Question 1

In the given figure reflection coefficient at load is

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Question 2

In the given figure the time taken by the wave to make one round trip is

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Question 3

In microwave networks, a uniform waveguide section behaves like which type of filter around its cutoff?

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Introduction / Context: Metallic waveguides support propagation only above their cutoff frequency. Below cutoff, fields are evanescent and power does not propagate. This frequency-selective behavior mirrors a classic filter response in RF design. Given Data / Assumptions: Uniform rectangular or circular waveguide section. Single-mode region around the dominant mode cutoff. Losses assumed small compared with cutoff effects. Concept / Approach: For a given mode, the propagation constant β is real for f > fc and imaginary for f < fc. Thus, the section 'passes' energy above fc and 'blocks' it below fc, which is the hallmark of a high-pass filter behavior. Step-by-Step Solution: Identify dominant mode with cutoff fc.For f < fc: β becomes imaginary; fields decay exponentially → no transmission.For f > fc: β is real; power propagates → transmission occurs.Therefore, the waveguide section functions as a high-pass filter. Verification / Alternative check: Plotting S21 versus frequency for an ideal waveguide shows near-zero transmission below fc and increasing transmission above fc, matching a high-pass characteristic. Why Other Options Are Wrong: Low-pass: Opposite of waveguide behavior. Band-pass / Band-stop: Would require two edges; a simple uniform guide exhibits a single cutoff edge. All-pass: Not true; magnitude and phase vary strongly around cutoff. Common Pitfalls: Confusing cavity resonators (band-pass) with uniform guides; overlooking that mode conversion or discontinuities can create additional filtering not present in a straight section. Final Answer: High-pass filter

Question 4

A transmission line has a load reflection coefficient Γ = 0.2 ∠45°. What is the standing-wave ratio (SWR) on the line?

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Introduction / Context: The standing-wave ratio (SWR) is a key performance metric of transmission line matching. It is computed directly from the magnitude of the reflection coefficient at the load. Given Data / Assumptions: Reflection coefficient Γ = 0.2 ∠45°. Only |Γ| affects SWR; the angle does not enter the SWR formula. Concept / Approach: The SWR on a line is SWR = (1 + |Γ|) / (1 − |Γ|). This follows from the ratio of maximum to minimum envelope of the standing wave formed by incident and reflected waves. Step-by-Step Solution: Compute magnitude: |Γ| = 0.2.Apply formula: SWR = (1 + 0.2) / (1 − 0.2).SWR = 1.2 / 0.8 = 1.5. Verification / Alternative check: As a quick check, |Γ| = 0 yields SWR = 1 (perfect match), and |Γ| = 0.2 should yield a modest mismatch above 1, consistent with 1.5. Why Other Options Are Wrong: 0.8, 1.1, 1.2: Do not satisfy the formula for |Γ| = 0.2. 2.0: Would require |Γ| = 1/3, not 0.2. Common Pitfalls: Using angle of Γ in SWR computation; mixing return loss (in dB) with SWR; forgetting to use magnitude only. Final Answer: 1.5

Question 5

Skin effect becomes more pronounced as the operating frequency increases. Is this statement accurate for practical conductors?

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Introduction / Context: Skin effect is the tendency of alternating current to concentrate near the surface of a conductor at higher frequencies. It is critical in RF, microwave, and high-speed digital design because it raises effective resistance and loss with frequency. Given Data / Assumptions: Good metallic conductors (e.g., copper) with finite conductivity. Frequency domain from kHz to GHz where classical skin effect applies. Concept / Approach: The skin depth δ is given by δ = sqrt(2 / (ω μ σ)), where ω is angular frequency, μ is permeability, and σ is conductivity. As frequency increases, δ decreases, crowding current toward the surface and increasing AC resistance. Step-by-Step Solution: Use δ = sqrt(2 / (ω μ σ)).As ω increases, δ decreases.Smaller δ means current flows in a thinner layer → higher effective resistance and more pronounced skin effect. Verification / Alternative check: Measured insertion loss of coaxial cables and PCB traces rises with frequency largely due to skin effect (and dielectric loss), confirming the theoretical trend. Why Other Options Are Wrong: False: Contradicts the frequency dependence of δ. Conditions about superconductors/plasma frequency/1 MHz are not relevant in ordinary conductor contexts. Common Pitfalls: Confusing skin effect with proximity effect; assuming plating does not help (silver plating can reduce RF loss by lowering surface resistance). Final Answer: True

