Q: Transformer ratios — in a step-down transformer, is the turns ratio Ns/Np less than 1? Assume turns ratio is defined as secondary turns divided by primary turns.

Introduction / Context: The turns ratio determines how a transformer changes voltage and current. A step down is used when a higher supply voltage must be reduced for a load, such as adapting mains to equipment levels. This item checks the sign and magnitude of the turns ratio for a step-down device. Given Data / Assumptions: Turns ratio defined as Ns/Np, where Ns is secondary turns and Np is primary turns. Ideal transformer assumptions for clarity. Steady sinusoidal operation. Concept / Approach: Voltage ratio follows Vs/Vp = Ns/Np. For a step-down transformer, Vs 1. Thus, a step-down device has fewer secondary turns than primary turns and consequently a turns ratio less than one under the given definition. Step-by-Step Solution: Start with Vs/Vp = Ns/Np.For step down, Vs/Vp 1, consistent with higher secondary current.Therefore, the statement is correct when ratio is defined as Ns/Np. Verification / Alternative check: Example: Np = 1000 turns, Ns = 100 turns → Ns/Np = 0.1. With Vp = 230 V, ideal Vs = 23 V, clearly a step down. Current is stepped up by 10 times ideally. Why Other Options Are Wrong: Marking “False” generally results from using the inverse definition (Np/Ns). Under that alternate definition, a step-down transformer has a ratio greater than one. The statement here explicitly uses Ns/Np. Common Pitfalls: Ambiguity in how ratio is quoted. Always specify whether you mean Ns/Np or Np/Ns when comparing devices to avoid misinterpretation in specifications and procurement. Final Answer: True

Q: Primary power vs. load change — behavior in an ideal transformer: If the load current on the secondary increases, does the real power drawn by the primary remain the same?

Introduction / Context: Transformers reflect load changes back to the primary. Understanding the dynamic between secondary load and primary input is crucial for power budgeting, fuse sizing, and efficiency calculations. If the secondary draws more power, the primary must supply more power (plus any losses) to maintain energy balance. Given Data / Assumptions: Ideal or high-efficiency transformer. Voltage ratio set by turns ratio; primary supply voltage remains nominal. No internal regulation inherent to the transformer itself. Concept / Approach: In steady state, conservation of power dictates P_primary ≈ P_secondary/efficiency. As the secondary current increases (for the same secondary voltage), secondary real power rises. Therefore, the primary real power must also rise proportionally (ignoring minor changes due to voltage regulation and losses). It cannot “remain the same.” Step-by-Step Solution: Let secondary power change by ΔP > 0 due to load current increase.Ideal transformer: Pp = Ps (no losses).Hence primary power must increase by the same ΔP.Therefore, the statement is false. Verification / Alternative check: Measure input current as the secondary load resistor is decreased. The primary current increases correspondingly, showing higher input wattage on a wattmeter. Why Other Options Are Wrong: “Correct” contradicts power balance. Qualifiers about step-down, unity power factor, or voltage regulation do not change the fundamental energy conservation requirement. Common Pitfalls: Assuming the transformer itself regulates power; in reality, it only transforms voltage/current ratios and transfers power demanded by the load. Final Answer: Incorrect

Question 1

Transformer Windings: Primary vs. Secondary Connections  Evaluate the statement: “The primary is the winding connected to the source, and the secondary is the winding connected to the load.”

Accepted Answer

Introduction / Context: Correctly naming transformer windings helps avoid wiring mistakes and data-sheet misinterpretation. The distinction between “primary” and “secondary” is based on connection, not on voltage magnitude or physical size of the winding. Given Data / Assumptions: A two-winding transformer operated in normal direction. Source side and load side can be swapped in some applications, but names follow connection. AC excitation and passive magnetic coupling apply. Concept / Approach: By convention, the winding connected to the input source is the primary, and the winding delivering power to the load is the secondary. Either winding may be higher or lower voltage depending on turns ratio and application (step-down or step-up). Naming does not imply voltage level. Step-by-Step Solution: Identify the source-connected winding → call it primary.Identify the load-connected winding → call it secondary.Apply turns ratio to determine voltage and current scaling but not the naming.Thus, the statement is consistent with standard practice and terminology. Verification / Alternative check: Datasheets label terminals “PRI” (primary) and “SEC” (secondary) with connection diagrams, confirming the naming depends on which side is tied to the source. Why Other Options Are Wrong: Restricting truth to step-down or center-tapped designs is unnecessary. The primary is not “always the higher-voltage winding”; step-up operation flips that condition. Common Pitfalls: Assuming the thicker wire is always the primary or that primary must be the higher voltage—both may vary with design constraints. Final Answer: True

Question 2

Center Tap (CT) Definition  Evaluate the statement: “A center tap (CT) is a connection at the midpoint of the secondary winding of a transformer.”

