Q: Impedance matching by transformer: What turns ratio is required to match an 80 Ω source to a 320 Ω load?

Introduction / Context: Impedance matching via a transformer uses the square law between turns ratio and impedance. Correct turns ratio minimizes reflections and maximizes power transfer in audio and RF systems alike. Given Data / Assumptions: Source resistance Rs = 80 Ω. Load resistance RL = 320 Ω. Ideal transformer; express turns ratio as Ns/Np. Concept / Approach: For ideal transformers, impedance scales as (Ns/Np)^2. To match, choose Ns/Np so that RL' = Rs, where RL' = RL * (Np/Ns)^2 when seen from the primary. Equivalently, (Ns/Np)^2 = RL / Rs. Step-by-Step Solution: (Ns/Np)^2 = RL / Rs = 320 / 80 = 4Ns/Np = sqrt(4) = 2 Verification / Alternative check: With Ns/Np = 2, a 320 Ω load reflects to Rs' = 320 / 4 = 80 Ω. This equals the source resistance, confirming matching. Why Other Options Are Wrong: 4, 20, 80: These would grossly mismatch the impedances. Common Pitfalls: Using the simple ratio RL/Rs rather than its square root for the turns ratio. Always take the square root when moving from impedance ratio to turns ratio. Final Answer: 2

Question 1

Loaded transformer basics: when the secondary voltage is one-fourth of the primary voltage, how does the secondary current compare to the primary current (ideal transformer)?

Accepted Answer

Introduction / Context: Transformers trade voltage for current while ideally conserving power (minus losses). Understanding the inverse current–voltage relationship underpins power supply design and impedance matching. Given Data / Assumptions: Secondary voltage Vs is one-fourth of primary voltage Vp. Ideal transformer assumption (no losses). Loaded condition (non-zero secondary current). Concept / Approach: For an ideal transformer, Vp/Vs = Np/Ns and Ip/Is = Ns/Np. Combining, Vs/Vp = Ns/Np, so Is/Ip = Np/Ns = Vp/Vs. If Vs = Vp/4, then Is = 4 * Ip, i.e., secondary current is four times the primary current. Step-by-Step Solution: Given Vs = Vp/4.Current ratio: Is/Ip = Vp/Vs = Vp/(Vp/4) = 4.Therefore Is = 4 * Ip (four times). Verification / Alternative check: Check ideal power: Pp ≈ Vp * Ip ≈ Vs * Is = (Vp/4) * (4Ip) = Vp * Ip, consistent with ideal conservation. Why Other Options Are Wrong: One-fourth the primary current / equal: contradicts inverse proportion for ideal transformers. Combined statement: logically inconsistent; cannot be both one-fourth and equal. Common Pitfalls: Mixing up turns and current ratios (current ratio is inverse of voltage ratio). Forgetting that losses slightly alter real values—here the question is ideal. Final Answer: four times the primary current

Question 2

Turns ratio for stepping 110 V AC down to 20 V AC: what transformer turns ratio is required (Np:Ns)?

Accepted Answer

Introduction / Context: Transformer design often begins with the desired voltage ratio. The primary-to-secondary turns ratio Np:Ns directly sets the output voltage under ideal conditions, a foundational concept for power converters and isolation transformers. Given Data / Assumptions: Input (primary) voltage Vp = 110 V AC. Desired output (secondary) voltage Vs = 20 V AC. Ideal transformer approximation (no losses). Concept / Approach: Ideal relation: Vp/Vs = Np/Ns. Rearranging gives the turns ratio directly. A step-down transformer requires more primary turns than secondary turns, so the ratio should be > 1. Step-by-Step Solution: Compute Np:Ns = Vp:Vs = 110:20.Simplify: 110/20 = 5.5.Therefore the turns ratio Np:Ns is 5.5:1. Verification / Alternative check: Back-check voltage: Vs = Vp / 5.5 = 110 / 5.5 = 20 V, as required. Why Other Options Are Wrong: 0.18 and 0.018: These are Ns:Np-like fractions, implying step-up, not step-down. 18: Would give Vs ≈ 110/18 ≈ 6.1 V, too low. 2.2: Would give Vs = 50 V, not 20 V. Common Pitfalls: Confusing Np:Ns with Ns:Np; always match the ratio to Vp:Vs for ideal cases. Ignoring regulation and load; those affect real outputs but not the ideal turns ratio. Final Answer: 5.5