Question 6

In a backward-wave oscillator (BWO), does the guided RF wave physically 'wind back and forth' along the line?

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Introduction / Context: A backward-wave oscillator (BWO) is a vacuum device where the phase velocity of the slow-wave structure supports a wave whose phase travels opposite to the direction of power flow. This enables oscillation under proper feedback conditions. Given Data / Assumptions: BWO with slow-wave structure (e.g., a helix or periodic waveguide). Electron beam interacts with a backward wave (negative phase velocity relative to the beam). Concept / Approach: 'Backward wave' refers to the direction of phase propagation relative to energy flow, not a literal 'winding back and forth' of energy or the field structure along the line. The energy still travels toward the output; it is the phase that appears to move backwards. Therefore, the statement that the wave 'winds back and forth' is inaccurate and misleading. Step-by-Step Solution: Recognize that in BWOs, v_phase and v_group have opposite directions.Phase traveling backward does not imply energy oscillates spatially back and forth.Hence the literal interpretation is false. Verification / Alternative check: Device analyses show negative dispersion slope (backward wave) with unidirectional energy transport; measured output confirms delivered power at the output port without 'shuttling' back and forth. Why Other Options Are Wrong: True (and qualifiers): Misinterpret the meaning of backward-wave as a physical zig-zag of energy. Depends on beam voltage: Beam voltage affects frequency and gain, not the fundamental definition of backward wave. Common Pitfalls: Equating negative phase velocity with negative energy flow; misreading cartoons of slow-wave structures as literal field paths. Final Answer: False

Question 7

At its design frequency, a quarter-wave open-circuited stub is equivalent to which reactive network at its input?

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Introduction / Context: Quarter-wave stubs are fundamental matching elements in RF/microwave design. An open-circuited λ/4 stub transforms boundary conditions and can mimic resonant L-C networks at the input, enabling compact impedance transformations. Given Data / Assumptions: Lossless, uniform transmission line stub. Electrical length l = λ/4 at the frequency of interest. Open circuit at the far end. Concept / Approach: The input impedance of an open-circuited stub is Z_in = −j Z0 cot(β l). At l = λ/4, βl = π/2, cot(π/2) = 0, so Z_in = 0 (a short). A zero input impedance at a single frequency corresponds to a series resonance (equivalent L and C in series producing a short at resonance). Thus, the open λ/4 stub acts like a series-resonant L-C at its input. Step-by-Step Solution: Use Z_in(open) = −j Z0 cot(βl).Set βl = π/2 (quarter wave) → cot(π/2) = 0.Therefore Z_in = 0 → input behaves as series resonance. Verification / Alternative check: A short-circuited λ/4 stub yields Z_in → ∞ (parallel resonance). By duality, the open-ended λ/4 stub must yield series resonance at the input, consistent with design tables. Why Other Options Are Wrong: Pure L or pure C: The input is a short at resonance, not purely inductive or capacitive. Parallel LC: That would correspond to a short-circuited λ/4 stub or open-circuited λ/2 behavior. Large resistor: Lossless stubs are reactive, not resistive (aside from matching transformations). Common Pitfalls: Confusing open and shorted stubs; forgetting that resonance type (series vs parallel) depends on termination. Final Answer: Inductance and capacitance in series (series resonant)

Question 8

Assertion–Reason: Artificial transmission lines are frequently used in laboratories. Reason: Such lines can emulate real lines and function as delay elements, attenuators, and filter networks. Which statement is correct?

