Question 1

Induced voltage from current slope: A 120 mH coil carries a current changing at 150 mA/s. What magnitude of voltage is induced across the coil?

Accepted Answer

Introduction / Context: Inductor voltage magnitude is proportional to the rate of change of current. Even relatively small slopes (in A/s) across moderate inductances generate measurable millivolt levels, which matter in precision analog design and sensor front-ends. Given Data / Assumptions: L = 120 mH = 0.12 H. di/dt = 150 mA/s = 0.15 A/s. We seek |v| = L * |di/dt|; polarity depends on Lenz's law and is not requested. Concept / Approach: Use v = L * di/dt with consistent SI units. Multiply henries by amperes per second to get volts directly. Step-by-Step Solution: v = 0.12 * 0.15 = 0.018 V.0.018 V = 18 mV. Verification / Alternative check: Order-of-magnitude: 0.1 H * 0.1 A/s = 0.01 V (10 mV). Our numbers are slightly larger, giving 18 mV, which is reasonable. Why Other Options Are Wrong: 180 mV: Off by a factor of 10. 1.8 mV and 1.25 mV: Too small; likely from arithmetic or prefix errors. Common Pitfalls: Mistaking mA/s for A/s or mH for H, causing 10× or 100× errors. Final Answer: 18 mV

Question 2

Unit comparison: An inductance of 0.08 µH is larger than which of the following quantities?

Accepted Answer

Introduction / Context: Comparing component values expressed with different SI prefixes can be tricky. This question reinforces careful conversion among henry (H), millihenry (mH), and microhenry (µH) to decide which figure is strictly smaller than 0.08 µH so that 0.08 µH is larger than it. Given Data / Assumptions: Reference value: 0.08 µH. Options provided in H and mH require conversion to a common base. We compare magnitudes only. Concept / Approach: Convert 0.08 µH to henries: 0.08 µH = 0.08 × 10^-6 H = 8 × 10^-8 H. Identify which option is less than 8 × 10^-8 H; that one is the value for which 0.08 µH is larger. Step-by-Step Solution: Target: 0.08 µH = 8e-8 H.Option checks: 0.0000008 H = 8e-7 H (larger; target is smaller, not larger than this).0.000000008 H = 8e-9 H (smaller; target 8e-8 H is larger).0.000008 H = 8e-6 H (much larger; target is smaller).0.00008 mH = 8e-8 H (equal; not larger). Verification / Alternative check: Convert all to µH: 0.000000008 H = 0.008 µH; indeed 0.08 µH > 0.008 µH by a factor of 10, confirming the selection. Why Other Options Are Wrong: 0.0000008 H and 0.000008 H: Both exceed 0.08 µH, so 0.08 µH is not larger than them. 0.00008 mH: Exactly equal to 0.08 µH; not “larger than.” Common Pitfalls: Confusing 10^-6 and 10^-9 prefixes or misplacing decimals when converting. Final Answer: 0.000000008 H

Question 3

Ohmmeter test on an inductor: An ohmmeter is placed across a coil and the pointer indicates approximately 0 Ω. What condition does this imply about the inductor?

Accepted Answer

Introduction / Context: Basic troubleshooting of inductors and transformer windings often starts with a simple resistance test. While inductors have low DC resistance, they should not read a true short (zero ohms). Interpreting ohmmeter readings correctly helps rapidly identify faults without specialized equipment. Given Data / Assumptions: Ohmmeter across a coil indicates ~0 Ω. We consider a typical small to medium coil with finite winding resistance. No other parallel paths are connected during the test. Concept / Approach: A healthy coil exhibits a small but non-zero DC resistance due to copper wire length. A reading of near 0 Ω indicates the winding is shorted (e.g., turns shorted together or terminals shorted), bypassing the normal length of wire and reducing measured resistance to nearly zero. Step-by-Step Solution: Interpretation: R_measured ≈ 0 Ω → electrical short across terminals.Healthy expectation: some ohms (or fractions) depending on gauge/turns; not 0 Ω.Therefore, the condition is “shorted.” Verification / Alternative check: Compare with “open” failure: an open winding reads infinite resistance. A “partly shorted” winding typically shows lower-than-normal but not zero resistance; performance under AC will also change (higher losses, overheating). Why Other Options Are Wrong: Open: Would read very high/infinite resistance, not zero. Good: A good coil should not read exactly zero; some resistance is expected. Partly shorted: Would not usually be exactly 0 Ω; meter reads small but non-zero. Common Pitfalls: Interpreting “very low” as “healthy”; check datasheet expected DCR values. Final Answer: shorted

Question 4

Transformer basics: What is true of a fixed transformer with respect to its turns ratio and adjustability?

