Question 1

In basic electromagnetism, when electrical energy is supplied to a coil (inductor), in which physical form is that energy primarily stored while current flows?

Accepted Answer

Introduction / Context: Inductors are everywhere in electrical and electronics engineering: power supplies, filters, motor drives, and radio-frequency circuits. A core idea is how inductors store and release energy. Unlike capacitors, which store energy in an electric field, coils store energy in a magnetic field produced by current. Understanding this difference helps with safe design (e.g., snubbers across coils) and accurate analysis of transient behavior. Given Data / Assumptions: A current-carrying coil (inductor) is energized by a source. We are interested in the intrinsic energy storage mechanism of the inductor. Ideal inductor assumptions for the concept; real parasitics do not change the fundamental storage mechanism. Concept / Approach: The energy stored by an inductor is associated with the magnetic field set up by current through its turns. As current increases, the magnetic field strengthens and energy increases; when current decreases, the collapsing field returns energy to the circuit. The well-known energy relation is E_L = 1/2 * L * I^2, where L is inductance in henries and I is current in amperes. Step-by-Step Solution: A current I flows through N turns, creating a magnetic field linking the coil.Magnetic energy density is proportional to the square of the field strength.Total energy stored in the inductor is E_L = 1/2 * L * I^2.Therefore, the energy is stored in the magnetic field surrounding the coil. Verification / Alternative check: When current through an energized coil is interrupted, a voltage spike appears that attempts to keep current flowing. This behavior is consistent with field energy being returned to the circuit as the magnetic field collapses (inductive kick). Snubber networks or flyback diodes are used to safely dissipate or recycle this energy. Why Other Options Are Wrong: Electrical/electrostatic fields describe capacitor energy storage, not inductor storage. “Force field” is not a precise engineering term here. Mechanical strain energy does not describe an inductor's normal operation. Common Pitfalls: Confusing the energy location (in the field) with dissipation in winding resistance. An ideal inductor stores but does not dissipate energy; real coils also have copper loss in series resistance. Final Answer: a magnetic field

Question 2

For a transformer with its secondary winding left open-circuited (no load connected), which statement best describes the secondary side condition under normal operation?

Accepted Answer

Introduction / Context: Transformers operate by magnetic coupling between primary and secondary windings. The behavior changes dramatically between loaded and unloaded (open-circuit) conditions. Knowing what happens on the secondary when it is not connected to a load is crucial for measurement, safety, and fault diagnosis. Given Data / Assumptions: Secondary winding is open-circuited (no load). Primary is energized at its rated voltage and frequency. Transformer is healthy and designed for the operating frequency. Concept / Approach: With the secondary open, the transformer still establishes a mutual magnetic flux and thus an induced secondary voltage proportional to the turns ratio. However, since the circuit is open, no continuous path exists for current on the secondary; hence, the steady-state secondary current is essentially zero (ignoring tiny capacitive leakage). The primary still draws a small magnetizing current to establish core flux and to cover core loss. Step-by-Step Solution: Apply Faraday’s law: an induced EMF appears on the secondary proportional to N_secondary.Check circuit completion: with open terminals, current cannot flow.Conclude: secondary current is zero in steady state; secondary voltage is present but no load current flows. Verification / Alternative check: Measuring with a voltmeter across an unloaded secondary shows the rated RMS voltage. Using an ammeter in series with a still-open terminal indicates no current flow. Primary input current is small (magnetizing and core-loss components). Why Other Options Are Wrong: “Secondary voltage increases” is generally false; it stays near nominal (slight rise possible in regulation, but not an increase trend as stated). “No secondary voltage” contradicts transformer action. “Secondary current increases” is the opposite of open-circuit behavior. “Secondary power equals primary power” is incorrect; with no load, real power transfer to the secondary is near zero aside from losses. Common Pitfalls: Confusing voltage presence with current flow; voltage can be present with zero current if the circuit is open. Final Answer: no secondary current flows

Question 3

Regarding transformer phase relationships, which statement is most accurate about the output phase of the secondary relative to the primary?

