Question 1

Assertion–Reason: The phenomenon of differential mobility is called the transferred electron effect. Gallium Arsenide (GaAs) exhibits this effect. Evaluate correctness of Assertion (A) and Reason (R).

Accepted Answer

Introduction / Context: This question explores semiconductor physics concepts behind the Gunn effect and transferred electron devices. Understanding differential mobility is important in high-frequency microwave device design. Given Data / Assumptions: Assertion: Differential mobility phenomenon is termed transferred electron effect. Reason: GaAs exhibits transferred electron effect. Concept / Approach: Transferred electron effect occurs in materials like GaAs where electrons move from a high-mobility valley to a low-mobility valley under strong electric fields. This leads to negative differential resistance, enabling oscillators. Step-by-Step Solution: Assertion is correct: definition of transferred electron effect.Reason is correct: GaAs is the classical material exhibiting this effect (basis of Gunn diodes).Reason also explains Assertion correctly, as GaAs is the experimental evidence of the phenomenon. Verification / Alternative check: Microwave engineering textbooks confirm GaAs shows Gunn oscillations due to transferred electron effect. Why Other Options Are Wrong: Any option denying either A or R contradicts semiconductor physics facts. Common Pitfalls: Confusing transferred electron effect with avalanche or tunneling effects. Final Answer: Both A and R are correct and R is correct explanation of A

Question 2

A transmission line is excited by a 100 V DC source. If the reflection coefficients at both ends are 1, what will happen to the oscillations on the line?

Accepted Answer

Introduction / Context: Transmission line theory considers wave reflections at mismatched terminations. Reflection coefficients determine how much energy is reflected back, influencing standing waves and oscillations. Given Data / Assumptions: DC source = 100 V applied. Reflection coefficient at both ends = 1 (perfect reflection). Concept / Approach: If both ends reflect 100% of the energy, no dissipation occurs. The wave will bounce indefinitely, sustaining oscillations unless loss mechanisms are introduced. Step-by-Step Solution: Reflection coefficient Γ = +1 means full constructive reflection.At source end and load end, energy reflects back entirely.Thus, energy is trapped → oscillations never decay. Verification / Alternative check: In practical systems, even small losses attenuate oscillations. But with ideal assumption (Γ = 1 at both ends), oscillations persist forever. Why Other Options Are Wrong: No oscillations: incorrect, reflection ensures oscillations.1 or 2: wrong, indefinite reflection occurs.Finite oscillations: contradicts perfect reflection assumption. Common Pitfalls: Forgetting that Γ = 1 means no energy is absorbed by load/source. Final Answer: the oscillations will continue indefinitely

Question 3

Is a reflex klystron oscillator essentially a low-power microwave device?

Accepted Answer

Introduction / Context: A reflex klystron oscillator is an early microwave vacuum tube that generates RF oscillations using velocity modulation and electron bunching. It is widely used in laboratories as a signal source at microwave frequencies, but its power level is relatively low compared with traveling-wave tubes (TWTs) or multi-cavity klystrons. Given Data / Assumptions: Device: Reflex klystron oscillator. Application: Microwave oscillator for test and measurement. Power levels: Generally a few milliwatts to hundreds of milliwatts. Concept / Approach: The reflex klystron uses a single resonant cavity and a repeller electrode to reflect electrons, creating velocity modulation and bunching. Due to its simple structure and single cavity, it cannot sustain high electron beam currents or high output powers. Hence, it is primarily classified as a low-power device. Step-by-Step Solution: Identify device: reflex klystron oscillator.Review operating principle: velocity modulation, repeller electrode, single cavity.Note limitation: low efficiency (10–20%), low beam current, limited to small power levels.Conclude: reflex klystron oscillator is essentially a low-power device. Verification / Alternative check: Textbooks confirm that reflex klystrons provide milliwatt-level power and are primarily used as local oscillators or test sources. Why Other Options Are Wrong: False: contradicts practical data, as power output is indeed low. Only in pulsed operation: even pulsed reflex klystrons are low power. Amplifier role: reflex klystrons are rarely used as amplifiers. Depends on cavity design: though design affects efficiency, the device class remains low power. Common Pitfalls: Confusing reflex klystrons with multi-cavity klystron amplifiers, which can produce high power. Final Answer: True

Question 4

A resistive microwave load with impedance ZL = 150 Ω is connected to a 50 Ω coaxial line. What will be the standing wave ratio (SWR) on the line?