Accepted Answer

Introduction / Context: Power supplies and signal interfaces frequently use transformers. Understanding the distinction between fixed and adjustable (tapped/variable) transformers ensures correct selection for voltage conversion and isolation tasks. Given Data / Assumptions: “Fixed transformer” means no user-adjustable taps or movable core/brush arrangements. Windings are predetermined at manufacture. Turns ratio sets the nominal voltage conversion. Concept / Approach: The output-to-input voltage ratio is determined by the turns ratio Ns/Np. In a fixed transformer, this ratio is set by the design and cannot be changed in service. Adjustable devices use multiple taps or autotransformer sliders; fixed ones do not. A center tap or multiple windings may or may not be present, but those features alone do not imply adjustability. Step-by-Step Solution: Define fixed unit: no user-selectable taps; constant turns ratio.Infer property: output voltage scales with a non-adjustable Ns/Np.Select the statement that emphasizes non-changeable turns ratio. Verification / Alternative check: Datasheets label transformers as “fixed” with a single secondary rating. Adjustable or “tapped” units explicitly list multiple ratios/terminals. Why Other Options Are Wrong: Center-tapped secondary / multiple windings / multiple-tapped secondary: These can exist, but do not define “fixed” unless taps are brought out for selection. The defining feature is the unchangeable ratio. Common Pitfalls: Equating “center-tapped” with “adjustable.” A center tap provides two equal halves but does not give arbitrary ratio control. Final Answer: turns ratios that cannot be changed

Question 5

Electromagnetic generation principle: Under which condition can a conductor (wire) generate an electromotive force (and hence electricity)?

Accepted Answer

Introduction / Context: Electromagnetic induction underlies generators, alternators, and many sensors. Recognizing the physical action that produces an induced voltage is essential for understanding energy conversion from mechanical motion to electrical energy. Given Data / Assumptions: A conductor is present in a magnetic field. We seek the condition that creates an induced electromotive force (emf). Quasi-static fields; standard Faraday/Lorentz laws apply. Concept / Approach: When a conductor cuts magnetic flux lines (i.e., moves through a magnetic field or the field varies with time around a stationary conductor), the magnetic flux linkage changes. By Faraday’s law, a changing flux linkage induces an emf: e = −dΦ/dt. Motion through a flux field is a classic case—used in rotating machines where coils sweep through the stator field to generate voltage. Step-by-Step Solution: Identify the requirement: change of flux linkage with the conductor.Produce that change: move the wire through a magnetic field (or vary the field around it).Conclude: a wire passing through a flux field (with relative motion or field change) generates emf. Verification / Alternative check: Right-hand rule or Lorentz force q(v × B) shows charge carriers experience a force when moving in a magnetic field, producing a potential difference across the conductor. Why Other Options Are Wrong: Carrying current: That is an effect of electricity, not a cause of generation. Wrapped as a coil: A coil increases induced voltage but still requires changing flux linkage. Neutral domains: Magnetic domains in a material do not by themselves generate electricity. Common Pitfalls: Assuming a coil alone generates electricity; without changing flux, no emf is induced. Final Answer: passing through a flux field

Question 6

Materials property: What is the (absolute) magnetic permeability of machine steel, expressed in henry per meter (H/m), from the typical reference values below?

Accepted Answer

Introduction / Context: Magnetic permeability μ characterizes how a material responds to magnetic fields and is central to inductor/transformer core design. Machine steel is a common ferromagnetic material with a permeability significantly higher than free space μ0 (≈ 1.256 × 10^−6 H/m) but much lower than specialized high-permeability alloys. Given Data / Assumptions: Choices represent absolute permeabilities in H/m. Typical machine steel has moderate relative permeability μr on the order of a few hundred. Operating in the linear region (unsaturated). Concept / Approach: Absolute permeability μ = μ0 * μr. If μr is a few hundred, μ is on the order of 10^−4 to 10^−3 H/m. Among the options, 5.65 × 10^−4 H/m aligns well with a μr around 450, consistent with “machine steel” rather than specialized silicon steel or high-μ alloys. Step-by-Step Solution: Estimate μr for machine steel ≈ 400–500 under modest flux.Compute order: μ ≈ μ0 * μr ≈ 1.256 × 10^−6 * 450 ≈ 5.65 × 10^−4 H/m.Select the closest tabulated value: 5.65 × 10^−4 H/m. Verification / Alternative check: Comparing with reference tables for common steels shows values in this range when unsaturated; much larger values would imply specialized cores (e.g., permalloy) rather than machine steel. Why Other Options Are Wrong: 1.1 × 10^−4: Too low (μr ≈ 87), not typical for machine steel. 6.9 × 10^−3 or 8.8 × 10^−3: Too high; such μ implies specialty alloys or measurement at very low flux densities. Common Pitfalls: Confusing absolute and relative permeability or assuming a constant μ regardless of flux density; real materials are nonlinear and frequency dependent. Final Answer: 5.65 × 10^−4

