns exceed primary turns). Determine whether this statement is correct.

Accepted Answer

Introduction / Context: Turns ratio is the defining parameter of an ideal transformer. A step-up transformer increases voltage from primary to secondary; correspondingly, the secondary has more turns than the primary, and the turns ratio N_s/N_p is greater than 1. Recognizing this relationship is key in power conversion and signal coupling. Given Data / Assumptions: Ideal transformer behavior. Step-up means V_s > V_p. Frequency within normal operating range; core not saturating. Concept / Approach: For an ideal transformer, V_s/V_p = N_s/N_p. Therefore, V_s > V_p requires N_s/N_p > 1. Current ratio inverts: I_s/I_p = N_p/N_s. This captures conservation of power (ignoring losses): V_p * I_p ≈ V_s * I_s. Step-by-Step Solution: Define step-up: secondary voltage exceeds primary.Use relation: V_s/V_p = N_s/N_p.Conclude N_s/N_p > 1 for step-up operation.Confirm current ratio reverses accordingly. Verification / Alternative check: Example: N_p = 100 turns, N_s = 500 turns gives N_s/N_p = 5 and V_s = 5 * V_p, which is a clear step-up case. Why Other Options Are Wrong: Incorrect / autotransformer only / 60 Hz only / laminated steel only: The turns-voltage relation is general and not restricted to a specific construction or frequency (within design limits). Common Pitfalls: Confusing turns ratio with current ratio; assuming core material alters the algebraic ratio rather than efficiency and saturation limits. Final Answer: Correct

Question 8

Inductor quality factor (Q): For a practical inductor modeled as an ideal L in series with winding resistance R, the quality factor Q at a given frequency is the ratio of inductive reactance to resistance, Q = X_L / R. Decide whether this statement …

Accepted Answer

Introduction / Context: Quality factor quantifies how reactive energy storage compares to resistive loss. For inductors using the common series R-L model, Q is a simple and widely used measure in filters, resonant tanks, and RF chokes. Given Data / Assumptions: Series model: L with series R (winding and equivalent losses). Operating at a single frequency f (since X_L depends on f). Inductive reactance X_L = 2 * pi * f * L. Concept / Approach: Define Q for a series inductor as Q = X_L / R. Larger Q means lower loss relative to reactance. At resonance in an RLC, higher Q leads to narrower bandwidth. This definition is frequency-dependent and is appropriate when resistance is small compared to reactance (high-Q devices). Step-by-Step Solution: Compute X_L = 2 * pi * f * L.Measure or estimate series resistance R.Form Q = X_L / R (dimensionless).Interpret: higher Q implies better energy storage vs loss. Verification / Alternative check: Compare two inductors at the same f and L but different R; the one with lower R has higher Q, behaving closer to an ideal inductor (sharper resonance, lower insertion loss). Why Other Options Are Wrong: Incorrect/DC only: At DC, X_L = 0, so Q is undefined/zero. The relation applies at AC. Core material only: Core affects losses and L, but Q still uses X_L/R. Common Pitfalls: Confusing series and parallel Q definitions; forgetting frequency dependence; ignoring additional core-loss components that can be folded into an effective R for small-signal models. Final Answer: Correct

Question 9

Dynamic behavior — do inductors “boost” current?: Inductors oppose rapid changes in current; they do not inherently amplify or “boost” current flow. Evaluate the correctness of the claim that an inductor will boost current.

Accepted Answer

Introduction / Context: Understanding how inductors influence current is essential in power converters, filters, and transient analysis. Inductors store energy in a magnetic field and resist changes in current according to v = L * di/dt. They are not active devices and cannot amplify current on their own. Given Data / Assumptions: Passive component: inductor with inductance L and some series resistance. Circuit context unspecified; consider general behavior. No active switching or control is implied. Concept / Approach: Inductors follow v = L * di/dt. To increase current, a positive voltage must be applied; the inductor then allows current to ramp at a rate di/dt = v/L. This is not “boosting” by the inductor; it is the source doing work while the inductor stores energy. In steady DC state after transients, an ideal inductor behaves like a short, passing whatever current the surrounding resistance allows, but it never amplifies current beyond what the source provides. Step-by-Step Solution: Note the governing relation: v = L * di/dt.Recognize that current rises only when source voltage is applied across the inductor.In steady state (DC), v ≈ 0 across an ideal inductor, so current is set by external resistance.Therefore, an inductor does not boost current; it shapes its rate of change. Verification / Alternative check: Examine a buck converter: although inductor current can exceed input current instantaneously due to energy transfer and duty-cycle control, the “boosting” action is from switching and storage, not from the inductor acting as an amplifier by itself. Why Other Options Are Wrong: Amplify current / ferrite-only / DC-only / wire-gauge-only: None convert a passive inductor into an active current booster; materials affect L and losses, not amplification. Common Pitfalls: Equating “passes more current in steady state” with amplification; misunderstanding that only active circuitry can provide gain; ignoring that inductors resist sudden current changes. Final Answer: Incorrect — an inductor does not boost current