Accepted Answer

Introduction / Context: Standing wave ratio (SWR) is a measure of impedance mismatch between a transmission line and its load. It indicates how much of the incident signal is reflected due to mismatch. Given Data / Assumptions: Line impedance Z0 = 50 Ω. Load impedance ZL = 150 Ω (purely resistive). Formula: SWR = (1 + |Γ|) / (1 − |Γ|). Reflection coefficient Γ = (ZL − Z0) / (ZL + Z0). Concept / Approach: SWR is always ≥ 1. If ZL = Z0, SWR = 1 (perfect match). If ZL differs from Z0, reflections occur, raising SWR. Step-by-Step Solution: Γ = (150 − 50) / (150 + 50) = 100 / 200 = 0.5.SWR = (1 + 0.5) / (1 − 0.5) = 1.5 / 0.5 = 3.Thus, the SWR = 3 (exactly). Verification / Alternative check: Measured VSWR in microwave labs for this ratio confirms the theoretical result of 3. Why Other Options Are Wrong: More than 3: incorrect, exact value is 3. Less than 3: mismatch is significant, not smaller. Either more than 3 or equal: introduces ambiguity not present in calculation. Exactly 2: would occur if ZL = 100 Ω, not 150 Ω. Common Pitfalls: Forgetting to take absolute value of Γ; misapplying formula. Final Answer: Equal to 3

Question 5

Assertion (A): A backward wave oscillator (BWO) has an internal positive feedback mechanism. Reason (R): Positive feedback is necessary for sustained oscillations. Which of the following is correct?

Accepted Answer

Introduction / Context: A backward wave oscillator (BWO) is a type of vacuum tube oscillator that uses an electron beam interacting with a slow-wave structure. It is important to understand the role of feedback in generating oscillations. Given Data / Assumptions: BWO principle: beam-wave interaction with backward wave propagation. Oscillations require feedback to reinforce the signal. Concept / Approach: In a BWO, internal positive feedback arises because the backward-traveling wave couples energy back to the electron beam. This feedback sustains oscillations. The reason statement correctly explains why oscillations persist: positive feedback is a fundamental requirement in all oscillators. Step-by-Step Solution: Assertion: BWO has internal positive feedback → true.Reason: Positive feedback necessary for oscillations → true.Link: The positive feedback in BWO explains its oscillation mechanism. Verification / Alternative check: Microwave device theory confirms BWOs operate on backward wave interaction providing feedback for self-oscillation. Why Other Options Are Wrong: Option B: The explanation is indeed correct. Option C or D: deny correct parts. Option E: both statements are not wrong. Common Pitfalls: Confusing feedback in BWO with external cavity feedback in klystrons. Final Answer: Both A and R are correct, and R is the correct explanation of A

Question 6

The velocity factor of a transmission line depends mainly on which property?

Accepted Answer

Introduction / Context: The velocity factor of a transmission line represents the ratio of signal propagation speed along the line to the speed of light in vacuum. It is critical for timing, phase delay, and wavelength calculations in RF and communication systems. Given Data / Assumptions: Transmission line filled with dielectric material. Velocity factor (VF) = v / c. v = velocity of wave in line, c = velocity of light in vacuum. Concept / Approach: The propagation speed in a line depends on the dielectric constant (relative permittivity) of the insulation material. Higher permittivity slows down the wave, reducing velocity factor. Temperature and skin effect mainly affect losses, not VF. Step-by-Step Solution: Formula: v = c / sqrt(εr).Thus, VF = 1 / sqrt(εr).Hence, velocity factor depends on εr of dielectric, not conductor losses or temperature. Verification / Alternative check: Cables with polyethylene dielectric (εr ≈ 2.25) have VF ≈ 0.67. Teflon (εr ≈ 2.1) → VF ≈ 0.69. These match practical data. Why Other Options Are Wrong: Temperature: affects loss, not speed directly. Skin effect: influences attenuation at high frequencies. None of the above: incorrect as permittivity is main factor. Magnetic permeability: usually close to 1 in dielectrics. Common Pitfalls: Confusing VF with attenuation factor; assuming conductor properties dominate propagation velocity. Final Answer: Relative permittivity of dielectric material

