Question 1

Definition in electromagnetics: What is electromagnetism in the context of a current-carrying conductor?

Accepted Answer

Introduction / Context: Electromagnetism connects electricity and magnetism. A foundational observation is that electric current flowing through a conductor produces a magnetic field encircling the wire. Recognizing this definition is essential for understanding motors, relays, inductors, and transmission lines. Given Data / Assumptions: Steady or time-varying current through a straight conductor. Classical field behavior (no quantum effects required for the concept). Right-hand rule applies to field direction around the conductor. Concept / Approach: By Ampere’s law and experimental evidence, a current I establishes a circumferential magnetic field H around the conductor. The field intensity is proportional to current and inversely related to distance from the wire. This effect underlies inductance and is different from induction caused by changing magnetic fields (Faraday’s law). Step-by-Step Solution: Identify the phenomenon: current → encircling magnetic field.Direction by right-hand rule: thumb points with current, curled fingers show field direction.Note the magnitude dependence: stronger current → stronger field.Recognize applications: electromagnets, solenoids, and magnetic actuators rely on this effect. Verification / Alternative check: Using a compass near a current-carrying wire demonstrates deflection that reverses with current direction, confirming the magnetic field produced by the current. Why Other Options Are Wrong: external voltage causing a magnetic field: Voltage alone does not create a magnetic field; current does. voltage caused by a magnetic field: That describes electromagnetic induction (emf) from changing flux, not the definition of electromagnetism here. current generated by an external magnetic field: That is induction again, requiring changing flux or motion. conversion of flux to heat: Not a definition; losses can cause heat but are not electromagnetism itself. Common Pitfalls: Confusing electromagnetism (field due to current) with electromagnetic induction (emf due to changing magnetic field). Ignoring field orientation; use the right-hand rule to avoid sign mistakes. Final Answer: the magnetic field generated around a conductor when a current passes through it

Question 2

Parallel inductors calculation: Two coils of 5 H and 100 mH are connected in parallel. What is the total equivalent inductance of the combination?

Accepted Answer

Introduction / Context: Combining inductors is common in filter design and energy storage networks. For inductors in parallel, the equivalent value decreases, analogously to parallel resistors. This question reinforces correct use of the parallel inductance formula and careful unit handling between henry and millihenry. Given Data / Assumptions: L1 = 5 H. L2 = 100 mH = 0.1 H. Coils are ideal and not magnetically coupled (no mutual inductance considered). Concept / Approach: For two uncoupled inductors in parallel, the equivalent inductance is L_eq = (L1 * L2) / (L1 + L2). This mirrors the product-over-sum form used for two resistors in parallel, but applies to inductors when mutual coupling is negligible. Step-by-Step Solution: Convert units: 100 mH = 0.1 H.Apply formula: L_eq = (5 * 0.1) / (5 + 0.1) H.Compute numerator: 0.5; denominator: 5.1 → L_eq = 0.5 / 5.1 ≈ 0.098039 H.Convert to mH: 0.098039 H ≈ 98.0 mH. Verification / Alternative check: Because one inductor (0.1 H) is much smaller than the other (5 H), the parallel combination should be slightly less than 0.1 H. The result 98 mH is consistent with this expectation. Why Other Options Are Wrong: 4.76 mH / 33.3 mH / 150.0 mH: Do not satisfy the parallel formula for the given values. 5.10 H: This is the series sum, not a parallel result. Common Pitfalls: Forgetting to convert millihenry to henry before calculation. Confusing series and parallel formulas or accidentally adding instead of using product-over-sum. Final Answer: 98.0 mH

Question 3

Faraday’s law — core idea: Faraday’s law primarily concerns which situation leading to an induced electromotive force (emf)?

Accepted Answer

Introduction / Context: Faraday’s law of electromagnetic induction explains how changing magnetic conditions produce an emf. This law underpins transformers, generators, and inductive sensors. Understanding the precise trigger for induction clarifies many practical designs. Given Data / Assumptions: Conductors or coils exposed to time-varying magnetic flux. Relative motion or changing field is present (dΦ/dt ≠ 0). Lenz’s law determines emf polarity opposing the change. Concept / Approach: Faraday’s law states that induced emf is proportional to the time rate of change of magnetic flux linkage. Practically, this occurs when a conductor cuts through magnetic field lines or when flux through a coil changes. The essential condition is changing flux linkage, commonly summarized as a magnetic field cutting a conductor (or equivalently, changing flux through a loop). Step-by-Step Solution: Identify mechanism: changing magnetic flux linkage → induced emf.Interpret “cutting a conductor”: relative motion of conductor and field or time-varying field causes changing linkage.Result: an emf and current (if closed path) are produced with polarity opposing the change (Lenz).Applications: generators, transformers, induction cookers, and pickup coils. Verification / Alternative check: Move a wire loop rapidly into or out of a magnetic field region; a measurable voltage appears only when the flux linkage changes, verifying Faraday’s law in practice. Why Other Options Are Wrong: a magnetic field in a coil / in a conductor: A static magnetic field with no change does not induce emf. hystersis: Misspelling of hysteresis; it concerns lagging magnetization and losses, not the induction condition. a static electric field: Not the Faraday induction mechanism. Common Pitfalls: Assuming a constant magnetic field is sufficient; the key is change in flux linkage. Confusing induction (emf due to changing flux) with electromagnetism (field due to current). Final Answer: a magnetic field cutting a conductor

