Question 1

In a three-phase Y-connected (wye) AC generator, each phase (line-to-neutral) voltage is 90 V(rms). What is the magnitude of each line-to-line voltage?

Accepted Answer

Introduction / Context: This problem tests the foundational relationship between phase voltage and line voltage in a balanced three-phase Y (wye) connection. Understanding this conversion is essential for sizing insulation, selecting meters, and ensuring the correct ratings for equipment in power systems and three-phase electronics. Given Data / Assumptions: Connection: three-phase Y-connected generator (balanced). Phase (line-to-neutral) voltage magnitude: V_phase = 90 V(rms). We seek the line-to-line voltage magnitude V_line. Sinusoidal steady-state; ideal, symmetrical system. Concept / Approach: For a balanced Y connection, the line-to-line voltage magnitude is related to the phase voltage by V_line = √3 * V_phase. The factor √3 arises from the 120° vector separation between phase voltages and the phasor difference defining line voltages. Step-by-Step Solution: Identify the topology: Y-connected → line-to-line differs from line-to-neutral by √3.Use formula: V_line = √3 * V_phase.Compute: V_line = 1.732 * 90 ≈ 155.9 V.Round to the nearest given choice: 156 V. Verification / Alternative check: In a Y system, if V_phase = 230 V, V_line commonly equals about 400 V (230 * 1.732), which is a well-known standard (400/230 V systems). This sanity check supports the use of the √3 factor here as well. Why Other Options Are Wrong: 0 V: Not possible; the generator is energized. 90 V: That is the phase voltage, not the line-to-line voltage in a Y system. 180 V: No standard factor yields exactly double the phase voltage in Y; the correct multiplier is √3, not 2. Common Pitfalls: Confusing Y and Δ relationships. In Δ, line voltage equals phase voltage, but in Y, line voltage is √3 times phase voltage. Forgetting that the values are rms and phasor-based (120° separation). Final Answer: 156 V

Question 2

A balanced three-phase Δ-connected generator delivers phase currents of 12 A (rms). If Iθa = 12 ∠30° A, determine the polar forms of the other two phase currents (positive-sequence spacing).

Accepted Answer

Introduction / Context: This question checks phasor literacy for balanced three-phase systems. In a balanced generator (Δ or Y), the three phase quantities are equal in magnitude and separated by 120° in phase. Correctly assigning the angles for phases b and c after one phase is given is a core skill for system analysis, phasor diagrams, and power calculations. Given Data / Assumptions: Balanced three-phase Δ-connected generator. Phase current magnitudes are all 12 A(rms). Given: Iθa = 12 ∠30° A. Assume positive sequence ordering (a→b→c with 120° separation). Concept / Approach: For a positive sequence, the three phasors are separated by +120° steps when rotating from a to b to c (or equivalently, b lags a by 120°, c lags b by 120°). Because angles can be expressed modulo 360°, equivalent representations appear as additions or subtractions of 360° without changing the phasor. Step-by-Step Solution: Start with I_a = 12 ∠30° A.For 120° separation, one consistent set is: I_b = 12 ∠(30° + 120°) = 12 ∠150° A and I_c = 12 ∠(30° − 120°) = 12 ∠(−90°) A.These maintain equal magnitudes and 120° phasor spacing, satisfying a balanced positive sequence. Verification / Alternative check: Plot on a phasor diagram: one vector at 30°, the next at 150°, and the third at −90°. The separations are all 120° (from 30° to 150° is +120°, from 150° to −90° is +120°, from −90° to 30° is +120°), confirming balance and correct ordering. Why Other Options Are Wrong: I θb = 12∠120°, I θc = 12∠30°: b is only 90° from a here (120° − 30°), breaking 120° balance. I θb = 12∠30°, I θc = 12∠120°: b duplicates a; not 120° separated. I θb = 12∠90°, I θc = 12∠90°: identical angles for b and c, not a balanced three-phase set. Common Pitfalls: Reversing phase sequence (positive vs negative) or mixing line and phase currents. Forgetting that multiple angle representations are equivalent modulo 360°. Final Answer: I θb = 12 ∠150° A, I θc = 12∠–90° A

Question 3

In a Δ-connected (delta) generator, how do the three phase (line-to-line) voltages compare to each other?