Question 7

A transmission line has a phase constant β = 29.8 rad/m. At a frequency of 1000 MHz, what will be the wavelength of the signal?

Accepted Answer

Introduction / Context: The phase constant β in a transmission line is related to the propagation constant and wavelength. Calculating wavelength from β is a common problem in RF and microwave engineering. Given Data / Assumptions: β = 29.8 rad/m. Frequency f = 1000 MHz = 1 GHz. Formula: β = 2π / λ. Concept / Approach: By definition, wavelength λ = 2π / β. Once β is known, λ can be calculated directly. Step-by-Step Solution: λ = 2π / β.= 6.283 / 29.8 ≈ 0.211 m.Thus wavelength = 0.21 m approximately. Verification / Alternative check: For 1 GHz in free space, λ ≈ 0.3 m. A phase constant of 29.8 rad/m corresponds to a slightly shorter wavelength due to dielectric or guide effects, consistent with 0.21 m. Why Other Options Are Wrong: 29.8 m, 2.98 m, 2.1 m: too large, do not satisfy β formula. 0.0298 m: too small, not matching calculation. Common Pitfalls: Using frequency-speed formula instead of β relation; ignoring 2π factor. Final Answer: 0.21 m

Question 8

In a klystron amplifier, the diagram used to represent the distance-time history of electrons is called what?

Accepted Answer

Introduction / Context: Klystron amplifiers are vacuum tube devices that amplify microwave signals using velocity modulation and bunching of electrons. Visualizing electron bunching is important for understanding their operation. Given Data / Assumptions: Device: klystron amplifier. Principle: velocity modulation → bunching → RF amplification. Concept / Approach: The bunching diagram plots electron position versus time, showing how initially uniform electron beams get grouped into bunches as they travel due to velocity modulation. This diagram is fundamental to analyzing klystron operation. Step-by-Step Solution: Electrons enter cavity → velocity modulated by RF.Different velocities cause electrons to drift closer or farther apart.Over distance, bunches form → represented in bunching diagram. Verification / Alternative check: All microwave engineering texts label this as a bunching diagram. Why Other Options Are Wrong: Apple gate: not a known term in klystron theory. Asynchronous: not standard for electron bunching. Velocity modulation diagram: refers to initial step, not full bunching visualization. Transit-time diagram: general term, not specific. Common Pitfalls: Confusing bunching diagram with velocity modulation graph. Final Answer: Bunching diagram

Question 9

In a circular waveguide, which of the following modes is the dominant mode of propagation?

Accepted Answer

Introduction / Context: Waveguides support various modes depending on frequency and geometry. The dominant mode is the one with the lowest cutoff frequency that propagates first when frequency increases. Given Data / Assumptions: Waveguide type: circular. Mode choices: TE and TM families. Concept / Approach: In rectangular waveguides, TE10 is dominant. In circular waveguides, the dominant mode is TE11, as it has the lowest cutoff frequency among supported modes. Step-by-Step Solution: Compare cutoff frequencies of modes.TE11 cutoff frequency is lowest in circular geometry.Therefore, TE11 is dominant mode. Verification / Alternative check: Standard microwave engineering reference confirms TE11 as dominant mode for circular waveguides. Why Other Options Are Wrong: TE01, TE20, TE21: higher cutoff frequencies. TM01: not the dominant one. Common Pitfalls: Assuming TE01 or TM01 is dominant due to numbering; in circular waveguides, TE11 is the actual dominant mode. Final Answer: TE11

Question 10

In a traveling-wave tube (TWT), does distributed interaction between the electron beam and the traveling wave take place?