Question 4

Inductive reactance formula — evaluate the statement: the magnitude of inductive reactance is given by XL = 2π f L for a linear inductor at frequency f (hertz) and inductance L (henry).

Accepted Answer

Introduction / Context: Inductive reactance quantifies how an inductor impedes alternating current due to the changing magnetic field. It plays a central role in AC analysis, filter design, resonance calculations, and impedance matching. The standard relationship ties reactance to frequency and inductance linearly in magnitude. Given Data / Assumptions: L is the inductance in henry (H). f is the signal frequency in hertz (Hz). The inductor is linear over the operating range (no core saturation, negligible parasitics). Steady-state sinusoidal analysis is assumed. Concept / Approach: For a sinusoidal steady state, the inductor voltage-current relation is v(t) = L * di/dt. In phasor form, this becomes V = j ω L * I, where ω = 2π f is the angular frequency. The impedance of the inductor is therefore Z_L = j ω L. The magnitude of this impedance is |Z_L| = ω L = 2π f L, which is by definition the inductive reactance, X_L. Thus, X_L grows linearly with frequency and inductance; at DC (f = 0), X_L = 0 Ω, behaving as a short in ideal form. Step-by-Step Solution: Write the time-domain relation: v = L * di/dt.Move to phasor domain: Z_L = j ω L with ω = 2π f.Take magnitude: X_L = |Z_L| = ω L = 2π f L.Interpretation: increasing f or L increases reactance proportionally. Verification / Alternative check: Example: L = 10 mH at f = 1 kHz → X_L = 2π * 1000 * 0.01 = about 62.8 Ω. A measurement with an LCR meter at 1 kHz would read a similar reactance (ignoring ESR and parasitics). Why Other Options Are Wrong: “Incorrect” contradicts the standard derivation from v = L di/dt.DC-only, iron-core-only, or kilohertz-only caveats are unnecessary; the formula applies broadly to linear inductors across frequencies where the lumped model holds. Common Pitfalls: Forgetting units (henry and hertz) leading to numeric errors; confusing reactance magnitude X_L with the complex impedance j ω L; neglecting nonidealities like winding resistance and self-resonance at very high frequencies. Final Answer: Correct

Question 5

Electromagnetism — definition check: Does self-inductance describe the property of a conductor (or coil) by which a changing current in itself induces an opposing electromotive force (emf) in the same circuit?

Accepted Answer

Introduction / Context: Self-inductance is a foundational idea in circuit theory and electromagnetics. Whenever the current through a conductor or coil changes, the associated magnetic field also changes. This changing flux links the same conductor and induces a voltage that opposes the change. Understanding this definition is crucial for analyzing RL transients, switch behavior, energy storage in magnetic fields, and practical issues such as inductive kickback in relays and motor windings. Given Data / Assumptions: Standard lumped-element model (coil modeled as an ideal inductor L in series with winding resistance). Quasi-static conditions where circuit dimensions are small relative to wavelengths of interest. No special core is required; self-inductance exists for air-core coils and even straight conductors (though the magnitude may be small). Concept / Approach: By Faraday’s law, induced emf magnitude is proportional to the time rate of change of linked magnetic flux. For a coil with N turns, v_L = L * di/dt, where L (henry) quantifies self-inductance. Lenz’s law gives the polarity: the induced emf acts to oppose the change in current. Thus, a rising current induces a counter-voltage that resists the rise; a falling current induces a voltage that tries to keep current flowing, leading to familiar spark/kickback phenomena if a path is suddenly opened. Step-by-Step Solution: State definition: self-inductance is the property of a circuit where its own changing current induces an emf in itself.Relate to formula: v_L = L * di/dt, showing proportionality to di/dt.Apply Lenz’s law for polarity: induced voltage opposes the change in current.Conclude: the stem statement correctly describes self-inductance. Verification / Alternative check: Observe an RL step response: when a DC source is suddenly applied, current does not jump to its final value; the inductor’s induced emf initially equals the source and then decays as current builds. This behavior confirms that the inductor generates an opposing emf due to its own changing current. Why Other Options Are Wrong: “Incorrect” contradicts Faraday–Lenz behavior. “Only at very high frequency,” “only in superconductors,” and “only with an iron core” are misconceptions; self-inductance exists broadly and does not require a core or extreme frequency, though L’s magnitude and parasitic effects may vary. Common Pitfalls: Confusing induced emf with power loss (ideal inductors store/return energy, they do not dissipate). Assuming a core is mandatory (air-core inductors are common in RF). Final Answer: Correct

Question 6

Electromagnetism — Lenz’s law comprehension: Does Lenz’s law state that an induced current (or induced emf) always acts to oppose the change in magnetic flux that produced it, thereby opposing the cause of motion or flux change?