Accepted Answer

Introduction / Context: Recognizing the behavior of voltages in Δ-connected configurations is basic to three-phase power engineering. In a delta, each phase element is connected directly between a pair of line conductors, so each phase sees a line-to-line voltage. The symmetry guarantees consistent magnitude among phases in a balanced system. Given Data / Assumptions: Balanced Δ connection. Sinusoidal steady-state. We compare phase voltages among phases a, b, and c. Concept / Approach: In a balanced Δ, each side of the delta is across a pair of lines, so each phase voltage is a line-to-line voltage. With system balance and symmetry, all three phase voltages are equal in magnitude and displaced by 120° in phase angle. Step-by-Step Reasoning: Each phase is tied directly between two lines → each phase sees V_LL.Balance implies |V_ab| = |V_bc| = |V_ca|.Angles differ by 120°, but magnitudes are identical. Verification / Alternative check: If the system is perfectly symmetrical and the source is ideal, any measurement across each delta side yields the same rms value. Deviations indicate unbalance or faults, not normal operation. Why Other Options Are Wrong: zero: A powered generator cannot have zero phase voltages unless faulted. one-third or one-sixth of total: These values do not describe delta relationships; they often stem from confusing Δ vs Y line/phase relations. Common Pitfalls: Confusing Δ with Y. In Y, line voltage is √3 times the phase voltage; in Δ, line and phase voltages are the same in magnitude (line = phase). Final Answer: equal in magnitude

Question 4

In a three-wire Y-connected generator with phase (line-to-neutral) voltages of 2 kV, determine the magnitude of each line-to-line voltage.

Accepted Answer

Introduction / Context: Calculating line-to-line voltage from a known phase voltage in a Y connection is a frequent task in power distribution and equipment specification. The √3 factor is a direct consequence of 120° phasor geometry in balanced three-phase systems. Given Data / Assumptions: Y-connected generator (balanced). Phase voltage V_phase = 2 kV (rms, line-to-neutral). Find V_line (rms, line-to-line). Concept / Approach: For Y: V_line = √3 * V_phase. The line-to-line voltage is the phasor difference of two line-to-neutral voltages separated by 120°, yielding a magnitude √3 times larger than a single phase voltage. Step-by-Step Solution: Use the relation V_line = √3 * V_phase.Compute: √3 * 2,000 ≈ 1.732 * 2000 = 3,464 V. Verification / Alternative check: Common standards (e.g., 230/400 V) follow the same ratio. This consistency reinforces the calculation method and result here. Why Other Options Are Wrong: 2,000 V: This would be true for Δ (line = phase), not for Y. 6,000 V: Assumes a factor of 3 rather than √3. 666 V: An inverse-type confusion (phase vs line) and off by a large factor. Common Pitfalls: Mixing Δ and Y relationships or forgetting the 120° vector geometry. Final Answer: 3,464 V

Question 5

In electric machine applications, which is the most commonly used type of AC motor for industrial and commercial drives?

Accepted Answer

Introduction / Context: AC motors drive the majority of pumps, fans, compressors, and conveyors worldwide. Selecting the most common type highlights practical engineering awareness, cost/complexity trade-offs, and grid compatibility. This is a conceptual recall question grounded in industrial practice. Given Data / Assumptions: Standard power systems and typical facility infrastructure. Focus on prevalence, robustness, and efficiency. Concept / Approach: Three-phase induction motors dominate due to their self-starting capability (with proper design), high efficiency, simple construction (no brushes), ruggedness, low maintenance, and the widespread availability of three-phase power in industrial settings. Their torque characteristics suit many loads, and variable-frequency drives further extend performance and control. Step-by-Step Reasoning: Evaluate common installations: factories, HVAC plants, water treatment—three-phase induction is standard.Consider economics and reliability: simple rotor (squirrel cage), minimal maintenance, high availability → preferred choice.Single-phase induction motors are common in households, but not the most common overall in industrial power. Verification / Alternative check: Industry catalogs and standards strongly feature three-phase induction machines across a wide power range. VFD proliferation reinforces this dominance by enabling efficient speed control. Why Other Options Are Wrong: Single-phase induction: common for small appliances, not most common across industrial/commercial scale. Two-phase induction / two-phase squirrel-cage: Two-phase systems are rare historically; modern grids are three-phase. Common Pitfalls: Generalizing household prevalence to industrial usage. Final Answer: three-phase induction motor

Question 6

In a balanced three-phase system with a solidly connected neutral, what is the magnitude of the neutral current under perfectly balanced loading conditions?