Question 3

Transformer reflected impedance: A 200 Ω load is connected across the secondary of a transformer whose turns ratio is 4. What value of load does the source 'see' reflected to the primary side?

Accepted Answer

Introduction / Context: This problem checks understanding of how a transformer reflects impedances from one side to the other. Being able to compute the reflected resistance is crucial when matching sources and loads for maximum power transfer and efficient design in power electronics and audio systems. Given Data / Assumptions: Load on secondary, RL(sec) = 200 Ω. Turns ratio is stated as 4 (interpreted here as Ns/Np = 4, a common convention in basic problems). Ideal transformer behavior; no winding resistance or core loss. Concept / Approach: The impedance reflection rule for an ideal transformer is: R_ref(primary) = RL(sec) * (Np/Ns)^2. If the given turns ratio is Ns/Np, then Np/Ns = 1 / (Ns/Np). A larger number of secondary turns steps the voltage up and the impedance down on the primary side by the square of the turns ratio inversion. Step-by-Step Solution: Given Ns/Np = 4 ⇒ Np/Ns = 1/4R_ref = RL * (Np/Ns)^2R_ref = 200 * (1/4)^2R_ref = 200 * 1/16 = 12.5 Ω Verification / Alternative check: If the transformer steps voltage up by 4, it steps current down by 4, and thus impedance is divided by 16 when seen from the primary. 200 Ω / 16 = 12.5 Ω, confirming the calculation. Why Other Options Are Wrong: 50 Ω: Only divides by 4, not by 16. 800 Ω: Would be the reflection if Np/Ns = 4, not Ns/Np = 4. 0 Ω: An ideal short is not created here. Common Pitfalls: Misinterpreting the turns ratio direction or forgetting that impedance reflection scales with the square of the turns ratio rather than the ratio itself. Final Answer: 12.5 Ω

Question 4

Transformer sizing: What kVA rating is required if the transformer must deliver up to 8 A at a secondary voltage of 2 kV?

Accepted Answer

Introduction / Context: Selecting a transformer requires estimating apparent power in volt-amperes. The kVA rating must meet or exceed the product of the maximum secondary voltage and current because transformers are rated in apparent power rather than real power due to varying power factor loads. Given Data / Assumptions: I_max = 8 A (secondary). V_secondary = 2 kV = 2000 V. Ideal transformer assumptions; rating is based on apparent power S = V * I. Concept / Approach: Apparent power S (VA) = V * I. Convert to kVA by dividing by 1000. This is independent of power factor for the purpose of transformer nameplate rating. Step-by-Step Solution: S = V * I = 2000 * 8S = 16,000 VAS = 16 kVA Verification / Alternative check: Quick check: 2 kV * 10 A would be 20 kVA, thus 8 A gives 16 kVA. The proportional reasoning matches the exact calculation. Why Other Options Are Wrong: 8 kVA or 4 kVA: Underrated; risk of overloading at full current. 0.25 kVA: Two orders of magnitude too small. Common Pitfalls: Confusing kW with kVA, or forgetting to convert kilovolts and kilovolt-amperes properly. Always compute apparent power for transformer sizing. Final Answer: 16 kVA

Question 5

Turns ratio application: A transformer has 110 V applied to the primary. If the turns ratio is 8 (secondary to primary), determine the secondary voltage.