Accepted Answer

Introduction / Context: A traveling-wave tube (TWT) is a broadband microwave amplifier. It amplifies signals by continuous interaction between an electron beam and an RF wave propagating along a slow-wave structure. Given Data / Assumptions: Device: TWT. Operation: distributed beam-wave interaction. Concept / Approach: Unlike klystrons (localized interaction), TWTs have distributed interaction along the length of the helix or coupled cavity structure, enabling wide bandwidth and high gain. Step-by-Step Solution: Electron beam injected into tube.RF signal propagates along helix (slow-wave structure).Electrons interact continuously along length, transferring energy. Verification / Alternative check: Engineering texts: TWTs are distributed amplifiers, unlike localized amplifiers. Why Other Options Are Wrong: False: contradicts principle of TWT. Only in pulsed mode: TWT works in CW as well. Backward wave operation: belongs to BWO, not TWT. Voltage dependency: interaction is continuous across operating range. Common Pitfalls: Confusing TWT with klystron's bunching interaction. Final Answer: True

Question 11

In a klystron amplifier, if the direct current (dc) electron velocity is v0 and the dc electron charge density is ρ0, then what is the expression for the dc beam current?

Accepted Answer

Introduction / Context: Beam current in electron devices depends on the product of charge density and velocity. This relation is fundamental in electron beam theory for microwave tubes. Given Data / Assumptions: Electron velocity = v0. Charge density = ρ0. Concept / Approach: Beam current density J = ρ * v. If beam cross-sectional area = 1 unit, then total current I = ρ0 * v0. This is the simplest form of the relation. Step-by-Step Solution: Current density J = ρ * v.Here ρ = charge density = ρ0.v = velocity = v0.Thus I = ρ0 * v0. Verification / Alternative check: All beam theory references confirm I = ρ0 * v0 * A, with A = 1 in this simplified form. Why Other Options Are Wrong: ρ0 / v0, v0 / ρ0, (v0 / ρ0)^2: incorrect dimensional analysis. ρ0 * v0^2: introduces extra power of velocity. Common Pitfalls: Forgetting beam area factor; misinterpreting density definition. Final Answer: ρ0 * v0

Question 12

Assertion (A): A coaxial line is a non-radiating transmission line. Reason (R): In a coaxial line, the electric and magnetic fields are confined to the region between the concentric conductors. Which statement is correct?

Accepted Answer

Introduction / Context: Coaxial transmission lines are widely used in RF applications because they confine electromagnetic fields and prevent radiation losses. The assertion-reason type question tests the understanding of why coaxial lines do not radiate. Given Data / Assumptions: Structure: inner and outer concentric conductors. Fields confined to dielectric between conductors. Concept / Approach: The fields in coaxial geometry cancel outside the line. The outer conductor shields the fields, confining them to the dielectric region. This explains why coaxial lines are non-radiating. Step-by-Step Solution: Assertion A: coaxial line is non-radiating → true.Reason R: fields confined between conductors → true.Link: confinement is exactly why no external radiation occurs. Verification / Alternative check: Coaxial cable is preferred for shielding-sensitive RF signals precisely for this reason. Why Other Options Are Wrong: Option B: R is indeed the correct explanation. Option C: R is not wrong. Option D: A is not wrong. Option E: Both are correct. Common Pitfalls: Confusing coaxial with two-wire lines that radiate due to unconfined fields. Final Answer: Both A and R are correct, and R is the correct explanation of A

Question 13

A lossless transmission line with characteristic impedance Z0 is terminated in a pure reactance equal to −j·Z0. What is the resulting Voltage Standing Wave Ratio (VSWR) on the line?

