Q: RL step response timing In a simple series RL circuit excited by a DC step, the inductor current rises toward its final value exponentially with time constant τ = L / R. After exactly two time constants have elapsed, the current reaches approximate…

Introduction / Context: Time constants are quick design tools for estimating how fast circuits respond. For RL and RC first-order systems, the exponential approach to steady state follows a consistent pattern that is widely memorized for 1τ, 2τ, 3τ, and so on. This question targets the numerical recall linked to two time constants. Given Data / Assumptions: Series RL with total resistance R and inductance L. Step input of DC voltage applied at t = 0. Ideal components; no saturation or parasitics. Concept / Approach: For a step from 0 to I_final, the current is i(t) = I_final * (1 − e^(−t/τ)), where τ = L / R. At t = 2τ, the fraction to final is 1 − e^(−2). Numerically, e^(−2) ≈ 0.1353, so the fraction is about 0.8647 or 86% of the final value. Step-by-Step Solution: Write i(t)/I_final = 1 − e^(−t/τ).Set t = 2τ → i(2τ)/I_final = 1 − e^(−2).Evaluate e^(−2) ≈ 0.1353.Compute percentage: (1 − 0.1353) * 100% ≈ 86%. Verification / Alternative check: Known milestones: at 1τ → 63%; at 2τ → 86%; at 3τ → 95%; at 4τ → 98%; at 5τ → 99%+. These standard approximations validate the calculation. Why Other Options Are Wrong: 63%: corresponds to 1τ, not 2τ.95%: corresponds roughly to 3τ.98%: corresponds roughly to 4τ.50%: would be near 0.69τ, not 2τ. Common Pitfalls: Mixing RC and RL forms (they share the same percentages); misremembering the milestone values; confusing current rise with voltage across the inductor (which decays). Final Answer: 86%

Question 1

Compute total inductive reactance — two uncoupled inductors are in series and driven at f = 1.0 kHz. If L1 = 0.30 H and L2 = 0.25 H, what is the total inductive reactance as seen by the source?

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Introduction / Context: Reactance depends on both element value and operating frequency. For inductors in series, the inductances add first; their combined reactance is then X_L = 2 * pi * f * L_total. This calculation is common when sizing drive stages or estimating filter impedances at test frequencies such as 1 kHz. Given Data / Assumptions: L1 = 0.30 H, L2 = 0.25 H, series and uncoupled. Frequency f = 1.0 kHz. Ideal inductors; we ignore resistance and coupling parasitics for this calculation. Concept / Approach: Total inductance for series inductors: L_total = L1 + L2. Then compute X_L = 2 * pi * f * L_total. Convert the answer to kilo-ohms for comparison with the options. This two-step process (sum L, then apply formula) is the reliable approach. Step-by-Step Solution: Sum inductances: L_total = 0.30 H + 0.25 H = 0.55 H.Compute X_L: X_L = 2 * pi * 1000 * 0.55.Numerical value: 2 * pi * 1000 ≈ 6283.19; multiply by 0.55 → ≈ 3455.8 Ω.Convert to kΩ: ≈ 3.456 kΩ ≈ 3.5 kΩ (rounded). Verification / Alternative check: Back-of-envelope: 2 * pi * 1k ≈ 6.28k; times 0.55 ≈ 3.45k Ω, consistent with the detailed computation. A quick SPICE AC analysis would report a magnitude near 3.46 kΩ at 1 kHz ignoring resistive losses. Why Other Options Are Wrong: ≈ 4.4 kΩ: Would correspond to L_total ≈ 0.70 H at 1 kHz; our L_total is 0.55 H. ≈ 16.3 kΩ or ≈ 23.3 kΩ: These imply multi-henry totals or much higher frequency; not the given values. Common Pitfalls: Forgetting to sum L before applying the formula; mixing angular frequency ω = 2 * pi * f with f; or unit errors (using mH without converting to H). Keep track of base units to avoid kilo-factor mistakes. Final Answer: ≈ 3.5 kΩ total inductive reactance at 1.0 kHz.

Question 2

Power terms in AC circuits — which type of power explicitly represents the power at a specific instant in time (the instantaneous rate of energy transfer), rather than the cycle-averaged value?

