Question 1

Inductive reactance and current: A sinewave voltage is applied across an inductor. If the frequency of the applied voltage is decreased, what happens to the current through the inductor (steady state)?

Accepted Answer

Introduction / Context:Inductors oppose changes in current via inductive reactance X_L, which depends on frequency. Understanding how current varies with frequency for a fixed applied voltage is essential in filter design, AC analysis, and impedance matching. Given Data / Assumptions: Pure inductor with inductance L. Applied sinusoidal voltage of fixed amplitude. Steady-state AC (no transients considered). Concept / Approach:Inductive reactance: X_L = 2 * π * f * L. Current magnitude with fixed voltage is I = V / X_L. Thus, as frequency f decreases, X_L decreases linearly, so current increases correspondingly. This inverse relationship holds for ideal inductors without series resistance considered. Step-by-Step Solution: Write reactance: X_L = 2πfL.Current magnitude: I = V / X_L = V / (2πfL).Decrease f → denominator decreases → current increases. Verification / Alternative check:Numerical example: Let V = 10 V, L = 10 mH. At 1 kHz: X_L ≈ 62.8 Ω → I ≈ 159 mA. At 100 Hz: X_L ≈ 6.28 Ω → I ≈ 1.59 A, showing the increase clearly (ideal case). Why Other Options Are Wrong: is decreased: Opposite of the reactance effect. does not change: Would only be true if X_L were constant, which it is not. momentarily goes to zero: Transient behavior is not part of steady-state AC analysis. Common Pitfalls: Confusing inductors with capacitors (where current increases with frequency in a different way). Final Answer:is increased

Question 2

RL time constant to near steady state: A 320 µH inductor is in series with a 3.3 kΩ resistor. Approximately how long does it take the current to build up to essentially its final value after a DC step (use about 5 time constants as 'full')?

Accepted Answer

Introduction / Context:In a first-order RL circuit driven by a DC step, the current approaches its final value exponentially with time constant τ = L / R. A common engineering rule is that about 5τ is effectively 'full' (over 99% of the final value). This question tests time-constant calculation and unit handling. Given Data / Assumptions: L = 320 µH = 320 × 10^-6 H. R = 3.3 kΩ = 3300 Ω. Use 5τ ≈ 5 * (L / R) for near-steady-state timing. Concept / Approach:Compute τ = L / R, then multiply by 5. Careful prefix conversion is crucial to avoid errors of 10^3 or 10^6 in the answer. Step-by-Step Solution: τ = L / R = (320 × 10^-6) / 3300 s ≈ 9.697 × 10^-8 s ≈ 0.09697 µs.Time to 'full': t_full ≈ 5τ ≈ 5 * 0.09697 µs ≈ 0.4848 µs.Rounded to the closest listed choice: 0.48 µs. Verification / Alternative check:At 4τ, response is ≈ 98.2%; at 5τ, ≈ 99.3%, which is typically considered essentially full in practical design. Why Other Options Are Wrong: 0.48 ms or 0.48 s or 48 s: These are orders of magnitude too large; they arise from mishandling µH/kΩ prefixes. Common Pitfalls: Treating microhenry as millihenry or forgetting to convert kilohms to ohms. Final Answer:0.48 µs

Question 3

Inductive reactance of a series–parallel inductor network: Two 10 H inductors are connected in parallel, and this parallel combination is then in series with a third 10 H inductor. What is the approximate total reactance at 7 kHz?