Accepted Answer

Introduction / Context: A center tap is common in power supplies and audio circuits. Understanding exactly what and where it is helps when wiring full-wave rectifiers, creating split supplies, or obtaining balanced signals from a single secondary. Given Data / Assumptions: Conventional transformer with a tapped secondary. “Midpoint” refers to equal turns on each side of the tap. Ideal, symmetrical winding for the definition. Concept / Approach: A center tap is a lead brought out from the halfway point of the secondary winding so that the turns are divided into two equal halves. This yields two equal but opposite-polarity voltages with respect to the CT reference, which is often used as a return or midpoint in rectifiers or dual-rail supplies. Step-by-Step Solution: Let the secondary have Ns turns from one end to the other.A CT connection is taken at Ns/2 turns, providing two segments of equal turns.Measured from CT to either end, the RMS voltage is approximately half of the full secondary voltage.Therefore, the CT is indeed the midpoint connection of the secondary winding. Verification / Alternative check: Check labeling on schematic symbols and transformer datasheets: the CT is identified as the middle lead, often used to create a pair of equal voltages for full-wave rectification. Why Other Options Are Wrong: Limiting to step-down or rectifier uses is unnecessary; CTs appear in many designs. Saying it is on the primary is incorrect by definition in this common context (although primary taps for voltage selection exist, a “CT” typically refers to the secondary midpoint in power supplies). Common Pitfalls: Confusing CT with multiple independent secondaries or with adjustable taps placed at non-midpoint locations. Final Answer: True

Question 3

Transformer fundamentals — if the secondary voltage is stepped up, is the secondary current stepped down (and vice versa)? Assume an ideal transformer with negligible losses and constant power transfer.

Accepted Answer

Introduction / Context: This statement checks a core idea of transformers used in power systems and electronics. In an ideal transformer, power on the primary side equals power on the secondary side, ignoring losses. Therefore, when voltage is changed by the turns ratio, current must change inversely so that the product of voltage and current stays approximately constant. Given Data / Assumptions: Ideal transformer operation with negligible copper and core losses. Turns ratio defined as Np:Ns where Np is primary turns and Ns is secondary turns. Sine wave steady state with no saturation. Concept / Approach: For an ideal transformer, Vp/Vs = Np/Ns and Ip/Is = Ns/Np. Combining the two relations gives Vp * Ip ≈ Vs * Is, which is constant power transfer in the ideal case. Hence, if the secondary voltage is stepped up, the secondary current must be stepped down proportionally so that power balance is satisfied. The reverse happens for a step down in voltage. Step-by-Step Solution: Voltage ratio: Vs = Vp * (Ns/Np).Current ratio: Is = Ip * (Np/Ns).Power balance: Vp * Ip ≈ Vs * Is.If Ns/Np > 1 (step up), then Is/Ip = Np/Ns < 1, so current is reduced; if Ns/Np < 1 (step down), current is increased. Verification / Alternative check: A numerical example makes the point. Suppose Vp = 120 V, Ip = 2 A, and Ns/Np = 2. Then Vs = 240 V and Is = 1 A, giving 240 V * 1 A = 240 W which closely matches the primary 120 V * 2 A = 240 W in the ideal case. The inverse relation between voltage and current is evident. Why Other Options Are Wrong: “False” contradicts the transformer current ratio derived from Faraday’s law and ideal power conservation. A higher secondary voltage without a lower secondary current would imply power creation in the transformer, which is not possible. Common Pitfalls: Confusing real devices with ideal behavior. Real transformers have copper loss, core loss, and limited regulation, but the inverse relation between voltage and current still describes the dominant behavior for design estimates. Final Answer: True

Question 4

Transformer ratios — in a step-down transformer, is the turns ratio Ns/Np less than 1? Assume turns ratio is defined as secondary turns divided by primary turns.

Accepted Answer

Introduction / Context: The turns ratio determines how a transformer changes voltage and current. A step down is used when a higher supply voltage must be reduced for a load, such as adapting mains to equipment levels. This item checks the sign and magnitude of the turns ratio for a step-down device. Given Data / Assumptions: Turns ratio defined as Ns/Np, where Ns is secondary turns and Np is primary turns. Ideal transformer assumptions for clarity. Steady sinusoidal operation. Concept / Approach: Voltage ratio follows Vs/Vp = Ns/Np. For a step-down transformer, Vs < Vp, which requires Ns/Np < 1. Current ratio is inverse: Is/Ip = Np/Ns > 1. Thus, a step-down device has fewer secondary turns than primary turns and consequently a turns ratio less than one under the given definition. Step-by-Step Solution: Start with Vs/Vp = Ns/Np.For step down, Vs/Vp < 1 → Ns/Np < 1.Current relation: Is/Ip = 1 / (Ns/Np) = Np/Ns > 1, consistent with higher secondary current.Therefore, the statement is correct when ratio is defined as Ns/Np. Verification / Alternative check: Example: Np = 1000 turns, Ns = 100 turns → Ns/Np = 0.1. With Vp = 230 V, ideal Vs = 23 V, clearly a step down. Current is stepped up by 10 times ideally. Why Other Options Are Wrong: Marking “False” generally results from using the inverse definition (Np/Ns). Under that alternate definition, a step-down transformer has a ratio greater than one. The statement here explicitly uses Ns/Np. Common Pitfalls: Ambiguity in how ratio is quoted. Always specify whether you mean Ns/Np or Np/Ns when comparing devices to avoid misinterpretation in specifications and procurement. Final Answer: True