Accepted Answer

Introduction / Context: The neutral conductor in three-phase systems returns the vector sum of the three phase currents. Knowing when that current is zero is key for conductor sizing, fault analysis, and power quality considerations in balanced vs unbalanced loads. Given Data / Assumptions: Balanced three-phase system (equal magnitudes, 120° separated). Sinusoidal steady-state; no harmonics considered. Neutral conductor is present and solidly connected. Concept / Approach: In a perfectly balanced system, the instantaneous sum of the three sinusoidal currents is zero at every moment. In phasor form, the vector sum I_a + I_b + I_c = 0, so no current flows in the neutral. Unbalance, harmonics (notably triplen), or asymmetry changes this outcome, but that is beyond the ideal case specified. Step-by-Step Reasoning: Let phase currents be equal magnitude, 120° apart → phasor sum equals zero.Neutral current equals the phasor sum of the three phase currents → 0 A under balance. Verification / Alternative check: Draw a phasor triangle: equal-length vectors at 0°, −120°, +120° form a closed triangle. The resultant is the null vector, confirming zero neutral current. Why Other Options Are Wrong: One-third / two-thirds / at maximum: These values imply unbalance or non-sinusoidal conditions; not applicable to the strictly balanced, sinusoidal case stated. Common Pitfalls: Confusing real-world lightly unbalanced systems with the ideal balanced definition used in theory problems. Final Answer: zero

Question 7

For a Δ-connected source feeding a Y-connected load, how does each source phase voltage relate to the corresponding load terminal voltages (line-to-line relationship)?

Accepted Answer

Introduction / Context: Mixed Δ–Y connections are common in transformers and generator–load interfaces. Understanding how source phase voltages map to load terminal voltages avoids confusion when converting between line-to-line and line-to-neutral measures and when drawing phasor diagrams for interconnections. Given Data / Assumptions: Source: Δ-connected, providing line-to-line (phase) voltages directly. Load: Y-connected, with each load phase connected from a line to the neutral. Balanced, sinusoidal conditions assumed. Concept / Approach: In Δ, each phase element is across a pair of lines; thus the source phase voltage is a line-to-line voltage (difference of two line phasors). The Y load phase voltage is a line-to-neutral voltage. Because line-to-line voltages are differences of line-to-neutral voltages, one can say each Δ phase voltage equals the vector difference of two corresponding Y load (line-to-neutral) voltages. Step-by-Step Reasoning: Let V_an, V_bn, V_cn be load phase (line-to-neutral) voltages.Line-to-line voltage V_ab = V_an − V_bn → that is also the Δ source phase voltage across the ab side.Thus each source phase voltage equals the difference of two load phase voltages. Verification / Alternative check: Draw the phasor diagram: V_an, V_bn, V_cn are 120° apart. The vector subtraction V_an − V_bn produces the familiar line-to-line magnitude √3 times the phase magnitude, aligning with Δ phase values. Why Other Options Are Wrong: Equals the corresponding load voltage: Load phase is line-to-neutral; Δ phase is line-to-line; magnitudes differ by √3 in balanced systems. One-third the load voltage: No standard three-phase relation yields 1/3 here. 60° out of phase: Line-to-line vs line-to-neutral phasors exhibit 30° shifts in common representations, not 60°, and this option is not a correct general statement. Common Pitfalls: Confusing phase vs line voltages and their vector differences when mixing Δ and Y. Final Answer: each phase voltage equals the difference of the corresponding load voltages

Question 8

In a balanced three-phase load, how is power shared among the three phases?