Accepted Answer

Introduction / Context: The turns ratio directly sets the voltage ratio for an ideal transformer. This problem checks comfort with scaling primary voltage to obtain the secondary voltage using the stated turns ratio convention. Given Data / Assumptions: Primary voltage Vp = 110 V. Turns ratio is 8, interpreted as Ns/Np = 8. Ideal transformer: Vs/Vp = Ns/Np. Concept / Approach: Use the ideal relation Vs = Vp * (Ns/Np). A turns ratio greater than 1 implies a step-up in voltage on the secondary. Step-by-Step Solution: Ns/Np = 8Vs = Vp * (Ns/Np)Vs = 110 * 8 = 880 V Verification / Alternative check: Since the ratio is 8:1, the secondary voltage must be eight times the primary. The numeric result 880 V is consistent with a simple step-up. Why Other Options Are Wrong: 8.8 V and 88 V: These would correspond to step-down scenarios or different ratio conventions. 8,800 V: Off by a factor of 10. Common Pitfalls: Misreading the direction of the turns ratio or mixing up primary and secondary positions in the ratio, which flips the result by a factor of 64 when squared for impedance or by 8 for voltage. Final Answer: 880 V

Question 6

Reflected resistance calculation: A transformer has 50 primary turns and 10 secondary turns. If the secondary load is 250 Ω, what resistance is reflected to the primary side?

Accepted Answer

Introduction / Context: Impedance reflection across ideal transformers is foundational for matching stages. When given explicit turn counts, you can compute the exact reflected resistance to the primary to analyze source loading and power transfer. Given Data / Assumptions: Np = 50 turns, Ns = 10 turns. RL(sec) = 250 Ω. Ideal transformer; R_ref(primary) = RL * (Np/Ns)^2. Concept / Approach: The impedance seen at the primary equals the secondary load multiplied by the square of the turns ratio from primary to secondary. A higher Np/Ns makes the primary appear more heavily loaded (higher reflected resistance). Step-by-Step Solution: Np/Ns = 50/10 = 5R_ref = RL * (Np/Ns)^2R_ref = 250 * 5^2 = 250 * 25R_ref = 6,250 Ω Verification / Alternative check: If this transformer were used to step voltage down by 5, it would step impedance up by 25 from the primary perspective; 250 Ω * 25 = 6,250 Ω matches the calculation. Why Other Options Are Wrong: 250 Ω: The actual secondary load, not the reflected primary value. 25 Ω: Would come from inverting the ratio incorrectly. 62,500 Ω: Off by 10x due to a powers-of-ten slip. Common Pitfalls: Forgetting the square on the turns ratio or swapping Ns and Np in the formula, which yields a drastically different value. Final Answer: 6,250 Ω

Question 7

Turns ratio needed to step 120 V AC up to 900 V AC using an ideal transformer.

Accepted Answer

Introduction / Context: Determining the turns ratio from desired input and output voltages is a routine transformer design task. The ideal transformer voltage ratio equals the turns ratio Ns/Np. Given Data / Assumptions: Input (primary) voltage Vp = 120 V AC. Desired output (secondary) voltage Vs = 900 V AC. Ideal transformer; Vs/Vp = Ns/Np. Concept / Approach: For ideal transformers: turns ratio r = Ns/Np = Vs/Vp. Compute the numeric ratio and choose the closest exact option. Step-by-Step Solution: r = Vs/Vp = 900 / 120r = 7.5 Verification / Alternative check: Check proportionality: 120 V × 7.5 = 900 V, confirming that a 7.5:1 step-up is required. Why Other Options Are Wrong: 75 or 750: Excessive step-up, yielding tens of kilovolts. 0.13: This would step down, not up. Common Pitfalls: Inverting the ratio or mixing up primary and secondary, which would give 0.133 instead of 7.5. Always compute Vs/Vp for Ns/Np when stepping up the voltage. Final Answer: 7.5

Question 8

Matching a 600 Ω amplifier to a 4 Ω speaker using an ideal coupling transformer: what turns ratio should be used for maximum power transfer?