Question 5

Ideal transformer polarity — dot convention statement: In an ideal transformer, when the primary and secondary dotted terminals are taken as the positive reference ends, the secondary voltage is in phase with the primary voltage.

Accepted Answer

Introduction / Context: Transformer polarity determines the phase relationship between primary and secondary voltages. The dot convention marks terminals that have the same instantaneous polarity, which is crucial when connecting multiple windings, building rectifiers, or summing secondary voltages without unintended cancellation. Given Data / Assumptions: Ideal transformer behavior (no losses, no leakage, infinite magnetizing inductance assumed for simplicity). Dots identify the corresponding winding ends. Phase comparison is with respect to the dotted ends as reference polarity. Concept / Approach: In the dot convention, an instantaneous positive voltage applied to the dotted end of the primary produces an instantaneous positive voltage at the dotted end of the secondary (scaled by turns ratio). Therefore, referenced at the dotted terminals, the voltages are in phase regardless of the magnitude change set by the turns ratio. Step-by-Step Solution: Identify dotted terminals on both windings.Apply a positive-going primary voltage at the dotted end.By definition, a positive secondary voltage appears at its dotted end simultaneously.Conclude: the voltages are in phase with respect to the dotted ends. Verification / Alternative check: Reversing either winding (switching the dot to the opposite end physically) introduces a 180° phase reversal. Practical tests with low AC safely confirm the indicated polarity via oscilloscope observation. Why Other Options Are Wrong: Polarity is independent of step-up/step-down status, line frequency within typical operation, and load type; these factors change magnitudes and currents, not the dot-defined phase relationship. Common Pitfalls: Confusing the sign of induced voltage due to winding direction, or ignoring the dots when paralleling or series-connecting secondaries, leading to inadvertent cancellation or shorting. Final Answer: Correct

Question 6

Transformer power conservation (ideal case): With 100% efficiency (no losses), the power transferred to the secondary equals the power drawn from the primary, independent of step-up or step-down operation.

Accepted Answer

Introduction / Context: Transformers transfer power between circuits via magnetic coupling. In the ideal case of 100% efficiency, the power into the primary equals the power delivered by the secondary. This principle underlies transformer ratings, turns-ratio design, and safety considerations in power systems and electronics. Given Data / Assumptions: Ideal (lossless) transformer: no copper loss, core loss, or leakage. Arbitrary turns ratio may step voltage up or down. Power factor may be anything; equality is for real power values. Concept / Approach: For an ideal transformer, Vp/Vs = Np/Ns and Ip/Is = Ns/Np. Apparent power balance gives Vp * Ip = Vs * Is. When no losses exist, input real power equals output real power even if voltage and current magnitudes change. The transformer trades current for voltage according to the turns ratio while conserving power. Step-by-Step Solution: Start with Vp/Vs = Np/Ns.Relate currents: Ip/Is = Ns/Np.Multiply: Vp * Ip = Vs * Is (apparent power equality).With 100% efficiency, real power in = real power out. Verification / Alternative check: Measure primary and secondary wattage with a true wattmeter under a resistive load; losses in practical units introduce small discrepancies but the ideal model remains an excellent approximation. Why Other Options Are Wrong: Limiting the statement to step-up or step-down is unnecessary. Unity power factor is not required for power equality; it affects currents and voltages but not ideal conservation. Common Pitfalls: Confusing apparent power S with real power P when power factor is not unity, and misinterpreting the roles of core and copper losses in non-ideal transformers. Final Answer: Correct

Question 7

Reflected load through a transformer — magnitude comparison: Is the equivalent (reflected) load seen at the primary always larger than the actual load connected to the secondary?