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Introduction / Context: Artificial (lumped) transmission lines are networks of inductors and capacitors designed to simulate the behavior of a distributed line. They are common in test benches, education, and prototyping where a physical long line is impractical. Given Data / Assumptions: Artificial line sections approximate real line parameters (L', C', R', G'). Lumped networks reproduce dispersion and delay over a band. Engineers use them as adjustable delay/attenuation/filter elements. Concept / Approach: By cascading L–C sections (e.g., constant-k, m-derived), one creates a ladder that mimics the characteristic impedance and propagation constant of a real line over a design band. This enables controlled group delay and magnitude shaping, making them valuable for lab applications. Step-by-Step Solution: Assertion: Artificial lines are frequently used → true.Reason: They emulate actual lines and provide delay, attenuation, and filtering → true.The reason explains precisely why they are used → correct explanation. Verification / Alternative check: Classic filter theory and delay line design (e.g., all-pass sections) show how lumped networks realize specified phase delay and amplitude control, confirming the stated applications. Why Other Options Are Wrong: Options B–E break the logical link or claim one of the statements is wrong; both are correct and causally related. Common Pitfalls: Assuming only distributed lines can provide delay; overlooking that artificial lines have finite bandwidth and tolerance-limited accuracy. Final Answer: Both A and R are correct and R is the correct explanation of A

Question 9

In a traveling-wave tube (TWT) amplifier, how does the amplitude of the resultant RF wave along the helix vary under small-signal operation?

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Introduction / Context: Traveling-wave tubes (TWTs) are broadband linear amplifiers. An electron beam interacts continuously with an RF wave on a slow-wave structure (e.g., helix), transferring kinetic energy to the wave and producing gain. Given Data / Assumptions: Small-signal regime (no saturation). Proper synchronism between electron beam and RF wave phase velocity. Neglecting distributed loss for the conceptual answer. Concept / Approach: In the linear region, the coupled-mode equations predict an exponential growth of the RF amplitude with distance along the helix, characterized by a spatial gain constant dependent on beam current, voltage, and circuit parameters. Step-by-Step Solution: Electron beam is velocity-modulated by the RF field on the helix.Bunching forms and exchanges energy with the wave.The wave amplitude grows approximately as exp(g * z), where g is the gain per unit length. Verification / Alternative check: Measured TWT gain (in dB) scales with length in small-signal operation, consistent with exponential growth until saturation effects limit further increase. Why Other Options Are Wrong: Linear increase: An oversimplification; the correct dependence is exponential in the small-signal model. Decrease constant: Contradicts amplification mechanism. Constant or oscillatory amplitude: Not true for an amplifying device in linear regime. Common Pitfalls: Confusing TWT amplifiers with oscillators; assuming uniform amplitude due to misinterpreting the helix as a passive line. Final Answer: It increases exponentially along the interaction region

Question 10

In a turnstile antenna, two half-wave resonant wire radiators are placed 90° apart in the same plane. How should they be excited to obtain the desired omnidirectional pattern and circular polarization?