Accepted Answer

Introduction / Context: Power terminology in AC and switched circuits can be confusing: instantaneous, real (average), reactive, and apparent power each capture different facets of energy flow. This question focuses on which term denotes the power at a specific instant in time, not averaged over a period or combined as an RMS product. Given Data / Assumptions: Time-varying voltage v(t) and current i(t). Focus on definitions, not on a particular waveform shape. Steady-state sinusoidal concepts are included but not required. Concept / Approach: Instantaneous power is defined as p(t) = v(t) * i(t). It represents the rate of energy transfer at each moment, and may be positive or negative depending on whether the element is absorbing or delivering energy at that instant. Real power P is the average of p(t) over a cycle and represents net energy converted to heat or work. Reactive power Q quantifies the amplitude of the oscillatory exchange of energy between reactive elements and the source (zero net over a cycle). Apparent power S = V_rms * I_rms summarizes the “size” of voltage and current without phase information (units VA) and bounds the vector sum P + jQ. Step-by-Step Solution: Recall definitions: p(t) = v(t) * i(t).Identify the “at an instant” descriptor: that is p(t), not an average.Match terminology: “Instantaneous power.”Select the option accordingly. Verification / Alternative check: For sinusoidal steady state: v(t) = V_m sin(ωt), i(t) = I_m sin(ωt − φ). Then p(t) = V_m I_m/2 * [cos φ − cos(2ωt − φ)]. The first term averages to P = V_rms I_rms cos φ (real power); the full expression is instantaneous power, varying within each cycle. Why Other Options Are Wrong: Apparent power: V_rms * I_rms; not time-instant specific. Reactive power: Represents energy exchange (vars), not instantaneous; average of the reactive component over a cycle is zero. Real power: Cycle-average of p(t), not the point-in-time value. Common Pitfalls: Assuming “real power” means the “real” momentary value; in power theory, “real” means average dissipated power. Keep p(t) (instantaneous) distinct from P (average), Q (reactive), and S (apparent). Final Answer: Instantaneous power represents the power at a particular point in time.

Question 3

Single-inductor reactance — in a circuit operating at f = 1.4 kHz, an inductor L2 has inductance 0.50 H. What is the inductive reactance of L2?

Accepted Answer

Introduction / Context: Inductive reactance quantifies how strongly an inductor resists changes in current at a given frequency. This calculation is central to filter design, impedance matching, and estimating current draw from AC sources. The problem asks for X_L of a single inductor at a specified frequency. Given Data / Assumptions: L2 = 0.50 H (ideal). Frequency f = 1.4 kHz. We ignore winding resistance and core losses for this calculation. Concept / Approach: The inductive reactance magnitude is X_L = 2 * pi * f * L. Substitute L = 0.50 H and f = 1.4 kHz to obtain a numerical value in ohms, then express in kilo-ohms for comparison with the options. Step-by-Step Solution: Write formula: X_L = 2 * pi * f * L.Insert numbers: X_L = 2 * pi * 1400 * 0.50.Compute: 2 * pi * 1400 ≈ 8796.46; multiply by 0.50 → ≈ 4398.23 Ω.Round: ≈ 4.40 kΩ (to two significant digits). Verification / Alternative check: Back-of-envelope: 2 * pi ≈ 6.28; 6.28 * 1.4k ≈ 8.79k; half of that (for 0.5 H) ≈ 4.40k Ω. A quick simulator would report essentially the same magnitude when plotting |Z_L| at 1.4 kHz. Why Other Options Are Wrong: ≈ 3.5 kΩ: Corresponds to about 0.40 H at 1.4 kHz or 0.55 H at 1.0 kHz; not our case. ≈ 16.3 kΩ or ≈ 23.3 kΩ: These would require multiple henries or much higher frequency. Common Pitfalls: Mixing f and angular frequency ω; remember ω = 2 * pi * f. Also, avoid unit slips (H vs mH) which can introduce thousand-fold errors. Final Answer: ≈ 4.4 kΩ is the inductive reactance of L2 at 1.4 kHz.

Question 4

Electromagnetic induction in electrical engineering  Electromagnetic induction is the physical process by which one form of energy creates the other when a magnetic field and an electric circuit interact. In standard terminology, electromagnetic ind…