Accepted Answer

Introduction / Context: The concept of reflected impedance allows us to analyze a transformer-coupled load from the source side. Designers choose turns ratios to “transform” impedances to desired levels. It is a common misconception that the reflected load is always larger than the physical load; the truth depends entirely on the turns ratio. Given Data / Assumptions: Ideal transformer with turns ratio a = Np/Ns. Load Z_L is connected across the secondary. Reflected impedance seen at the primary is Z_in = a^2 * Z_L. Concept / Approach: Because Z_in = (Np/Ns)^2 * Z_L, the input impedance scales by the square of the turns ratio. If Np > Ns (step-down voltage), a > 1 and Z_in > Z_L. If Np < Ns (step-up voltage), a < 1 and Z_in < Z_L. Therefore, the reflected load can be larger, equal, or smaller, depending on the chosen ratio. Step-by-Step Solution: Define a = Np/Ns.Compute Z_in = a^2 * Z_L.Case a > 1 ⇒ Z_in > Z_L; case a = 1 ⇒ Z_in = Z_L; case a < 1 ⇒ Z_in < Z_L.Conclude that “always larger” is false. Verification / Alternative check: Example: Np = 100, Ns = 200 ⇒ a = 0.5. For Z_L = 100 Ω, Z_in = 0.25 * 100 = 25 Ω, clearly smaller than Z_L. Why Other Options Are Wrong: “Correct” contradicts the square-law scaling. Statements limited to specific frequencies ignore that the impedance transformation is frequency-independent in the ideal model (though Z_L itself may be frequency-dependent). Common Pitfalls: Forgetting the square on the turns ratio and assuming a linear scaling; mixing up primary and secondary labeling when applying a^2. Final Answer: Incorrect

Question 8

Primary power vs. load change — behavior in an ideal transformer: If the load current on the secondary increases, does the real power drawn by the primary remain the same?

Accepted Answer

Introduction / Context: Transformers reflect load changes back to the primary. Understanding the dynamic between secondary load and primary input is crucial for power budgeting, fuse sizing, and efficiency calculations. If the secondary draws more power, the primary must supply more power (plus any losses) to maintain energy balance. Given Data / Assumptions: Ideal or high-efficiency transformer. Voltage ratio set by turns ratio; primary supply voltage remains nominal. No internal regulation inherent to the transformer itself. Concept / Approach: In steady state, conservation of power dictates P_primary ≈ P_secondary/efficiency. As the secondary current increases (for the same secondary voltage), secondary real power rises. Therefore, the primary real power must also rise proportionally (ignoring minor changes due to voltage regulation and losses). It cannot “remain the same.” Step-by-Step Solution: Let secondary power change by ΔP > 0 due to load current increase.Ideal transformer: Pp = Ps (no losses).Hence primary power must increase by the same ΔP.Therefore, the statement is false. Verification / Alternative check: Measure input current as the secondary load resistor is decreased. The primary current increases correspondingly, showing higher input wattage on a wattmeter. Why Other Options Are Wrong: “Correct” contradicts power balance. Qualifiers about step-down, unity power factor, or voltage regulation do not change the fundamental energy conservation requirement. Common Pitfalls: Assuming the transformer itself regulates power; in reality, it only transforms voltage/current ratios and transfers power demanded by the load. Final Answer: Incorrect

Question 9

Center-tapped secondary — phase relationship of the two half-windings: In a center-tapped transformer, the two secondary end-to-center-tap voltages have equal magnitude and are 180° out of phase with respect to the center tap.

Accepted Answer

Introduction / Context: Center-tapped secondaries are common in full-wave rectifiers and dual-polarity supplies. The phase between the two half-windings determines how diodes conduct in rectifier circuits and how outputs can be combined to generate positive and negative rails relative to the center tap. Given Data / Assumptions: Ideal transformer with a center-tapped secondary winding. The center tap is taken as the reference (0 V). Symmetrical construction so both halves have equal turns. Concept / Approach: Because the two secondary halves are wound in opposite directions relative to the center reference, their instantaneous polarities are opposite. At any instant, one end is at +V with respect to the center tap while the other is at –V of equal magnitude. This phase opposition is 180° when referenced to the center tap, a key requirement for classic full-wave rectifier operation using two diodes and a center tap. Step-by-Step Solution: Take the center tap as 0 V.At an instant when one outer end is +V, the other is –V of the same magnitude.Thus the two voltages are equal in amplitude and opposite in phase.Conclusion: 180° out of phase with respect to the center tap. Verification / Alternative check: Oscilloscope traces show two sine waves of equal amplitude mirrored about 0 V when probing each end to the center tap. Dot markings also confirm the opposite instantaneous polarities. Why Other Options Are Wrong: The relationship does not depend on step-up/down ratio, load, or line frequency, provided symmetry and proper phasing are maintained. Common Pitfalls: Referencing end-to-end voltage instead of each end to center tap; miswiring that swaps dot orientation causing unexpected phasing. Final Answer: Correct

Question 10

Open-circuit secondary — primary current behavior: If the secondary winding of a transformer is open-circuited, the primary current is nearly zero (only the small magnetizing current flows).