Accepted Answer

Introduction / Context: Transformers are widely used to match a high-impedance source to a low-impedance load. Correct turns ratio selection ensures the load is reflected as the source's impedance, maximizing power transfer and efficiency. Given Data / Assumptions: Source resistance (amplifier output) Rs = 600 Ω. Load (speaker) RL = 4 Ω. Ideal transformer and standard matching condition: R_ref at source = Rs. Concept / Approach: For an ideal transformer, reflected resistance from secondary to primary is R_ref = RL * (Np/Ns)^2. Maximum power transfer requires R_ref = Rs. Express turns ratio as r = Ns/Np per the answer set. Step-by-Step Solution: Set RL * (Np/Ns)^2 = Rs(Np/Ns)^2 = Rs / RL = 600 / 4 = 150Np/Ns = sqrt(150) ≈ 12.247Hence Ns/Np = 1 / 12.247 ≈ 0.0817Closest option: 0.08 Verification / Alternative check: Using Ns/Np = 0.08 implies Np/Ns ≈ 12.5, giving R_ref ≈ 4 * 12.5^2 = 4 * 156.25 = 625 Ω, close to 600 Ω and consistent with a rounded choice. Why Other Options Are Wrong: 8 or 80: These would be very large Ns/Np and would not reflect 4 Ω up to about 600 Ω. 0.8: Too large; would reflect far less than 600 Ω. Common Pitfalls: Mixing up Ns/Np and Np/Ns or forgetting the square relationship between turns ratio and impedance reflection leads to large errors. Final Answer: 0.08

Question 9

Compute mutual inductance: Given coupling coefficient k = 0.65, L1 = 2 µH, and L2 = 5 µH, find the mutual inductance M.

Accepted Answer

Introduction / Context: Mutual inductance quantifies how strongly two inductors couple magnetically. It is central to transformer action and coupled inductor behavior in filters and converters. Given Data / Assumptions: k = 0.65 (dimensionless coupling coefficient). L1 = 2 µH, L2 = 5 µH. Ideal linear conditions; standard mutual inductance formula applies. Concept / Approach: The mutual inductance is M = k * sqrt(L1 * L2). Use consistent units (here, microhenries) and take the square root carefully. Step-by-Step Solution: Compute product: L1 * L2 = 2 * 5 = 10 (µH^2)sqrt(10) ≈ 3.1623 µHM = 0.65 * 3.1623 ≈ 2.055 µHRounded to listed options: 2 µH Verification / Alternative check: Because k < 1, M must be less than sqrt(L1 * L2) ≈ 3.162 µH; 2 µH fits this bound and the calculation. Why Other Options Are Wrong: 4 µH or 8 µH: Exceed sqrt(L1 L2) without even applying k, impossible with k ≤ 1. 2 mH: Wrong unit and magnitude by 10^3. Common Pitfalls: Forgetting to multiply by k or mixing units (mH vs µH). Always verify that M ≤ sqrt(L1 L2). Final Answer: 2 µH

Question 10

Impedance matching by transformer: What turns ratio is required to match an 80 Ω source to a 320 Ω load?

Accepted Answer

Introduction / Context: Impedance matching via a transformer uses the square law between turns ratio and impedance. Correct turns ratio minimizes reflections and maximizes power transfer in audio and RF systems alike. Given Data / Assumptions: Source resistance Rs = 80 Ω. Load resistance RL = 320 Ω. Ideal transformer; express turns ratio as Ns/Np. Concept / Approach: For ideal transformers, impedance scales as (Ns/Np)^2. To match, choose Ns/Np so that RL' = Rs, where RL' = RL * (Np/Ns)^2 when seen from the primary. Equivalently, (Ns/Np)^2 = RL / Rs. Step-by-Step Solution: (Ns/Np)^2 = RL / Rs = 320 / 80 = 4Ns/Np = sqrt(4) = 2 Verification / Alternative check: With Ns/Np = 2, a 320 Ω load reflects to Rs' = 320 / 4 = 80 Ω. This equals the source resistance, confirming matching. Why Other Options Are Wrong: 4, 20, 80: These would grossly mismatch the impedances. Common Pitfalls: Using the simple ratio RL/Rs rather than its square root for the turns ratio. Always take the square root when moving from impedance ratio to turns ratio. Final Answer: 2

Question 11

Fundamental function of a transformer: identify the correct statement about what a transformer does.