Accepted Answer

Introduction / Context: Electromagnetic induction underpins generators, transformers, inductors, and many sensors. It explains how changing magnetic fields induce voltages and currents in conductors. Understanding the “from–to” relationship (what causes what) is foundational for both power and electronics domains. Given Data / Assumptions: A conductor or coil is present. A magnetic field linked to that conductor changes with time (by motion or field variation). We consider idealized behavior (no saturation, negligible losses) to focus on the principle. Concept / Approach: Faraday’s law states the induced electromotive force in a coil is proportional to the negative rate of change of magnetic flux through it: e_induced = −N * dΦ/dt. In plain words, a changing magnetic field “through” a circuit produces (induces) an electrical voltage; if a closed path exists, current flows. Lenz’s law gives the polarity that opposes the change causing it. Therefore, electromagnetic induction is the generation of electricity (voltage/current) from magnetism (changing flux). Step-by-Step Solution: Identify the cause: varying magnetic flux linkage (by motion, varying current elsewhere, or both).Relate to effect: an induced voltage appears across the conductor or coil terminals.Conclude: the process converts magnetic-field change into electrical energy transfer (electricity).Apply to devices: generators (mechanical → magnetic change → electricity) and transformers (AC in primary → changing flux → induced voltage in secondary). Verification / Alternative check: Rotate a coil in a steady magnetic field: Φ varies sinusoidally with angle, creating a sinusoidal induced voltage. Alternatively, keep the coil still but vary the magnetic field (e.g., AC in a primary winding) and observe induced secondary voltage. Both confirm “electricity from magnetism.” Why Other Options Are Wrong: magnetism, electricity: describes the inverse (how electromagnets arise from current), not induction.electricity, electricity: omits the magnetic intermediary; that is conduction or amplification, not induction.magnetism, magnetism: self-referential; no energy conversion pathway.mechanical force, voltage: mechanical energy can create induction via motion, but the immediate cause in EM induction is changing magnetism, not force directly. Common Pitfalls: Confusing induction (changing flux → voltage) with the magnetic field created by current (current → magnetism). Also, assuming static magnetic fields induce voltage; it is the change (dΦ/dt) that matters. Final Answer: electricity, magnetism

Question 5

Inductor behavior with alternating current  The opposition offered by an ideal inductor to alternating current (AC) is quantified by inductive reactance. Inductive reactance is directly proportional to which parameter in steady-state sinusoidal anal…

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Introduction / Context: Inductors behave differently under DC and AC. At DC an ideal inductor looks like a short (after transients), but in AC it resists current changes. The measure of this opposition is called inductive reactance, a frequency-dependent quantity essential for filters, matching networks, and resonant circuits. Given Data / Assumptions: Linear inductor with inductance L (henry). Sinusoidal steady-state at frequency f (hertz). Parasitics like winding resistance and self-capacitance are neglected. Concept / Approach: The inductor voltage-current relationship is v = L * di/dt. Under sinusoidal excitation, i and v are sinusoidal; in phasor form, Z_L = j * 2π * f * L. The magnitude of this impedance is X_L = 2π * f * L. Hence, X_L scales linearly with frequency f and also with inductance L. When f increases, X_L increases proportionally; when f decreases toward zero, X_L tends to zero. Step-by-Step Solution: Write Z_L = j * 2π * f * L.Take magnitude: X_L = |Z_L| = 2π * f * L.Observe proportionality: X_L ∝ f (and ∝ L).Conclude: the inductor’s opposition to AC is directly proportional to frequency. Verification / Alternative check: Measure current through a fixed inductor driven by a fixed-amplitude source at two frequencies, f1 < f2. Since I ≈ V / X_L, current at f2 is smaller by the ratio f2/f1, confirming linear dependence on frequency. Why Other Options Are Wrong: resistance: resistors yield frequency-independent opposition (for ideal parts).applied voltage amplitude: changing V alters current but does not define X_L.wire length only: geometry affects L, but X_L’s direct proportionality is to f (and to L, not merely length).capacitance: belongs to capacitive reactance (X_C = 1 / (2π f C)), not inductive behavior. Common Pitfalls: Confusing X_L (reactance magnitude) with the complex j operator; forgetting that at very high frequency, parasitics can shift behavior from ideal. Final Answer: frequency

Question 6

Energy and polarity in an inductor during a decreasing current  When the current through an ideal inductor is forced to decrease, what happens to the inductor’s magnetic field and to the instantaneous voltage polarity across the inductor terminals?

Accepted Answer

Introduction / Context: Inductors store energy in magnetic fields. The way they respond to changes in current is central to switching power supplies, flyback circuits, and transient suppression. Understanding field behavior and voltage polarity during current decay prevents design errors and device stress. Given Data / Assumptions: Ideal inductor with inductance L. Current is decreasing (di/dt < 0). We adopt the passive sign convention: the indicated terminal voltage is positive for current entering the labeled positive terminal. Concept / Approach: The inductor relation is v = L * di/dt. If di/dt is negative (current decreasing), v becomes negative with respect to the passive sign convention, meaning the inductor’s polarity flips to oppose the decrease. Simultaneously, the magnetic field energy W = 0.5 * L * I^2 must diminish as current falls, so the field collapses, returning energy to the circuit. Lenz’s law captures both effects: the induced voltage polarity always opposes the change in current. Step-by-Step Solution: Identify di/dt < 0 for a falling current.Compute voltage sign from v = L * di/dt → inductor voltage reverses relative to the previous conduction polarity.Relate energy to current: as I decreases, W = 0.5 * L * I^2 decreases, so the magnetic field collapses.Conclude: field collapses; voltage polarity reverses to keep current flowing. Verification / Alternative check: Open an inductive circuit: a high-voltage spike of opposite polarity appears, often clamped by a flyback diode. The spike’s polarity is consistent with maintaining current flow briefly despite the interruption. Why Other Options Are Wrong: expands, reverses: field expands only when current rises.collapses, remains the same / expands, remains the same: contradict v = L * di/dt; polarity must change if di/dt changes sign.remains unchanged, polarity random: ignores electromagnetic laws; behavior is deterministic. Common Pitfalls: Forgetting sign conventions; assuming the inductor “blocks” changes entirely—instead, it resists them by generating an opposing voltage. Final Answer: collapses, reverses