Accepted Answer

Introduction / Context:Series–parallel combinations of inductors reduce to an equivalent inductance that determines the overall reactance X_L = 2π f L_eq. Accurate reduction and unit conversion are core skills for AC network analysis at audio and RF frequencies. Given Data / Assumptions: L1 = 10 H in parallel with L2 = 10 H. That parallel pair is in series with L3 = 10 H. Frequency f = 7 kHz; ideal inductors (no resistance considered). Concept / Approach:Two equal inductors in parallel produce L_par = L/2. Then add the series inductor to get L_eq. Finally compute X_L = 2π f L_eq and convert to kilohms for comparison to options. Step-by-Step Solution: Parallel of two 10 H: L_par = (10 * 10)/(10 + 10) = 5 H.Equivalent inductance: L_eq = L_par + 10 H = 5 + 10 = 15 H.Reactance: X_L = 2π f L_eq = 2π * 7000 * 15 ≈ 659,734 Ω ≈ 660 kΩ. Verification / Alternative check:Reasonableness: A very large inductance at several kilohertz should yield hundreds of kilohms of reactance; the computed value aligns with that intuition. Why Other Options Are Wrong: 66 kΩ: A factor-of-10 error from dropping a zero in the frequency or inductance. 219 kΩ or 1.3 MΩ: Do not match 2πfL with L_eq = 15 H at 7 kHz. Common Pitfalls: Mistaking parallel of equal inductors as 2L (it is L/2). Dropping units when converting to kilohms. Final Answer:660 kΩ

Question 4

Factors that decrease inductance of an iron-core coil: Which action will definitely reduce the inductance value of a given coil (all else equal)?

Accepted Answer

Introduction / Context:The inductance L of a coil depends on geometry (number of turns, cross-sectional area, length) and core permeability. Designers adjust these parameters to reach target L-values for filters, chokes, and transformers. Knowing how each factor influences L prevents trial-and-error mistakes. Given Data / Assumptions: Iron-core coil; compare the effect of changing geometry or core material. Standard approximate relation: L ∝ μ * N^2 * A / l, where μ is core permeability, N is turns, A is core area, and l is magnetic path length. All else (unlisted variables) held constant when evaluating each option independently. Concept / Approach:From L ∝ μ * N^2 * A / l, increasing μ, increasing N, or increasing A will increase inductance; increasing l will decrease inductance. Removing the iron core effectively reduces μ toward μ0 (air), which substantially decreases L compared to iron-core values. Step-by-Step Solution: Assess μ: removing iron → μ drops → L decreases.Assess N: increasing turns → N^2 rises → L increases (not a decrease).Assess A: increasing area → L increases (not a decrease).Assess material: inserting higher-μ core → μ rises → L increases (not a decrease). Verification / Alternative check:Common lab observation: swapping an iron core for air reduces measured inductance dramatically, confirming μ’s dominant influence on L. Why Other Options Are Wrong: increasing number of turns: Raises inductance via N^2 scaling. inserting higher-permeability core: Raises μ, thus increases L. increasing cross-sectional area: Increases A, raising L. Common Pitfalls: Thinking that shortening length always decreases L; here length is not the option. The presented incorrect options all increase L. Final Answer:removing the iron core from the coil

Question 5

Inductors in parallel and series with AC source: L1 = 25 µH in parallel with L2 = 50 µH, then in series with L3 = 20 µH, driven by 60 V rms at 300 kHz. What is the total rms current drawn?

Accepted Answer

Introduction / Context:Analyzing combinations of inductors with an AC source requires computing the equivalent inductance and then using inductive reactance to find the current. This problem reinforces parallel–series reduction for inductors and conversion from inductance to reactance at a specified frequency. Given Data / Assumptions: L1 = 25 µH, L2 = 50 µH in parallel. L3 = 20 µH in series with the parallel pair. Source: 60 V rms, f = 300 kHz (sinusoidal). Ideal inductors; no resistance; steady-state AC. Concept / Approach: For two inductors in parallel: L_p = (L1 * L2) / (L1 + L2). The series combination is L_total = L_p + L3. Inductive reactance is X_L = 2 * π * f * L_total. The current magnitude is I_rms = V_rms / X_L. Step-by-Step Solution: Compute L_p = (25 µH * 50 µH) / (25 µH + 50 µH) = 1250 / 75 µH ≈ 16.67 µH.L_total = 16.67 µH + 20 µH ≈ 36.67 µH.X_L = 2 * π * 300,000 * 36.67 × 10^-6 ≈ 69.1 Ω.I_rms = 60 / 69.1 ≈ 0.868 A ≈ 870 mA. Verification / Alternative check: Order-of-magnitude check: tens of microhenries at hundreds of kilohertz yields reactance on the order of tens of ohms; 60 V over ~70 Ω gives about 0.85 A, consistent. Why Other Options Are Wrong: 87 mA and 337 mA underestimate the current (would require much higher reactance). 87 A is unrealistically high for these values. Common Pitfalls: Forgetting that inductors in parallel combine like resistors in parallel; mixing microhenry units; using f instead of 2πf for reactance. Final Answer: 870 mA

