Question 1

Two separate coils are available to build a step-up transformer. Coil 1 has 100 more turns than Coil 2. To achieve a step-up action (higher secondary voltage), which coil should serve as the primary?

Accepted Answer

Introduction / Context: Transformer step-up or step-down behavior depends directly on the turns ratio between secondary and primary windings. Choosing which physical coil to use for primary or secondary determines whether the output voltage increases or decreases compared to the input. Given Data / Assumptions: Coil 1 has more turns than Coil 2 (by 100 turns). We desire a step-up transformer: V_secondary > V_primary. Ideal transformer assumptions: negligible losses and leakage. Concept / Approach: The ideal transformer ratio is V_secondary / V_primary = N_secondary / N_primary. To step up voltage, N_secondary must exceed N_primary. Since Coil 1 has more turns, it should be used as the secondary, leaving the fewer-turn coil (Coil 2) as the primary. Step-by-Step Solution: Let N1 = turns of Coil 1, N2 = turns of Coil 2, with N1 = N2 + 100.For step-up, choose N_secondary > N_primary.Assign N_secondary = N1 and N_primary = N2 to satisfy N_secondary > N_primary.Therefore, Coil 2 should be the primary. Verification / Alternative check: Example: if N1 = 600 and N2 = 500, then V_secondary / V_primary = 600/500 = 1.2, a 20% step-up. Reversing roles would give a step-down. Why Other Options Are Wrong: Claiming identical windings are required is incorrect; any ratio other than 1:1 can step voltages. Using Coil 1 as primary produces step-down behavior, not step-up. “Too small to matter” is false; even modest differences change output voltage. Turns count directly affects voltage ratio; it cannot be ignored. Common Pitfalls: Confusing which side should have more turns for step-up. Always place the higher-turns winding on the secondary side to raise voltage. Final Answer: Coil 2 should be the primary.

Question 2

For a long, tightly wound coil (solenoid-like) with magnetic core, how does its inductance vary with basic physical parameters, all else held constant?

Accepted Answer

Introduction / Context: Inductance quantifies a coil’s ability to store magnetic energy for a given current. Designers adjust turns, core materials, and geometry to achieve target inductance values for filters, chokes, and resonant circuits. Knowing the correct proportionalities prevents costly trial-and-error. Given Data / Assumptions: Long, uniformly wound coil with a linear magnetic core (no saturation). Fringing and leakage are neglected for first-order analysis. We compare qualitative dependences (increase/decrease), not exact numbers. Concept / Approach: A widely used approximate relation is L ∝ μ * N^2 * A / l, where μ is permeability, N is turns count, A is cross-sectional area, and l is magnetic path length. Thus, inductance grows with permeability, with area, and with the square of the turns, and it decreases with longer magnetic paths. Step-by-Step Solution: Start from the proportionality: L ∝ μ * N^2 * A / l.Analyze parameter effects: increasing N strongly raises L because of the N^2 factor.Increasing length l lowers L; increasing area A raises L; higher μ raises L.Therefore, the statement “directly proportional to the number of turns squared” is the most accurate among the options. Verification / Alternative check: Doubling turns approximately quadruples inductance if μ, A, and l remain constant. This square-law behavior is well established in inductor design references and verified in practical winding experiments. Why Other Options Are Wrong: Direct proportionality to length is incorrect; L falls as length increases. Inverse proportionality to area and to permeability are both wrong; L increases with A and μ. Independence from core material is false; μ of the core is a dominant factor. Common Pitfalls: Assuming L increases linearly with turns; the correct dependence is N^2 for tightly coupled coils. Final Answer: directly proportional to the number of turns squared

Question 3

Which electromechanical device is best described as a coil that produces linear motion of a movable iron plunger or core when energized?