Accepted Answer

Introduction / Context: This question tests standing-wave behavior on a lossless transmission line when the load is purely reactive. Understanding how the reflection coefficient magnitude relates to VSWR is core to RF and microwave engineering. Given Data / Assumptions: Lossless line with real Z0. Load ZL = −j·Z0 (pure reactance, magnitude equal to Z0). VSWR is defined using the magnitude of the reflection coefficient |Γ|. Concept / Approach: The reflection coefficient at the load is Γ = (ZL − Z0) / (ZL + Z0). For a purely reactive load, |Γ| = 1 because the load cannot absorb average power; it only stores and returns energy. VSWR = (1 + |Γ|) / (1 − |Γ|). Step-by-Step Solution: Let ZL = −j·Z0.Γ = (−j·Z0 − Z0) / (−j·Z0 + Z0) = (−1 − j) / (1 − j).|−1 − j| = √(1 + 1) = √2 and |1 − j| = √2 ⇒ |Γ| = 1.VSWR = (1 + 1) / (1 − 1) = 2 / 0 → infinity. Verification / Alternative check: If a load is purely reactive, average real power at the load is zero; thus, all incident power is reflected (|Γ| = 1), which forces VSWR to be unbounded (infinite). Why Other Options Are Wrong: 2 or 10 or 5: These correspond to |Γ| strictly less than 1; not possible with a pure reactance on a lossless line.1: VSWR = 1 only for a perfect match (ZL = Z0), not for a reactive load. Common Pitfalls: Confusing the sign or value of the reactance with resistive matching; any purely reactive termination on a lossless line yields |Γ| = 1. Final Answer: infinity

Question 14

In a directional coupler, which relationship between coupling, isolation, and directivity (in dB) is correct?

Accepted Answer

Introduction / Context: Directional couplers are four-port passive microwave components characterized by coupling (how much power is sampled), isolation (how much the isolated port is suppressed), and directivity (ability to discriminate direction). Given Data / Assumptions: All quantities are in decibels. Standard coupler port labeling and definitions apply. Concept / Approach: Directivity (in dB) is defined as isolation (in dB) minus coupling (in dB). Rearranged: isolation = coupling + directivity. This identity is fundamental in coupler specifications and cascaded design calculations. Step-by-Step Solution: Start from D (dB) = Iso (dB) − Cpl (dB).Rearrange to Iso (dB) = Cpl (dB) + D (dB).Hence, isolation equals coupling plus directivity (in dB). Verification / Alternative check: Manufacturers’ datasheets present these three figures of merit; calculating any one from the other two follows this identity. Why Other Options Are Wrong: Statements that add isolation and directivity to get coupling—or multiply terms—contradict the standard definition. Common Pitfalls: Confusing linear ratios with dB; the add/subtract rule applies only in dB. Final Answer: isolation (dB) equals coupling + directivity

Question 15

The operating principle of a backward-wave oscillator (BWO) is most similar to which microwave device?

Accepted Answer

Introduction / Context: Backward-wave oscillators (BWOs) are microwave sources that generate oscillations by interaction of an electron beam with a slow-wave structure supporting backward-wave propagation (negative group velocity). Given Data / Assumptions: Comparison is about the nature of oscillation and feedback mechanism. Reflex klystron is a one-cavity oscillator relying on electron bunching and feedback via transit-time. Concept / Approach: Like the reflex klystron, a BWO is an oscillator (not merely an amplifier) that uses an internal feedback mechanism tied to electron transit time and wave interaction. While a BWO uses a slow-wave structure (as in TWTs), its operation is as a self-excited oscillator akin to a reflex klystron rather than a linear amplifier. Step-by-Step Solution: BWO supports backward-wave mode on a slow-wave structure.Electron beam couples energy into the backward wave, sustaining oscillations.This self-oscillatory behavior is conceptually closer to a reflex klystron oscillator than to a simple amplifier. Verification / Alternative check: Microwave textbooks classify BWO as an oscillator often contrasted with reflex klystrons due to negative resistance / feedback mechanisms. Why Other Options Are Wrong: Klystron amplifier / TWT: primarily linear amplifiers requiring input drive.Magnetron / gyrotron: different interaction mechanisms (crossed-field, cyclotron resonance). Common Pitfalls: Assuming similarity to TWT because of the slow-wave structure; the key distinction is oscillator vs. amplifier. Final Answer: reflex klystron oscillator

Question 16

In microstrip transmission lines used on printed circuit boards, can the RF/microwave signal path be redirected simply by changing (routing/turning) the centre strip geometry?