Accepted Answer

Introduction / Context: In many transformer applications—such as multi-secondary power supplies, audio coupling, and feedback networks—the instantaneous polarity (phase) of the secondary relative to the primary matters. Designers use this knowledge to connect windings series-aiding or series-opposing and to ensure correct rectifier phasing. Given Data / Assumptions: Ideal transformer model for conceptual clarity. Standard dot notation on schematics denotes corresponding ends of windings. Supply frequency remains constant and appropriate for the core. Concept / Approach: The phase of the induced secondary voltage is determined by the winding sense (direction of turns) relative to the primary. If the dotted ends are treated as corresponding ends, a positive rise at the dotted primary end produces a simultaneous positive rise at the dotted secondary end. Thus, phase can be 0 degrees (in phase) or 180 degrees (out of phase) depending on how the windings are oriented or connected; it is not intrinsically set by frequency. Step-by-Step Solution: Identify polarity dots on the schematic.If primary dotted terminal becomes positive relative to its undotted terminal, the secondary dotted terminal becomes positive relative to its undotted terminal at the same instant.Therefore, output phase depends on the winding direction/connection. Verification / Alternative check: Using an oscilloscope and a low-voltage test, probe primary and secondary dotted ends simultaneously. The waveforms rise together when the dots correspond to in-phase ends. Why Other Options Are Wrong: Always in phase or always out of phase is an overgeneralization; either can occur based on winding orientation. Dependence on frequency for phase inversion is incorrect; frequency affects magnitude and core behavior, not the instantaneous polarity mapping. Random phase is false; dot notation removes ambiguity. Common Pitfalls: Ignoring dot notation and assuming schematic placement (left/right) implies phase. Only the dots or documented lead designation define the relationship. Final Answer: dependent on the direction of the primary and secondary windings

Question 4

In real-world components, which characteristics does a practical inductor exhibit in addition to ideal inductance?

Accepted Answer

Introduction / Context: No manufactured inductor behaves as an ideal element. Practical coils exhibit parasitic resistance and capacitance that shape frequency response, losses, and self-resonance. Recognizing these non-idealities is essential for RF design, switching power supplies, and precision filtering. Given Data / Assumptions: Wound inductor with finite wire resistivity and inter-turn spacing. Operation over a range of frequencies where parasitics become significant. Linear operation (no core saturation) for the conceptual discussion. Concept / Approach: A practical inductor can be modeled as an ideal inductance L in series with a winding resistance R_w (copper loss), and in parallel (or distributed) with a capacitance C_p due to inter-turn and layer-to-layer capacitances. At low frequency, R_w causes I^2*R loss; at high frequency, C_p can resonate with L, creating a self-resonant frequency beyond which the device behaves capacitively. Step-by-Step Solution: Identify ideal property: inductance L stores energy in a magnetic field.Add series resistance: R_w represents copper loss and skin/proximity effects at higher frequencies.Add parasitic capacitance: C_p arises from the proximity of turns and insulation.Conclude: a practical inductor exhibits all of these characteristics simultaneously. Verification / Alternative check: Impedance analyzer sweeps show rising inductive reactance up to a peak at self-resonance, after which the inductor appears capacitive—evidence of C_p. DC resistance measurements reveal R_w. Why Other Options Are Wrong: Choosing only one property ignores well-documented parasitics present in every coil. Claiming “only inductance with zero loss” contradicts reality; wire has resistance and dielectrics form capacitance. Common Pitfalls: Using an inductor near or above its self-resonant frequency without accounting for the capacitive behavior. Always check datasheet curves. Final Answer: all of the above

Question 5

Impedance matching with a transformer: What turns ratio is required to match two 16 Ω loads connected in parallel (effective 8 Ω) to an output impedance of 22.2 Ω on the other side?

Accepted Answer

Introduction / Context: Transformers are commonly used to match a source impedance to a load impedance to maximize power transfer or meet amplifier stability requirements. The impedance seen at one winding is the actual load scaled by the square of the turns ratio. Selecting the correct ratio prevents power loss, distortion, and overheating. Given Data / Assumptions: Two 16 Ω loads in parallel → Z_load = 8 Ω. Desired matched impedance on the other side is 22.2 Ω. Ideal transformer approximation (no losses, perfect coupling). Concept / Approach: For an ideal transformer, the impedance referred from secondary to primary obeys Z_primary = (N_primary / N_secondary)^2 * Z_secondary. Let a = N_primary : N_secondary. We need the reflected impedance to equal 22.2 Ω when the real load is 8 Ω. Step-by-Step Solution: Compute equivalent load: Z_s = 16 ∥ 16 = (16*16)/(16+16) = 8 Ω.Set matching condition: Z_p = a^2 * Z_s = 22.2 Ω.Solve for a: a^2 = 22.2 / 8 = 2.775.a = sqrt(2.775) ≈ 1.666 → required turns ratio ≈ 1.67:1 (primary:secondary). Verification / Alternative check: Reflect 8 Ω through a = 1.67: Z_p ≈ (1.67)^2 * 8 ≈ 2.7889 * 8 ≈ 22.3 Ω, essentially the target 22.2 Ω within rounding. This confirms the selection. Why Other Options Are Wrong: 1.38:1 and 0.72:1 yield reflected impedances significantly different from 22.2 Ω. 0.60:1 steps the impedance down instead of up. 2.22:1 would reflect 8 Ω to ≈ 39.5 Ω, far from the requirement. Common Pitfalls: Forgetting that impedance scales with the square of the turns ratio, not linearly. Also, be careful to combine parallel loads correctly before reflection. Final Answer: 1.67:1

Question 6

When current through an inductor begins to change, it cannot change instantaneously. Which fundamental effect prevents an instantaneous change in inductor current?