Accepted Answer

Introduction / Context: Understanding primary current under different load conditions is crucial for safe testing and operation of transformers. With the secondary open, the transformer is unloaded; therefore, only the magnetizing current (to establish core flux and cover core losses) flows in the primary. Given Data / Assumptions: Idealized view: no load on the secondary, so no real power transfer. Real transformers: a small magnetizing current and small core-loss current still exist. Frequency within normal operating range. Concept / Approach: When the secondary is open, there is no path for load current. The primary draws just enough current to magnetize the core and supply core losses (hysteresis and eddy currents). This current is typically a few percent of full-load current. Thus, compared to loaded operation, the primary current is “nearly zero.” Step-by-Step Solution: Open secondary ⇒ I_secondary ≈ 0.No reflected load ⇒ input impedance is high.Primary current reduces to magnetizing + core-loss components.Therefore, I_primary is small relative to rated current. Verification / Alternative check: Clamp-meter measurement on a small power transformer with open secondary typically shows 2–10% of rated primary current, verifying the “nearly zero” characterization for power transfer purposes. Why Other Options Are Wrong: Step-up/step-down status, exact frequency, or core lamination do not change the fundamental behavior; only the magnitude of magnetizing/core-loss current varies slightly. Common Pitfalls: Expecting literally zero current; in practice, some current is always required to sustain core flux and cover losses. Final Answer: Correct

Question 11

Voltage–current trade in an ideal transformer: If the secondary voltage is stepped up by a certain ratio, the secondary current is stepped down by the same ratio (inverse of the turns ratio).

Accepted Answer

Introduction / Context: Transformers exchange voltage for current according to their turns ratio. This principle allows transmission at high voltage and low current to reduce I^2R losses, and then stepping down voltage for safe utilization while increasing current. Recognizing the inverse relationship prevents overcurrent conditions and ensures correct component sizing. Given Data / Assumptions: Ideal transformer with turns ratio a = Np/Ns. Voltage ratio: Vs/Vp = Ns/Np. Current ratio: Is/Ip = Np/Ns (inverse of voltage ratio). Concept / Approach: Power conservation in the ideal case implies Vp * Ip = Vs * Is. Combining this with the voltage ratio immediately yields the inverse current ratio. Therefore, if voltage is stepped up by factor k, current is stepped down by factor k, preserving power (neglecting losses). Step-by-Step Solution: Let k = Ns/Np (voltage step-up factor).Then Vs = k * Vp and Is = (1/k) * Ip.Thus, increasing voltage reduces current by the same factor.This is independent of load type in the ideal model. Verification / Alternative check: Example: 1:10 step-up. If Vp = 10 V and Ip = 1 A, then Vs = 100 V and Is ≈ 0.1 A (neglecting losses), confirming the inverse relationship. Why Other Options Are Wrong: Load power factor and transformer size do not alter the ratio identity; they affect real vs. reactive power and efficiency but not the ideal ratio relationships. Common Pitfalls: Confusing primary vs. secondary labels or using the turns ratio directly for current rather than its reciprocal. Final Answer: Correct

Question 12

Transformer basics — turns ratio check: A transformer has 700 turns on the primary winding and 35 turns on the secondary winding. Determine whether the stated turns ratio of 20:1 (primary:secondary) is correct, based on the definition turns ratio = …

Accepted Answer

Introduction / Context: Transformers are characterized by their turns ratio, defined as the number of primary turns to the number of secondary turns. This ratio directly sets the ideal voltage ratio and relates to the current ratio inversely. The question asks you to verify whether a primary of 700 turns and a secondary of 35 turns corresponds to a 20:1 turns ratio. Given Data / Assumptions: Primary turns Np = 700. Secondary turns Ns = 35. Standard definition of turns ratio: a = Np:Ns. Assume conventional winding definitions; no need for frequency or load to compute the ratio. Concept / Approach: The turns ratio is purely a geometric/winding count relationship. For ideal transformers, Vp/Vs = Np/Ns and Is/Ip = Np/Ns. Determining the turns ratio requires only the turn counts, not operating conditions. Step-by-Step Solution: Compute a = Np / Ns = 700 / 35.700 / 35 = 20.Therefore, a = 20:1 (primary:secondary).The given statement “turns ratio is 20:1” is correct. Verification / Alternative check: If the transformer is ideal, a = Vp/Vs. A 20:1 turns ratio would step 120 V to 6 V, or 240 V to 12 V, which are common secondary voltages, supporting the plausibility of the computed ratio. Why Other Options Are Wrong: Incorrect: The arithmetic clearly shows 700:35 reduces to 20:1. Correct only for ideal transformers: The turns ratio itself is independent of ideality; ideality affects voltage regulation and losses, not the count ratio. Indeterminate without frequency: Frequency does not change the turns ratio calculation. Common Pitfalls: Confusing turns ratio with impedance ratio (which is the square of the turns ratio). Mixing up primary and secondary order when expressing the ratio. Final Answer: Correct

Question 13

Condition for maximum power transfer — statement check: In a linear DC network, maximum power is transferred to the load when the load resistance equals the Thevenin (source) resistance seen from the load terminals. Evaluate this statement.