Question 7

RL decay timing: For a series RL circuit with L = 20 H and R = 10 Ω, what is the approximate time required for a 0.4 A inductor current to decay essentially to zero after a switch opens?

Accepted Answer

Introduction / Context: Inductor current decays exponentially after the driving source is removed. Engineers often estimate a “practical zero” using the time constant concept to plan protection networks (snubbers), decide measurement windows, or ensure safe discharge intervals in power circuits. Given Data / Assumptions: Series RL path with L = 20 H, R = 10 Ω. Initial current I0 = 0.4 A (any nonzero value decays with the same time constant). Looking for “complete” or “practical” decay (≈ 1% or less of I0). Concept / Approach: The RL time constant is τ = L / R. Current decays as i(t) = I0 * e^(−t/τ). A common practical rule: after 5τ, i(t) ≈ 0.0067 * I0 (less than 1%), often treated as “essentially zero.” Step-by-Step Solution: Compute τ: τ = L / R = 20 H / 10 Ω = 2 s.Estimate complete decay time: t ≈ 5 * τ = 5 * 2 s = 10 s.Therefore, the current falls to near zero in about 10 seconds. Verification / Alternative check: Exact check: at t = 10 s, i(10) = 0.4 * e^(−10/2) = 0.4 * e^(−5) ≈ 0.4 * 0.0067 ≈ 2.7 mA, effectively zero for most purposes. Why Other Options Are Wrong: 2 s (1τ): Leaves about 36.8% of I0; far from zero. 4 s (2τ): Leaves about 13.5% of I0. 8 s (4τ): Leaves about 1.8% of I0; closer, but the standard rule of thumb uses 5τ. Common Pitfalls: Confusing τ with full decay time; τ is just the characteristic time, not the endpoint. Also, forgetting that the initial current magnitude does not change τ. Final Answer: 10 seconds

Question 8

Definition of electromagnetic induction: Electromagnetic induction is most accurately described as the generation of which quantity from which cause?

Accepted Answer

Introduction / Context: Electromagnetic induction is the backbone of electric power generation and many sensing technologies. Understanding the definition clarifies why rotating machinery, moving magnets, or time-varying fields can produce usable electrical power. Given Data / Assumptions: We focus on the cause–effect relationship identified by Faraday’s law. “Electricity” here means induced voltage/current in a conductor. Magnetism refers to magnetic fields or changing magnetic flux. Concept / Approach: Faraday’s law states that a changing magnetic flux through a loop induces an electromotive force around the loop. Thus, motion of a conductor within a magnetic field or variation of the field over time generates an electrical response. This is succinctly “electricity from magnetism.” Step-by-Step Solution: Recognize that induced emf is proportional to −dΦ/dt.Create dΦ/dt by moving conductors or varying fields.Conclude that the phenomenon produces electrical energy/signals originating from magnetic effects. Verification / Alternative check: In generators, a coil rotates in a static magnetic field, changing flux → induced sinusoidal emf. In transformers, a time-varying primary current creates a changing flux in the core → induced voltage in the secondary (no mechanical motion required). Why Other Options Are Wrong: Electricity from electricity: That is conduction or amplification, not induction. Magnetism from electricity: That is the inverse effect (Ampère's law), not induction. Magnetism from magnetism: Tautological and not a generation mechanism. Common Pitfalls: Confusing electromagnetic induction with electromagnetic radiation or with the magnetic field produced by a current (the latter is the opposite direction of causality). Final Answer: electricity from magnetism

Question 9

Magnetic field due to current: Which statement correctly describes the magnetic field polarity behavior around a conductor for DC versus AC current?