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Introduction / Context: The turnstile antenna is widely used in VHF/UHF broadcasting, satellite beacons, and telemetry. It consists of two orthogonal half-wave dipoles intended to produce an omnidirectional horizontal pattern and circular (or near-circular) polarization when properly excited. Given Data / Assumptions: Two half-wave wire radiators are mounted at right angles (90°) in the same plane. We want the typical turnstile radiation pattern and polarization behavior. Free-space operation and ideal symmetry are assumed. Concept / Approach: To obtain circular polarization from two orthogonal linear radiators, the fields must have equal magnitudes and a phase quadrature of ±90°. In the turnstile, one dipole is fed with a 90° phase shift relative to the other so that the vector sum of their fields rotates uniformly over time, yielding circular polarization broadside to the plane of the elements and an approximately omnidirectional pattern in the horizontal plane when combined with a reflector or phasing structure. Step-by-Step Solution: Let E1 be the field from dipole 1 and E2 from dipole 2, mounted orthogonally.Set |E1| = |E2| by equal current magnitudes (balanced feeding).Impose phase(E2) − phase(E1) = ±90° (phase quadrature).The instantaneous vector sum E = E1 + jE2 traces a circle in time, giving circular polarization and the characteristic turnstile pattern. Verification / Alternative check: If the dipoles are fed in phase (0°) or in anti-phase (180°), the polarization is linear and the azimuth pattern becomes figure-eight like for individual elements, defeating the classic turnstile goal. Only ±90° quadrature with equal amplitudes yields circular polarization in the principal directions. Why Other Options Are Wrong: In phase / in phase opposition / 45°: These do not produce true circular polarization; the field remains linear or elliptical with undesired nulls.180° (listed as 'in phase opposition'): also linear and degrades omnidirectionality. Common Pitfalls: Feeding equal amplitude but forgetting the 90° phase shift; or applying 90° without maintaining equal amplitudes—both ruin circularity. Final Answer: 90° out of phase with each other

Question 11

In a backward-wave oscillator (BWO), which adjustment most effectively tunes the oscillation frequency?

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Introduction / Context: A backward-wave oscillator (BWO) is a vacuum-electron microwave source using a slow-wave structure that supports a backward wave (negative group velocity). Frequency control is crucial for agile sources and swept measurements. Given Data / Assumptions: BWO oscillation relies on interaction between the electron beam and a backward wave on a slow-wave structure (often a helix or coupled cavities). Beam voltage sets electron velocity; beam current primarily influences output power and starting conditions. Concept / Approach: The synchronism condition determining the oscillation frequency depends on matching the electron beam's axial velocity with the slow-wave phase velocity. Changing the beam voltage changes the electron velocity and thus shifts the phase synchronism point, directly tuning the oscillation frequency. Beam current affects gain and saturation but has far less direct leverage on frequency. Step-by-Step Solution: Electron beam velocity v ≈ √(2eVb / me), where Vb is beam voltage.Phase synchronism requires ω / β ≈ v / η (η is structure constant), linking frequency ω to Vb.Therefore, sweeping Vb forces ω to shift to maintain synchronism, providing practical frequency tuning. Verification / Alternative check: Manufacturer data for BWOs show wideband tuning via anode (beam) voltage sweeping; the beam current control is mainly used to set output power and avoid over-loading, not to tune frequency precisely. Why Other Options Are Wrong: Beam current alone: alters amplitude/efficiency, minimal frequency shift.Thermionic emission or collector potential: secondary effects, not the principal tuning handle.Combined minor tweaks still rely primarily on beam-voltage control. Common Pitfalls: Confusing BWO with reflex klystron repeller-voltage tuning; both change electron transit but in BWO the primary, smooth tuning is via beam voltage. Final Answer: Varying the voltage that controls the electron beam velocity

Question 12

Assertion–Reason (Transmission Lines): A matched load has impedance equal to the line’s characteristic impedance because a matched termination absorbs all incident power without reflection.

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Introduction / Context: Matching in transmission lines eliminates reflections, maximizes power transfer, and ensures flat frequency response. Understanding why ZL = Z0 is fundamental in RF design. Given Data / Assumptions: Lossless or low-loss line with characteristic impedance Z0. Load ZL can be chosen for matching. Reflection coefficient Γ = (ZL − Z0) / (ZL + Z0). Concept / Approach: When ZL = Z0, Γ = 0 and no power is reflected; all incident power is absorbed (and ideally delivered to the load). The definition of a matched termination is precisely that it equals Z0 and thus absorbs all incident power, providing the causal link from the Reason to the Assertion. Step-by-Step Solution: Set ZL = Z0 ⇒ Γ = 0.VSWR = (1 + |Γ|) / (1 − |Γ|) = 1 (perfect match).No standing waves form; all incident power is delivered to the matched load. Verification / Alternative check: Time-domain view: a wave reaching a matched load sees no discontinuity and thus no reflection; frequency-domain view: the impedance transformation along the line terminates in Z0 with flat response. Why Other Options Are Wrong: If R stated were false or unrelated, we would not obtain Γ = 0 or VSWR = 1.Only the given Reason directly explains why the Assertion holds in standard line theory. Common Pitfalls: Confusing “maximum power transfer” Thevenin matching in lumped circuits with transmission-line matching; here, ZL must equal Z0 at the operating frequency. Final Answer: Both A and R are correct and R is correct explanation of A