Accepted Answer

Introduction / Context: Coils can convert electrical energy into mechanical motion in several ways. Identifying the correct device by its construction and motion is important for applications such as valves, latches, and actuators. Given Data / Assumptions: The device has a wound coil. It contains a movable iron core or plunger. When energized, it produces a linear push–pull motion. Concept / Approach: A solenoid consists of a coil and a ferromagnetic plunger that is drawn into the coil when current flows, producing linear motion. Relays also use coils but operate contacts via a pivoted armature; reed switches are magnetically actuated contacts sealed in glass and lack a moving iron plunger as described. An “armature” is a component inside machines and relays, not a complete device. Voice coil actuators are similar in function but are built differently and usually do not use an iron plunger; they employ a coil in a magnetic gap for precise motion. Step-by-Step Solution: Match description: coil + movable iron core = solenoid.Exclude relay: coil actuates contacts via lever/armature, not a plunging core as the main motion.Exclude reed switch: no iron plunger; uses magnetic fields to close reed contacts.Exclude armature: that is a component, not the full device. Verification / Alternative check: Component catalogs list solenoids specifically for linear actuation with plunger strokes and force ratings, confirming the definition. Why Other Options Are Wrong: Reed switch: passive magnetically actuated contacts; no plunger motion. Relay: primarily an electrical switch; movement is of an armature, not a sliding iron core inside the coil as the actuator itself. Armature: not the complete device. Voice coil actuator: typically lacks an iron plunger and uses different geometry. Common Pitfalls: Calling any coil-driven motion a relay. Relays switch contacts; solenoids create linear force and displacement. Final Answer: solenoid

Question 4

In magnetic units, one weber equals how many lines of magnetic flux in the older cgs system, and which term names a “value of ten to the eighth power magnetic lines of force”?

Accepted Answer

Introduction / Context: Magnetic quantities are expressed in SI and in older cgs-based units. Engineers often encounter legacy documentation that mentions “lines of force” or “maxwells.” Converting between systems prevents confusion when comparing specifications or reading vintage texts. Given Data / Assumptions: We are dealing with magnetic flux (not flux density). SI unit of magnetic flux is the weber (Wb). Older cgs unit is the maxwell (Mx), colloquially “lines of force.” Concept / Approach: The exact conversion is 1 weber = 10^8 maxwells. Thus, a “value of ten to the eighth power magnetic lines of force” corresponds to precisely one weber of flux. Remember that flux density is measured in tesla (SI), where 1 tesla = 1 weber per square meter. Step-by-Step Solution: Identify the quantity: total magnetic flux, not field intensity.Recall conversion: 1 Wb = 10^8 Mx.Map the phrase “ten to the eighth power lines” to 1 Wb.Choose the term: weber. Verification / Alternative check: Handbooks and standards list 1 Wb = 10^8 Mx; dimensional analysis confirms flux density (T) relates as T = Wb/m^2, not as a count of lines. Why Other Options Are Wrong: Field and flux are generic words; the specific SI unit asked is weber. Coulomb is an electric charge unit, not magnetic flux. Tesla measures flux density, not total flux. Common Pitfalls: Confusing flux (Wb) with flux density (T). Always distinguish total flux from flux per area. Final Answer: weber (1 Wb = 10^8 maxwells)

Question 5

As electric current flows through a straight conductor, which fundamental electromagnetic effect always accompanies it?

Accepted Answer

Introduction / Context: Electromagnetism links electricity and magnetism: moving charges create magnetic fields, and changing magnetic fields induce electric fields. This fundamental relationship underlies motors, generators, inductors, and countless electromagnetic devices. Given Data / Assumptions: A steady current flows through a long, straight conductor. Space around the conductor is otherwise empty and linear. We are interested in qualitative effects, not exact field strengths. Concept / Approach: A current produces a magnetic field that forms concentric circles around the conductor, with direction given by the right-hand rule: thumb points in the direction of conventional current, fingers curl in the direction of the magnetic field. The magnitude at distance r is proportional to current I and inversely proportional to r (for an ideal long straight conductor). Step-by-Step Solution: State the principle: current-carrying conductors generate surrounding magnetic fields.Describe geometry: field lines are circular and centered on the wire.Determine direction: use the right-hand rule for conventional current.Infer applications: basis for inductance, magnetic forces, and electromagnetic actuation. Verification / Alternative check: Use a small compass near a powered wire: the compass deflects when current flows, indicating a magnetic field. Turning current off removes the deflection. Why Other Options Are Wrong: “Aids the current” is vague and not a defined physical effect. “Wire tries to point north” anthropomorphizes behavior; only a permanent magnet aligns with Earth’s field. An electrostatic field does not necessarily oppose current; static fields arise from charge distributions, not from steady current alone. Heat is produced due to resistance, but magnetic effects are always present with current flow. Common Pitfalls: Confusing heat generation (I^2 * R) with the magnetic field. Both occur simultaneously; the magnetic field is a fundamental outcome of moving charges. Final Answer: a magnetic field is developed around it

Question 6

In a 1 kHz series RL circuit, the resistor R and the inductive reactance XL are both 1000 Ω. What is the magnitude of the total impedance seen by the source?