Accepted Answer

Introduction / Context: Understanding the core purpose of a transformer is essential for any study of power systems and electronics. Transformers rely on electromagnetic induction, which inherently requires alternating magnetic fields, and therefore alternating currents and voltages. Given Data / Assumptions: Standard iron-core or air-core transformer behavior. No active circuitry inside the transformer; it is a passive device. Concept / Approach: Transformers work by mutual induction: a changing current in the primary creates a changing magnetic flux that induces a voltage in the secondary. A changing flux requires AC excitation. Thus, transformers cannot convert AC to DC or DC to AC by themselves, nor do they affect DC voltages except transiently at switching moments. Step-by-Step Reasoning: AC in primary ⇒ changing flux ⇒ induced AC in secondaryVoltage ratio equals turns ratioTherefore a transformer steps AC voltage up or down while ideally conserving apparent power Verification / Alternative check: Apply steady DC to a transformer: after a brief transient, there is no changing flux and thus no sustained induced secondary voltage. This confirms that only AC is effectively transformed. Why Other Options Are Wrong: changes ac to dc / changes dc to ac: These are rectification or inversion functions performed by active devices, not a passive transformer alone. steps up or down dc voltages: Not possible in steady state without switching. Common Pitfalls: Confusing transformer operation with that of rectifiers or inverters. A transformer can be part of those systems but is not sufficient to perform conversion by itself. Final Answer: steps up or down ac voltages

Question 12

Safety consideration for power transformers: the primary winding of a power transformer should always be what?

Accepted Answer

Introduction / Context: Power transformers interface directly with mains or other energy sources. Proper protection is essential to prevent faults, overheating, and fire hazards. The question targets basic safety practice for transformer primaries. Given Data / Assumptions: Standard power transformer connected to a supply. Goal: protect wiring and transformer from overcurrent conditions. Concept / Approach: A fuse or circuit breaker on the primary side provides overcurrent protection. During faults such as shorts or severe overloads, the protective device opens the circuit to prevent damage. This is a fundamental code-compliant design practice in power electronics and electrical installations. Step-by-Step Reasoning: Place a fuse in series with the primaryAbnormal current ⇒ fuse opens ⇒ transformer de-energizedEquipment and conductors remain protected Why Other Options Are Wrong: open: An open primary is not a protective measure; it simply means not connected to operate. shorted: Dangerous and destructive; never short a primary. switched: A switch provides manual control but not automatic overcurrent protection. Common Pitfalls: Relying solely on secondary-side fusing while omitting primary protection. Both sides may require protection depending on codes and design, but primary fusing is essential. Final Answer: fused

Question 13

Transformer basics: A transformer's primary is connected to a 60 V AC source, its secondary feeds a 330 Ω load, and the turns ratio is 3:1 (primary:secondary). What is the secondary RMS voltage delivered by the transformer?