Question 13

Identify the expression that is not a traveling electromagnetic wave in one spatial dimension.

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Introduction / Context: Traveling waves vary in both space and time with a specific phase relationship. Recognizing traveling versus standing or static patterns is a key skill in electromagnetics and transmission-line theory. Given Data / Assumptions: Expressions are scalar field examples along x. β is phase constant, ω is angular frequency, Em is amplitude. Concept / Approach: A one-dimensional traveling wave has dependence on (βx ± ωt); any expression missing time or space dependence cannot represent a traveling wave. Purely spatial (or purely temporal) dependence corresponds to static distributions (or uniform oscillations everywhere) rather than propagation. Step-by-Step Solution: Forms like sin(βx − ωt), cos(βx − ωt), sin(ωt − βx) clearly include both x and t with a constant phase velocity v = ω / β.The term sin(βx) has no time dependence; it is a spatial standing pattern or static profile, not a propagating wave.Therefore the non-traveling expression is e = Em sin(βx). Verification / Alternative check: For a traveling wave, points of constant phase satisfy βx − ωt = constant ⇒ x = (ω/β)t + constant, i.e., motion in x over time. No such relation exists for sin(βx) alone. Why Other Options Are Wrong: All other options contain (βx ± ωt), representing waves moving in ±x. Both sine and cosine forms are equivalent representations shifted in phase. Common Pitfalls: Assuming sin(ωt) alone is a traveling wave; it has no spatial dependence and thus no propagation. Final Answer: e = Em sin(βx)

Question 14

In typical high-frequency transmission-line modeling (RLCG model), which primary distributed parameter is often negligible under many practical conditions?

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Introduction / Context: Distributed-parameter models describe lines using R, L, C, and G per unit length. Understanding which parameters dominate guides low-loss approximations and design choices across frequency bands. Given Data / Assumptions: Dielectric is low-loss (tan δ small). Conductor losses and skin effect may be present but manageable. Typical RF/microwave coaxial, microstrip, or twin-lead environments. Concept / Approach: At many RF frequencies with quality dielectrics, shunt conductance G (which models dielectric leakage) is very small compared to susceptance from C and reactance from L. Hence, the well-known 'low-loss line' approximation treats R and G as small compared to ωL and ωC, with G often the smallest of the four. Step-by-Step Solution: Telegrapher parameters: R (Ω/m), L (H/m), C (F/m), G (S/m).Dielectric loss modeled by G is minimal for high-quality, dry dielectrics; tan δ ≈ ωC/G is low.Thus, for many practical lines, G is negligible first; sometimes R is also small but not as negligible as G in many PCB and coax cases. Verification / Alternative check: Attenuation constant α ≈ (R/2)√(C/L) + (G/2)√(L/C); for low-loss lines, the G-term is frequently orders of magnitude smaller than the R-term. Why Other Options Are Wrong: L and C define the line's phase velocity and characteristic impedance and are never negligible.R can be small, but conductor loss is often significant compared with G, especially with skin effect. Common Pitfalls: Assuming G is always zero; humidity or lossy dielectrics can increase G at low or very high frequencies. Final Answer: G (shunt conductance)

Question 15

Which microwave device employs a helix slow-wave structure to achieve wideband interaction with an electron beam?