Accepted Answer

Introduction / Context: Microstrip lines are planar transmission lines consisting of a conducting strip over a dielectric substrate with a ground plane on the other side. They are ubiquitous in RF PCBs for routing signals. Given Data / Assumptions: Standard microstrip on dielectric (FR-4, Rogers, etc.). Controlled impedance achieved via strip width, substrate height, and dielectric constant. Concept / Approach: Because microstrip is a planar structure, the signal follows the metal strip. Changing the strip's path (turns, bends, meanders) redirects the guided wave, provided impedance is controlled (corner mitering, bend radius) to minimize reflections and radiation. Step-by-Step Solution: Routing the centre strip defines current path and fringing fields.Bends and tees re-direct the guided wave if designed with proper impedance control.Thus, altering the strip geometry does redirect the signal path on the board. Verification / Alternative check: EM simulators and impedance calculators for microstrip confirm field confinement to the vicinity of the strip and ground reference, following the trace path. Why Other Options Are Wrong: 'False' options ignore practical PCB RF designs where routing is exactly how signals are guided. Common Pitfalls: Ignoring added discontinuity effects at sharp bends; proper mitering or arcs are used to control impedance. Final Answer: True

Question 17

Estimate the approximate half-power beamwidth (HPBW) in degrees for a 1 m diameter parabolic antenna operating at 10 GHz.

Accepted Answer

Introduction / Context: Parabolic reflectors concentrate energy into a narrow main lobe. A common quick estimate for HPBW is used by antenna engineers for link budgeting and pointing accuracy. Given Data / Assumptions: Antenna diameter D = 1 m. Frequency f = 10 GHz ⇒ wavelength λ ≈ 0.03 m. Rule-of-thumb for HPBW in degrees: HPBW ≈ 70 * (λ / D). Concept / Approach: The HPBW of a well-illuminated parabolic reflector is inversely proportional to aperture size and directly proportional to wavelength. Quick formulas give a practical estimate, adequate for exam problems and early design. Step-by-Step Solution: Compute λ = c / f ≈ 3×10^8 / 10×10^9 = 0.03 m.HPBW ≈ 70 * (λ / D) = 70 * (0.03 / 1) = 2.1 degrees.Among the given discrete choices, 1° is the nearest practical choice (far closer than 5°). Verification / Alternative check: Some texts use 58–70 factor depending on illumination efficiency; with 58 the HPBW ≈ 1.74°, still closer to 1° than to 5°. Why Other Options Are Wrong: 5°, 50°, 100°: too wide for a 1 m dish at 10 GHz.0.2°: unrealistically narrow given the aperture and frequency. Common Pitfalls: Confusing first-null beamwidth (about 2× HPBW) with HPBW, leading to larger estimates. Final Answer: 1°

Question 18

Do both IMPATT and TRAPATT microwave diodes rely on the avalanche effect as part of their operation?