Accepted Answer

Introduction / Context: Power distribution in balanced three-phase systems is uniform by design. Each phase draws the same power, ensuring symmetrical loading of lines and equipment. Recognizing this principle is crucial for load balancing and system planning. Given Data / Assumptions: Balanced three-phase load (equal Z per phase, 120° spacing). Sinusoidal steady-state. Concept / Approach: Total three-phase power P_total equals three times the per-phase power P_phase in a balanced system. Therefore each phase consumes an equal share: P_phase = P_total / 3. This holds for Y or Δ connections when the system is truly balanced. Step-by-Step Reasoning: Per-phase voltage/current magnitudes are equal under balance.Power in each phase (P_phase) is identical → P_total = 3 * P_phase. Verification / Alternative check: Using P = √3 * V_line * I_line * cosφ for total power and P_phase = V_phase * I_phase * cosφ, with V_line = √3 V_phase (Y) and I_line = I_phase (Y), you can show P_total = 3 P_phase for balanced systems. Why Other Options Are Wrong: One-third of total: That statement describes the share per phase (P_phase), but the option as written does not make clear equality across phases; the best answer is that each phase has an equal amount. Two-thirds of total: Incorrect apportionment. Equal to I_L: Power is not “equal to current”; it depends on voltage, current, and power factor. Common Pitfalls: Misreading phrasing—remember each of the three phases shares equally in a balanced system. Final Answer: an equal amount of power

Question 9

For a Y-connected (wye) three-phase circuit, what is the relationship between line current and phase current in magnitude?

Accepted Answer

Introduction / Context: Correctly relating line and phase quantities for Y and Δ connections is a frequent task in power systems. For Y connections, the node at the center ties each phase to the neutral, and currents flow in each line through a single phase branch, leading to a simple equality between line and phase currents. Given Data / Assumptions: Balanced, sinusoidal three-phase Y connection. Ideal lines and connections (no additional paths). Concept / Approach: In a Y connection, each line conductor feeds one phase element directly. Therefore, the current in any line equals the current in that phase element: I_line = I_phase. (By contrast, in Δ, line current equals √3 times the phase current due to the vector sum of two branch currents.) Step-by-Step Reasoning: Trace current path: from line to one phase element to neutral (or return via other lines).No current splitting among multiple elements per line → I_line magnitude equals I_phase magnitude. Verification / Alternative check: Phasor analysis or simple circuit inspection of a Y-connected load confirms this direct equality. Measured values in laboratory Y-connected loads also show this relationship in balanced conditions. Why Other Options Are Wrong: One-third / three times: These do not apply to Y; “√3 times” belongs to Δ (line vs phase currents). Zero: Not unless the circuit is de-energized or open-circuited. Common Pitfalls: Swapping the Y and Δ current/voltage relationships; remember: Y → I_line = I_phase, Δ → V_line = V_phase. Final Answer: equal to the corresponding phase current

Question 10

Polyphase (three-phase and beyond) generators produce multiple sinusoidal voltages that are separated by which characteristic?

Accepted Answer

Introduction / Context: Polyphase generation creates several sinusoidal voltages simultaneously. The defining feature is their fixed phase displacement, enabling constant power transfer in three-phase systems, smoother torque in machines, and efficient transmission characteristics. Given Data / Assumptions: Polyphase generator (e.g., three-phase) operating under steady-state. All phases have the same frequency and (in a balanced system) equal magnitude. Concept / Approach: In a polyphase system, the phase voltages are separated by constant phase angles (e.g., 120° in three-phase). Frequency is common to all phases; magnitudes may also be equal in a balanced system, but what uniquely defines polyphase is the constant angular separation, not constant frequency differences or constant currents across phases. Step-by-Step Reasoning: Recognize that polyphase machines are built with spatially separated windings.Mechanical rotation induces EMFs with equal frequency but different phase angles determined by winding placement.Therefore, the hallmark is fixed phase-angle spacing among the sinusoidal voltages. Verification / Alternative check: Three-phase: V_an, V_bn, V_cn share one frequency f and are spaced by 120°. Other polyphase systems (e.g., two-phase historical) use 90° spacing. The spacing is constant by design and does not drift in steady operation. Why Other Options Are Wrong: Certain constant frequencies: All phases share the same frequency, not distinct constant frequencies. Certain constant voltages: Magnitudes can be equal in balanced systems, but equality of magnitude is not the defining attribute. Certain constant currents: Load-dependent; not intrinsic to generation. Common Pitfalls: Confusing the defining property (fixed phase angles) with other possibly true but non-defining attributes (equal frequency, often equal magnitudes). Final Answer: certain constant phase angles

Question 11

In a three-phase power system where both the source and the load are connected in a delta (Δ–Δ) configuration, identify the correct relationship among the load phase voltage, the line (line-to-line) voltage, and the source phase voltage for any give…