Question 10

DC through an inductor — what limits it?: In a DC circuit after transients, the current through an ideal inductor is limited by the total series resistance in the loop (source resistance plus any resistors). Decide whether this statement is correct.

Accepted Answer

Introduction / Context: Inductors behave differently during transients and in steady-state. In DC steady-state, an ideal inductor has zero voltage across it (di/dt = 0), acting as a short circuit. Therefore, the loop's resistances determine the final current value, a key point in power supply and relay driver design. Given Data / Assumptions: DC excitation long after any switching event (steady-state). Inductor treated as ideal (no series resistance). The circuit includes finite series resistance: R_total. Concept / Approach: In steady DC, di/dt = 0 implies v_L = L * 0 = 0. KVL then reduces the loop to the source voltage dropping entirely across resistive elements, so I_DC = V_source / R_total. The inductance value L influenced the transient ramp rate but not the final DC current (for an ideal inductor). Step-by-Step Solution: Write v_L = L * di/dt; at steady DC, v_L = 0.Apply KVL: V_source = I * R_total.Solve for I: I = V_source / R_total.Conclude the inductor does not limit DC current; resistance does. Verification / Alternative check: Measure current after a long time in a series RL step response; the final current equals V/R_total, independent of L (though the time constant tau = L/R_total set the rise time). Why Other Options Are Wrong: Incorrect / iron core / depends solely on L / current cap at 1 A: Core or L affects dynamics and saturation, not the ideal DC final value; the final current is set by resistance. Common Pitfalls: Confusing the time constant (which uses L) with the final steady-state current; overlooking small but present winding resistance in real inductors that should be included in R_total. Final Answer: Correct

Question 11

Transformers and DC: Can a conventional transformer step up or step down a DC voltage directly (without switching or chopping)? Choose the correct evaluation.

Accepted Answer

Introduction / Context: Transformers rely on changing magnetic flux to induce voltage according to Faraday’s law. A steady DC produces no changing flux, so no secondary voltage is induced once transients die out. Understanding this limitation is critical in power conversion and explains why DC-DC converters first switch DC to AC-like waveforms before using a transformer. Given Data / Assumptions: Conventional iron/ferrite-core transformer, not a special motor-generator or electronic converter. Input is steady DC, not a pulsed or switched waveform. No internal electronics present in the transformer itself. Concept / Approach: Induced voltage e = -N * dΦ/dt. With constant DC and a static core, dΦ/dt = 0 after initial transient, so e_secondary ≈ 0. Applying DC continuously risks saturating the core and overheating windings due to magnetizing current, offering no useful voltage transformation in steady state. Step-by-Step Solution: Apply DC: initial transient may cause a brief change in flux.After steady state, flux is constant; dΦ/dt = 0, so induced secondary voltage is ≈ 0.Meanwhile, primary draws DC magnetizing current that can overheat the transformer.Therefore a transformer cannot step DC directly; switching is required. Verification / Alternative check: DC-DC converters (e.g., flyback, forward) first chop DC into a pulsating waveform at kHz frequencies, then use a transformer; the presence of switching confirms the necessity of changing flux for induction. Why Other Options Are Wrong: Works for DC / more turns / laminated cores / wire gauge: None overcome the need for time-varying flux. Turns ratio matters only with AC or changing waveforms. Common Pitfalls: Assuming “more turns” guarantees step-up without considering Faraday’s law; forgetting core saturation and heating when DC is applied directly. Final Answer: Incorrect — conventional transformers do not transform steady DC

Question 12

Phase relation in a parallel RL circuit: In AC analysis, does the inductor branch current lead or lag the resistor branch current in a parallel RL network? Evaluate the statement: “Inductor current leads the resistor current.”