Question 6

Series inductance calculation: Three inductors of 15 mH each are connected in series. What is the total inductance?

Accepted Answer

Introduction / Context:Inductors in series add directly, similar to resistors in series. This provides a simple way to obtain a higher total inductance for filters, chokes, and timing networks. Given Data / Assumptions: Three inductors, each 15 mH. All three are connected in series. Ideal components with negligible coupling. Concept / Approach: For series inductors (with negligible mutual coupling), L_total = L1 + L2 + L3. Units must be consistent (millihenry here). Step-by-Step Solution: L_total = 15 mH + 15 mH + 15 mH = 45 mH. Verification / Alternative check: If series inductors had coupling, mutual inductance M terms would affect the result. With ideal, un-coupled coils, simple addition applies. Why Other Options Are Wrong: 30 mH implies only two coils added. 15 mH implies one coil. 5 mH is unrelated to series addition. Common Pitfalls: Confusing series with parallel behavior; ignoring mutual coupling in real layouts (not part of this ideal problem). Final Answer: 45 mH

Question 7

Parallel inductance: Four coils of 75 µH, 40 µH, 25 µH, and 15 µH are connected in parallel. Estimate the total inductance.

Accepted Answer

Introduction / Context:Parallel inductors combine like resistors in parallel: the reciprocal of the equivalent inductance equals the sum of the reciprocals. This configuration reduces inductance and increases current-handling capability. Given Data / Assumptions: L1 = 75 µH, L2 = 40 µH, L3 = 25 µH, L4 = 15 µH. All four inductors are in parallel. Ideal components; negligible coupling and series resistance. Concept / Approach: Use 1/L_eq = Σ(1/L_i). After summing reciprocals, invert to find L_eq. Keep units in microhenries throughout to avoid conversion errors. Step-by-Step Solution: Compute reciprocals: 1/75 ≈ 0.01333, 1/40 = 0.025, 1/25 = 0.04, 1/15 ≈ 0.06667 (all in 1/µH).Sum: 0.01333 + 0.025 + 0.04 + 0.06667 = 0.145.Invert: L_eq = 1 / 0.145 ≈ 6.8966 µH ≈ 6.9 µH. Verification / Alternative check: L_eq must be less than the smallest individual value (15 µH). The computed 6.9 µH satisfies this sanity check. Why Other Options Are Wrong: 14 µH is too large for a four-branch parallel. 2.2 µH would require much larger reciprocal sum. 155 µH incorrectly suggests series addition. Common Pitfalls: Arithmetic mistakes in reciprocals; forgetting to invert the final sum; mixing µH with mH. Final Answer: 6.9 µH

Question 8

RL step response: A 220 Ω resistor and a 60 mH inductor in series are connected to a 20 V DC source via a switch. What is the inductor current two time constants after closing the switch?