Accepted Answer

Introduction / Context: This problem tests series RL impedance calculation. In AC circuit analysis, impedance combines resistance and reactance vectorially. Because resistor voltage is in phase with current while inductor voltage leads current by 90 degrees, the total magnitude is found using the Pythagorean relationship, not simple arithmetic addition. Given Data / Assumptions: Frequency f = 1 kHz (value does not affect the result once XL is given). R = 1000 Ω (purely resistive). XL = 1000 Ω (inductive reactance). Series connection of R and L (same current through both). Concept / Approach: Impedance of a series RL network is Z = R + jXL. The magnitude is |Z| = sqrt(R^2 + XL^2). The phase angle is theta = arctan(XL / R), though only the magnitude is requested here. Since R and XL are equal, the triangle is isosceles, and |Z| will be R * sqrt(2). Step-by-Step Solution: Compute magnitude: |Z| = sqrt(R^2 + XL^2)Insert values: |Z| = sqrt(1000^2 + 1000^2)|Z| = sqrt(1,000,000 + 1,000,000) = sqrt(2,000,000)|Z| ≈ 1414 Ω (since 1000 * sqrt(2) ≈ 1414.21) Verification / Alternative check: Because R = XL, the phase angle is arctan(1) = 45 degrees. A 45-degree impedance triangle has hypotenuse R * sqrt(2), confirming the numeric result without a calculator. Why Other Options Are Wrong: 500 Ω and 707 Ω: These correspond to parallel combinations or RMS-like misapplications, not series magnitude. 1000 Ω: Ignores the inductive reactance. 2000 Ω: Simple arithmetic sum, not vector sum, so incorrect. Common Pitfalls: Adding R and XL directly instead of using the square root of the sum of squares. Confusing magnitude with phase; both are important, but only the magnitude is asked. Final Answer: 1414 Ω

Question 7

A single-phase transformer is rated 10 kVA, 1000/500 V, 60 Hz. What is the maximum current on the 1000 V side at full apparent power (neglecting losses)?

Accepted Answer

Introduction / Context: Transformer nameplates specify apparent power (kVA) and rated voltages on each winding. At full load, the maximum current on a given side is determined by I = S / V on that side. This question checks quick conversion from nameplate data to rated current, a routine design and sizing task. Given Data / Assumptions: Rated apparent power S = 10 kVA = 10,000 VA. Primary (source) side voltage considered: 1000 V. Ideal transformer assumptions (ignore losses and power factor for rating current). Concept / Approach: For a single-phase transformer, I_rated (on a given winding) = S_rated / V_rated (of that winding). The kVA rating applies to both sides; voltage differs, so the current differs correspondingly to keep S the same (S = V * I). Step-by-Step Solution: Compute current on 1000 V side: I1 = S / V1I1 = 10,000 VA / 1000 V = 10 ATherefore, the maximum rated current on the 1000 V side is 10 A. Verification / Alternative check: On the 500 V side, rated current would be I2 = 10,000 / 500 = 20 A. The ampere-turns balance through the turns ratio, consistent with ideal transformer behavior. Why Other Options Are Wrong: 2 A, 5 A, 7.5 A: Too small for 10 kVA at 1000 V. 20 A: That is the rated current on the 500 V side, not the 1000 V side. Common Pitfalls: Confusing primary and secondary sides; always divide the kVA by the side voltage in question. Mixing up kW and kVA; rated current is based on kVA regardless of power factor. Final Answer: 10 A

Question 8

In a simple AC generator, as the armature passes the 270° position of rotation within a uniform magnetic field, what is the instantaneous polarity/magnitude of the induced voltage?

Accepted Answer

Introduction / Context: An elementary AC generator produces a sinusoidal voltage as the armature (or coil) rotates in a magnetic field. Understanding how the induced voltage relates to angular position helps visualize phase, polarity, and peak occurrences within a cycle. Given Data / Assumptions: Uniform rotation with angular position measured from the zero-crossing reference where induced voltage follows a sine function. Ideal sinusoidal behavior: v(θ) ∝ sin(θ). Neglect effects like armature reaction and nonuniform fields. Concept / Approach: If v(θ) is sinusoidal, then key angles are: 0° → 0 V, 90° → +Vmax, 180° → 0 V, 270° → −Vmax, and 360° → 0 V (completing the cycle). Thus, at 270°, the induced voltage reaches its most negative peak, indicating both maximum magnitude and negative polarity. Step-by-Step Solution: Model: v(θ) = Vmax * sin(θ).Evaluate at θ = 270°: sin(270°) = sin(3π/2) = −1.Therefore, v(270°) = −Vmax, i.e., the maximum negative value.This corresponds to the coil plane and flux cutting rate aligning for maximum magnitude with negative sign based on the chosen reference. Verification / Alternative check: Plotting one cycle of a sine wave or recalling standard unit circle values confirms the mapping of angles to peaks and zero crossings, validating the 270° result as −Vmax. Why Other Options Are Wrong: Between maximum negative and zero / between zero and maximum positive: These describe intermediate magnitudes, not the peak at 270°. At maximum positive: Occurs at 90°, not 270°. Zero volts: Occurs at 0°, 180°, and 360°. Common Pitfalls: Confusing mechanical position reference; ensure the sine reference starts at 0° = zero crossing. Mixing sine and cosine references; cosine would shift all angles by 90°. Final Answer: at maximum negative