Accepted Answer

Introduction / Context: IMPATT (IMPact Avalanche Transit-Time) and TRAPATT (TRApped Plasma Avalanche Triggered Transit) diodes are negative-resistance microwave devices used in oscillators and amplifiers. Given Data / Assumptions: Both devices use avalanche breakdown to initiate carrier multiplication. Transit-time effects provide the necessary phase delay for negative resistance. Concept / Approach: In IMPATT, carriers generated by avalanche are swept across the drift region, producing a 180° current–voltage phase shift. TRAPATT also starts with avalanche, but operates with a trapped plasma domain and typically exhibits higher efficiency and lower operating frequency than IMPATT. Step-by-Step Solution: Avalanche multiplication → carrier surge.Transit-time through drift region → phase delay leading to negative resistance.Both devices depend on the avalanche trigger to start the process. Verification / Alternative check: Standard microwave device references categorize both as avalanche transit-time devices, differing in domain dynamics and efficiency. Why Other Options Are Wrong: Statements that only one is avalanche-based are incorrect.Frequency limits (e.g., 'only above 100 GHz') are irrelevant to the avalanche principle. Common Pitfalls: Confusing Gunn diodes (transferred electron effect) with avalanche devices. Final Answer: True

Question 19

In manufacturing, which process is typically used to fabricate a microstrip transmission line on a dielectric substrate?

Accepted Answer

Introduction / Context: Microstrip lines are planar RF interconnects realized on PCB or hybrid substrates. Their fabrication parallels printed circuit board processes to define controlled-impedance traces. Given Data / Assumptions: Standard copper cladding on dielectric (e.g., FR-4, Rogers). Patterning via photolithography and chemical etching. Concept / Approach: The umbrella term 'printed circuit technique' covers imaging the trace (photoresist), UV exposure through a mask, development, and chemical etching—often called photo-etching as a step in the printed circuit process. In industry, one describes the overall approach as the printed circuit technique. Step-by-Step Solution: Start with copper-clad substrate.Apply photoresist, expose pattern, develop.Etch away unwanted copper to leave the microstrip line. Verification / Alternative check: RF PCB houses implement microstrip/CPW structures using standard PCB photolithography and etching flows. Why Other Options Are Wrong: Oxidation/cladding/electroplating alone are materials steps, not the full patterning/fabrication method for the line geometry.'Photo etching' is a step, but the accepted holistic process name is 'printed circuit technique'. Common Pitfalls: Equating a single step (etching) with the entire manufacturing process nomenclature. Final Answer: printed circuit technique

Question 20

In practical transmission engineering, what is the approximate loss angle (dielectric loss tangent angle) of a good-quality insulated cable?

Accepted Answer

Introduction / Context: The loss angle (often discussed via tan delta) quantifies dielectric losses in insulated cables. A small loss angle indicates that the dielectric behaves almost ideally capacitive with minimal energy dissipation per cycle, which is desired in high-quality power and communication cables. Given Data / Assumptions: Term: 'loss angle' refers to the angle between the current vector and the ideal quadrature (purely capacitive) current in a lossy dielectric. 'Good quality cable' implies low dielectric loss material (e.g., polyethylene, XLPE, PTFE). Typical power or RF usage where dielectric heating should be minimal. Concept / Approach: The dielectric loss tangent (tan δ) is the ratio of the resistive (loss) component to the reactive (capacitive) component of current. The loss angle δ is small for good dielectrics. Converting between tan δ and degrees, practical values for premium cable dielectrics correspond to loss angles well below a few degrees, commonly near 1° or less at the operating frequency. Step-by-Step Solution: Identify that a 'good' dielectric has tan δ ≪ 1.Small tan δ means δ (in radians) ≈ tan δ; thus δ is only a few degrees at most.Industry data for quality cable dielectrics (e.g., XLPE, PTFE) show tan δ in the range of 10^-4 to 10^-3 at power frequencies, corresponding to a very small angle.Hence, an order-of-magnitude correct choice is about 1°, not tens of degrees. Verification / Alternative check: Manufacturers specify dissipation factor (tan δ) values; translating these to an angle confirms a very small δ, supporting the ~1° choice for a quick, standards-style question. Why Other Options Are Wrong: 30°/70°/90°: Such large angles imply enormous dielectric loss, inconsistent with 'good quality' insulation. 15°: Still far too large for premium cable dielectric behavior. Common Pitfalls: Confusing loss angle with phase angle of current in an ideal capacitor; assuming higher operating temperature automatically implies large δ for good materials. Final Answer: 1°

Microwave Communication Questions