Accepted Answer

Introduction / Context: This problem checks fundamental transformer relations: how the turns ratio links primary voltage to secondary voltage under AC excitation. Correctly interpreting a ratio written as primary:secondary is essential for estimating output levels and sizing loads. Given Data / Assumptions: Primary voltage, Vp = 60 V (AC, RMS). Turns ratio Np:Ns = 3:1 (step-down). Secondary drives a 330 Ω load; assume ideal transformer and negligible voltage drop inside windings for the voltage calculation. Concept / Approach: In an ideal transformer, voltage ratio equals turns ratio: Vp / Vs = Np / Ns. Therefore Vs = Vp * (Ns / Np). With a 3:1 step-down, the secondary has one-third the primary voltage. Step-by-Step Solution: Turns ratio: Np:Ns = 3:1 → Ns / Np = 1 / 3.Compute secondary voltage: Vs = Vp * (Ns / Np) = 60 * (1 / 3) = 20 V.Thus the secondary supplies approximately 20 V RMS to the 330 Ω load. Verification / Alternative check: If the transformer were ideal, load current would be Is = Vs / RL = 20 / 330 ≈ 0.0606 A, and apparent power on secondary ≈ 1.21 VA. The primary would reflect the same apparent power (neglecting losses), consistent with step-down behavior. Why Other Options Are Wrong: 18 V: Not equal to one-third of 60 V. 2 V: Off by a factor of 10; misreading ratio as 30:1. 180 V: Would be a step-up, opposite of 3:1 primary:secondary. 12 V: Would correspond to a 5:1 step-down from 60 V. Common Pitfalls: Inverting the ratio (treating 3:1 as secondary:primary). Mixing peak and RMS quantities; here, all values are RMS. Final Answer: 20 V

Question 14

Transformer design goal: What primary voltage is required if a transformer with turns ratio Np/Ns = 0.1 must produce a 9 V secondary output?

Accepted Answer

Introduction / Context: This question tests your ability to manipulate the transformer voltage–turns relation when the turns ratio is given as a single number Np/Ns, not as a colon pair. It is common in datasheets and calculations to express the ratio this way. Given Data / Assumptions: Turns ratio r = Np/Ns = 0.1. Desired secondary voltage Vs = 9 V (RMS). Ideal transformer; ignore losses and regulation. Concept / Approach: Voltage ratio equals turns ratio: Vp / Vs = Np / Ns. Rearranging gives Vp = Vs * (Np/Ns). When the turns ratio is less than 1, the transformer is step-up (secondary has more turns than primary), so the required primary voltage is lower than the secondary voltage. Step-by-Step Solution: Use Vp / Vs = Np / Ns = 0.1.Solve for Vp: Vp = Vs * 0.1 = 9 * 0.1 = 0.9 V.Therefore, only 0.9 V RMS on the primary is needed to obtain 9 V RMS on the secondary. Verification / Alternative check: If Vp = 0.9 V and Np/Ns = 0.1, then Vs = Vp / (Np/Ns) = 0.9 / 0.1 = 9 V, confirming consistency. Why Other Options Are Wrong: 90 V and 900 V: These imply step-down assumptions; they invert the ratio. 9 V: Would only be correct if Np = Ns (ratio 1). 18 V: Unrelated to the required 9 V with the given ratio. Common Pitfalls: Confusing Np/Ns with Ns/Np; always match Vp/Vs = Np/Ns. Forgetting the problem states Np/Ns = 0.1, not Ns/Np. Final Answer: 0.9 V

Question 15

Turn counting: A transformer has 400 primary turns and 2,000 secondary turns. What is the turns ratio Np:Ns (expressed as a number Np/Ns)?

Accepted Answer

Introduction / Context: Accurately computing the turns ratio is foundational for predicting output voltage and reflected impedances. This problem reinforces the simple division that defines Np/Ns in a transformer. Given Data / Assumptions: Np = 400 turns. Ns = 2,000 turns. Ideal transformer conventions. Concept / Approach: The turns ratio expressed as a single number is Np/Ns. Use direct division of the primary turns by the secondary turns. Step-by-Step Solution: Compute r = Np/Ns = 400 / 2000.400 / 2000 = 0.2.Therefore, the turns ratio (as a number) is 0.2. Verification / Alternative check: As a colon ratio, this is 1:5 (since 0.2 = 1/5). That means the transformer is step-up by 5 in voltage when driven on the primary. Why Other Options Are Wrong: 0.4: Would correspond to 400/1000, not the given turns. 5 or 25: Those are Ns/Np magnitudes or unrelated; the question asks Np/Ns. 2: Not supported by the given turn counts. Common Pitfalls: Accidentally computing Ns/Np instead of Np/Ns. Forgetting to simplify the fraction correctly. Final Answer: 0.2

Question 16

Coupling effects: A transformer has a turns ratio of 1:1 and a coupling coefficient k = 0.85. If 2 V AC is applied to the primary, what RMS voltage appears at the secondary?