Accepted Answer

Introduction / Context: Magnetic fields arise from electric currents. Their polarity and time variation depend on whether the current is direct (DC) or alternating (AC). This concept is essential in understanding inductors, transformers, and electromagnetic interference (EMI). Given Data / Assumptions: Long straight conductor reference. Right-hand rule defines field direction for conventional current. Current may be constant (DC) or time-varying (AC). Concept / Approach: For DC, the current direction is constant, so the magnetic field has a fixed polarity and constant magnitude (neglecting edge effects). For AC, current reverses direction periodically, so the field polarity reverses as well and its magnitude varies sinusoidally (or with the waveform shape). Therefore only DC yields a fixed polarity field. Step-by-Step Solution: Apply right-hand rule for DC: thumb = current direction; fingers wrap to show fixed field circulation.Consider AC: current changes sign → field polarity flips every half-cycle.Therefore select the DC fixed polarity statement as correct. Verification / Alternative check: Observation with a compass near a DC-carrying wire shows a steady deflection; with AC, the compass needle vibrates or averages to near zero due to rapid reversals. Why Other Options Are Wrong: AC causes fixed polarity: False; polarity alternates. DC causes alternating field: False; polarity is fixed for unidirectional current. Combined statement: Reverses the true behaviors for AC and DC. Common Pitfalls: Equating “fixed polarity” with “constant magnitude.” In DC, both are constant; in AC, both polarity and magnitude vary with time. Final Answer: A DC current causes a fixed polarity magnetic field.

Question 10

Wave-shaping with RL networks: With a square-wave input applied to a series RL differentiator and the output taken across the inductor, what output waveform shape is observed?

Accepted Answer

Introduction / Context: First-order RL and RC networks are used for simple differentiation and integration of waveforms. Knowing where the output is taken (across R or across L) determines whether you see sharp spikes, exponential edges, or low-pass behavior. This is crucial in pulse circuits, timing, and EMI considerations. Given Data / Assumptions: Series RL circuit driven by an ideal square wave. Output measured across the inductor L. Differentiator condition: time constant small compared to the square-wave period. Concept / Approach: Voltage across an inductor is v_L = L * di/dt. When the input steps at each square-wave transition, current attempts to change from one steady value to another exponentially, so di/dt is largest at the transition and then decays toward zero. Consequently, v_L exhibits pulses that start at a high value and decay exponentially back to zero each half-cycle—an exponential-shaped pulse at each edge. Step-by-Step Solution: At a rising edge: input steps upward → current begins rising → v_L initially large then decays exponentially.Between edges: current nearly steady → di/dt ≈ 0 → v_L ≈ 0.At a falling edge: input steps downward → current begins decreasing → v_L reverses polarity with exponential decay back to zero. Verification / Alternative check: Time constant τ = L/R. If τ ≪ half-period, pulses are narrow compared to the flat portions. Oscilloscope traces clearly show exponentially decaying lobes of alternating polarity. Why Other Options Are Wrong: dc: Only after a long time in a pure inductorless circuit; here the inductor voltage averages to zero between edges. Short time-duration pulses: True for output across R (current spikes), but across L the shape is explicitly exponential. Only a phase shift: RL differentiators produce transient wave-shapes, not mere sinusoidal phase shifts. Common Pitfalls: Confusing where the output is taken; swapping R and L changes the observed waveform. Ensure the differentiator condition (small τ) to get distinct pulses. Final Answer: exponential-shaped pulses

Question 11

Transformer operation at no-load: When no load is connected to the secondary winding (open-circuit secondary), what happens to the primary current and the secondary voltage of an ideal power transformer?