Accepted Answer

Introduction / Context: Lenz’s law determines the direction (polarity) of induced currents and voltages in electromagnetic induction. It is a qualitative statement that ensures energy conservation and aligns with Faraday’s law. Engineers rely on it when predicting back-emf in motors, inductive kickback in coils, eddy current braking, and transformer polarities. Given Data / Assumptions: Changing magnetic flux through a conducting loop or coil. Faraday’s induction principle applies: magnitude of induced emf is proportional to rate of change of linked flux. Conventional passive sign convention and standard circuit orientation. Concept / Approach: Lenz’s law: the induced emf produces currents whose magnetic field opposes the original change in flux. If flux increases, the induced current creates an opposing field to reduce the increase. If flux decreases, the induced current creates a field that tries to maintain it, resisting the decrease. This opposition manifests as mechanical forces resisting motion (e.g., in generators) or electrical voltages resisting current change (e.g., in inductors). Step-by-Step Solution: Identify the cause: changing flux linkage through a loop.Apply Faraday–Lenz: induced emf polarity chosen so that its effect opposes the change in flux.Infer current direction: right-hand/clock rules yield the loop direction that creates the opposing field.Conclude: the statement matches Lenz’s law precisely. Verification / Alternative check: Consider moving a magnet toward a loop: the induced current direction creates a magnetic field repelling the magnet (opposing the approach). Pulling the magnet away flips the current to attract it (opposing separation). This physical opposition is a direct demonstration of Lenz’s law. Why Other Options Are Wrong: “Incorrect” conflicts with the definition. Limiting it to DC, to increases only, or to ferromagnetic cores is false—Lenz’s law applies generally to any changing flux linkage in conductive loops. Common Pitfalls: Confusing “oppose the change” with “oppose the flux.” It opposes the change in flux, not necessarily the flux itself. Final Answer: Correct

Question 7

Circuit laws — applicability: Does Kirchhoff’s Voltage Law (KVL) fail to apply to a simple series R–L circuit driven by a source in lumped-element analysis?

Accepted Answer

Introduction / Context: Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of voltage drops around any closed loop is zero at every instant. This law is widely used in analyzing RL, RC, and RLC circuits for both transient and steady-state conditions. The claim that KVL does not apply to series RL circuits is a misconception and is corrected here. Given Data / Assumptions: Lumped-element model (component dimensions small vs. wavelength; negligible radiation). Ideal components: resistor R, inductor L, and a voltage source. Quasi-static field conditions without distributed transmission-line effects. Concept / Approach: KVL emerges from electrostatic field conservativeness in lumped circuits and is valid instant-by-instant. In a series RL loop driven by v_s(t), the loop equation is v_s(t) − v_R(t) − v_L(t) = 0, where v_R(t) = i(t) * R and v_L(t) = L * di/dt. This is the standard starting point for deriving the RL differential equation and its step response. Step-by-Step Solution: Write KVL for the loop: v_s(t) = v_R(t) + v_L(t).Substitute element laws: v_R(t) = iR, v_L(t) = Ldi/dt.Obtain RL DE: Ldi/dt + Ri = v_s(t).Conclude: KVL holds at all times under lumped conditions. Verification / Alternative check: Measure with an oscilloscope: the instantaneous source voltage equals the instantaneous sum of resistor and inductor voltages. Simulations (SPICE) also confirm v_s = v_R + v_L for arbitrary inputs (steps, sinusoids, pulses). Why Other Options Are Wrong: Limiting KVL to resistive circuits, steady state, or low frequency is incorrect within lumped-element validity. Only when circuits become distributed (high-frequency transmission lines, significant radiation) do we need full Maxwell’s equations instead of simple KVL. Common Pitfalls: Confusing instantaneous addition with RMS magnitudes; in AC, phasor magnitudes do not add arithmetically, but the time-domain KVL always holds. Final Answer: Incorrect

Question 8

Inductors in series (no coupling): For inductances connected in series with negligible mutual coupling, is the total inductance LT equal to L1 + L2 + L3 + … ?