Accepted Answer

Introduction / Context: Real transformers do not couple all magnetic flux perfectly. The coupling coefficient 0 ≤ k ≤ 1 captures how effectively primary flux links the secondary. This question tests applying k to estimate secondary voltage when the ideal turns ratio alone would suggest equality. Given Data / Assumptions: Turns ratio Np:Ns = 1:1. Coupling coefficient k = 0.85. Primary RMS voltage Vp = 2 V. Assume negligible resistance and leakage beyond k's effect on induced voltage. Concept / Approach: With imperfect coupling, the induced secondary voltage magnitude is approximately Vs ≈ k * (Ns/Np) * Vp. With Ns/Np = 1, Vs ≈ k * Vp = 0.85 * 2 V. Step-by-Step Solution: Compute Vs ≈ k * Vp = 0.85 * 2 = 1.7 V.Because k < 1, the secondary voltage is lower than the ideal 1:1 value. Verification / Alternative check: In the ideal case (k = 1), Vs would be 2 V. Reducing k to 0.85 scales the induced magnitude to 85% of ideal, consistent with 1.7 V. Why Other Options Are Wrong: 0.85 V: Would correspond to applying k twice or using Vp = 1 V. 1 V: Not supported by the given k and Vp. 0 V: Only true for DC excitation or open primary; not here. 2 V: Would require perfect coupling (k = 1). Common Pitfalls: Ignoring k and assuming ideal behavior. Confusing k with efficiency; k relates to flux linkage, not directly to power loss. Final Answer: 1.7 V

Question 17

Transformers and DC excitation: The primary of a transformer is connected to a 6 V battery (DC). Turns ratio is 1:3 and the secondary load RL = 100 Ω. What steady-state voltage is across the load?

Accepted Answer

Introduction / Context: Transformers require changing magnetic flux to induce secondary voltage. A pure DC source does not sustain changing flux after the initial transient, so understanding steady-state behavior prevents design errors and overheating scenarios. Given Data / Assumptions: DC source: 6 V battery on the primary. Turns ratio 1:3 (primary:secondary). Secondary load RL = 100 Ω. Steady-state condition after any switching transient; idealized core behavior for concept. Concept / Approach: Induced voltage is proportional to time rate of change of flux (Faraday’s law). With DC, after the initial connection transient, dΦ/dt → 0, so the induced secondary voltage tends to 0 V. The transformer core may saturate if DC persists, but the key answer is the steady-state secondary voltage. Step-by-Step Solution: Apply DC to the primary → transient flux change → momentary secondary voltage.After settling, dΦ/dt = 0 → induced secondary voltage Vs = 0 V.Therefore, RL sees 0 V in steady state. Verification / Alternative check: Using the transformer equation V = N * dΦ/dt qualitatively: with constant current and constant flux (ignoring saturation), the time derivative is zero, hence no induced EMF. Why Other Options Are Wrong: 6 V or 18 V or 2 V: These presume AC action or a lasting voltage ratio; not applicable with DC at steady state. 0.6 V: No mechanism produces this steady value without time-varying flux. Common Pitfalls: Assuming the turns ratio applies to DC the same way as AC. Ignoring that transformers are inductive devices needing changing current. Final Answer: 0 V

Question 18

Transformer loss accounting: With 120 W into the primary and 8.5 W lost to winding resistance, what output power is delivered to the load (neglect other losses)?