Accepted Answer

Introduction / Context:In a first-order RL circuit, current grows exponentially toward a final value after a DC step is applied. The time constant τ governs how quickly the current approaches steady state. This question practices using τ to find the current at a specified multiple of τ. Given Data / Assumptions: R = 220 Ω, L = 60 mH, V = 20 V DC. Initial current i(0) = 0 A (switch just closed). Find i(t) at t = 2τ, where τ = L / R. Concept / Approach: For a rising RL current: i(t) = I_final * (1 − e^(−t/τ)), where I_final = V / R. Compute τ, then evaluate i at t = 2τ. Step-by-Step Solution: τ = L / R = 0.06 H / 220 Ω ≈ 0.0002727 s.I_final = V / R = 20 / 220 = 0.09091 A ≈ 90.9 mA.At t = 2τ: i(2τ) = I_final * (1 − e^(−2)) ≈ 90.9 mA * 0.8647 ≈ 78.6 mA.Rounded to the nearest listed value: 78 mA. Verification / Alternative check: At t = 3τ the current would be ~95% of final (≈ 86.4 mA); at 2τ we expect ≈ 86% of final, matching ≈ 78–79 mA. Why Other Options Are Wrong: 91 mA is the final current, not the value at 2τ. 57 mA corresponds to an earlier time. 400 mA exceeds the DC limit V/R. Common Pitfalls: Using e^(−tR/L) but mixing R and L units; forgetting I_final; rounding too early. Final Answer: 78 mA

Question 9

DC coil current: An 18 V supply is connected across a coil with winding resistance 180 Ω (ignore inductance at steady state). What is the current?

Accepted Answer

Introduction / Context:Once steady state is reached in a DC circuit, an inductor behaves as a short circuit and only its winding resistance limits current. This is a direct application of Ohm's law. Given Data / Assumptions: Supply voltage V = 18 V (DC). Winding resistance R = 180 Ω. Steady state, so inductive reactance is zero. Concept / Approach: Use Ohm's law I = V / R at steady state. The inductive element does not impede DC after transients decay. Step-by-Step Solution: I = 18 V / 180 Ω = 0.1 A = 100 mA. Verification / Alternative check: If the transient were considered, current would rise exponentially from 0 A to 100 mA with time constant τ = L/R. At t → ∞, the value is 100 mA as computed. Why Other Options Are Wrong: 1 mA and 10 mA are too small by factors of 100 and 10 respectively. 1 A would require only 18 Ω of resistance, not 180 Ω. Common Pitfalls: Applying AC reactance to a DC steady-state question; forgetting that inductors pass DC after transients. Final Answer: 100 mA

Question 10

Parallel inductors check: Inductors of 5 mH, 4.3 mH, and 0.6 mH are connected in parallel. Qualitatively, the total inductance will be…

Accepted Answer

Introduction / Context:Understanding how inductors combine helps in quick sanity checks without full calculations. For parallel inductors, the equivalent inductance is always less than the smallest branch value, analogous to resistors in parallel. Given Data / Assumptions: L1 = 5 mH, L2 = 4.3 mH, L3 = 0.6 mH. All inductors are in parallel. Assume ideal behavior and no coupling. Concept / Approach: The rule for parallel inductors is 1/L_eq = Σ(1/L_i). Because reciprocals add, L_eq must be less than the smallest L_i. Here, the smallest is 0.6 mH; therefore, L_eq < 0.6 mH. Step-by-Step Solution: Recognize smallest branch: 0.6 mH.Since 1/L_eq = 1/5 + 1/4.3 + 1/0.6, the right side is positive and greater than 1/0.6.Therefore, L_eq < 0.6 mH (no exact arithmetic needed for this qualitative question). Verification / Alternative check: Numerically, 1/L_eq ≈ 0.2 + 0.2326 + 1.6667 ≈ 2.099, so L_eq ≈ 0.476 mH, confirming it is less than 0.6 mH. Why Other Options Are Wrong: Values near 9.9 mH suggest series addition, not parallel. 'Greater than 5 mH' contradicts the parallel rule. Common Pitfalls: Mixing up series and parallel rules; forgetting that parallel reduces the equivalent inductance. Final Answer: less than 0.6 mH

Question 11

Energy in an inductor: As the current in an inductor decreases, what happens to the energy stored in its magnetic field?