Accepted Answer

Introduction / Context: Just as resistances in series add, inductances in series also add when there is negligible mutual coupling. This rule is used to realize a desired inductance by stacking multiple coils and is fundamental in filter and timing networks. However, coupling between coils can change the result, so the assumption matters. Given Data / Assumptions: Series connection of inductors carrying the same current. Negligible mutual inductance (coils are physically separated or oriented to minimize coupling). Lumped-element operation. Concept / Approach: For series inductors with the same current, total flux linkages sum, yielding an equivalent inductance equal to the arithmetic sum of individual inductances: L_T = L1 + L2 + .... If there is coupling, cross terms involving mutual inductance M appear (L_T = L1 + L2 ± 2M for two coils), which can increase or decrease L_T depending on relative winding sense. Step-by-Step Solution: Assume same current I flows through all series inductors.Voltage across each: v_k = L_k * di/dt.Total series voltage: v_T = sum(v_k) = (sum L_k) * di/dt.Therefore L_T = sum L_k when coupling is negligible. Verification / Alternative check: Measure with an LCR meter: two inductors 10 mH and 15 mH well separated read about 25 mH in series. Bringing them close and aligned changes the reading due to mutual M—evidence that the simple sum assumes negligible coupling. Why Other Options Are Wrong: Requiring an iron core, identical values, or DC operation is unnecessary. The relation is general under the stated assumption. “Incorrect” ignores standard series behavior. Common Pitfalls: Ignoring coupling effects; in compact layouts, mutual inductance can be significant and must be accounted for in precision designs. Final Answer: Correct

Question 9

Series connection effect: If multiple inductors are connected in series with negligible mutual coupling, does the total inductance increase compared to any single inductor?

Accepted Answer

Introduction / Context: Understanding how inductors combine is crucial for filter design, power electronics, and RF circuits. In series, each inductor contributes an induced voltage proportional to L * di/dt. The contributions add, producing a larger equivalent inductance than any individual element—assuming negligible mutual coupling. Given Data / Assumptions: Series connection; same current flows through each inductor. Negligible mutual inductance between coils. Ideal lumped components. Concept / Approach: The total voltage across series inductors is v_T = (L1 + L2 + …) * di/dt. Equating to v_T = L_T * di/dt yields L_T = L1 + L2 + …. Because each L_k is positive, L_T is necessarily greater than any single L_k. If coupling exists, M terms can modify L_T up or down depending on winding orientation; the “increase” claim is assured under negligible coupling. Step-by-Step Solution: Write each drop: v_k = L_k * di/dt.Sum drops: v_T = (sum L_k) * di/dt.Define L_T by v_T = L_T * di/dt.Conclude L_T = sum L_k > any single L_k. Verification / Alternative check: Practical check with two separated coils on a bench shows the LCR meter reading increases when placed in series compared to either coil alone. Why Other Options Are Wrong: “Incorrect” conflicts with basic series behavior. Constraints like equal values, ferrite cores, or temperature conditions are not required for the fundamental addition rule without coupling. Common Pitfalls: Overlooking mutual coupling in dense layouts; coupling can reduce or increase L_T, so layout and orientation matter. Final Answer: Correct

Question 10

RL transient — time constants: Is it accurate to say that current in an inductor “reaches its maximum value” exactly after five time constants in a first-order RL step response?

Accepted Answer

Introduction / Context: First-order RL circuits respond to a DC step with an exponential current rise toward a final value. Engineers often use “time constants” to estimate when the response is essentially complete. However, exponential functions are asymptotic; they approach the final value but reach it only at infinite time. Thus, “exactly at five time constants” is not correct, though 5τ is a common practical benchmark (~99.3%). Given Data / Assumptions: Series RL circuit driven by a DC step. Time constant τ = L / R for an RL network. Ideal inductor and resistor. Concept / Approach: The current response is i(t) = I_final * (1 − e^(−t/τ)). At t = τ, the response is ~63.2%; at 2τ, ~86.5%; 3τ, ~95.0%; 4τ, ~98.2%; 5τ, ~99.3%. The current never truly equals I_final at any finite t; it only gets arbitrarily close as t increases. Therefore, the claim that it “reaches its maximum” at 5τ is false in the strict sense. Step-by-Step Solution: Use RL step formula: i(t) = I_final * (1 − e^(−t/τ)).Evaluate at t = 5τ: i(5τ) ≈ 0.993 * I_final.Recognize asymptotic behavior: equality i(t) = I_final requires t → ∞.Conclude: the statement is inaccurate as written. Verification / Alternative check: Oscilloscope measurements show the current trace flattening after several τ but never mathematically touching the final value. Designers often treat 5τ as “essentially final” to within about 0.7% error. Why Other Options Are Wrong: “Correct” contradicts exponential behavior. Restrictions about core type, R = 0, or waveform do not change the asymptotic nature for the DC step case. Common Pitfalls: Equating “practically complete” with “exactly complete.” Always state tolerances when using time-constant heuristics. Final Answer: Incorrect

Question 11

Fields and forces — definitions: Is “magnetic force” the correct term for the cause that produces a magnetic field, or is a magnetic field produced by electric currents and changing electric fields while magnetic force acts on charges in a field?