Accepted Answer

Introduction / Context: The maximum power transfer principle is a staple of circuit design, particularly in instrumentation, audio, and communications. The practical question is: what relationship between the source and load ensures that the load receives the greatest possible power for a given source? Given Data / Assumptions: Linear DC or low-frequency network modeled by a Thevenin source (Vth in series with Rth). Load is purely resistive with resistance RL. No reactive components or frequency-dependent matching in this simplified statement. Concept / Approach: For a Thevenin source, power in the load is P = (Vth^2 * RL) / (Rth + RL)^2. Differentiating with respect to RL and setting the derivative to zero yields the optimum at RL = Rth. This result maximizes load power but does not maximize efficiency (which is 50% at the optimum). Step-by-Step Solution: Start with P(RL) = (Vth^2 * RL) / (Rth + RL)^2.Differentiate: dP/dRL = 0 occurs when Rth = RL.Check second derivative or compare values to confirm it is a maximum.Conclusion: Maximum power transfer occurs at RL = Rth (for DC resistive networks). Verification / Alternative check: Choose Vth and Rth, plot P versus RL; the peak appears at RL = Rth. In AC with complex impedances, the generalized condition is ZL = Zth* (complex conjugate). Why Other Options Are Wrong: Incorrect: Conflicts with the calculus optimization result. Applies only at resonance: Resonance is unrelated to the DC resistive case; in AC, conjugate matching is the general rule. Requires RL = 0: A short circuit yields zero load power (all power dissipates in the source resistance). Common Pitfalls: Confusing maximum power transfer with maximum efficiency. Forgetting that in AC networks with reactance, matching is to the complex conjugate of the source impedance. Final Answer: Correct

Question 14

Mutual inductance geometry — concept check: For mutual inductance to occur between two coils, must the coils be oriented at right angles to each other, or is coupling maximized when their magnetic axes are aligned?

Accepted Answer

Introduction / Context: Mutual inductance quantifies how effectively a changing current in one coil induces a voltage in another. Coil orientation and core material are key design variables in transformers, inductive sensors, and wireless power systems. Given Data / Assumptions: Two coils placed in proximity, potentially with a magnetic core. Mutual inductance M depends on flux linkage from the primary to the secondary. Coupling coefficient k ranges from 0 (no coupling) to 1 (perfect coupling). Concept / Approach: Mutual inductance M = k * sqrt(L1 * L2). The coupling coefficient k is maximized when the magnetic axes of the coils are aligned so that the primary’s flux directly links the secondary. When coils are oriented at right angles (orthogonal), the linked flux is minimized, and k trends toward zero. Step-by-Step Solution: Consider coil axes: parallel alignment → maximum shared flux; right-angle alignment → minimal shared flux.Flux linkage drives induced voltage: e2 = M * di1/dt; with small k at right angles, e2 is small.Therefore, mutual inductance does not require right-angle orientation and is actually minimized in that orientation.Conclusion: The statement requiring right-angle placement is false. Verification / Alternative check: Transformer design universally aligns windings concentrically on a common core to maximize coupling; cross-axis placement is used when intentional low coupling is desired (e.g., some pickup coils). Why Other Options Are Wrong: Correct: Would assert an incorrect geometrical condition. High-frequency or nonmagnetic qualifiers do not invert the fundamental orientation effect: aligned axes produce more coupling than orthogonal axes. Common Pitfalls: Confusing electromagnetic interference reduction strategies (orthogonal placement reduces coupling) with the requirement for mutual inductance. Assuming any proximity guarantees strong coupling. Final Answer: Incorrect

Question 15

Impedance matching with a transformer — compute required turns ratio: Find the primary-to-secondary turns ratio needed so that a 5 kΩ source (to be matched) properly drives an 8 Ω loudspeaker. Use the relationship (Np/Ns)^2 = Zp/Zs and report the ra…

Accepted Answer

Introduction / Context: Audio and RF systems often need to match a relatively high source impedance to a low speaker or antenna impedance. Transformers provide impedance transformation where the impedance ratio equals the square of the turns ratio. Given Data / Assumptions: Desired match: Zp = 5 kΩ (5000 Ω) to Zs = 8 Ω. Ideal transformer assumption for first-pass calculation (no leakage, no losses). Definition: (Np/Ns)^2 = Zp/Zs. Concept / Approach: Impedance seen from the primary is related to the load by Z_reflected = (Np/Ns)^2 * Z_load. To make the source “see” 5 kΩ, choose Np/Ns so that (Np/Ns)^2 = 5000/8. Step-by-Step Solution: Compute the ratio: 5000 / 8 = 625.Take the square root: sqrt(625) = 25.Therefore, Np/Ns = 25, so the required turns ratio is 25:1 (primary:secondary).This ratio matches 5 kΩ to 8 Ω under ideal conditions. Verification / Alternative check: If Np/Ns = 25, then Z_reflected = 25^2 * 8 = 625 * 8 = 5000 Ω, exactly the target source impedance. This confirms the computation. Why Other Options Are Wrong: 5:1 and 8:1 yield impedance ratios of 25 and 64, far from 625. 625:1 produces an absurd impedance ratio of 390,625, which is not appropriate here. Common Pitfalls: Forgetting to square the turns ratio when converting impedances. Mixing up the direction (primary vs secondary) and thereby inverting the ratio. Ignoring non-idealities (leakage, winding resistance), which slightly alter real-world matching but not the fundamental ratio. Final Answer: 25:1