Accepted Answer

Introduction / Context: A core concept in transformer theory is understanding behavior at no-load, i.e., when the secondary is open-circuited. Technicians frequently measure voltages and currents under this condition to verify turns ratio, excitation current, and core losses. Recognizing what stays the same and what changes prevents misdiagnosis of “faults” that are actually normal behaviors. Given Data / Assumptions: The secondary winding is open (no external load is connected). The transformer is considered ideal for first-order reasoning (leakage drops and copper loss neglected). Rated primary voltage is applied at the primary terminals. Concept / Approach: With an open secondary, there is no load current in the secondary. However, an ideal transformer still requires a small primary current, called magnetizing current, to establish alternating flux in the core and to supply core (hysteresis and eddy) losses. Because the secondary is open, essentially no I*Z drops occur in the windings, so the secondary terminal voltage is very close to the ideal turns-ratio value (V2 ≈ V1 * N2/N1). Step-by-Step Solution: Secondary open → I_secondary ≈ 0 A.Primary still draws magnetizing current I_m to create core flux and supply core losses.Copper drops are negligible at no-load, so V_secondary ≈ induced emf per turn × N2 → near rated turns-ratio value.Therefore: primary current is small but non-zero; secondary voltage remains approximately nominal. Verification / Alternative check: Practical measurement with a voltmeter on an unloaded secondary shows close to nameplate secondary voltage. A clamp meter on the primary shows a small current (typically a few percent of full-load current) confirming non-zero magnetizing draw. Why Other Options Are Wrong: Primary current becomes zero and the secondary voltage becomes zero: incorrect; magnetizing current never vanishes and secondary voltage does not collapse at no-load.Primary current becomes inversely proportional to primary impedance and the secondary voltage collapses: this confuses general Ohm’s law with transformer excitation behavior; the secondary does not collapse at no-load.Secondary voltage proportional to primary impedance: secondary voltage is set primarily by turns ratio, not the primary impedance. Common Pitfalls: Assuming “no load” means “no current at all.” In reality, excitation current flows even with an open secondary. Also, small regulation effects can make the no-load secondary voltage slightly higher than the full-load value due to the absence of copper drops. Final Answer: Primary current is small but non-zero (magnetizing current) and the secondary voltage remains approximately the rated value set by the turns ratio

Question 12

Lenz’s law in action: When the current in a coil is interrupted and its magnetic field collapses, what is the direction of the induced current relative to the original current?

Accepted Answer

Introduction / Context: Lenz’s law governs the polarity of induced emf in inductors and transformers. Understanding the direction of induced current during switching events is vital for selecting snubbers, flyback diodes, and protection components in power electronics and relay/solenoid driving circuits. Given Data / Assumptions: A coil (inductor) carries current that is suddenly reduced or interrupted. The magnetic field stored around the inductor collapses. No specific circuit is given, but the polarity/direction question is general. Concept / Approach: Lenz’s law states that the induced emf always acts to oppose the change in current that produced it. When the source current is removed, the inductor “tries” to keep current flowing in the same direction. Hence the induced current will continue in the original direction for as long as energy remains in the magnetic field, frequently producing a large voltage spike to maintain that current through any available path. Step-by-Step Solution: Initial condition: coil carries current I in a given direction.Source is opened → di/dt is negative (current is attempting to drop quickly).Induced emf polarity is such that it sustains current in the same direction to oppose the decrease.Therefore, induced current direction is the same as the original current direction. Verification / Alternative check: Observe diode-flyback behavior: with a diode placed across a coil (cathode to positive side), the current freewheels in the same direction through the diode path when the driving switch opens—experimental confirmation of Lenz’s law. Why Other Options Are Wrong: Go to zero: current decays but does not instantly drop to zero; the inductor resists instantaneous change.Aid counter emf buildup (without direction): incomplete; the crucial point is the same-direction current to oppose the change.Opposite direction: would accelerate the change rather than oppose it, violating Lenz’s law. Common Pitfalls: Confusing steady-state current direction with induced current polarity during transient events. Remember, inductors resist change in current, not current itself. Final Answer: be in the same direction

Question 13

Inductor turn-off behavior: When an inductor’s magnetic field collapses after the driving source is removed, what immediate electrical effect is observed at the inductor terminals?

Accepted Answer

Introduction / Context: When current through an inductor is interrupted, the stored magnetic energy must dissipate. The mechanism by which energy leaves the inductor involves a transient voltage often called flyback or counter-emf. Recognizing this behavior is crucial for protecting switches and semiconductor devices from overvoltage stress. Given Data / Assumptions: An inductor has stored energy E = 0.5 * L * I^2. The driving source is removed or the circuit is opened. We observe terminal behavior immediately after turn-off. Concept / Approach: Inductors oppose changes in current. If current attempts to drop rapidly to zero, the inductor develops whatever voltage is necessary (within circuit limits) to keep current flowing briefly. This produces a high counter-emf spike whose polarity sustains the former current direction, potentially arcing across switches or overstressing transistors unless a clamp (diode, snubber, TVS) is provided. Step-by-Step Solution: Stored energy: E = 0.5 * L * I^2 resides in the magnetic field.At turn-off, di/dt is large and negative.Inductor voltage: v_L = L * di/dt; a large magnitude results when di/dt is large.Hence a high-voltage spike (counter-emf) appears across the inductor terminals. Verification / Alternative check: Oscilloscope measurements of relay coils or buck converter inductors during switch turn-off show sharp voltage spikes. Adding a flyback diode limits the spike near one diode drop, confirming the underlying behavior. Why Other Options Are Wrong: Current doubles instantaneously: inductors prevent instantaneous current change; it cannot double.A new larger field expands: the field collapses; it does not expand.No voltage because source is gone: induced voltage is produced by changing flux, not by the source after turn-off. Common Pitfalls: Underestimating the magnitude of the spike and omitting protective components. Always consider clamp paths to control v_L during turn-off. Final Answer: A large voltage is induced (a high counter-emf spike appears) across the inductor