Accepted Answer

Introduction / Context:Inductors store energy in a magnetic field proportional to the square of current. Understanding how energy changes with current is essential for analyzing transients, flyback behavior, and energy transfer in converters. Given Data / Assumptions: Inductor with inductance L, carrying current I. Current is decreasing in magnitude. No exotic effects (e.g., core hysteresis) are considered. Concept / Approach: The stored energy is W = (1/2) * L * I^2. Because I^2 is the determining factor, any decrease in |I| reduces W. If I tends to zero, energy tends to zero; the inductor releases energy back into the circuit. Step-by-Step Solution: Write: W = 0.5 * L * I^2.If current changes from I1 to I2 with |I2| < |I1|, then I2^2 < I1^2.Therefore, W2 = 0.5 * L * I2^2 < W1 = 0.5 * L * I1^2 → energy decreases. Verification / Alternative check: Consider an RL decay: i(t) = I0 * e^(−t/τ). Then W(t) = 0.5 * L * I0^2 * e^(−2t/τ), clearly decreasing over time. Why Other Options Are Wrong: 'Increases' contradicts the I^2 dependence. 'Remains the same' would require constant current. 'Doubles' would need current magnitude to increase by √2. Common Pitfalls: Confusing stored energy with instantaneous voltage across the inductor; energy depends on current, not directly on voltage. Final Answer: decreases

Question 12

Inductor initial condition: An inductor, a 1 kΩ resistor, and a switch are in series across a 6 V DC source. Exactly at the moment the switch closes, what is the voltage across the inductor?

Accepted Answer

Introduction / Context:Inductors oppose sudden changes in current. At the instant a DC source is applied, the current through an ideal inductor cannot change instantaneously, which determines the initial voltage across it. Given Data / Assumptions: Series circuit: DC source 6 V, resistor 1 kΩ, ideal inductor. Switch just closed (t = 0+). Initial current is zero at the instant of closing. Concept / Approach: At t = 0+, the inductor current is still 0 A, so the resistor drop is i * R = 0 V. KVL requires the inductor to drop the entire source voltage initially. More formally, v_L = L * di/dt, which is initially equal to the source since di/dt is finite and the resistor drop is zero. Step-by-Step Solution: At t = 0+: i = 0 A ⇒ v_R = i * R = 0 V.KVL: v_source = v_R + v_L ⇒ 6 V = 0 + v_L.Thus, v_L(0+) = 6 V. Verification / Alternative check: As time increases, current rises toward 6 mA (6 V / 1 kΩ), and the inductor voltage decays to 0 V, consistent with first-order RL behavior. Why Other Options Are Wrong: 0 V would imply instantaneous current rise. 12 V exceeds the source; no such surge exists in this ideal DC step. 4 V would require a simultaneous 2 V resistor drop at t = 0+, which contradicts i = 0 A. Common Pitfalls: Confusing capacitor and inductor initial conditions; an inductor initially behaves like an open circuit for DC steps. Final Answer: 6 V

Question 13

Series RL at changing frequency: In a series circuit with an inductor and a resistor, the frequency is initially set so that XL = R. If the frequency is increased, which relationship becomes true for the voltage drops?

Accepted Answer

Introduction / Context:In a series RL circuit, the same current flows through both elements, and the voltage drops are proportional to their impedances. Changing frequency alters the inductor's reactance, which in turn changes the relative drops. Given Data / Assumptions: Series RL; initial condition XL = R. Frequency is increased from that starting point. Sinusoidal steady state with ideal components. Concept / Approach: Inductive reactance XL = 2 * π * f * L. Increasing f increases XL. With the same current I in series, VL = I * XL and VR = I * R. If XL becomes greater than R, then VL exceeds VR. Step-by-Step Solution: Start: XL = R ⇒ VL = VR.Increase f ⇒ XL increases, while R stays constant.Therefore, I is common, and VL = I * XL grows beyond VR = I * R. Verification / Alternative check: Impedance magnitude Z = √(R^2 + XL^2) increases; current decreases, but the ratio VL/VR = XL/R increases > 1, so VL dominates. Why Other Options Are Wrong: VL = VR holds only at the specific frequency where XL = R. VR ≥ VL or VL < VR would require XL ≤ R, not true after increasing f. Common Pitfalls: Assuming current increase with frequency; in RL, higher frequency actually increases impedance and reduces current, but the ratio of drops still favors the inductor. Final Answer: VL > VR

Question 14

Unit conversion for inductance: A 240 µH inductor is equivalent to which value expressed in millihenries (mH) or henries (H)?