Question 9

On a B–H (magnetization) curve, the magnetizing force H is defined as the ampere-turns per unit length of the magnetic path. Which expression correctly represents H and its SI unit?

Accepted Answer

Introduction / Context: The B–H curve relates magnetic flux density B to magnetizing force H. Correctly defining H is fundamental in magnetic circuit analysis, core design, and understanding hysteresis. H quantifies the magnetizing effort produced by current-carrying turns along a magnetic path of a certain length. Given Data / Assumptions: N turns carrying current I create magnetomotive force (MMF) equal to N * I ampere-turns. Magnetic path length is l (meters). Homogeneous path assumption for definition (local values can vary in complex cores). Concept / Approach: Magnetizing force H is the MMF per unit length of magnetic path: H = MMF / l = (N * I) / l. Its SI unit is A/m (ampere per meter). This field intensity is the cause, while B (flux density) is the effect through the material relationship B = μ * H (with μ = μ0 * μr). Step-by-Step Solution: Compute MMF: MMF = N * I (ampere-turns).Normalize by path length: H = (N * I) / l.Identify SI units: A/m.Relate to B: In free space, B = μ0 * H ⇒ H = B / μ0, consistent with units A/m. Verification / Alternative check: Dimensional analysis: N is dimensionless, I in A, l in m, so H has dimensions A/m, confirming the expression. Electromagnetic textbooks consistently define H using MMF per length. Why Other Options Are Wrong: H = I × N (AT): That is MMF, not H; it lacks division by length. AT per weber: Not the definition of H; mixes unrelated quantities. Wb/m^2 (tesla): That is B, not H. H = B / μ0 (A/m): This is correct only in free space as a derived relation, not the general defining expression requested. Common Pitfalls: Confusing B and H or their units (tesla vs A/m). Forgetting path length in practical cores; air gaps change effective l and H required. Final Answer: H = (N * I) / l, units A/m

Question 10

When an electromagnet (ferromagnetic core with coil) reaches its maximum achievable flux density for the applied magnetizing force, the material is said to be at what condition?

Accepted Answer

Introduction / Context: Saturation is a key concept in magnetic circuits, transformers, and inductors. It limits how much additional flux density B a core can support even as magnetizing force H increases. Designers must recognize saturation to avoid distortion, overheating, and loss of inductance in power electronics and magnetics. Given Data / Assumptions: Ferromagnetic core with coil producing MMF = N * I. B–H curve exhibits a knee beyond which B increases slowly despite large increases in H. Quasi-static operation (ignore dynamic effects like eddy currents for the concept). Concept / Approach: As H increases, domains align and B rises rapidly at first. Near the knee of the B–H curve, most domains are aligned; further increases in H yield minimal increases in B. This condition is called magnetic saturation. Operation deep into saturation causes high magnetizing current and severe nonlinearity. Step-by-Step Explanation: Observe B–H curve: Identify linear region and the knee.At the knee and beyond, incremental permeability drops significantly.Core no longer stores additional magnetic energy efficiently; B approaches a practical maximum.Thus, the electromagnet is said to be at saturation. Verification / Alternative check: Inductors exhibit reduced incremental inductance L = dΦ/di when saturated. Measurement of magnetizing current spikes or audible transformer buzz under heavy load also indicate saturation onset. Why Other Options Are Wrong: In field condition / at field strength: Vague phrasing, not the established term. Inhibited: Not a magnetics term in this context. In remanence: Refers to residual flux when H is removed, not maximum driven state. Common Pitfalls: Confusing saturation with remanence or coercivity, which are hysteresis-related but different. Ignoring temperature; saturation flux density decreases with heat. Final Answer: at saturation

Question 11

Four primary factors determine the inductance of a coil: number of turns, coil length, coil cross-sectional area, and which material property of the core?