Accepted Answer

Introduction / Context: Inductors resist changes in current. This time-dependent behavior governs the shape of transients in RL circuits, the operation of switching converters, and the design of snubbers. The physical reason is the voltage that the inductor itself produces whenever its current changes. Given Data / Assumptions: An inductor with inductance L is part of a circuit where current is changing. We are considering short time intervals around the change. Core remains unsaturated; linear behavior assumed. Concept / Approach: An inductor obeys v_L = L * di/dt. A rapid attempted change in current (large di/dt) generates a large induced voltage whose polarity opposes the change (Lenz’s law). This self-induced or “counter” emf fights the change, limiting di/dt to a finite value set by circuit conditions. Step-by-Step Solution: Consider a sudden step in applied voltage across L.Inductor produces v_L that maintains current continuity: di/dt = v_L / L.As di/dt increases, v_L increases in opposing polarity, preventing an instantaneous jump in current.Thus, the counter emf of the coil is the effect that limits the rate of change of current. Verification / Alternative check: In an RL step response, current follows i(t) = I_final * (1 - e^(-t/τ)), where τ = L/R, clearly not instantaneous. The exponential rise or decay arises from the inductor’s induced voltage opposing current change. Why Other Options Are Wrong: Fixed resistance affects steady-state current and time constant but does not forbid instantaneous change by itself. Applied emf drives current; it does not limit change. Eddy currents and dielectric absorption are secondary effects and not the primary cause of current continuity in L. Common Pitfalls: Thinking an inductor “blocks DC.” It does not block steady DC; it resists changes in current. After transients settle, an ideal inductor behaves like a short circuit. Final Answer: the counter emf of the coil

Question 7

Parallel RL at low power factor: What is the overall phase angle (in degrees) for a 24 Vac parallel RL circuit with R = 45 Ω and XL = 1100 Ω?

Accepted Answer

Introduction / Context: For parallel RL circuits, the source voltage is common, while currents through the resistor and inductor differ in phase. The overall current lags the voltage slightly when the inductor branch susceptance is small relative to conductance. Computing the phase angle from admittance is a standard AC analysis task. Given Data / Assumptions: R = 45 Ω, XL = 1100 Ω at 24 Vac (RMS). Ideal components; frequency implicit in XL value. We seek the phase angle θ of total current relative to voltage. Concept / Approach: Use admittance: Y = G + jB, where G = 1/R and B = −1/XL (inductive susceptance is negative). The phase angle of the total current is θ = arctan(B/G). For an inductive parallel circuit, θ is negative (current lags). We typically report the magnitude for comparison with given options. Step-by-Step Solution: Compute conductance: G = 1/R = 1/45 ≈ 0.02222 S.Compute susceptance: B = −1/XL = −1/1100 ≈ −0.0009091 S.Find angle: θ = arctan(B/G) = arctan(−0.0009091 / 0.02222) ≈ −2.34 degrees.Magnitude ≈ 2.34 degrees → closest option is 2.300 degrees (lagging). Verification / Alternative check: Compute power factor: pf = cos(|θ|) ≈ cos(2.34°) ≈ 0.9992, indicating a very small reactive component, consistent with a large XL in parallel. Why Other Options Are Wrong: 0.001° and 89.900° imply negligible or extreme phase shifts not matching B/G. 87.600° is characteristic of nearly pure reactance, not this case. 45.000° would require G = |B|, which is not true here. Common Pitfalls: Confusing series and parallel formulas. In parallel, compute with admittances (G and B); in series, use impedances (R and X) and their phasor angle. Final Answer: 2.300 degrees

Question 8

On schematic diagrams, two vertical transformer windings with dashed parallel lines drawn between them typically indicate what type of core?