Accepted Answer

Introduction / Context: Precise terminology matters. A magnetic field is produced by electric currents (moving charges) and changing electric fields; the field then exerts a magnetic force on moving charges or magnetic dipoles. Saying “magnetic force produces the magnetic field” reverses cause and effect and confuses core concepts of electromagnetism and Lorentz force. Given Data / Assumptions: Classical electromagnetism framework. Moving charges and time-varying electric fields as sources of magnetic fields. Lorentz force law linking fields to forces on charges. Concept / Approach: Maxwell’s equations indicate that currents and changing electric fields create magnetic fields. The Lorentz force F on a charge q moving with velocity v in a magnetic field B is F = q * v × B (magnetic component). Thus, “magnetic force” is an effect of a pre-existing magnetic field acting on charges; it is not the agent that creates B. Permanent magnets produce fields due to atomic currents (electron spin/orbit) at the microscopic level—not because “force” is creating the field. Step-by-Step Solution: Identify source: currents and changing E-fields create B-fields.Define effect: magnetic force acts on moving charges/dipoles within B.Resolve the statement: it incorrectly claims force creates the field.Conclude: the statement is incorrect. Verification / Alternative check: In a straight wire with DC current, a circular B-field forms around the wire (right-hand rule). The force on a nearby test charge appears only because the B-field already exists due to current; turning off current removes the field and the magnetic component of force. Why Other Options Are Wrong: “Correct” misstates causality. Context qualifiers (permanent magnets, stationary charges, cryogenic temperatures) do not change the definitions. Common Pitfalls: Confusing “field sources” with “forces caused by fields.” Keeping these roles separate prevents conceptual errors in device analysis (motors, generators, magnetic storage). Final Answer: Incorrect

Question 12

Power transfer via transformers — matching condition: Is maximum power transfer obtained when the primary winding’s impedance equals the secondary winding’s impedance, regardless of source and load, or when the load reflected to the primary matches …

Accepted Answer

Introduction / Context: Transformers are used to match impedances between a source and a load. The classic maximum power transfer theorem states that a load receives maximum power when it equals the complex conjugate of the source’s Thevenin impedance (for purely resistive sources, equal resistance). In transformer-coupled systems, this condition is met when the load, reflected to the primary through the turns ratio, matches the source impedance—not when the primary and secondary impedances are arbitrarily made equal to each other. Given Data / Assumptions: Linear, ideal transformer model for first-order reasoning. Source characterized by a Thevenin impedance Z_S and Thevenin voltage V_S. Load Z_L connected to the transformer secondary. Concept / Approach: Reflected impedance: Z_reflected_to_primary = (N_P/N_S)^2 * Z_L. Maximum power transfer condition: Z_reflected_to_primary ≈ Z_S (or Z_S* for complex matching). Therefore, the engineering task is to choose the turns ratio such that the load, when reflected, equals the source impedance. Simply equating “primary impedance” and “secondary impedance” is meaningless because they exist on different sides and are related by the square of the turns ratio; equality between them does not guarantee matching to the source. Step-by-Step Solution: Start from maximum power transfer: Z_load_seen_by_source should ≈ Z_source.Reflect Z_L to primary: Z_L′ = (N_P/N_S)^2 * Z_L.Set Z_L′ ≈ Z_S to satisfy the condition.Conclude the original statement (primary equals secondary) is not the correct criterion. Verification / Alternative check: Example: source 50 Ω, load 200 Ω. Choose N_P/N_S = sqrt(50/200) = 1/2 so that the 200 Ω load reflects as 50 Ω; maximum transfer occurs. Primary and secondary impedances are not “equal” in absolute terms; they are related by the square of turns ratio. Why Other Options Are Wrong: “Correct” contradicts the theorem. Conditions like 1:1 transformers, resonance, or ideal sources do not change the fundamental matching requirement. Common Pitfalls: Assuming equal impedances on both windings ensures maximum power. The correct condition is source matching to the load as seen through the transformer. Final Answer: Incorrect

Question 13

Inductance dependence on geometry: Is the inductance of a coil directly proportional to its cross-sectional area (all else equal)?