Accepted Answer

Introduction / Context:Engineering problems often require converting units cleanly. Inductance is commonly expressed in henries (H), millihenries (mH), and microhenries (µH). Knowing the decimal relationships avoids scale errors during design or analysis. Given Data / Assumptions: L = 240 µH. 1 mH = 1000 µH; 1 H = 1000 mH. We seek an equivalent expression. Concept / Approach: Use powers-of-ten conversions: micro (10^-6), milli (10^-3). Thus, 240 µH * (1 mH / 1000 µH) = 0.240 mH. Converting to henries would give 0.000240 H. Step-by-Step Solution: Convert µH → mH: 240 / 1000 = 0.240 mH.Optionally, mH → H: 0.240 / 1000 = 0.000240 H.Choose the matching statement from the options. Verification / Alternative check: Reverse check: 0.240 mH = 0.240 * 1000 µH = 240 µH, confirming equivalence. Why Other Options Are Wrong: 240 mH is 1000× too large. 0.000240 mH is 1000× too small. 240 H is six orders of magnitude larger than the original value. Common Pitfalls: Misplacing decimal points; confusing micro with milli prefixes; forgetting that 1 mH = 1000 µH. Final Answer: 0.240 mH inductor

Question 15

Series inductors with geometric progression: Five inductors are connected in series. The smallest inductance is 8 µH, and each subsequent inductor has twice the inductance of the previous one (connected in ascending order). What is the total series …

Accepted Answer

Introduction / Context:Questions on series inductors often test whether learners remember that series inductances add directly, just like series resistances. This problem adds a twist by specifying a geometric progression (each value is twice the previous), which requires careful listing of the individual inductances before summing them. Such skills are fundamental in filter design, chokes, and energy storage planning in power electronics. Given Data / Assumptions: Five inductors connected in series. Smallest inductance L1 = 8 µH. Each subsequent inductor is twice the previous: L2 = 2L1, L3 = 2L2, etc. Ideal components; mutual coupling ignored (no coupling coefficient specified). Concept / Approach:For inductors in series (without mutual coupling), the equivalent inductance is the arithmetic sum of individual inductances: L_total = L1 + L2 + L3 + L4 + L5. With a ratio of 2, the list is straightforward to generate and then sum. Keeping consistent microhenry units avoids conversion mistakes. Step-by-Step Solution: L1 = 8 µH.L2 = 28 = 16 µH.L3 = 216 = 32 µH.L4 = 232 = 64 µH.L5 = 264 = 128 µH.L_total = 8 + 16 + 32 + 64 + 128 = 248 µH. Verification / Alternative check:Recognize the geometric-series pattern: sum of n terms with first term a and ratio r is S_n = a * (r^n − 1)/(r − 1). Here a = 8 µH, r = 2, n = 5: S_5 = 8*(2^5 − 1)/(2 − 1) = 8*(32 − 1) = 8*31 = 248 µH, matching the direct sum. Why Other Options Are Wrong: 64 µH or 32 µH: These are intermediate single inductor values, not the series total. 8 H: Unit and magnitude error; result must remain in µH for these values and is far smaller than 1 H. Common Pitfalls: Forgetting that inductances add in series even when values double. Mistaking geometric progression with exponential growth of the total; it still sums linearly. Final Answer:248 µH

Question 16

Coil construction: Which change definitely increases the DC winding resistance of a coil (ignore core effects on resistance)?