Accepted Answer

Introduction / Context: Inductance quantifies a coil's ability to store magnetic energy. For a single-layer solenoid-style inductor, physical dimensions and core material dominate L. Understanding which parameters matter most helps when adjusting inductance for filters, chokes, and transformers. Given Data / Assumptions: Single coil with a magnetic path that largely follows the core. Quasi-static magnetic behavior without strong leakage. Classical approximation for solenoid inductance applies. Concept / Approach: Approximate inductance for a solenoid: L ∝ (μ * N^2 * A) / l, where μ is permeability of the core material, N is the number of turns, A is cross-sectional area, and l is magnetic path length. Thus, in addition to turns, length, and area, the key independent material property is permeability μ = μ0 * μr. Step-by-Step Explanation: Identify geometric factors: N (turns), A (area), l (length).Include material factor: μ, which scales magnetic flux density for a given H.Increasing μ (e.g., adding an iron or ferrite core) increases L dramatically.Hence, the fourth factor is permeability. Verification / Alternative check: Empirical measurements show that swapping an air core (μr ≈ 1) for a ferrite core (μr >> 1) increases inductance by roughly μr times, aligning with the proportional relation above. Why Other Options Are Wrong: Reluctance: Inversely related to μ and geometry; not an independent design parameter of the coil itself. Counter emf: A result of inductance (L * di/dt), not a determining factor. Coefficient of coupling: Pertains to mutual inductance between coils, not self-inductance of a single coil. Wire resistivity: Affects losses, not L directly. Common Pitfalls: Confusing permeability (material property) with reluctance (circuit property). Overlooking air gaps that reduce effective μ and L. Final Answer: permeability

Question 12

To deliberately decrease the inductance of an existing coil (without changing core material), which geometric change would be effective?

Accepted Answer

Introduction / Context: Adjusting a coil's inductance can be done by altering geometry or materials. When core material must remain unchanged, geometric manipulations are the lever. This question checks which change specifically reduces L for a given coil and core material. Given Data / Assumptions: Core material is fixed (permeability unchanged). We can modify turn spacing, area, or number of turns. Quasi-static solenoid approximations apply. Concept / Approach: For a solenoid-like coil, L ∝ (μ * N^2 * A) / l and is also influenced by turn proximity and leakage. Increasing spacing between turns reduces mutual coupling between adjacent turns and can effectively lower the coil's inductance compared to tightly wound turns. Increasing N or A would increase L; increasing μ would also increase L. Step-by-Step Explanation: Hold μ constant; consider geometry effects.Increase turn spacing → reduce effective coupling and increase mean length → lower L.Contrast with other changes: more turns (N↑) or larger area (A↑) both raise L.Therefore, increasing spacing between turns decreases inductance. Verification / Alternative check: Practical RF inductors use spaced turns or basket weave to reduce self-capacitance and inductance; simulations and measurements confirm reduced L versus tightly wound coils with the same N. Why Other Options Are Wrong: Increase turns: L rises approximately with N^2. Increase permeability: L increases in proportion to μ. Increase area: L increases in proportion to A. Remove the core: While this can reduce L, it violates the stated constraint of not changing the core material; the most direct geometric change among the listed options is spacing. Common Pitfalls: Assuming all geometric enlargements increase L; spacing reduces coupling and can lower L. Overlooking that core changes dominate L; here core is held constant by assumption. Final Answer: increase the spacing between turns

Question 13

A 24 V peak sinusoidal source drives a 900 µH inductor at f = 22 kHz. What is the reactive power (inductive VAR) drawn by the inductor?