Accepted Answer

Introduction / Context: The inductance of a coil depends on its geometry and the magnetic properties of the core. For a simple solenoid model, L increases with the square of the number of turns, with permeability, and with cross-sectional area, and decreases with magnetic path length. Recognizing these dependencies helps engineers tune inductors for filters, chokes, and energy storage components. Given Data / Assumptions: Simple solenoid formula context: L ∝ μ * N^2 * A / l. Constant turns N, path length l, and permeability μ for comparison. Negligible fringing and parasitics for first-order reasoning. Concept / Approach: Holding N, μ, and l constant, doubling the cross-sectional area A doubles the inductance L. Physically, a larger area allows more magnetic flux for the same magnetizing force, increasing energy storage in the magnetic field for a given current. Practical cores (ferrite, powdered iron) obey this trend within their linear regions. Step-by-Step Solution: Start with proportional form: L ∝ μ * N^2 * A / l.Fix μ, N, l; vary A.Observe: increasing A increases L linearly.Conclude: the statement is correct. Verification / Alternative check: Compare two inductors identical except for core cross-section: the device with larger A measures a higher inductance on an LCR meter. Magnetic circuit modeling likewise shows flux Φ ∝ MMF / Reluctance, with Reluctance ∝ l / (μ * A); larger A reduces reluctance and increases L. Why Other Options Are Wrong: Claiming dependence only with iron cores is false; air-core coils also follow this proportionality. Frequency and winding layering do affect parasitics (ESR/ESL distribution) but not this first-order geometric proportionality. Common Pitfalls: Ignoring saturation and non-linear μ in high-flux conditions; in those regimes, proportionality can deviate. For first-order, small-signal analysis, the relationship holds well. Final Answer: Correct

Question 14

DC circuits — energy dissipation sources: Is it true that in a direct-current (DC) circuit the only conversion of electrical energy to heat occurs in the inductance of a coil?

Accepted Answer

Introduction / Context: Power dissipation in circuits is tied to resistance (or other lossy mechanisms), not ideal reactance. In DC conditions, an ideal inductor behaves as a short circuit with zero average power dissipation; practical heat arises in its winding resistance and other resistive components. Understanding where heat comes from helps with thermal design, efficiency, and reliability. Given Data / Assumptions: DC steady state after any transients have decayed. Inductor modeled as ideal L in series with winding resistance R_w. Other circuit elements may include resistors and semiconductors with conduction losses. Concept / Approach: Instantaneous power p = v * i. For an ideal inductor at steady DC, di/dt = 0, so v_L = L * di/dt = 0; therefore, p_L = v_L * i = 0. Real heating occurs in resistive parts: P = I^2 * R_w in windings, I^2 * R in series resistors, and conduction/switching losses in semiconductors. During transients, energy is temporarily stored or released by the inductor (E = 0.5 * L * I^2), but ideal inductors do not dissipate that energy—resistive paths do. Step-by-Step Solution: At steady DC, di/dt = 0 → v_L = 0.Power in ideal L: p = v_L * i = 0 → no heating from ideal inductance.Identify heat sources: P = I^2 * R_w in windings and other resistive elements.Conclude the statement is false; heating is due to resistance, not ideal inductance. Verification / Alternative check: Measure coil temperature with and without current: heating correlates with I^2 * R_w. If the same coil had zero resistance (superconducting), DC heating would vanish despite finite inductance, confirming that resistance—not inductance—causes heat. Why Other Options Are Wrong: “Correct” contradicts power relations. Transient-only and core-type qualifiers miss the steady-state point. Superconductors illustrate the opposite: inductance remains, but DC heating disappears because R ≈ 0. Common Pitfalls: Confusing reactive energy exchange with dissipation; only resistive elements dissipate real power in ideal analyses. Final Answer: Incorrect

Question 15

RL filters in basic AC electronics: depending on where the output is taken (across R or across L) in a simple series RL network driven by a sinusoidal source, the resulting filter can behave as a high-pass or as a low-pass. Confirm or reject this st…

Accepted Answer

Introduction / Context: Passive filters built from a single resistor and a single inductor are common first-order building blocks in AC electronics. Whether such an RL network passes low frequencies or high frequencies depends not just on the parts, but on where the output is measured. This item checks your understanding of output placement and frequency response. Given Data / Assumptions: Series RL network excited by an AC voltage source. Ideal components unless otherwise noted (L with no series loss, R purely resistive). Output can be taken either across the resistor or across the inductor. Small-signal linear operation; no saturation effects. Concept / Approach: The frequency-dependent term is the inductive reactance: XL = 2 * π * f * L. At low frequency, XL is small; at high frequency, XL is large. The voltage division between R and XL determines the output. If the output is taken across R, low frequencies appear mostly across R (since XL is small), so the network behaves as a low-pass. If the output is taken across L, high frequencies develop a larger drop across L (since XL grows with f), so the network behaves as a high-pass. Step-by-Step Solution: 1) Write the divider ratio: V_R = Vin * (R / sqrt(R^2 + XL^2)); V_L = Vin * (XL / sqrt(R^2 + XL^2)).2) Evaluate low f: XL → small ⇒ V_R ≈ Vin (low-pass), V_L ≈ small.3) Evaluate high f: XL → large ⇒ V_R ≈ small (attenuated), V_L ≈ Vin (high-pass).4) Conclude that output placement determines whether the RL network is low-pass or high-pass. Verification / Alternative check: Plot magnitude responses of V_R/Vin and V_L/Vin versus frequency. Each is complementary around the corner frequency fc = R / (2 * π * L). Why Other Options Are Wrong: Incorrect: contradicts the clear divider behavior described above. Applies only at resonance: a single RL has no resonance; resonance needs at least L and C. Valid only for ideal components: small losses slightly alter fc and slope but not the pass/stop nature. Common Pitfalls: Confusing RL behavior with RC rules; forgetting that the measured node (R or L) defines which band is passed; assuming “filter type” is fixed by component types alone. Final Answer: Correct

Question 16

Ideal transformers and phasing: without specifying dot orientation or reference terminals, is it valid to state that “the primary and secondary are in phase”?