Accepted Answer

Introduction / Context: Schematic symbols communicate not just connectivity but also construction details. For transformers and inductors, the presence or absence of core lines conveys whether the device is air-cored or uses a ferromagnetic core, which affects inductance, losses, and frequency range. Given Data / Assumptions: Two vertical coils are drawn facing each other (transformer symbol). Dashed lines are shown between the windings. Standard drafting conventions used in many textbooks and datasheets. Concept / Approach: When a core is indicated between windings, it signifies magnetic material coupling the coils. In many basic schematic conventions, drawing parallel lines (often dashed or solid) between coils denotes a laminated iron (steel) core transformer, whereas no core lines implies air core. Specific CAD libraries may differentiate ferrite with variations, but the essential meaning is “magnetic core present,” commonly laminated iron in power-frequency contexts. Step-by-Step Solution: Identify the two parallel lines between windings as a core symbol.Associate this with laminated iron cores found in mains-frequency transformers.Therefore, dashed lines between the coils typically indicate an iron core transformer. Verification / Alternative check: Consult common schematic symbol references: air-core transformers are drawn without core lines; adding core lines indicates a ferromagnetic core, usually iron for power applications. Why Other Options Are Wrong: Air core would show no core lines at all. “Nonmetallic core” is ambiguous; ferrite is nonmetallic but is usually identified more specifically and by context (high frequency). Ferrite cores are often used at high frequency; many symbol sets still simply show core lines without distinguishing material. Common Pitfalls: Assuming the dashed style always encodes material type (iron vs ferrite). Unless the legend specifies otherwise, interpret lines between windings as “has ferromagnetic core,” which in power circuits typically means laminated iron. Final Answer: an iron core

Question 9

According to Lenz’s law, as a magnetic field collapses in a coil, what is the polarity of the induced voltage relative to the original cause (the force creating the field)?

Accepted Answer

Introduction / Context: Lenz’s law is the sign convention of electromagnetic induction. It dictates how induced voltages and currents oppose the changes that produced them. This principle explains flyback voltages, back-EMF in motors, and the self-protective behavior of inductors. Given Data / Assumptions: A magnetic field linked to a coil is collapsing (flux decreasing). The original “force” is the cause of the field (source current or magnetizing action). We are considering the polarity of the induced voltage. Concept / Approach: Faraday’s law says induced EMF is proportional to the negative rate of change of flux linkage. Lenz’s law interprets the negative sign: the induced voltage drives a current whose magnetic field opposes the change in original flux. During collapse, the induced voltage polarity acts to keep current flowing in the same direction, thereby opposing the decrease in flux. Step-by-Step Solution: Original field decreases → coil responds with induced EMF.By Lenz’s law, this EMF produces a magnetic effect opposing the decrease.Therefore, the induced voltage polarity is opposite to the cause that created the field.Practically, this appears as a “kick” of opposite polarity when current is interrupted. Verification / Alternative check: Observe a relay coil without a flyback diode: interrupting current produces a high-voltage spike of opposite polarity across the coil, consistent with opposition to the change. Why Other Options Are Wrong: Independence or randomness contradicts Faraday–Lenz law. “Identical” would reinforce the change, violating energy conservation. “Only if stationary” is irrelevant; induction arises from changing flux. Common Pitfalls: Confusing “opposes the change” with “opposes the original quantity.” The induced effect tries to keep flux from changing, not always to cancel the original field steady value. Final Answer: opposite to the force creating it

Question 10

Transformer design: If 120 V (RMS) is applied to a 2400-turn primary, how many turns are required on the secondary to obtain 7.5 V (RMS) at the secondary?

Accepted Answer

Introduction / Context: The transformer turns ratio directly sets the voltage ratio between windings in the ideal case. Designing low-voltage secondaries (for logic supplies, heaters, or instrumentation) requires correctly scaling turns from a known primary. Given Data / Assumptions: Primary voltage Vp = 120 V RMS. Primary turns Np = 2400 turns. Desired secondary voltage Vs = 7.5 V RMS. Ideal transformer approximation (no regulation drop). Concept / Approach: In an ideal transformer, Vs/Vp = Ns/Np. Therefore, Ns = Np * (Vs/Vp). This linear proportionality allows immediate calculation of required secondary turns for a target voltage. Step-by-Step Solution: Compute ratio: Vs/Vp = 7.5 / 120 = 0.0625.Apply to turns: Ns = 2400 * 0.0625 = 150 turns.Thus, 150 turns on the secondary will yield approximately 7.5 V RMS. Verification / Alternative check: Reverse-check: Ns/Np = 150/2400 = 1/16; applying to voltage gives Vp/16 = 120/16 = 7.5 V. Agreement confirms the result. Why Other Options Are Wrong: 75 turns would produce 3.75 V; 900 turns would produce 45 V; 1920 turns would give 96 V; 15 turns would give 0.75 V. None match the target 7.5 V. Common Pitfalls: Confusing turns ratio with impedance ratio. Voltages scale linearly with turns; impedances scale with the square of the turns ratio. Final Answer: 150 turns

Question 11

In a practical loaded transformer operating from an alternating-current (AC) supply, how does the primary current respond when the secondary load current increases, so that mutual induction maintains nearly constant core flux?