Accepted Answer

Introduction / Context: Reactive power quantifies the non-dissipative exchange of energy between reactive components and the source. For a pure inductor, average real power is zero, but reactive power Q is nonzero and depends on the RMS voltage and inductive reactance XL. This question requires converting peak to RMS and applying the correct formula for Q. Given Data / Assumptions: Vp = 24 V (sinusoidal peak). f = 22 kHz, L = 900 µH. Ideal inductor (no resistance), sinusoidal steady state. Concept / Approach: Reactive power for a purely inductive load is Q = Vrms^2 / XL, where XL = 2 * π * f * L. Also Vrms = Vp / sqrt(2) for a sine wave. Compute XL, then compute Q. Units are VAR (volt-ampere reactive). Step-by-Step Solution: Vrms = 24 / sqrt(2) ≈ 16.97 V.XL = 2 * π * 22,000 * 900e-6 ≈ 124.4 Ω.Q = Vrms^2 / XL = (16.97^2) / 124.4 ≈ 288.0 / 124.4 ≈ 2.31 VAR.Rounded to two significant figures: Q ≈ 2.3 VAR. Verification / Alternative check: Compute current magnitude: I = Vrms / XL ≈ 16.97 / 124.4 ≈ 0.136 A. Then Q = Vrms * I ≈ 16.97 * 0.136 ≈ 2.31 VAR, consistent with the first method. Why Other Options Are Wrong: 4.6 VAR and 9.3 VAR: Would correspond to either double or quadruple the correct current/voltage relationships. 0 VAR: Only true for zero reactive components; a pure inductor at AC has nonzero Q. 1.1 VAR: Roughly half the correct value, inconsistent with calculations. Common Pitfalls: Using peak voltage directly in Q = V^2 / XL; must use RMS. Confusing real power P with reactive power Q; for an ideal inductor, P = 0 while Q ≠ 0. Final Answer: 2.3 VAR

Question 14

An inductor is in series with a voltage source, a switch, and a resistor. At the instant the switch is first closed (t = 0+), the inductor behaves as which effective element?

Accepted Answer

Introduction / Context: Inductors resist changes in current. When a previously unenergized inductor is suddenly connected in series to a source via a switch, the immediate, transient behavior is governed by di/dt limitations. Recognizing the instant-after behavior is critical in transient analysis and in understanding why inductors protect against inrush currents. Given Data / Assumptions: Initial inductor current iL(0−) = 0 A (switch open for a long time beforehand). Ideal inductor (no series resistance), ideal switch, and DC source. Observation is at t = 0+ (the instant just after closure). Concept / Approach: For an inductor, vL = L * di/dt. Current through an inductor cannot change instantaneously; iL(0+) = iL(0−). With iL initially zero, the inductor initially allows no current (iL remains 0 at t = 0+), acting like an open circuit at that instant. Over time, current rises exponentially toward its steady-state value with time constant τ = L / R, and in steady state a DC inductor behaves like a short (neglecting winding resistance). Step-by-Step Explanation: Before closing: iL = 0 A (de-energized).At t = 0+: iL cannot jump, so still 0 A.Zero current implies the path is effectively open at that moment.As time progresses, iL increases and the inductor's effective impedance falls toward 0 Ω in DC steady state. Verification / Alternative check: Transient solution for RL: i(t) = (V/R) * (1 − e^(−t/τ)). At t = 0, i(0) = 0, confirming no instantaneous current and open-circuit behavior initially. Why Other Options Are Wrong: A short: Describes steady-state DC behavior after a long time, not the instant after switching. A resistor: Inductor impedance is frequency-dependent; at the instant of switching it is effectively infinite for DC current. A voltage or current source: Those are active elements; the inductor is passive and does not act as an ideal source at t = 0+. Common Pitfalls: Assuming inductors instantly pass current like resistors; they do not. Confusing initial and steady-state behaviors of reactive components. Final Answer: an open

Question 15

RL circuits — When an inductance value (in henry) is divided by the circuit’s resistance (in ohms), what engineering quantity do you obtain for a first-order RL network?

Accepted Answer

Introduction: The ratio of inductance to resistance, written as L / R, is a cornerstone parameter for first-order RL circuits. It sets how quickly current builds up or decays when a step voltage is applied or removed, directly determining transient speed in power electronics, filters, and sensor interfaces. Given Data / Assumptions: First-order series RL network with inductance L (henry) and resistance R (ohm). Excited by a step or switched DC voltage source. Linear, time-invariant components; no saturation or temperature drift considered. Concept / Approach: The natural response of an RL circuit follows an exponential with time constant tau = L / R. This constant determines how fast current approaches its final steady value. After approximately 5 * tau, current is essentially settled. The same tau governs decay after de-energizing the inductor. Step-by-Step Solution: Define time constant: tau = L / R (units: second, since henry/ohm reduces to second).Charging (rise) with step input: i(t) = I_final * (1 − e^(−t/tau)).Discharging (decay) after opening the source: i(t) = I_initial * e^(−t/tau).Interpretation: larger L or smaller R → larger tau → slower dynamics. Verification / Alternative check: Dimensional analysis: 1 H = 1 V·s/A and 1 Ω = 1 V/A, so H/Ω = (V·s/A)/(V/A) = s, confirming a time measure. Simulation or oscilloscope traces of RL steps validate that 63.2% of the final current is reached at t = tau. Why Other Options Are Wrong: counter emf value: That is an instantaneous induced voltage, not a constant and not L/R. induced voltage amplitude: Depends on di/dt and turns; not equal to L/R. quality factor: Q involves reactance/resistance at a frequency; not simply L/R in seconds. phase angle at resonance: Resonance pertains to RLC; an RL alone does not resonate. Common Pitfalls: Confusing L/R (seconds) with X_L = 2 * pi * f * L (ohms). Assuming tau gives exact settling time; it only characterizes exponential rate (use multiples of tau for design). Final Answer: rise or decay time constant