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Introduction / Context: Slow-wave structures reduce the phase velocity of RF waves to interact efficiently with electron beams. Recognizing device architectures is essential in microwave engineering. Given Data / Assumptions: Helix supports a slow traveling wave over a broad bandwidth. Electron beam interacts continuously along the helix for amplification. Concept / Approach: The traveling-wave tube (TWT) uses a helical slow-wave structure to achieve near-constant phase velocity over a wide frequency range, enabling broad bandwidth gain. By contrast, klystrons use resonant cavities (buncher, catcher) and do not rely on a helix. Step-by-Step Solution: Identify devices employing helices: TWT is the standard example.Klystrons use discrete cavities and drift spaces; not helix structures.Therefore, the correct choice is TWT. Verification / Alternative check: Any microwave textbook shows cross-sections: TWT with helix winding around the beam tunnel; klystrons with cavities. Why Other Options Are Wrong: Klystron amplifier/oscillator: cavity-based, narrower bandwidth than helix-TWT.Magnetron: crossed-field device with resonant cavities; no helix. Common Pitfalls: Assuming 'klystron' generically uses any slow-wave structure; only specific special designs (not standard klystrons) would. Final Answer: TWT (traveling-wave tube)

Question 16

In a rectangular waveguide, which TM mode exhibits the lowest cutoff frequency (i.e., the dominant TM mode)?

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Introduction / Context: Rectangular waveguides support TE and TM modes with cutoff frequencies determined by the guide dimensions and mode indices m and n. Identifying the dominant TM mode is a classic exam task. Given Data / Assumptions: For TM modes, both m and n must be nonzero (no TMm0 or TM0n). Cutoff frequency: fc(mn) = (c/2) * √[(m/a)^2 + (n/b)^2] for TMmn. Dimensions a ≥ b for standard guides. Concept / Approach: Because TMm0 or TM0n are not permitted (Ez must vanish at the walls), the lowest indices allowed are m = 1 and n = 1. Thus TM11 is the TM mode with the lowest possible cutoff, making it the dominant TM mode (though overall guide dominant is TE10). Step-by-Step Solution: Disallow TM10 and TM01 by boundary conditions (one index cannot be zero).Evaluate smallest valid pair ⇒ (m, n) = (1, 1).Hence, TM11 is the lowest-cutoff TM mode. Verification / Alternative check: Textbook tables list TE10 as the overall dominant mode and TM11 as the dominant TM mode for a ≥ b rectangular guides. Why Other Options Are Wrong: TM10, TM01: invalid for TM due to boundary conditions.TM21: higher indices → higher cutoff than TM11. Common Pitfalls: Mistaking the dominant guide mode (TE10) for dominant TM; they are different questions. Final Answer: TM11

Question 17

True or False: The RF impedance level of IMPATT diodes is generally lower than that of Gunn diodes.

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Introduction / Context: Gunn and IMPATT diodes are common solid-state microwave sources but operate on different physical principles and present different device impedances to resonant circuits or transmission lines. Given Data / Assumptions: Typical Gunn diode impedances are relatively low (a few to a few tens of ohms). Typical IMPATT diode impedances are higher (tens to hundreds of ohms), depending on design and frequency. Concept / Approach: Gunn diodes (transferred electron effect) are commonly matched using low-impedance resonant circuits, while IMPATT diodes (avalanche transit-time) exhibit higher negative resistance magnitude and effective impedance, thus often require different matching networks. Therefore, the statement that IMPATT has “lower” impedance than Gunn is incorrect. Step-by-Step Solution: Recall typical ranges: Z_Gunn ≈ few Ω to ~20 Ω; Z_IMPATT ≈ ~20 Ω to a few hundred Ω.Compare: Z_IMPATT > Z_Gunn in most practical designs.Hence the assertion “IMPATT impedance is generally lower than Gunn” is false. Verification / Alternative check: Device application notes and oscillator design examples routinely show low-Q, low-Z resonators for Gunns, whereas IMPATT oscillators use higher-Z matching networks or resonators. Why Other Options Are Wrong: “True” variants contradict widely observed device characteristics.“Cannot be determined” ignores broad empirical ranges known to practitioners. Common Pitfalls: Confusing the higher RF power capability of IMPATT in some bands with lower impedance; these are independent attributes. Final Answer: False

Question 18

In a three-cavity klystron amplifier, in which cavities are RF oscillations (RF fields) present and excited during normal operation?