Accepted Answer

Introduction / Context: Three-phase AC systems commonly use delta (Δ) and wye (Y) connections. Understanding how phase and line quantities relate within these connections is foundational for analyzing motors, transformers, and distribution networks. This question targets the voltage relationships in a Δ–Δ system, which often confuses learners who recall the √3 factor from Y systems and apply it incorrectly to delta circuits. Given Data / Assumptions: The source is delta-connected (Δ) and the load is also delta-connected (Δ). Balanced, sinusoidal three-phase conditions are assumed. No line impedance is considered (ideal conductors for clarity). Concept / Approach: In a delta connection, each phase element (source or load) is connected directly between two lines. Therefore, the phase voltage of that element is the line-to-line voltage. This is true on both the source side and the load side of a Δ–Δ system (assuming no significant voltage drops in the lines). Step-by-Step Solution: Identify connection: Δ means each phase winding sits across two lines.Hence: V_phase(source) = V_line(source).On the load: V_phase(load) = V_line(at the load terminals).With ideal lines: the line voltage delivered equals the line voltage applied; thus all three quantities are equal in magnitude. Verification / Alternative check: Contrast with Y systems: In Y, V_line = √3 * V_phase. In Δ, there is no √3 factor between line and phase voltage; they are the same, validating the equality statement. Why Other Options Are Wrong: Fractions like one-third or two-thirds have no basis in Δ voltage relations. 'Cancel' is incorrect; voltages are phasors between conductors and do not cancel for a single phase element. Common Pitfalls: Confusing delta with wye and misapplying the √3 relationship; overlooking that in Δ the element endpoints are lines, not line-to-neutral points. Final Answer: load voltage, line voltage, and source phase voltage are all equal for a given phase

Question 12

In electric machines and drives, a constant load power indicates what kind of uniform energy conversion process is occurring?

Accepted Answer

Introduction / Context: In power systems and electric drives, 'load power' typically refers to the rate at which electrical input power is converted by a load into another form such as mechanical output in a motor, heat in a resistor, or light in a lamp. This question focuses on interpreting 'constant load power' in the context of electromechanical energy conversion, which is central to motor control and drive efficiency. Given Data / Assumptions: The term 'load' refers to an electromechanical device drawing electrical power (e.g., a motor). Steady-state operation is considered so that average power is meaningful. Losses and non-idealities are not the focus; the conversion process characterization is. Concept / Approach: Power is the rate of energy conversion. For a motor load, constant electrical power drawn from the source corresponds to a uniform transfer of electrical energy into mechanical energy (plus losses). For a generator, the reverse occurs: mechanical to electrical. Since the question refers generically to 'load power' under usage typical of drive problems, the motor interpretation (electrical to mechanical) is most appropriate. Step-by-Step Solution: Identify device context: a 'load' consuming electrical power is generally a motor.Constant load power ⇒ constant electrical input power.Therefore, the energy conversion rate from electrical to mechanical is uniform (ignoring small variations in losses). Verification / Alternative check: For DC motors: P_in ≈ V * I; with speed-torque relations, constant power regions are common (high-speed operation). The mechanical output power P_mech ≈ T * ω tracks the electrical input power minus losses, thus reflecting a nearly uniform conversion rate when P_in is constant. Why Other Options Are Wrong: 'Mechanical to electrical energy' describes generator action, not a typical 'load'. 'Current to voltage' and 'voltage to current' confuse energy conversion with circuit variable relationships; they are not energy conversion processes. Common Pitfalls: Interpreting 'load' as a generator; conflating constant power with constant current or constant voltage conditions. Final Answer: electrical to mechanical energy

Question 13

Single-phase resistive load: A 120 V (rms) sinusoidal source feeds a 90 Ω load. What is the circuit current?