Accepted Answer

Introduction / Context: Transformer voltage phasing is essential whenever windings are interconnected, rectified, or paralleled. The relative instantaneous polarity of primary and secondary voltages is not inherently “always in phase” or “always out of phase.” It is defined by the dot convention and how terminals are referenced. This question probes that concept. Given Data / Assumptions: Ideal transformer model (no loss, infinite magnetizing inductance for analysis). No explicit dot markings or terminal reference given in the statement. Sinusoidal steady-state assumed. Concept / Approach: For an ideal transformer with dots on one end of each winding, the instantaneous polarities at the dotted ends are in phase: a rising voltage applied to the dotted primary end induces a voltage of the same instantaneous polarity at the dotted secondary end. If one observes opposite (dotted to undotted) terminals, the measured voltages appear 180 degrees out of phase. Therefore, the blanket statement “primary and secondary are in phase” lacks required context and is not universally valid. Step-by-Step Solution: 1) Use the ideal relation: V2 / V1 = N2 / N1, with polarities assigned by the dot convention.2) If measurements are taken at dotted ends simultaneously, voltages are in phase; at dotted vs undotted ends, they are inverted.3) Because the statement specifies neither reference ends nor dot orientation, it cannot be assumed true. Verification / Alternative check: Manufacturer schematics always include dot marks or explicit polarity marks to avoid ambiguity when building multiwinding supplies or audio crossovers. Why Other Options Are Wrong: Correct: ignores the need to specify terminals/dots. Depends only on turns ratio: ratio sets magnitude, not phase. True only for step-down: phasing is not tied to whether N2 < N1 or N2 > N1. Common Pitfalls: Assuming “step-down equals in phase”; forgetting that scope probe reference selection can flip apparent phase; overlooking the dot convention entirely. Final Answer: Incorrect

Question 17

Transformer loading: does the secondary current primarily depend on the secondary voltage and the connected load resistance (I2 ≈ V2 / Rload in steady state)?

Accepted Answer

Introduction / Context: In practical transformer applications, the secondary current is set by the load connected to the secondary and the secondary voltage delivered by the turns ratio. Understanding this relationship underpins power supply sizing, regulation expectations, and thermal design. Given Data / Assumptions: Ideal transformer approximation: V2 = (N2/N1) * V1. Sinusoidal steady-state operation. Load modeled as a resistance Rload (no significant reactance). Negligible winding resistance and leakage for first-order reasoning. Concept / Approach: With an ideal source, the secondary presents a Thevenin-like supply of magnitude V2. The current through a resistive load is I2 = V2 / Rload. The primary current adjusts automatically (by reflected impedance) to support the delivered secondary power, with I1 ≈ (N2/N1) * I2 for an ideal device, neglecting magnetizing component. Step-by-Step Solution: 1) Compute secondary voltage from the turns ratio.2) Use Ohm’s law on the load: I2 = V2 / Rload.3) Recognize that the transformer reflects the load to the primary: Rreflected = (N1/N2)^2 * Rload.4) The source supplies the required VA; the secondary current is dictated by V2 and Rload. Verification / Alternative check: Replace the secondary and load by their primary-reflected equivalents; the same current results after transforming back, confirming consistency. Why Other Options Are Wrong: Incorrect: contradicts Ohm’s law at the secondary terminals. Only true for autotransformers or saturated cores: both claims are unrelated to the linear, ideal behavior described. Common Pitfalls: Confusing magnetizing current (small and mostly reactive) with load current; assuming secondary current is “fixed by the transformer” rather than by the load. Final Answer: Correct

Question 18

RL transient behavior: in a first-order inductive (RL) circuit, does the time constant depend on the circuit resistance?