Accepted Answer

Introduction / Context:A coil’s DC winding resistance affects copper loss and temperature rise. Designers control this resistance by choosing conductor gauge and number of turns. Understanding which parameters increase resistance is key for balancing efficiency, size, and cost in inductors, transformers, and electromagnets. Given Data / Assumptions: We focus on DC winding resistance only (not reactance). Material of the wire is copper unless otherwise stated; resistivity is constant for this comparison. Core changes do not alter the copper’s ohmic resistance. Concept / Approach:Wire resistance R = ρ * (ℓ / A), where ρ is resistivity, ℓ is conductor length, and A is cross-sectional area. Increasing turns increases the length ℓ, raising R. Using thinner wire decreases A, also raising R. Core material does not appear in this formula and thus does not affect winding resistance directly. Step-by-Step Solution: If turns ↑ → length ℓ ↑ → R ↑ (direct proportionality).If wire thinner → area A ↓ → R ↑ (inverse proportionality).Changing the magnetic core alters inductance (via permeability) but not the copper’s DC ohmic resistance. Verification / Alternative check:Practical test: measuring coil resistance with an ohmmeter before/after adding turns or switching to a smaller gauge consistently shows higher resistance; swapping the core alone leaves ohmic resistance unchanged. Why Other Options Are Wrong: Changing the core material: affects inductance and losses like hysteresis/eddy currents, not winding DC resistance. Using thinner wire only or increasing turns only: each individually increases R, but the most complete correct statement is that either change will increase R. Common Pitfalls: Confusing inductance changes with resistance changes. Ignoring the strong effect of conductor cross-section on R. Final Answer:increasing the number of turns or using thinner wire

Question 17

Inductor energy dependence on current: When the current through an inductor is reduced to half its original value, how does the stored magnetic energy change?

Accepted Answer

Introduction / Context:Energy storage in inductors underpins power conversion, snubbers, and resonant circuits. The relationship between current and stored energy is quadratic, not linear, and mixing these up leads to major design errors. This question checks that you remember the exact current-squared dependence. Given Data / Assumptions: Ideal inductor with inductance L (constant). Initial current I; then current is reduced to I/2. No losses considered during the change for the purpose of comparing stored energy values. Concept / Approach:The magnetic energy stored in an inductor is W = 0.5 * L * I^2. If current changes by a factor k, the energy scales by k^2. Halving current (k = 0.5) therefore reduces energy by (0.5)^2 = 0.25, i.e., one quarter of the original. Step-by-Step Solution: Initial energy: W1 = 0.5 * L * I^2.New current: I2 = I/2.New energy: W2 = 0.5 * L * (I/2)^2 = 0.5 * L * (I^2 / 4) = (1/4) * (0.5 * L * I^2) = W1 / 4. Verification / Alternative check:Pick L = 1 H, I = 2 A for a quick check: W1 = 0.514 = 2 J. With I/2 = 1 A, W2 = 0.511 = 0.5 J = 2/4, confirming the quartering effect. Why Other Options Are Wrong: Quadruples or doubles: These imply increasing, not reducing, energy. Does not change: Contradicts the I^2 dependence; energy must change when current changes. Common Pitfalls: Assuming linear energy-current relation instead of quadratic. Confusing inductive energy with capacitor energy (which also uses a squared term, but in voltage). Final Answer:is quartered

Question 18

Magnitude comparison: An inductance of 0.43 mH is smaller than which of the following values?