Question 16

Relays — Inside an electromechanical relay, which component physically moves when the coil is energized by the electromagnet?

Accepted Answer

Introduction: Electromechanical relays convert electrical control signals into mechanical motion to open or close power circuits. Knowing the internal parts and their roles helps in selecting, wiring, and troubleshooting relays in automation and protection systems. Given Data / Assumptions: A standard electromagnetic relay with coil, armature, return spring, and contacts. Energizing the coil creates a magnetic field. We are identifying the mechanically moving element actuated by the field. Concept / Approach: When the coil is powered, magnetic flux draws a ferromagnetic lever called the armature toward the core. The armature’s motion transfers force to the contact set (moving contacts), changing their state (NO closes, NC opens). The spring returns the armature when the coil is de-energized. Step-by-Step Solution: Coil energized → magnetic field established.Magnetic attraction pulls the armature toward the core/pole piece.Armature movement actuates the contact mechanism.De-energize coil → spring restores original position. Verification / Alternative check: Datasheets and teardown diagrams label the moving ferrous lever as the armature; contacts are stationary and moving blades but are not themselves the driving member. Why Other Options Are Wrong: conductor: Generic; everything contains conductors but they are not the moving actuator. contacts: Their state changes, but they are driven by the armature. solenoid: A solenoid is a type of actuator; in a relay the moving piece is specifically the armature linked to the contacts. commutator: Part of DC machines, not relays. Common Pitfalls: Confusing the coil (electromagnet) with the moving part; the coil is fixed. Assuming contacts are the primary moving component; they follow the armature. Final Answer: armature

Question 17

Inductors and transient behavior — During changes in current, the ability of a conductor/coil to generate a counter-emf within itself is called:

Accepted Answer

Introduction: Inductors resist changes in current by creating an induced voltage that opposes the change. Distinguishing the phenomenon from the resulting voltage is crucial for precise terminology in circuits, power supplies, and signal conditioning. Given Data / Assumptions: Single coil or conductor with changing current. No external secondary coil is required (not mutual induction). Linear operation assumed for clarity. Concept / Approach: According to Faraday’s law, a changing current produces a changing magnetic field, which links the same conductor and induces a voltage opposing the change (Lenz’s law). The ability or property is termed self-induction; the induced opposing voltage is often called back emf. The unit of inductance (henry) quantifies how much voltage appears for a given di/dt but is not the effect itself. Step-by-Step Solution: A time-varying current i(t) in a coil produces magnetic flux Φ(t).Change of flux dΦ/dt links the same turns, inducing v_L = −L * di/dt.This opposition to current change is the phenomenon of self-induction.The observed voltage is called back emf; L is measured in henrys. Verification / Alternative check: Opening a current through a relay coil causes a high voltage spike (back emf) due to self-induction, commonly clamped by a flyback diode—textbook evidence of the phenomenon. Why Other Options Are Wrong: back emf: Names the voltage result, not the property. a henry: Unit of inductance, not the effect. impedance: General AC opposition (includes R and X), not specifically the self-inductive mechanism. mutual coupling: Requires two coils; question is about a single conductor. Common Pitfalls: Using “back emf” and “self-induction” interchangeably; keep property vs. outcome distinct. Assuming only coils exhibit self-induction; any conductor carrying changing current does, though coils increase the effect. Final Answer: self-induction

Question 18

Inductance determinants — Which set of properties primarily controls the inductance value of a coil (assume a uniform core and tightly wound turns)?