Accepted Answer

Introduction / Context: Transformers rely on mutual induction: a changing current in the primary creates a changing magnetic flux in the core, which induces voltage in the secondary. When a load is connected, current flows in the secondary and affects the primary via the core. This question tests your understanding of how primary current automatically adjusts to changing secondary load to keep the core flux approximately constant in steady operation. Given Data / Assumptions: Ideal transformer approximation (small losses). Constant AC primary voltage. Magnetizing current is comparatively small. Secondary load current increases due to added load. Concept / Approach: The secondary load current produces a demagnetizing magnetomotive force (MMF) that tends to reduce core flux. To counteract this and maintain the flux set by the applied primary voltage, the primary current increases. The additional component of primary current (the load component) creates an opposing MMF that balances the secondary's effect, keeping net flux nearly unchanged. Step-by-Step Solution: Primary applied voltage fixes core flux waveform approximately (neglecting saturation).Secondary load current introduces a demagnetizing MMF: N2 * I2.Transformer action draws extra primary current I1(load) so that N1 * I1(load) ≈ N2 * I2 (ampere-turns balance).Total primary current = magnetizing current + load component; thus it increases with secondary load. Verification / Alternative check: Using power balance for an ideal transformer: V1 * I1 ≈ V2 * I2. If I2 increases at fixed V2, then I1 must increase at fixed V1, confirming the qualitative behavior. Why Other Options Are Wrong: Primary decrease when secondary increases: Opposite of ampere-turns balance. Primary remains fixed: Violates power and MMF balance. “Secondary decrease causes primary increase”: Also contrary to balance; less load means less primary current. Common Pitfalls: Confusing magnetizing current (relatively small) with total primary current. Forgetting that the applied voltage, not the load, primarily sets the flux; current adjusts to maintain it. Final Answer: primary current is increased by an increase in secondary current

Question 12

In power-supply filtering applications, where should an inductor be placed relative to the load, and does it aid or oppose rapid changes in current through the load?

Accepted Answer

Introduction / Context: Inductors are often used in power supplies as smoothing elements. Their fundamental behavior is to resist sudden changes in current, which makes them useful in reducing ripple current reaching the load after rectification. This question explores correct placement and qualitative behavior of an inductor used as a filter choke. Given Data / Assumptions: DC output with superimposed ripple from a rectifier. Inductor used as a choke in a filter network (e.g., L filter or LC filter). Load expects relatively steady current. Concept / Approach: For a series choke filter, the inductor is placed in series with the load. Because vL = L * di/dt, the inductor develops a voltage opposing rapid changes in current. High-frequency ripple tries to change current quickly, but the inductor resists, thereby smoothing current. Capacitors, conversely, are commonly placed in shunt to ground to bypass ripple voltage. Step-by-Step Solution: Place L in series: current to the load must pass through L.Ripple attempts rapid current change; inductor generates opposing voltage.Result: reduced ripple current through the load, improved smoothing.Often paired with shunt capacitors to form LC or π filters for better attenuation. Verification / Alternative check: Frequency-domain view: inductor impedance XL = 2 * π * f * L increases with frequency, so higher-frequency ripple components see greater impedance and are attenuated when L is in series with the load. Why Other Options Are Wrong: Parallel placements described here are capacitor-like roles, not inductive choke roles. “Aids current changes”: The opposite of an inductor’s behavior; it opposes di/dt. Shorting ripple across the rectifier is unsafe and incorrect for an inductor. Common Pitfalls: Confusing capacitor (shunt) and inductor (series) filter positions. Assuming the inductor reduces voltage ripple directly; it primarily smooths current which, with capacitors, yields lower voltage ripple. Final Answer: placed in series with the load and opposes any current changes

Question 13

In electromagnetic coupling between two circuits, what is the correct term for the effect where magnetic flux produced by one circuit links and cuts the conductors of the other circuit, inducing a voltage?