Question 14

Transformer fundamentals: A voltage is induced in the secondary winding primarily due to which physical action?

Accepted Answer

Introduction / Context: Transformers transfer energy via magnetic coupling. Understanding what actually induces voltage in the secondary clarifies how turns ratio, flux, and core behavior relate and helps avoid common misconceptions about “voltage being transferred” directly through wires. Given Data / Assumptions: Ideal transformer model for explanation. Alternating voltage is applied to the primary winding. Core provides high mutual coupling so most primary flux links the secondary. Concept / Approach: Applying AC to the primary creates a time-varying flux in the core. According to Faraday’s law, any coil linked by changing flux experiences an induced emf proportional to the rate of change of flux and the number of turns. Thus, the secondary voltage is induced by the primary-generated mutual flux that cuts the secondary turns. The turns ratio determines the magnitude relationship V2/V1 ≈ N2/N1 but is not itself the mechanism; the mechanism is changing flux linkage. Step-by-Step Solution: Primary AC current → time-varying core flux φ(t).Secondary turns are linked by φ(t) → e2 = N2 * dφ/dt.Therefore, V_secondary arises from mutual flux set by the primary.Magnitude relationship is governed by the turns ratio, but induction is due to dφ/dt. Verification / Alternative check: Open-circuit test: with the secondary open, a voltmeter still measures near-rated voltage on the secondary—evidence that induction is caused by primary-driven flux, not by load current. Why Other Options Are Wrong: Secondary magnetic field acting alone: without primary excitation, there is no changing flux to induce voltage.Primary turns ratio as a cause: the ratio sets the proportionality but does not cause induction.Secondary counter-emf independent of the primary: secondary emf is induced by the primary-created flux. Common Pitfalls: Equating turns ratio to mechanism; always attribute induction to changing flux as per Faraday’s law. Final Answer: the alternating magnetic field produced by the primary that links the secondary (mutual flux in the core)

Question 15

Frequency dependence of inductors: How does inductive reactance behave as the excitation frequency changes in a purely inductive circuit?

Accepted Answer

Introduction / Context: Inductive reactance determines how strongly an inductor resists AC. Designers use this dependence to create filters, chokes, and impedance-matching networks that behave differently over frequency. Recognizing the trend with frequency is fundamental to AC analysis. Given Data / Assumptions: Ideal inductor with inductance L. Sinusoidal steady-state excitation at frequency f. No significant parasitics considered. Concept / Approach: Inductive reactance is defined by XL = 2 * pi * f * L. Thus, for fixed inductance L, XL grows linearly with frequency. Higher frequency means larger reactance, which reduces current for a given applied AC voltage (I = V / XL in a purely inductive branch). Step-by-Step Solution: Formula: XL = 2 * pi * f * L.Holding L constant, compare f1 and f2: if f2 > f1 then XL2 > XL1.Therefore, as frequency increases, XL increases.Conversely, lowering frequency reduces XL proportionally. Verification / Alternative check: Plot XL versus f; the graph is a straight line through the origin with slope 2 * pi * L, confirming linear dependence. Why Other Options Are Wrong: Decrease with frequency: opposite of the formula.Independent from frequency: contradicts XL expression.Depend on XC: capacitive reactance is unrelated to the definition of XL. Common Pitfalls: Mixing up XL and XC. Remember XC = 1 / (2 * pi * f * C) decreases with frequency, whereas XL increases. Final Answer: increase with frequency

Question 16

Quality factor to reactance: If a coil has a quality factor Q = 60 and a winding resistance of 5 Ω at a given frequency, what is its inductive reactance at that frequency?