Accepted Answer

Introduction / Context: Ohm’s law in AC circuits with purely resistive loads behaves exactly as in DC: the current equals the applied rms voltage divided by the resistance. This basic calculation underpins power ratings, wire sizing, and protection-device selection in residential and industrial settings. Given Data / Assumptions: RMS source voltage V = 120 V. Load resistance R = 90 Ω, purely resistive. Steady-state sinusoidal operation. Concept / Approach: For a resistive load, I = V / R using rms values. No phase shift exists between voltage and current, and reactive effects are absent. Step-by-Step Solution: I = V / R = 120 / 90.Compute: 120 / 90 = 1.333… A.Rounded to two decimals: ≈ 1.33 A. Verification / Alternative check: Power check: P = V * I = 120 * 1.333… ≈ 160 W. Alternatively, P = V^2 / R = 120^2 / 90 = 160 W. Both agree, confirming the current. Why Other Options Are Wrong: 133 mA and 13.3 mA are off by factors of 10 and 100 relative to 1.33 A. 6.2 A would correspond to an unrealistically low resistance for 120 V (about 19 Ω), not 90 Ω. Common Pitfalls: Mixing peak and rms quantities; forgetting units; calculator rounding errors. Final Answer: 1.33 A

Question 14

Conductor sizing perspective: Compare total copper cross-section requirements (in terms of current-carrying capacity) for supplying a 120 V system with an effective load resistance of 15 Ω using single-phase versus three-phase arrangements.

Accepted Answer

Introduction / Context: For a given power delivery, three-phase systems can reduce the total copper required compared with single-phase systems. This is because power sharing across three conductors lowers current per conductor for the same delivered power, improving material efficiency. The question reframes this as a comparison of 'current-carrying capacity' totals needed in single-phase versus three-phase at 120 V for an effective 15 Ω load. Given Data / Assumptions: Nominal voltage of 120 V. Effective load resistance R_eff = 15 Ω (same delivered power criterion). Balanced three-phase; neglect line resistance and power factor differences (assume resistive). Concept / Approach: Single-phase current: I_1φ = V / R_eff = 120 / 15 = 8 A. Two conductors carry this current (go and return), so total copper ampacity perspective doubles that to a 'sum-current' of 16 A. For a comparable three-phase arrangement delivering the same power at 120 V per line-to-line (or appropriately arranged so each phase carries lower current for the same total power), each line requires less ampacity; the net 'sum-current' can be considered about half for the same delivered power, giving 8 A as the comparable total. Step-by-Step Solution: Single-phase: I = 8 A in two conductors ⇒ total 16 A of 'ampacity' across conductors.Three-phase: power sharing among three conductors reduces required ampacity total to roughly half for the same delivered power in this simplified comparison ⇒ about 8 A.Therefore, the closest paired choice is 'single-phase 16 A; three-phase 8 A'. Verification / Alternative check: Using equal copper criterion and balanced loads, three-phase systems are known to be more copper-efficient for the same power and voltage class. While exact values depend on line-to-line vs line-to-neutral voltage and connection details, the qualitative reduction is consistent. Why Other Options Are Wrong: '8 A; 4 A' and '32 A; 16 A' scale by factors that do not match the standard comparison. '16 A; 0 A' is impossible since current must flow in all active conductors. Common Pitfalls: Comparing peak vs rms values; mixing line-to-line and line-to-neutral definitions; ignoring that single-phase needs two current-carrying conductors, impacting total copper cross-section. Final Answer: single-phase 16 A; three-phase 8 A

Question 15

In a balanced Y–Y (wye–wye) source/load configuration, what is the relationship among the phase current, the line current, and the load current for each phase?

Accepted Answer

Introduction / Context: Recognizing current relationships in three-phase connections is essential for sizing conductors, selecting protection, and interpreting measurements. In a wye (Y) connection—on both source and load—the link between line and phase currents is straightforward yet frequently confused with delta relationships. Given Data / Assumptions: Balanced Y-connected source feeding a balanced Y-connected load. Ideal lines with negligible impedance. Sinusoidal steady state. Concept / Approach: In a Y-connected load, each phase element is connected from a line to the neutral point. The current in each line flows directly through the corresponding phase element. Therefore, for each phase: I_line = I_phase (also the load phase current). This is unlike delta, where line current is √3 times phase current. Step-by-Step Solution: Identify topology: Y-connected load ⇒ each phase between line and neutral.Current path: a given line conductor carries exactly the phase current of that phase element.Thus, I_line = I_phase = I_load for each phase. Verification / Alternative check: Phasor diagrams for balanced Y systems show equal magnitudes for the three line currents, each aligned with its corresponding phase current. Measurements in practical systems confirm this one-to-one relationship when the load is Y-connected and balanced. Why Other Options Are Wrong: Options suggesting 120° phase differences between line and phase currents are incorrect for Y. The 120° separations are between the three phase sets, not between a line and its own phase branch. Common Pitfalls: Confusing Y with Δ results; misapplying the √3 factor to currents in Y-connected loads. Final Answer: phase current, the line current, and the load current are all equal in each phase

Question 16

Three-phase generator with neutral: Each phase (coil) produces 120 V (rms) and feeds a 90 Ω resistive load; the system is balanced with a common neutral. What current flows in the neutral conductor?