Question 16

Induction terminology — name the concept: Identify the correct term for the phenomenon in which a changing current in one coil induces a voltage in a nearby second coil.

Accepted Answer

Introduction / Context: Transformers, wireless power transfer, and many sensing applications rely on one circuit inducing a response in another via magnetic fields. Knowing the correct terminology helps you read datasheets and design notes accurately. Given Data / Assumptions: Two separate coils placed such that magnetic flux from one can link the other. Current in the first coil is time varying, producing a changing magnetic field. The second coil experiences a changing flux linkage and therefore an induced voltage. Concept / Approach: The induced voltage in the second coil due to current change in the first is described by mutual inductance M. The coupling coefficient k (0 to 1) quantifies how much of the first coil’s flux links the second. Self-inductance L, by contrast, refers to voltage induced in the same coil that carries the changing current. Step-by-Step Solution: A changing current i1 in coil 1 creates a changing flux.A portion of that flux links coil 2, creating e2 = M * di1/dt.This cross-coupling is called mutual inductance.Therefore, the correct term is “Mutual inductance.” Verification / Alternative check: Transformers are practical demonstrations: primary current changes induce secondary voltages proportional to the mutual inductance (through the turns ratio framework). Why Other Options Are Wrong: Electrical isolation: a benefit of transformer design but not the induction mechanism’s name. Self-inductance: refers to a coil inducing voltage in itself from its own current change. Coefficient of coupling: a dimensionless parameter k describing linkage efficiency, not the fundamental phenomenon’s name. Common Pitfalls: Confusing mutual inductance with simple capacitive coupling. Assuming mutual inductance requires physical contact; it is through the magnetic field. Final Answer: Mutual inductance

Question 17

Step-up vs step-down — what changes on the secondary? For a step-up transformer (higher secondary voltage than primary), indicate what quantity increases at the secondary side in an ideal transformer.

Accepted Answer

Introduction / Context: Transformers trade voltage and current while conserving power (ideally). A step-up transformer produces a higher voltage at the secondary than at the primary, which has practical applications in power distribution and certain electronics. Given Data / Assumptions: Ideal transformer (no copper losses, no core losses, no leakage). Turns ratio a = Np/Ns < 1 for a step-up (Ns > Np). Power balance approximates to Vp * Ip ≈ Vs * Is (ideal). Concept / Approach: Voltage ratio equals turns ratio: Vs/Vp = Ns/Np. For a step-up transformer, Ns > Np, so Vs > Vp. To keep power roughly constant, current goes the other way: Is < Ip. Power does not increase; transformers do not create power, they only transform it. Step-by-Step Solution: Recognize “step-up”: Ns > Np → Vs/Vp = Ns/Np > 1 → Vs increases.Current relation: Is/Ip = Np/Ns < 1 → secondary current decreases.Ideal power: Ps ≈ Pp; real transformers have small losses.Thus, the quantity that increases at the secondary is voltage. Verification / Alternative check: Example: Np = 100, Ns = 1000. If Vp = 12 V, Vs ≈ 120 V (voltage up), while current scales down by 10x. Why Other Options Are Wrong: Increase the current: False; current decreases when voltage steps up. Increase the power: Power is not increased by an ideal transformer; it is transferred with losses only. Decrease the voltage: Opposite of a step-up function. Common Pitfalls: Assuming transformers “boost power.” Forgetting that higher voltage implies lower current for similar power transfer. Final Answer: increase the voltage

Question 18

Transformer nonideal effect — name the phenomenon: Identify the effect that occurs when some of the primary’s magnetic flux does not link the core path to the secondary and instead travels through surrounding air paths.