Accepted Answer

Introduction / Context: Practical transformers exhibit copper (I^2R) losses in windings and core losses. This question focuses on simple power balance when a single loss component is known, a common step in efficiency and thermal calculations. Given Data / Assumptions: Primary input power Pin = 120 W. Winding copper loss Pcu = 8.5 W (given). Neglect all other losses (e.g., core loss), so remaining power is transferred to the load. Concept / Approach: By conservation of power: Pout ≈ Pin - Losses. Here only one loss term is specified, so subtract it directly from the input power to estimate output power to the load terminals. Step-by-Step Solution: Compute Pout = Pin - Pcu = 120 - 8.5 = 111.5 W.Therefore, the load receives approximately 111.5 W. Verification / Alternative check: If we expressed efficiency η = Pout / Pin = 111.5 / 120 ≈ 92.9%, which is a plausible transformer efficiency figure depending on size and design. Why Other Options Are Wrong: 14.1 W or 0 W: Ignore the stated power flow and losses. 1,020 W: Exceeds input power; violates conservation. 118.5 W: Would imply only 1.5 W loss, contradicting the given 8.5 W. Common Pitfalls: Adding losses instead of subtracting from input power. Confusing efficiency percentage with watts. Final Answer: 111.5 W

Question 19

Step-down calculation: A transformer has a 110 V primary and a 15:1 turns ratio (primary:secondary). With RL = 120 Ω, what is the approximate load voltage?

Accepted Answer

Introduction / Context: Here we apply the turns ratio to obtain the secondary voltage of a step-down transformer under load. For most practical cases with modest current and low winding resistance, the ideal ratio gives a good approximation of the load voltage. Given Data / Assumptions: Vp = 110 V (RMS). Turns ratio Np:Ns = 15:1. Load RL = 120 Ω; assume ideal transformer behavior for voltage estimation. Concept / Approach: For an ideal transformer, Vs = Vp * (Ns / Np). A 15:1 primary:secondary ratio means one-fifteenth of the primary voltage appears on the secondary winding. Step-by-Step Solution: Compute Ns/Np = 1 / 15.Vs = 110 * (1 / 15) ≈ 7.33 V.Rounded to one decimal place, ≈ 7.3 V across RL. Verification / Alternative check: Secondary current would be Is ≈ Vs / RL ≈ 7.33 / 120 ≈ 61 mA. Reflected to primary, Ip ≈ Is / 15 ≈ 4.1 mA; apparent power ≈ 0.45 VA—small, justifying negligible regulation for this estimate. Why Other Options Are Wrong: 73 V: Off by a factor of 10. 88 V and 880 V: Do not follow the 15:1 step-down relation. 11 V: Would correspond to 10:1, not 15:1. Common Pitfalls: Mistaking 15:1 as secondary:primary, which would predict a step-up. Using peak instead of RMS values. Final Answer: 7.3 V

Question 20

Ideal transformer power transfer: If 25 W is applied to the primary of an ideal transformer with turns ratio 10, how much power is delivered to the secondary load?

Accepted Answer

Introduction / Context: While the turns ratio changes voltage and current, an ideal transformer conserves power (ignoring losses). This question emphasizes that power in equals power out under ideal conditions, regardless of the ratio value itself. Given Data / Assumptions: Primary input power Pin = 25 W. Transformer is ideal (no copper or core loss). Turns ratio given but not needed for power equality. Concept / Approach: Ideal transformer property: Psecondary ≈ Pprimary. The turns ratio scales voltage and current inversely so that their product remains the same (within small rounding). Step-by-Step Solution: Recognize ideal behavior: Pout = Pin.Therefore Pout = 25 W. Verification / Alternative check: If voltage is stepped up by 10, current is stepped down by 10, keeping V * I constant, thus preserving power ideally. Why Other Options Are Wrong: 0 W: Would imply open circuit or total loss—contrary to the prompt. 250 W or 2.5 W: Misuse the ratio on power instead of on voltage/current. 50 W: Would require ≥50% efficiency gain; not applicable. Common Pitfalls: Multiplying or dividing power directly by the turns ratio; only V and I scale with turns. Final Answer: 25 W

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