Accepted Answer

Introduction / Context: When two circuits are magnetically coupled, changes in current in the first generate magnetic flux that links the second. This linkage causes an induced voltage in the second circuit according to Faraday’s law. The phenomenon is fundamental to transformers, inductive sensors, and many communication systems. Given Data / Assumptions: Two nearby coils or conductors. Time-varying current in the primary circuit. Magnetic flux from the primary links the secondary. Concept / Approach: Mutual inductance M quantifies the amount of induced voltage in one circuit due to a time-varying current in the other: v2 = M * di1/dt (sign depending on polarity). High mutual inductance occurs with tight coupling, high permeability cores, and geometries that maximize shared flux. Step-by-Step Explanation: Primary current changes → magnetic flux varies in the core/space.Portion of this flux links the secondary turns.Linked flux changing in time induces voltage in the secondary (Faraday’s law).This cross-circuit effect is called mutual inductance. Verification / Alternative check: In a transformer, open-circuit secondary voltage scales with turns ratio. The very existence of a predictable induced secondary voltage demonstrates strong mutual inductance between windings on a common core. Why Other Options Are Wrong: Self-inductance: Induced voltage within the same circuit due to its own current change. Rectification: Conversion of AC to DC using nonlinear devices. Magnetism: A broad term, not the specific coupling mechanism. Eddy-coupling: Not a standard term; eddy currents are loss mechanisms, not the desired coupling. Common Pitfalls: Confusing mutual inductance with leakage flux (flux that does not link the other circuit). Ignoring orientation and spacing; poor alignment reduces M strongly. Final Answer: mutual inductance

Question 14

Faraday’s law of electromagnetic induction: which statement best captures the law's quantitative essence regarding induced electromotive force (emf)?

Accepted Answer

Introduction / Context: Faraday’s law states how changing magnetic fields induce voltages in circuits. It provides the quantitative relationship used in generators, transformers, and inductive sensors. Understanding the dependence on the rate of change of flux is essential for designing efficient electromagnetic devices. Given Data / Assumptions: Conductor or coil experiencing changing magnetic flux. Time-varying magnetic field or relative motion between conductor and field. Negligible parasitics for conceptual clarity. Concept / Approach: Faraday’s law in magnitude: emf = dΦ/dt for a single turn (more generally, emf = N * dΦ/dt for N turns). The induced voltage scales with how fast flux linkages change, whether by varying field strength, changing area/orientation, or relative motion. Lenz’s law gives the direction, opposing the change, but Faraday’s law sets the magnitude relation. Step-by-Step Explanation: Identify the flux linkage: λ = N * Φ.Differentiate with respect to time: emf = dλ/dt = N * dΦ/dt.Therefore, faster flux change → larger induced emf.Waveforms with higher frequency or steeper edges induce higher voltages for the same flux swing. Verification / Alternative check: In generators, increasing rotational speed increases dΦ/dt, boosting output voltage. In transformers, higher frequency for the same core flux swing increases induced voltage per turn, consistent with Faraday’s law. Why Other Options Are Wrong: Opposition/aiding effects refer to Lenz’s law (direction), not Faraday’s magnitude relation. “Related to current direction” is incomplete and often incorrect; current need not flow in an open circuit for emf to exist. “Depends only on field strength” ignores the crucial time derivative. Common Pitfalls: Mixing Faraday’s quantitative law with Lenz’s directional rule. Ignoring that changing geometry (angle/area) also changes flux linkage, not just B magnitude. Final Answer: emf depends on the rate of cutting flux

Question 15

For which purpose is a transformer fundamentally used in electrical engineering practice—considering energy transfer between circuits of the same nature?

Accepted Answer

Introduction / Context: Transformers are passive electromagnetic devices used pervasively in power grids and electronics. Their core function is to transfer energy between circuits via magnetic coupling, typically changing voltage and current levels while (ideally) conserving apparent power and frequency. Given Data / Assumptions: Alternating current (AC) source to establish a time-varying flux. Two windings magnetically coupled on a common core. Neglecting losses for fundamental description. Concept / Approach: Only a time-varying current can induce a continuous voltage in the secondary; thus, transformers operate with AC (or time-varying waveforms). The device couples energy from one AC circuit to another, enabling isolation, voltage transformation, and impedance matching. Step-by-Step Explanation: Primary AC establishes alternating core flux.This changing flux links the secondary and induces an AC voltage proportional to turns ratio.Load on the secondary draws current; primary current adjusts to supply the power magnetically.Therefore, transformers couple AC-to-AC energy; they are not DC energy converters without additional circuitry (e.g., choppers/inverters). Verification / Alternative check: If a DC source is applied to an ideal transformer, after transients no changing flux exists, and no sustained secondary voltage is induced—further confirming the AC-only nature for steady transfer. Why Other Options Are Wrong: DC-to-DC, AC-to-DC, DC-to-AC transfers require active conversion circuits; a transformer alone cannot perform these functions. “Convert chemical energy” describes batteries/fuel cells, unrelated to transformers. Common Pitfalls: Assuming a transformer “steps DC voltage”—not possible without switching. Overlooking isolation and impedance matching roles in addition to voltage transformation. Final Answer: to couple electric energy from one ac circuit to another

Question 16

In magnetic field terminology, what is the correct SI unit for magnetic flux density B (i.e., webers per square meter)?