Accepted Answer

Introduction / Context: Real transformers deviate from the ideal due to losses and imperfect coupling. Understanding each nonideal effect is essential for predicting voltage regulation, efficiency, and bandwidth. Given Data / Assumptions: Primary and secondary windings share a core but not all flux is perfectly confined. Some field lines escape the core and loop through air or structural parts. We classify this behavior among nonideal effects. Concept / Approach: Magnetic flux leakage refers to the portion of magnetic flux generated by the primary that fails to couple to the secondary. Leakage effectively inserts series inductances and reduces the coupling coefficient k, lowering mutual inductance and altering transient response. Step-by-Step Solution: Primary creates magnetizing flux; in ideal case, nearly all links the secondary.In practice, some flux escapes the core path → air paths with much higher reluctance.This uncoupled portion is “leakage flux,” causing leakage inductance.Therefore, the named effect is magnetic flux leakage. Verification / Alternative check: Designers reduce leakage by interleaving windings, using high-permeability cores, and minimizing air gaps. Measurement of leakage inductance in equivalent circuits confirms the effect quantitatively. Why Other Options Are Wrong: Hysteresis loss: Energy lost per cycle in the core due to magnetization reversal; not uncoupled flux per se. Winding resistance: Copper loss (I^2*R), unrelated to flux paths. Winding capacitance: Parasitic capacitive effects between turns/layers; not flux leakage. Common Pitfalls: Confusing leakage flux with fringing flux at air gaps; both involve air paths but have different implications. Equating leakage with core loss; they are distinct. Final Answer: Magnetic flux leakage

Question 19

Terminology — what is the practice of making a load appear equal to a source’s internal resistance? Choose the term commonly used in circuits and systems when the load is adjusted (often via a matching network or transformer) so that it equals the s…

Accepted Answer

Introduction / Context: Interfacing a source with a load is a universal task in electronics. Designers aim either for maximum power transfer, maximum voltage transfer, or minimum reflection (in RF). The process of adjusting networks so the load “looks like” the desired value at the interface has a standard name. Given Data / Assumptions: We consider the interface at a given frequency band. Matching may be accomplished by resistive pads, transformers, or reactive L/C networks. The goal is to present the source with its preferred load value. Concept / Approach: Impedance matching is the general term for making the load impedance equal (or conjugate-equal in AC) to the source impedance at the interface. One objective of matching is maximum power transfer (in DC/resistive networks, RL = Rs), but in RF lines it also minimizes reflections (return loss) and standing waves. Step-by-Step Solution: Identify design goal: make the load appear equal to the source impedance.Choose a network (e.g., transformer with specific turns ratio, L-match, Pi, T) to transform the actual load to the required value.Verify with Z_in at the source port: Z_in ≈ Z_source (or Z_in ≈ Z_source* in AC).Therefore, the practice is called impedance matching. Verification / Alternative check: Transmission line theory: Γ = (ZL − Z0)/(ZL + Z0). If ZL = Z0 (a matching condition), Γ = 0 and power is delivered with no reflection, confirming the matching concept. Why Other Options Are Wrong: Maximum power transfer: A design objective, not the general term for the act itself. Reflected resistance and reflected load: Terms used with transformers to describe how a load appears at another winding; they are not the broad process name. Common Pitfalls: Equating “impedance matching” strictly with resistive equality in DC; in AC, the proper condition is conjugate matching. Confusing the goal (maximum power) with the method (matching). Final Answer: impedance matching

Question 20

Transformer design for high coupling: To achieve a high coefficient of coupling k (i.e., strong magnetic linkage between windings), fixed transformers are typically wound on a __________ so that the primary and secondary share the same magnetic path.

Accepted Answer

Introduction / Context: A transformer transfers energy from primary to secondary by magnetic coupling. The fraction of magnetic flux from the primary that links the secondary is captured by the coefficient of coupling k, which ideally approaches 1. Understanding how construction affects k is essential for minimizing leakage inductance and improving regulation in power and signal transformers. Given Data / Assumptions: The transformer is a fixed (not air-gapped tunable) unit. Goal: maximize flux linkage between windings. Windings and core style can be selected to enhance coupling. Concept / Approach: Magnetic coupling improves when both windings share the same concentrated magnetic path. Winding the primary and secondary on a common core leg (or interleaving them) ensures most primary flux passes through the secondary window, raising k and reducing leakage. Construction choices like core material and lamination style (EI, CI, toroidal) matter, but the key design idea is a common core path for both windings. Step-by-Step Solution: Define coupling: k increases as shared flux increases.Shared flux increases when windings occupy the same magnetic circuit.A “common core” arrangement gives both windings the same flux path.Therefore the best generic answer is “common core.” Verification / Alternative check: Interleaved windings on a common core leg (including toroidal and EI with both windings on the center leg) are standard practice for high coupling, confirming the principle independent of a specific lamination shape. Why Other Options Are Wrong: EI core / CI core: These identify lamination shapes, not the coupling principle. Either can be common or not, depending on winding layout. Iron core: A material choice that increases permeability but does not, by itself, guarantee high k without shared path. Aluminum core: Non-magnetic; would not support efficient coupling. Common Pitfalls: Equating core material or alphabet shape (EI, CI) with coupling quality; ignoring winding placement and interleaving which dominate k. Final Answer: common core

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