Accepted Answer

Introduction / Context: Phase relationships determine how currents add in AC circuits. In a parallel RL, the resistor branch current is in phase with the voltage, while the inductor branch current lags the voltage by 90 degrees (for an ideal inductor). This directly sets the phase relation between the two branch currents. Given Data / Assumptions: Sinusoidal steady-state AC. Ideal components: resistor R and inductor L in parallel across the same voltage. Reference voltage is the same across both branches. Concept / Approach: For a resistor: i_R is in phase with v. For an inductor: v = L * di_L/dt, which leads to i_L lagging v by 90 degrees. Since v is common to both branches, i_R (in phase with v) inherently leads i_L by 90 degrees; equivalently, i_L lags i_R by 90 degrees. Therefore, the statement that the inductor current leads the resistor current is incorrect. Step-by-Step Solution: Set the reference: v(t) is common across branches.Resistor: i_R(t) = v(t)/R (0-degree phase shift).Inductor: i_L lags v by 90 degrees in the ideal model.Conclude i_L lags i_R; it does not lead. Verification / Alternative check: Phasor diagram: draw I_R along the positive real axis; I_L along negative imaginary axis. The total current is the vector sum, confirming lag of the inductor branch current. Why Other Options Are Wrong: Leads at low frequency / depends only on Q / only if R = X_L: None reverse the fundamental 90-degree lag for the inductor branch in the ideal case. Common Pitfalls: Confusing series RL behavior with parallel; mixing the inductor’s voltage-current relation with the capacitor’s, where current leads voltage. Final Answer: Incorrect — inductor current lags the resistor current

Question 13

Transformer terminology — primary vs secondary: The primary winding of a transformer is the winding connected to the source, while the secondary is connected to the load. Assess the claim: “The primary winding is the side connected to the load.”

Accepted Answer

Introduction / Context: Clear terminology prevents wiring errors. By definition, the primary winding is connected to the driving source; the secondary winding delivers power to the load. Some contexts label windings by function (input vs output) rather than voltage magnitude (step-up or step-down), but “primary-to-source” remains the standard convention. Given Data / Assumptions: Conventional two-winding transformer. Primary: driven by a source; secondary: supplies a load. No unusual reconfiguration such as backfeeding unless explicitly stated. Concept / Approach: Primary and secondary roles are determined by which side is energized by the external source. Even if a transformer is used in reverse (backfed), the side currently connected to the source is, by function, acting as the primary for that use. Therefore, the claim that the primary is connected to the load is contrary to standard terminology. Step-by-Step Solution: Define “primary”: winding connected to the source.Define “secondary”: winding from which power is taken to the load.Apply definitions to any use case; names follow function.Hence the statement in the stem is incorrect. Verification / Alternative check: Datasheets label windings as primary (input) and secondary (output). Installation instructions always treat the source-connected side as primary, avoiding confusion during wiring and protection sizing. Why Other Options Are Wrong: Primary connected to load: Opposite to the standard definition. True only for step-down / depends on turns ratio / autotransformers: The naming is independent of ratio or construction; it is based on which side the source energizes. Common Pitfalls: Assuming “higher-voltage side” equals primary always; in backfeeding scenarios, roles can swap functionally — but the side tied to the source remains the primary for that setup. Final Answer: Incorrect — primary connects to the source

Question 14

Inductors in series — two ideal inductors are connected end-to-end (series). If L1 = 0.20 mH and L2 = 0.40 mH, what is the total equivalent inductance seen by the source?