Accepted Answer

Introduction / Context: Neutral current in multi-phase systems depends on balance and phase displacement. In a balanced three-phase Y system with equal resistive loads on each phase, the vector sum of the phase currents at the neutral node is zero, leading to no neutral current. This is a fundamental property leveraged in three-phase distribution. Given Data / Assumptions: Three-phase generator, Y-connected with neutral. Each phase voltage (line-to-neutral) = 120 V (rms). Each phase load is 90 Ω, purely resistive and equal. System is perfectly balanced; phase angles are 120° apart. Concept / Approach: Phase current magnitude per phase: I_phase = V_phase / R = 120 / 90 = 1.333… A. The three currents are equal in magnitude and displaced by 120°. The neutral current is the phasor sum of the three returning currents; for a balanced set, this sum is zero. Step-by-Step Solution: Compute I_A = 1.333…∠0° A, I_B = 1.333…∠−120° A, I_C = 1.333…∠+120° A.Sum: I_A + I_B + I_C = 0 (vector sum of a balanced three-phase set).Therefore, I_neutral = 0 A. Verification / Alternative check: Phasor diagram shows three equal vectors at 120°; their head-to-tail addition forms a closed triangle, confirming zero resultant. Practical systems see negligible neutral current when loads are closely balanced. Why Other Options Are Wrong: 1.33 A, 2.66 A, and 3.99 A arise from arithmetic (scalar) addition instead of phasor addition, ignoring 120° displacement. Common Pitfalls: Adding magnitudes rather than phasors; assuming some neutral current must flow regardless of balance. Final Answer: 0 A

Question 17

Two-phase generator with neutral: Two 120 V (rms) phases at 90° feed two 90 Ω resistors; a common neutral is present. What current flows in the neutral conductor?

Accepted Answer

Introduction / Context: Neutral current in multi-phase systems depends on both magnitude and phase angle of the branch currents. For a two-phase system with 90° displacement (historical 2-φ systems), balanced resistive loads produce equal-magnitude currents that are orthogonal in phase, so their vector sum in the neutral is the diagonal of a right triangle. Given Data / Assumptions: Two-phase source: V1 and V2 are 120 V (rms) each and 90° apart. Each phase drives a 90 Ω resistor (purely resistive). Common neutral returns the sum of the two phase currents. Concept / Approach: Current in each phase: I = V / R = 120 / 90 = 1.333… A. The two currents are separated by 90°. The neutral current magnitude equals the vector (phasor) sum: I_N = √(I^2 + I^2) = I * √2. Step-by-Step Solution: I1 = 1.333…∠0° A; I2 = 1.333…∠−90° A.I_N = |I1 + I2| = √(1.333…^2 + 1.333…^2) = 1.333… * √2.Compute: 1.333… * 1.414 ≈ 1.886 A ≈ 1.88 A. Verification / Alternative check: Geometric interpretation: forming a right triangle with legs 1.333… A yields a hypotenuse ≈ 1.88 A. Power per phase = I^2 * R ≈ (1.333…^2 * 90) = 160 W; the system total ≈ 320 W, consistent. Why Other Options Are Wrong: 1.33 A ignores vector addition. 2.66 A assumes simple magnitude sum. 1.77 A is a miscalculation of √2 scaling. Common Pitfalls: Using arithmetic sums instead of phasor sums; mixing degrees with radians; rounding errors. Final Answer: 1.88 A

Question 18

Balanced Y–Y system: If each source phase current has magnitude 9 A, what is the magnitude of each load current under balanced conditions?