Accepted Answer

Introduction / Context:Unit conversion agility (µH, mH, H) is indispensable when comparing component values from different datasheets or suppliers. This problem asks you to identify which listed quantity is strictly larger than 0.43 mH, because only then can we say 0.43 mH is smaller than that quantity. Given Data / Assumptions: Reference value: 0.43 mH. Options include microhenry, millihenry, and henry forms. Accurate unit conversion is required. Concept / Approach:Convert each option to the same unit (e.g., mH) for clean comparison. Remember: 1 H = 1000 mH and 1 mH = 1000 µH. Equality must not be mistaken for “smaller than.” Step-by-Step Solution: 430 µH = 0.430 mH = 0.43 mH (equal, not larger).0.00043 H = 0.43 mH (equal, not larger).43 µH = 0.043 mH (smaller than 0.43 mH).0.05 H = 50 mH (much larger than 0.43 mH). Verification / Alternative check:Compare orders of magnitude: 0.43 mH is far below 1 H. Converting 0.05 H to mH gives 50 mH, which clearly exceeds 0.43 mH, validating the choice. Why Other Options Are Wrong: 430 µH and 0.00043 H: These are exactly equal to 0.43 mH, not larger. 43 µH: Smaller than 0.43 mH, so the statement “0.43 mH is smaller than 43 µH” is false. Common Pitfalls: Mistaking equality for greater-than or less-than. Dropping or misplacing decimal points when converting between H, mH, and µH. Final Answer:0.05 H

Question 19

Series combination of inductors: A 2 mH, a 3.3 mH, and a 0.2 mH inductor are connected in series. What is the total inductance of the series path?

Accepted Answer

Introduction / Context:For uncoupled inductors, series connection leads to simple arithmetic addition of inductances. This is analogous to series resistors and is frequently used to realize non-standard inductance values by combining available parts. Given Data / Assumptions: L1 = 2 mH, L2 = 3.3 mH, L3 = 0.2 mH. Series connection; mutual coupling neglected. All values expressed in millihenries for easy arithmetic. Concept / Approach:Use L_total(series) = L1 + L2 + L3. No conversions are needed since all are in mH. Ensure you do not mistakenly compute a parallel equivalent. Step-by-Step Solution: L_total = 2 + 3.3 + 0.2 = 5.5 mH. Verification / Alternative check:If mutual coupling were present with tight coupling (not given here), results could change due to induced voltages. Since the problem omits coupling, pure addition applies—consistent with standard textbook treatment. Why Other Options Are Wrong: 55 mH: Off by a factor of 10. less than 5.5 mH / less than 0.2 mH: Incorrect because series values add; they cannot yield less than the largest single inductor. Common Pitfalls: Accidentally treating the connection as parallel and applying reciprocal formulas. Final Answer:5.5 mH

Question 20

Inductor voltage law: For a coil with L = 8 µH, if di/dt = 20 mA/µs, what is the induced voltage across the coil (v = L * di/dt)?

Accepted Answer

Introduction / Context:The defining inductor relation v = L * di/dt directly links voltage to how quickly current changes. Mastering unit conversions (µH, mA/µs) is critical for translating lab measurements into correct voltage estimates, especially in switching converters and transient analysis. Given Data / Assumptions: L = 8 µH = 8 × 10^-6 H. di/dt = 20 mA/µs = 20 × 10^-3 A per 10^-6 s = 2.0 × 10^4 A/s. Ideal inductor; polarity per Lenz's law is not required for magnitude. Concept / Approach:Apply the formula v = L * di/dt using consistent SI units. Convert microhenry to henry and milliampere-per-microsecond to ampere-per-second before multiplying to avoid powers-of-ten errors. Step-by-Step Solution: Convert L: 8 µH = 8e-6 H.Convert di/dt: 20 mA/µs = 0.02 A / 1e-6 s = 2.0e4 A/s.Compute v: v = 8e-6 * 2.0e4 = 1.6e-1 V = 0.16 V = 160 mV. Verification / Alternative check:Scale check: 10 µH * 10 mA/µs → 0.1 V. Our L and di/dt are close, giving ≈0.16 V, which is consistent by proportionality. Why Other Options Are Wrong: 16 mV or 1.6 mV or 2.5 mV: These miss factors of 10^1–10^2 due to unit-conversion mistakes. Common Pitfalls: Treating mA/µs as mA/s or forgetting to convert µH to H. Final Answer:160 mV

Inductors Questions

Questions

Practice smarter