Accepted Answer

Introduction / Context: The quality factor Q of an inductor quantifies how reactive energy storage compares to resistive loss at a particular frequency. It is frequently used in RF design, filters, and resonant circuits to predict selectivity and loss. Converting Q and winding resistance to reactance is a common quick calculation. Given Data / Assumptions: Coil winding resistance R = 5 Ω (series model). Quality factor Q = XL / R = 60 at the operating frequency. We seek the inductive reactance XL at that frequency. Concept / Approach: For a series R–L model at a specific frequency, Q = XL / R. Rearranging gives XL = Q * R. This relation holds at the specified frequency because Q is frequency-dependent via XL = 2 * pi * f * L, but R is typically considered the effective series resistance at that same frequency. Step-by-Step Solution: Use Q definition: Q = XL / R.Solve for XL: XL = Q * R.Substitute values: XL = 60 * 5 Ω = 300 Ω.Therefore, the inductive reactance is 300 Ω. Verification / Alternative check: If L were known, XL = 2 * pi * f * L could be equated to 300 Ω to back-calculate the operating frequency or inductance—consistent with the same Q value. Why Other Options Are Wrong: 0.083 Ω: reciprocal-style error; not from Q * R.12 Ω or 30 Ω: result from incorrect multiplication or mixing R and XL. Common Pitfalls: Confusing series Q with parallel Q or using DC resistance instead of effective AC resistance. Always use values specified for the operating frequency. Final Answer: 300 Ω

Question 17

Time-constant control in RL networks: Which action will decrease (reduce) the time constant τ of a simple series RL circuit, where τ = L / R?

Accepted Answer

Introduction / Context: The time constant τ characterizes how fast current and voltage waveforms rise or decay in first-order circuits. For the series RL network, τ = L / R. Designers often need to speed up or slow down responses—for instance, reducing relay kick times or shaping edges in pulse networks—by adjusting L or R appropriately. Given Data / Assumptions: Simple single-loop series RL circuit. Time constant τ = L / R (L in henries, R in ohms). Objective: reduce τ (achieve a faster response). Concept / Approach: Because τ = L / R, there are two direct levers: decrease L or increase R. Increasing R reduces τ (faster decay and rise to steady state), while increasing L increases τ (slower response). Changing input amplitude does not affect τ, and swapping component positions in a single loop has no effect on τ. Step-by-Step Solution: Formula: τ = L / R.To reduce τ, either lower L or raise R.Option selected: add series resistance → R_total increases → τ decreases.Therefore, the circuit responds faster (shorter transient time). Verification / Alternative check: Simulate with R = 100 Ω, L = 10 mH → τ = 0.01 / 100 = 0.0001 s. Add 100 Ω in series → R = 200 Ω → τ = 0.01 / 200 = 0.00005 s, exactly half, confirming the effect. Why Other Options Are Wrong: Adding an inductor in series: increases L, increasing τ (slower response).Decreasing input amplitude: scales voltages/currents but does not change τ.Exchanging component positions: in a series loop, ordering does not change τ. Common Pitfalls: Trying to reduce τ by putting a resistor in parallel with the series resistor; that actually lowers equivalent R and increases τ. Always check whether the modification raises or lowers the series resistance seen by the inductor. Final Answer: Adding an extra resistor in series to increase the total resistance

Question 18

Protection devices: A magnetic (electromagnetic) circuit breaker trips and opens the circuit to protect equipment under which electrical condition?

Accepted Answer

Introduction / Context: Magnetic circuit breakers and overload relays protect wiring and equipment from damage due to excessive current. Knowing the parameter that causes tripping helps technicians select the correct protective device and diagnose nuisance trips versus genuine faults. Given Data / Assumptions: A magnetic (electromagnetic) breaker uses a solenoid or magnetic actuator. Trip action is determined by the magnitude of current. Thermal or electronic elements may supplement magnetic action, but the trigger remains overcurrent. Concept / Approach: Magnetic trip elements respond nearly instantaneously to high overcurrent (short-circuit) conditions because magnetic force is proportional to current. Thermal elements respond to overloads over time using bimetal heating. In either case, the triggering quantity is current, not voltage, and certainly not a normal open or proper operation condition. Step-by-Step Solution: Fault causes current to exceed breaker rating (e.g., short to ground).Magnetic field in trip solenoid increases rapidly with current.Actuator trips the mechanism and opens contacts.Circuit is interrupted, protecting downstream equipment and conductors. Verification / Alternative check: Breaker curves (time–current characteristics) show trip times primarily as a function of current magnitude, confirming the overcurrent basis. Why Other Options Are Wrong: Voltage exceeded: overvoltage protection is typically provided by surge protectors or MOVs, not magnetic breakers.An open in equipment: an open reduces current, not increases it.Operating properly: no protective device should trip in normal conditions. Common Pitfalls: Confusing voltage surges with overcurrent events. While surges can cause high currents, the breaker trips on the current, not directly on the voltage magnitude. Final Answer: current is exceeded