Question 7

RL step response timing  In a simple series RL circuit excited by a DC step, the inductor current rises toward its final value exponentially with time constant τ = L / R. After exactly two time constants have elapsed, the current reaches approximate…

Accepted Answer

Introduction / Context: Time constants are quick design tools for estimating how fast circuits respond. For RL and RC first-order systems, the exponential approach to steady state follows a consistent pattern that is widely memorized for 1τ, 2τ, 3τ, and so on. This question targets the numerical recall linked to two time constants. Given Data / Assumptions: Series RL with total resistance R and inductance L. Step input of DC voltage applied at t = 0. Ideal components; no saturation or parasitics. Concept / Approach: For a step from 0 to I_final, the current is i(t) = I_final * (1 − e^(−t/τ)), where τ = L / R. At t = 2τ, the fraction to final is 1 − e^(−2). Numerically, e^(−2) ≈ 0.1353, so the fraction is about 0.8647 or 86% of the final value. Step-by-Step Solution: Write i(t)/I_final = 1 − e^(−t/τ).Set t = 2τ → i(2τ)/I_final = 1 − e^(−2).Evaluate e^(−2) ≈ 0.1353.Compute percentage: (1 − 0.1353) * 100% ≈ 86%. Verification / Alternative check: Known milestones: at 1τ → 63%; at 2τ → 86%; at 3τ → 95%; at 4τ → 98%; at 5τ → 99%+. These standard approximations validate the calculation. Why Other Options Are Wrong: 63%: corresponds to 1τ, not 2τ.95%: corresponds roughly to 3τ.98%: corresponds roughly to 4τ.50%: would be near 0.69τ, not 2τ. Common Pitfalls: Mixing RC and RL forms (they share the same percentages); misremembering the milestone values; confusing current rise with voltage across the inductor (which decays). Final Answer: 86%

Question 8

Filter behavior of classic building blocks  Consider the standard small-signal approximations for first-order networks. An RL integrator (output taken appropriately so the circuit performs time integration) behaves as what type of simple filter, and…

Accepted Answer

Introduction / Context: First-order RC and RL circuits can act as crude differentiators and integrators when operated within limits (small or large time-constant assumptions relative to signal period). Recognizing their corresponding filter “shapes” helps when sketching Bode plots or predicting pass/stop behavior. Given Data / Assumptions: Ideal passive components. Small-signal, linear operation near the intended use regime. Standard topologies: RL integrator output across R; RC differentiator output across R in series with C (or equivalent forms yielding the classic responses). Concept / Approach: An RL integrator is commonly realized with a series RL input and output taken across the resistor. Its transfer magnitude is |Vout/Vin| = R / √(R^2 + (ωL)^2), which rolls off with frequency and passes low frequencies better than high—this is low-pass behavior. An RC differentiator is often a series RC with output across the resistor; its high-frequency gain approaches that of the divider, while low frequencies are attenuated, giving a high-pass response. When used strictly as integrator/differentiator, additional constraints on frequency range are imposed so that the intended operation dominates. Step-by-Step Solution: Write RL integrator transfer: H_RL(ω) = R / (R + jωL) → low-pass magnitude trend.Write RC differentiator transfer: H_RC(ω) = jωRC / (1 + jωRC) when output is across R → high-pass magnitude trend.Relate time-domain roles: integration emphasizes low-frequency content; differentiation emphasizes high-frequency content.Conclude mapping: RL integrator → low-pass; RC differentiator → high-pass. Verification / Alternative check: Bode slopes: both are first-order with 20 dB/decade slope. The RL integrator magnitude decreases with frequency (low-pass), while the RC differentiator magnitude increases with frequency up to the asymptote (high-pass). Why Other Options Are Wrong: high-pass, low-pass: reverses the correct pairing.low-pass, low-pass / high-pass, high-pass / band-pass, band-stop: do not match the canonical behaviors for these elementary blocks. Common Pitfalls: Confusing whether the output is taken across R or L/C; the terminal choice flips the filter type. Also, forgetting that “integrator” and “differentiator” descriptions are valid only over a restricted frequency band. Final Answer: low-pass, high-pass

Inductors Questions