Accepted Answer

Introduction / Context: Magnetic quantities have distinct symbols and SI units. Confusing magnetic flux, flux density, and field intensity leads to errors in analysis and design. Flux density B measures how much magnetic flux passes per unit area, a key parameter for core sizing and saturation considerations. Given Data / Assumptions: B denotes magnetic flux density. SI base relation: B = Φ / A, with Φ in webers (Wb) and A in square meters (m^2). Standard SI unit naming conventions apply. Concept / Approach: Because B = Φ/A, its unit is weber per square meter. The SI derived unit name for Wb/m^2 is the tesla (T). Other related quantities include H (magnetic field strength) in A/m and Φ (flux) in webers. Non-SI cgs units like gauss are sometimes used (1 T = 10,000 gauss). Step-by-Step Explanation: Start with definition: B = Φ / A.Units: Φ in Wb, A in m^2, so B in Wb/m^2.By SI naming, Wb/m^2 is called tesla (T).Therefore, the correct unit is tesla. Verification / Alternative check: Dimensional analysis and standard references list B's SI unit as tesla. Also, the relationship B = μ * H yields units T = (H/m) * (A/m) = N/(A·m), consistent with tesla's alternative forms. Why Other Options Are Wrong: Magnetomotive force (MMF) is measured in ampere-turns, not B. Weber is the unit of flux Φ, not flux density. Maxwell is a cgs unit of flux, not SI and not B. Ampere per meter is the unit of H (field intensity), not B. Common Pitfalls: Mixing B and H or their units (tesla vs ampere per meter). Confusing flux (Wb) with flux density (T). Final Answer: a tesla

Question 17

In ferromagnetic materials, the observed time-dependent lag of magnetic flux density B relative to magnetizing force H under alternating excitation is known as what phenomenon?

Accepted Answer

Introduction / Context: When ferromagnetic materials are subjected to alternating magnetizing force, the relationship between B and H is not single-valued. Instead, a loop forms in the B–H plane because B lags H. This behavior is central to understanding core losses and dynamic magnetics in transformers, inductors, and motors. Given Data / Assumptions: Alternating magnetizing force H(t). Ferromagnetic core material with domain dynamics. Quasi-periodic steady state. Concept / Approach: The lag between B and H is called hysteresis. The B–H trajectory forms a closed loop over a cycle; the enclosed area equals energy lost per unit volume per cycle (hysteresis loss). Coercivity (the field required to reduce B to zero after saturation) and remanence (residual B when H returns to zero) are parameters describing hysteresis behavior. Step-by-Step Explanation: Apply alternating H → domains rotate and switch with delay.B does not instantaneously follow H → phase and state-dependent lag arises.Plot B vs H over a cycle → a loop appears; its area corresponds to hysteresis energy loss.This phenomenon is termed hysteresis. Verification / Alternative check: Measuring core loss vs frequency shows a component roughly proportional to frequency (hysteresis) and another to frequency squared (eddy current), confirming distinct mechanisms. Why Other Options Are Wrong: Eddy current: Circulating currents causing I^2R losses; not the B–H lag itself. Saturation: The knee where B increases slowly with H; not a lag term. Coercivity: A specific hysteresis parameter, not the entire lag phenomenon. Paramagnetism: Weak, linear magnetization without hysteresis loop. Common Pitfalls: Equating all core losses to hysteresis; eddy currents also contribute significantly. Confusing coercivity or remanence (points on the loop) with the loop phenomenon itself. Final Answer: hysteresis

Question 18

Using the definition of the volt from electromagnetic induction, the time required for one weber (1 Wb) of magnetic flux to cut a conductor and induce one volt (1 V) is equal to what duration?