Accepted Answer

Introduction: Designing inductors requires understanding which geometric and material factors set the inductance. This knowledge guides core selection and winding specifications for filters, chokes, and transformers. Given Data / Assumptions: Coil wound uniformly on a magnetic path with effective length l and cross-sectional area A. Permeability μ depends on the core material (air, iron, ferrite, etc.). Skin and proximity effects ignored for the inductance value itself. Concept / Approach: A simplified form shows L proportional to μ * N^2 * A / l. Thus inductance increases with higher permeability, more turns, and larger core area, and decreases with longer magnetic path length. Wire type (gauge) affects resistance and current rating, not the magnetostatic inductance directly. Step-by-Step Solution: Identify controlling variables: μ (core material), N (turns), A (core area), l (path length).Relate to L: L ∝ μ * N^2 * A / l.Evaluate distractors: “type of wire” and “length of wire” are not primary determinants of L (except insofar as they alter geometry).Hence, the correct set is μ, N, A, and l. Verification / Alternative check: Inductor design calculators and core datasheets apply exactly these variables, confirming the proportionality and practical design steps for target inductance values. Why Other Options Are Wrong: Options with “type of wire” or “length of wire” emphasize resistance/temperature rise, not inductance. Irrelevant attributes (color, solder type) do not affect L. Common Pitfalls: Confusing DC resistance effects with inductance; both matter but are distinct. Ignoring fringing and air-gap effects in gapped cores that reduce μ and L. Final Answer: Type of core material, number of turns, cross-sectional area of core, and length of core

Question 19

Transformer basics — If a transformer has a turns ratio of 5:1 defined as Ns:Np = 5:1, what does this imply about the relative number of turns?

Accepted Answer

Introduction: Turns ratio definitions must be explicit. The common convention is to state the ratio as Ns:Np (secondary to primary) or V2:V1 for ideal transformers. Clarity prevents design and measurement mistakes in step-up vs. step-down applications. Given Data / Assumptions: Turns ratio is stated as 5:1 and explicitly defined Ns:Np = 5:1. Ideal transformer behavior for conceptual understanding. Same frequency on both windings; negligible losses. Concept / Approach: In an ideal transformer, V2/V1 = Ns/Np and I2/I1 = Np/Ns. Therefore, Ns:Np = 5:1 means five times as many turns on the secondary as on the primary. This configuration is a step-up transformer for voltage (and step-down for current). Step-by-Step Solution: Given Ns:Np = 5:1 → Ns = 5 * Np.Voltage relationship: V2/V1 = Ns/Np = 5 → V2 is 5 times V1 (ideal).Current relationship: I2/I1 = Np/Ns = 1/5 → secondary current is one-fifth of primary current.Thus, more turns on the secondary than on the primary. Verification / Alternative check: Practical measurements with open-circuit secondary show V2 approximately five times V1 for this ratio, confirming the turns relationship. Why Other Options Are Wrong: more primary turns: Opposite of the defined ratio. more secondary current: Inverse is true; current is reduced when turns and voltage are stepped up. source/load impedance statements: Not directly implied by turns ratio alone without context. equal turns: Contradicts the 5:1 statement. Common Pitfalls: Confusing Ns:Np with Np:Ns; always specify orientation. Ignoring that stepping up voltage steps down current for similar power. Final Answer: more secondary turns than primary turns

Question 20

Transformer operation — Which statement correctly describes what a transformer will have or maintain in normal operation?

Accepted Answer

Introduction: A transformer is a static electromagnetic device that transfers energy between circuits via mutual induction. Understanding which characteristics are intrinsic vs. situational prevents confusion in specifications and testing. Given Data / Assumptions: Conventional power or signal transformer with two windings. Same frequency present on both windings under ideal conditions. No commutation or moving parts. Concept / Approach: Every transformer, by definition, has at least two windings (primary and secondary). Currents and power depend on load conditions; frequency is the same on both sides (neglecting tiny phase shifts), since the magnetic coupling does not change the input frequency. Step-by-Step Solution: Identify essential structure: primary winding + secondary winding around a magnetic core (or air core).Operation: applying AC to the primary induces AC in the secondary of the same frequency.Power transfer: primary input power ≈ secondary output power + losses; there is no mechanical conversion.Therefore, the guaranteed characteristic is having primary and secondary windings. Verification / Alternative check: Datasheets always specify turns for both windings; frequency on secondary equals primary’s frequency in steady state. Why Other Options Are Wrong: primary and secondary current: Secondary current can be near zero at no load. primary wattage producing secondary current: Wording is misleading; power transfer depends on load and losses. different frequencies: Incorrect; frequency is preserved. commutator: Transformers have no rotating parts. Common Pitfalls: Assuming secondary current exists without load; only magnetizing current flows then. Thinking a transformer can change frequency; it cannot. Final Answer: primary and secondary windings

Inductors Questions