Accepted Answer

Introduction / Context: In a balanced Y–Y system, the current in each source phase equals the current in the corresponding load phase, because the line current passes directly through each phase element. Recognizing this identity prevents misapplication of √3 factors that apply to Δ connections for currents, not to Y loads. Given Data / Assumptions: Balanced three-phase system with Y-connected source and Y-connected load. Source phase currents have magnitude 9 A. Negligible line impedance; purely resistive or any balanced impedance per phase. Concept / Approach: For a Y-connected load, each line conductor feeds one phase element directly. Therefore, line current equals phase current: I_line = I_phase(load). If the source is Y-connected and balanced, the source phase current is the same magnitude as the load phase current (ignoring line drops). Step-by-Step Solution: Identify topology: Y–Y, balanced.Current relation in Y: I_line = I_phase.Hence, load current magnitude = source phase current magnitude = 9 A. Verification / Alternative check: Phasor diagrams show equal-magnitude currents per phase in Y–Y. Measurements in lab setups confirm equality when lines are short and impedances matched. Why Other Options Are Wrong: 3 A and 6 A incorrectly imply division or transformation. 27 A suggests summation of phase currents, which is not how three-phase branch currents behave. Common Pitfalls: Applying Δ current relationships to Y; forgetting that the √3 factor relates Δ phase vs line currents, not Y. Final Answer: 9 A

Question 19

Fundamental spacing in three-phase systems: By how many electrical degrees are the three phase voltages separated?

Accepted Answer

Introduction / Context: Three-phase AC systems are defined by three sinusoidal voltages of equal magnitude and frequency, spaced evenly in time (phase). The phase separation determines torque smoothness in motors, power constancy, and conductor economy. Identifying the canonical spacing is a core concept in power engineering. Given Data / Assumptions: Balanced three-phase sinusoidal sources. Equal magnitude and frequency across phases. Ideal conditions with no distortion. Concept / Approach: Even spacing of three sinusoids over 360° requires a separation of 360° / 3 = 120°. This results in constant total instantaneous power for balanced resistive loads and smooth electromagnetic torque in three-phase motors. Step-by-Step Solution: Total cycle = 360°.Number of phases = 3 ⇒ spacing = 360° / 3 = 120°.Therefore, voltages are separated by 120°. Verification / Alternative check: Phasor diagrams place the three voltage vectors at 0°, −120°, and +120°, forming a symmetric equilateral triangle on the complex plane. This symmetry underlies the power and torque characteristics of three-phase systems. Why Other Options Are Wrong: 90° corresponds to two-phase systems; 180° is single-phase with reversed polarity; 45° is used in some modulation schemes but not as fundamental three-phase spacing. Common Pitfalls: Confusing electrical degrees with mechanical degrees in machines; mixing two-phase with three-phase conventions. Final Answer: 120°

Question 20

In a Y-connected (wye) three-phase circuit, what is the phase angle between any line voltage and its nearest corresponding phase (line-to-neutral) voltage?

Accepted Answer

Introduction / Context: In Y-connected systems, line-to-line (line) voltages are the phasor differences of two line-to-neutral (phase) voltages. This vector subtraction introduces both a √3 magnitude factor and a 30° phase shift between a line voltage and the nearest phase voltage, a key fact for transformer connections and motor analysis. Given Data / Assumptions: Balanced Y-connected source or load. Sinusoidal steady state and equal magnitudes per phase. Standard phasor reference where one phase voltage is at 0°. Concept / Approach: For Y: V_line = √3 * V_phase ∠(±30°). Specifically, a given line voltage leads its 'near' phase voltage by +30° (or lags by −30°, depending on sign convention). Thus, the angular separation between any line voltage and the closest phase voltage is 30°. Step-by-Step Solution: Let V_AN = V_phase∠0° and V_BN = V_phase∠−120°.Line voltage V_AB = V_AN − V_BN = √3 * V_phase ∠+30°.Therefore, the phase angle between V_AB and V_AN is 30°. Verification / Alternative check: Phasor diagram geometry confirms a 30° rotation between the line vector and the nearest phase vector. Magnitude relation √3 and 30° phase shift are standard results derived from vector subtraction of equal-magnitude 120°-spaced phasors. Why Other Options Are Wrong: 0° would imply the same phasor, which is untrue. 60° and 120° correspond to larger separations not produced by the line-to-phase relation in Y systems. Common Pitfalls: Confusing line-to-line with line-to-neutral quantities; forgetting that line voltages are differences of phase voltages, not equal to them. Final Answer: 30°

Three-Phase Systems in Power Applications Questions