Accepted Answer

Introduction / Context: Series combinations of inductors appear in filter chokes, input EMI networks, and energy-storage chains. Knowing how to combine inductances quickly is a bread-and-butter skill in electronics. This problem asks for the equivalent inductance when two ideal inductors are connected in series, using values that add neatly to 0.60 mH. Given Data / Assumptions: L1 = 0.20 mH (ideal; no winding resistance or coupling losses). L2 = 0.40 mH (ideal). Series connection; mutual coupling is negligible (k ≈ 0) unless otherwise stated. Small-signal or DC equivalent; we are only aggregating L values, not reactances at a specific frequency. Concept / Approach: For uncoupled inductors in series, total inductance is the sum of individual inductances: L_total = L1 + L2. This mirrors resistors in series. If there were significant mutual coupling, a ±2M term would appear (M is mutual inductance), but in most layout situations with separated coils or orthogonal cores, M is small and the simple sum applies. Step-by-Step Solution: Identify topology: series connection of L1 and L2.Apply rule: L_total = L1 + L2.Compute: L_total = 0.20 mH + 0.40 mH = 0.60 mH.Report with units consistent to mH. Verification / Alternative check: At a test frequency f, the net reactance is X_L = 2 * pi * f * L_total. If each coil has reactances X1 and X2 measured separately, you will observe X_total ≈ X1 + X2, confirming the series rule in practice (ignoring small parasitics and coupling). Why Other Options Are Wrong: 0.40 mH or 0.20 mH: These correspond to individual values, not the sum. 0.80 mH: Would require an additional 0.20 mH or constructive coupling that adds effective inductance; not the given case. Common Pitfalls: Confusing series and parallel formulas (parallel uses reciprocals), forgetting that tight coupling can modify the sum, and mixing units (µH vs mH). Always confirm units before adding. Final Answer: 0.60 mH is the total inductance for 0.20 mH and 0.40 mH in series (uncoupled).

Question 15

Inductor cores — what does the term “air-core inductor” most accurately imply about the material inside the coil former?

Accepted Answer

Introduction / Context: Cores determine an inductor’s inductance, quality factor, saturation behavior, and frequency performance. Designers must distinguish “air-core” from ferrite or powdered-iron cores to choose the right part for RF, power, or signal applications. This question clarifies what “air-core inductor” actually means. Given Data / Assumptions: An inductor is a coil of wire wound around some medium (core or former). “Air-core” appears on datasheets for RF chokes and high-frequency coils. No specific frequency is mandated; the concept is general. Concept / Approach: An air-core inductor has no ferromagnetic material (no ferrite, steel, or iron powder) within the coil. The “core” is literally air or a non-magnetic support like PTFE, ceramic, fiberglass, or plastic. Without a magnetic path of high permeability, inductance per turn is lower than with magnetic cores, but linearity is excellent and there is no magnetic saturation. Losses due to hysteresis are eliminated, making air-core designs ideal for high-frequency RF circuits and for applications requiring very high linearity or low distortion. Step-by-Step Solution: Interpret the phrase “air-core.”Recognize that it excludes ferromagnetic materials.Conclude the coil surrounds air or a non-magnetic former only.Choose the option that explicitly states this. Verification / Alternative check: Compare datasheets: air-core inductors show relatively low inductance for given size and turns, high self-resonant frequency, and minimal core losses; ferrite or powdered-iron cores show higher inductance density but may introduce saturation and frequency-dependent losses. Why Other Options Are Wrong: Laminated steel with air gaps: Describes power transformers or large chokes, not air-core. Iron slug tuning: That is an adjustable iron-core or ferrite-core inductor. Ferrite rod: Clearly a magnetic core, the opposite of air-core. Common Pitfalls: Assuming “air-core” refers to gaps in iron cores; it does not. Also, mistaking the plastic bobbin as a “core” with magnetic effects; non-magnetic formers contribute negligible permeability. Final Answer: It uses no ferromagnetic core; the coil surrounds air or a non-magnetic former.

Question 16

Coil parasitics — the inherent DC resistance of an inductor's wire winding is called what?

Accepted Answer

Introduction / Context: Real inductors are not ideal. In addition to inductance, they exhibit losses and parasitic elements. One important parameter is the ohmic resistance of the copper (or aluminum) wire itself, which affects efficiency, Q factor, and temperature rise. This question asks for the standard term used for that inherent resistance. Given Data / Assumptions: The element is a coil of conductive wire. We are discussing its DC property (not frequency-dependent skin or proximity effects). Terminology should match common datasheet language. Concept / Approach: The resistance of the wire making up the coil, measured at DC, is called winding resistance (sometimes R_w or DCR for “DC resistance”). It is separate from the frequency-dependent increase in effective resistance caused by skin effect and proximity effect, often described as AC resistance or ESR in the context of power magnetics. Winding resistance contributes I^2 * R losses and lowers Q = X_L / R_total at a given frequency. Step-by-Step Solution: Identify the physical phenomenon: ohmic resistance of the wire at DC.Select the standard term used by manufacturers and textbooks.Answer: “winding resistance.”Note the distinction from reactance (frequency-dependent imaginary part). Verification / Alternative check: Check typical inductor datasheets: “Inductance L,” “DCR (mΩ),” “Q at frequency,” and “SRF” are commonly listed. DCR equals the winding resistance at room temperature and rises with conductor temperature. Why Other Options Are Wrong: Permeability: A property of core material, not conductive loss. Reactance: Imaginary opposition X_L = 2 * pi * f * L, not a real loss. AC resistance: A frequency-dependent effective resistance; the stem asks for inherent (DC) resistance. Common Pitfalls: Equating ESR/AC resistance with the DC winding resistance; ESR typically exceeds DCR at higher frequencies due to skin and proximity effects. Final Answer: winding resistance is the inherent DC resistance of the coil wire.