Accepted Answer

Introduction / Context: Time constant governs how quickly currents and voltages rise or decay during switching events in first-order circuits. For RL networks, the time constant directly affects relay delays, inrush current, and filter settling. This question asks whether resistance influences that time constant. Given Data / Assumptions: Series RL driven by a step input. Inductor L and total effective series resistance R (including source and winding resistance). Linear operation without core saturation. Concept / Approach: The RL time constant is τ = L / R (seconds). Larger resistance speeds up decay (smaller τ), while larger inductance slows it down (larger τ). This emerges from the differential equation: L * di/dt + R * i = V_step, whose homogeneous solution is i(t) = I_final * (1 − exp(−t/τ)) for a rise or i(t) = I_initial * exp(−t/τ) for a decay. Step-by-Step Solution: 1) Write the KVL for the series RL under a step: L * di/dt + R * i = V.2) Identify the coefficient of i: R / L sets the exponential rate.3) Conclude τ = L / R and note its units: henry/ohm = second.4) Interpret physically: more resistance dissipates energy faster, so the inductor current reaches the new steady state sooner. Verification / Alternative check: Check limits: as R → 0, τ → ∞ (current changes very slowly); as R → ∞, τ → 0 (current cannot build, response is fast). Why Other Options Are Wrong: Incorrect or “depends only on inductance”: both ignore the R term that sets the decay rate. Depends only on source voltage: voltage scales final current, not the exponential time constant. Common Pitfalls: Confusing RL (τ = L/R) with RC (τ = R*C); forgetting that any series resistance adds to R, including winding and source resistances. Final Answer: Correct

Question 19

Comparing passives in practice: in common electronics applications (especially at board and IC level), is it generally accurate to say inductors are less versatile than capacitors?

Accepted Answer

Introduction / Context: Both capacitors and inductors are fundamental energy-storage elements, yet their practical usage breadth differs in typical electronic products. This item targets a qualitative comparison frequently taught in introductory courses and observed in mainstream PCB and IC design. Given Data / Assumptions: “Versatility” refers to the variety of roles in common circuits: coupling, decoupling, filtering, timing, energy storage, impedance shaping. Practical constraints such as size, cost, DC resistance, and electromagnetic interference are considered. Focus on low-to-mid frequency analog/digital electronics rather than specialized high-power or magnetics domains. Concept / Approach: Capacitors are broadly used for power-supply decoupling, AC coupling, timing (RC), oscillator networks, sample-and-hold, and DC blocking, and they integrate well on silicon. Inductors, while essential in switch-mode power supplies, RF matching, and chokes, are bulkier, harder to integrate, have parasitic resistance and core losses, and can radiate/receive EMI. Consequently, across a wide range of everyday designs, capacitors play more roles and are easier to deploy, which underpins the teaching shorthand that inductors are “less versatile.” Step-by-Step Solution: 1) List common capacitor roles: decoupling, filtering (with R or L), timing, coupling, integrators, sample-and-hold.2) List common inductor roles: SMPS energy storage, RF tanks/matching, EMI suppression, chokes.3) Compare integration: capacitors (especially small) can be on-chip; inductors usually off-chip or large.4) Infer generality: capacitors appear in more contexts across consumer/embedded designs. Verification / Alternative check: Survey any digital board’s BOM: capacitors dominate counts for decoupling and timing; inductors are fewer, typically tied to power stages or EMI filters. Why Other Options Are Wrong: Incorrect: overlooks pervasive capacitor use and inductor constraints. Only at RF or only for ideal parts: the observation spans many frequency regimes and real components. Common Pitfalls: Interpreting “less versatile” as “less important”; conflating specialization (power magnetics, RF inductors) with overall breadth of use. Final Answer: Correct

Question 20

Faraday–Lenz principle: does the induced voltage polarity in a coil oppose the change in current that created it?

Accepted Answer

Introduction / Context: Induction underlies inductors, transformers, and many sensors. Faraday’s law quantifies induced electromotive force (emf), and Lenz’s law specifies its polarity. Together they explain why an inductor “resists” changes in current by developing a voltage that counteracts the change. Given Data / Assumptions: Single coil carrying a time-varying current. Magnetic coupling is localized to the coil’s own field (self-inductance). Linear operation without core saturation. Concept / Approach: Faraday: induced emf magnitude is proportional to the time rate of change of magnetic flux linkage. For an inductor, v_L = L * di/dt. Lenz: the induced voltage polarity is such that the resulting current opposes the change in flux that produced it. Hence, if the current attempts to increase, the inductor produces a voltage that drops opposite to the source; if the current attempts to decrease, the inductor generates voltage that tries to keep current flowing in the same direction. Step-by-Step Solution: 1) Write the inductor law: v_L = L * di/dt.2) Consider di/dt > 0 (current rising): v_L polarity opposes the rise, requiring extra source voltage.3) Consider di/dt < 0 (current falling): v_L reverses polarity to sustain current, often causing flyback.4) Conclude that the induced voltage always opposes the causative change. Verification / Alternative check: Scope a switching inductor: during turn-off, the node voltage spikes of opposite polarity appear, consistent with v_L = L * di/dt and Lenz’s law. Why Other Options Are Wrong: Incorrect: contradicts fundamental induction behavior. “True only for DC” or “only with iron core”: the law holds for AC and DC transients and does not require ferromagnetic cores. Common Pitfalls: Confusing “opposes current” with “blocks DC” (an inductor passes DC once steady state is reached); misreading polarity during switching. Final Answer: Correct

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