Question 19

Changing flux through a coil: When a permanent magnet enters a coil and a voltage is induced across the coil terminals, the phenomenon is referred to as what?

Accepted Answer

Introduction / Context: Whenever magnetic flux linking a coil changes with time, an emf is induced across the coil according to Faraday’s law. This is the basic principle behind generators, pickups, and many sensors. A magnet moving into a coil is a classic illustration used in physics and electrical engineering courses. Given Data / Assumptions: A permanent magnet is moved into a stationary coil. The magnetic flux linking the coil’s turns changes with time. We observe an induced voltage across the coil terminals. Concept / Approach: The induced emf magnitude is proportional to the rate of change of flux linkage: e = − N * dΦ/dt, where N is the number of turns and Φ is the flux per turn. The phrase “flux linkage” emphasizes that it is the linkage (N * Φ) changing in time that causes the emf. While terms like “counter emf” may apply to coils opposing changes in their own current, the more precise description for a moving magnet scenario is changing flux linkage producing electromagnetic induction. Step-by-Step Solution: Move magnet into coil → Φ through turns increases with time.Flux linkage λ = N * Φ therefore changes with time (dλ/dt ≠ 0).By Faraday’s law, an emf is induced: e = − dλ/dt.Hence, the observed voltage is due to changing flux linkage. Verification / Alternative check: Reverse the direction of motion or magnet polarity and observe the measured polarity reversal of the induced voltage—consistent with the sign in Faraday’s law and Lenz’s law. Why Other Options Are Wrong: An alternation: a vague term; not the specific phenomenon.Self-inductance: pertains to a coil’s own current change, not a magnet moving nearby.Counter emf: general term for emf opposing current change; less precise here than flux linkage induced by external motion. Common Pitfalls: Equating all induced voltages with “back emf.” Reserve “self-inductance” and “counter emf” for current-change scenarios within the same winding; use “changing flux linkage” for external magnetic changes. Final Answer: flux linkage (change in flux linkage causing electromagnetic induction)

Question 20

Definition check: In basic electricity and magnetism, what is meant by electromagnetism?

Accepted Answer

Introduction / Context: Electromagnetism unifies electric and magnetic phenomena. For practical technicians, the key takeaway is that electric current creates magnetic fields, and changing magnetic fields can induce electric fields—principles used in motors, generators, relays, and transformers. Given Data / Assumptions: We seek a concise definition applicable to basic circuits and devices. Context includes wires, coils, and magnetic fields generated by current. Permanent magnets are not the primary focus. Concept / Approach: When current flows through a conductor, a magnetic field forms in concentric circles around it (right-hand rule). Coiling the conductor concentrates the field, creating an electromagnet. Conversely, Faraday’s and Lenz’s laws explain how changing magnetic fields induce voltages in conductors. The term “electromagnetism” in this introductory context refers to magnetism produced by electric current and the interplay between electric and magnetic fields. Step-by-Step Solution: Apply current to a straight wire → magnetic field appears around it.Form the wire into a coil → field lines add, strengthening the magnetism.Introduce a ferromagnetic core → further concentrates and strengthens the field (electromagnet).Thus, electromagnetism is the magnetic action resulting from electric current. Verification / Alternative check: Demonstrations with a compass near a current-carrying wire or the operation of a relay (coil energizes, armature pulls in) visibly confirm the production of magnetic effects by electrical current. Why Other Options Are Wrong: Permanent magnet alone: that is magnetostatics of permanent materials, not electromagnetism from current.Interaction between permanent and artificial magnet: describes magnet-to-magnet behavior, not current-generated fields.“Current in the coil” only states a quantity, not the magnetic action it causes. Common Pitfalls: Thinking electromagnetism requires coils only; even a straight conductor exhibits magnetic effects. Coils and cores simply enhance and shape those fields. Final Answer: the magnetic field action associated with a current-carrying conductor (electric current producing magnetism)

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