Accepted Answer

Introduction / Context: The SI definition of the volt ties electrical potential to the rate of change of magnetic flux linkage. One volt equals one weber per second when induced by electromagnetic induction. This question checks recognition of that fundamental unit relationship. Given Data / Assumptions: Flux linkage is changing uniformly. Induced emf due to cutting or linking flux (Faraday’s law). Single-turn equivalence for simplicity. Concept / Approach: Faraday’s law (magnitude): emf = dΦ/dt for a single turn. If emf = 1 V and the total change in flux is 1 Wb, then the time required is t such that 1 V = 1 Wb / t, giving t = 1 s. Rotational “revolutions” do not directly define voltage unless related to flux change per unit time. Step-by-Step Solution: Set emf = 1 V.Let ΔΦ = 1 Wb over interval t.Then 1 = 1 / t → t = 1 s.Therefore, the required time is one second. Verification / Alternative check: In rotating machines, increasing speed reduces the time for a given flux change, increasing induced voltage proportionally—consistent with 1 V = 1 Wb/s as the base relationship. Why Other Options Are Wrong: Revolutions are mechanical counts, not time units; voltage depends on flux change rate, not merely turns. One millisecond or one minute contradicts the 1 Wb per 1 V requirement unless emf differs from 1 V. Common Pitfalls: Confusing webers (flux) with tesla (flux density) and seconds (time). Forgetting the turn count N scales emf; here we reference 1 turn equivalence for clarity. Final Answer: one second

Question 19

An ideal transformer has a 4:1 turns ratio (primary:secondary). If the measured input (primary) power is 5 W under load, what is the output (secondary) power delivered to the load (neglecting losses)?

Accepted Answer

Introduction / Context: For an ideal transformer, power in equals power out (P_in ≈ P_out), aside from small losses. While the turns ratio changes voltage and current levels, it does not create or destroy power. This question checks your understanding of power conservation across a transformer. Given Data / Assumptions: Turns ratio Np:Ns = 4:1. Measured primary (input) power = 5 W. Neglect core and copper losses (ideal device). Concept / Approach: Ideal transformer relations: V2/V1 = Ns/Np and I2/I1 = Np/Ns. Apparent power is conserved: P2 = V2 * I2 = V1 * I1 = P1. Therefore, the secondary power equals the primary power when losses are negligible. Step-by-Step Solution: Given P_primary = 5 W.Assuming ideal behavior: P_secondary = P_primary.Thus, P_secondary = 5 W. Verification / Alternative check: If the secondary voltage is V2 = V1 * (Ns/Np) = V1/4, then I2 = I1 * (Np/Ns) = 4 * I1. Multiplying gives V2 * I2 = (V1/4) * (4 * I1) = V1 * I1 = P1, confirming power equality independent of ratio. Why Other Options Are Wrong: 1.25 W or 20 mW: Incorrect scaling; power is not reduced by the turns ratio in an ideal transformer. 9.5 W: Would imply power gain without an external source—violates energy conservation. 1.25 pW: Physically nonsensical given the stated input power. Common Pitfalls: Mistaking voltage/current scaling for power scaling; turns ratio redistributes voltage and current while conserving power. Ignoring real-world losses; though small, they slightly reduce output below input in practice. Final Answer: 5.0 W

Question 20

Inductance unit check in circuit theory: If an inductor experiences a current change of 1 ampere per second and the induced voltage across it is 1 volt, this corresponds to which standard unit value?

Accepted Answer

Introduction / Context: When studying inductors in basic circuit theory, a key relationship is the voltage induced across an inductor due to a changing current. Identifying the correct unit from the voltage–current rate-of-change relationship helps cement the definition of the henry, the SI unit of inductance. Given Data / Assumptions: Induced voltage across the inductor: v = 1 V. Rate of change of current: di/dt = 1 A/s. Standard inductor behavior with linear inductance and no core saturation. Concept / Approach: The governing formula for an ideal inductor is v = L * di/dt. Rearranging gives L = v / (di/dt). Substituting the provided values immediately yields the inductance in henrys. The henry is defined so that 1 H produces 1 V when current changes at 1 A/s. Step-by-Step Solution: Start with v = L * di/dt.Solve for L: L = v / (di/dt).Insert values: L = 1 V / (1 A/s) = 1 H.Therefore, the condition describes exactly one henry of inductance. Verification / Alternative check: Dimensional analysis confirms consistency: volt = henry * ampere/second → henry = volt / (ampere/second) = voltsecond/ampere, which is the SI definition of 1 H. Why Other Options Are Wrong: a lenz: Not a unit; Lenz’s law states polarity of induced emf. an ohm: Unit of resistance; relates voltage and current directly (V = IR), not di/dt. a farad: Unit of capacitance defined by i = C * dv/dt, not v = L * di/dt. a weber: Unit of magnetic flux; not the unit for inductance. Common Pitfalls: Confusing the inductor formula with the capacitor formula and swapping roles of current and voltage derivatives. Mixing up henry with weber or tesla; those quantify flux and flux density, not inductance itself. Final Answer: a henry

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