Question 17

Inductor dynamics — which quantity in an ideal inductor cannot change instantaneously at the exact moment a switching event occurs?

Accepted Answer

Introduction / Context: Inductors store energy in magnetic fields. Their fundamental behavior is governed by v = L * di/dt. This directly constrains how rapidly current through an inductor can change, which is crucial in switching converters, motor drives, and transient analysis. The question asks which quantity cannot exhibit an instantaneous jump in an ideal inductor. Given Data / Assumptions: Ideal inductor with inductance L, no series resistance. Possibility of step changes in applied voltage (switching). We consider the instantaneous behavior at t = 0+ when a switch toggles. Concept / Approach: From v = L * di/dt, the derivative of current is finite for finite voltage. An instantaneous jump in current would require infinite di/dt, which in turn would require infinite voltage — impossible in real circuits and not permitted in idealized linear models. Therefore, in an inductor, current is continuous. Voltage, however, can change abruptly as circuit conditions change because it is proportional to the rate of change of current, not to the current itself. Step-by-Step Solution: Recall the constitutive relation: v(t) = L * di(t)/dt.Assume a finite voltage step is applied; di/dt becomes finite.Hence i(t) changes with slope, not a jump; i(t) is continuous at switching instants.Conclude that the inductor current cannot change instantaneously. Verification / Alternative check: Observe RL step response: i(t) = I_final * (1 − exp(−t/τ)) + i(0) * exp(−t/τ) with τ = L/R. At t = 0+, the exponential term assures continuity of current, matching lab waveforms. Why Other Options Are Wrong: Voltage: It can jump depending on circuit constraints, especially in series with switches or when di/dt changes sign. Instantaneous power: p(t) = v(t) * i(t) can change abruptly since v may step. Magnetic field orientation: Orientation is determined by current direction; while direction can reverse over time, the instantaneous change would again imply a current discontinuity — the governing restriction is on current. Common Pitfalls: Confusing capacitor and inductor continuity laws (capacitor voltage is continuous; inductor current is continuous). Mixing energy storage variables leads to wrong transient predictions. Final Answer: current cannot change instantaneously in an inductor.

Question 18

Solving for an unknown series inductor — three uncoupled inductors are in series. If L1 = 120 mH, L2 = 70 mH, and the measured total inductance is 340 mH, what is the value of L3?

Accepted Answer

Introduction / Context: Determining an unknown component value from a measured total is common in repair and prototyping. With series inductors that are not magnetically coupled, the total inductance is just the arithmetic sum. Given two known values and the total, the third can be obtained by subtraction. This mirrors the way series resistors add. Given Data / Assumptions: L1 = 120 mH, L2 = 70 mH, L3 = ? Total series inductance L_total = 340 mH. Inductors are uncoupled (negligible mutual inductance). Ideal components for the computation; parasitics do not affect the sum appreciably. Concept / Approach: For uncoupled inductors in series, L_total = L1 + L2 + L3. Solve for the unknown: L3 = L_total − (L1 + L2). If coils were tightly coupled, the ±2M term could alter the equivalent inductance; however, in most practical series aggregates with spacing or orthogonal orientation, M is small and can be ignored for first-order calculations. Step-by-Step Solution: Write the sum: L_total = L1 + L2 + L3.Rearrange: L3 = L_total − (L1 + L2).Compute: L3 = 340 mH − (120 mH + 70 mH) = 340 mH − 190 mH = 150 mH.Choose the matching option. Verification / Alternative check: Measure reactance at a test frequency f: X_L,total ≈ 2 * pi * f * 0.340 H. Separately summing X_L of L1, L2, and the computed L3 will yield the same total within measurement tolerance, confirming the calculated L3. Why Other Options Are Wrong: 220 mH, 295 mH, 315 mH: These do not satisfy L1 + L2 + L3 = 340 mH and would give totals that are too large. Common Pitfalls: Accidentally using the parallel formula (reciprocals) or mixing units (mH vs H). Always check that the configuration is truly series and that units are consistent. Final Answer: 150 mH for L3.

Question 19

First-order transient rule of thumb — after exactly one time constant (τ) in a first-order RL or RC response, by what percentage has the transient changed, and what percentage remains?

Accepted Answer

Introduction / Context: First-order circuits (RL and RC) exhibit exponential responses to steps. Engineers routinely estimate settling using the time constant τ. A classic benchmark: after one τ, the exponential has moved 63.2% of the way from its initial value toward its final value, leaving 36.8% of the excursion to go. This aids quick design checks without a calculator. Given Data / Assumptions: First-order system with step input; no oscillation. Time constant τ = L/R for RL or τ = R*C for RC. We are asking for normalized percentages independent of the specific values. Concept / Approach: Standard exponential form: y(t) = y_final + (y_initial − y_final) * exp(−t/τ). At t = τ, exp(−1) ≈ 0.3679. Therefore, the remaining difference from final is 36.79% (≈ 36.8%), and the amount of change completed is 1 − 0.3679 ≈ 0.6321 or 63.21% (≈ 63.2%). These percentages hold for rising or falling responses (signs invert but magnitudes match). Step-by-Step Solution: Write normalized response: normalized_remaining = exp(−t/τ).Evaluate at t = τ: exp(−1) ≈ 0.3679 → 36.8% remaining.Compute changed percentage: 100% − 36.8% = 63.2% changed.Select the matching option. Verification / Alternative check: At t = 2τ, remaining ≈ 13.5%; at t = 3τ, ≈ 5.0%; at t = 5τ, ≈ 0.67%. These checkpoints form the common “1τ = 63.2%” mnemonic and the “5τ ≈ settled” rule. Why Other Options Are Wrong: 36.8% changed / 63.2% remaining: Reversed. 13.5% or 86.5% pairs: Those correspond to t = 2τ (13.5% remaining) and its complement, not 1τ. Common Pitfalls: Confusing “remaining” with “changed,” or mixing the percentages for 1τ and 2τ. Always derive from exp(−t/τ) to avoid memory slips. Final Answer: 63.2% changed, 36.8% remaining after one time constant.

Question 20

Opposition to AC in an inductor — the specific term for an inductor’s frequency-dependent opposition to alternating current is what?

Accepted Answer

Introduction / Context: While resistors oppose current independent of frequency (ideally), inductors and capacitors oppose AC in a frequency-dependent manner. Using precise terminology helps avoid confusion when calculating circuit behavior. This question asks for the name of an inductor’s AC opposition. Given Data / Assumptions: Linear time-invariant circuit elements. Sinusoidal steady-state analysis at a single frequency f. Ideal inductor with reactance proportional to frequency. Concept / Approach: The inductor’s AC opposition is called inductive reactance, symbol X_L, given by X_L = 2 * pi * f * L (in ohms). “Impedance” Z is the vector sum of resistance and reactance (Z = R + jX). “Reluctance” is a magnetic circuit term for opposition to magnetic flux (analogous to resistance but in the magnetic domain). “Resistance” alone denotes the real (lossy) part, not the frequency-dependent imaginary part provided by the inductor. Step-by-Step Solution: Identify the specific phenomenon: frequency-dependent, imaginary opposition.Recall formula: X_L = 2 * pi * f * L.Name that quantity: inductive reactance.Select the option “reactance.” Verification / Alternative check: Phasor calculations: the inductor’s impedance is Z_L = jX_L = j * (2 * pi * f * L). The magnitude |Z_L| equals X_L, confirming that “reactance” is the proper term for the opposition’s magnitude. Why Other Options Are Wrong: Impedance: Too general; includes both resistance and reactance. Reluctance: Wrong domain; magnetic circuits, not electrical AC circuits. Resistance: Real power-dissipative term, not the reactive opposition unique to inductors/capacitors. Common Pitfalls: Using “impedance” and “reactance” interchangeably. Remember: impedance is the complex total; reactance is the imaginary component (X_L or X_C). Final Answer: reactance is the correct term.

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