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Introduction / Context: Klystron amplifiers use cavity resonators and electron-beam bunching to achieve high-gain microwave amplification. Understanding where RF fields exist helps clarify the amplifier mechanism. Given Data / Assumptions: Three-cavity topology: input (buncher), intermediate (idler), and output (catcher). Small-signal excitation at the input; power extracted at the output. Concept / Approach: The input cavity is driven by the external RF signal, creating velocity modulation in the beam. After drifting and bunching (aided by the idler cavity in three-cavity designs), the bunched beam deposits RF energy into the output cavity, where oscillatory fields are sustained and power is coupled out. Therefore, oscillations (RF fields) are excited in both the input and output cavities during normal operation. Step-by-Step Solution: External RF excites the input cavity → velocity modulation.Beam bunches in drift region; idler cavity improves bunching and gain stability.Bunched beam induces oscillations in the output cavity → RF power output. Verification / Alternative check: Oscilloscope or network analyzer measurements (via weak probes) show RF fields in both input and output cavities; the idler cavity is tuned but typically not externally driven. Why Other Options Are Wrong: Input only or output only ignores the dual-cavity nature of excitation and extraction.Intermediate only is incorrect; it provides additional gain/tuning but is not the sole locus of oscillations.“Purely DC” is incorrect; RF cavities define the device. Common Pitfalls: Conflating amplifier klystrons with klystron oscillators; amplifiers require an input signal but still sustain resonant RF fields in cavities. Final Answer: both (a) and (b)

Question 19

Which transmission line configuration is effectively non-radiating in normal operation?

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Introduction / Context: Transmission lines are intended to guide energy, not radiate it. Line geometry determines field confinement and susceptibility to radiation losses, interference, and EMC issues. Given Data / Assumptions: Coaxial line: inner conductor, dielectric, outer shielding conductor (return), with TEM mode confined between conductors. Open two-wire line: parallel conductors exposed to free space, supporting external fields. Concept / Approach: Coaxial cables are designed so that the electromagnetic field is almost entirely confined within the dielectric between inner and outer conductors; external radiation is negligible when the cable is properly terminated and shield integrity is maintained. Open two-wire lines, by contrast, have fringing fields in free space and can radiate, especially when unbalanced or near discontinuities. Step-by-Step Solution: Consider field lines: coax TEM fields reside between inner/outer conductors.The outer conductor serves as a shield, preventing radiation and pickup.Therefore, coaxial line is effectively non-radiating; open two-wire line is not. Verification / Alternative check: EMC measurements show far less radiated emission from coax than from open two-wire lines for the same guided power, assuming correct terminations and integrity. Why Other Options Are Wrong: Open two-wire: exposed fields → potential radiation, particularly at bends/mismatches.“Both” or “None” misstate basic field confinement differences. Common Pitfalls: Damaged or poorly shielded coax can leak; the statement assumes proper construction and terminations. Final Answer: Coaxial line

Question 20

In a power or communication cable, how does the charging reactive power (MVAR) change as the operating frequency is increased?

Accepted Answer

Introduction / Context: Cables behave predominantly like capacitors. The reactive power drawn due to their capacitance is commonly referred to as charging MVAR. Understanding how this charging requirement varies with frequency is essential in power system planning and in high-frequency communications links using long coaxial or underground cables. Given Data / Assumptions: Cable acts as a distributed capacitance with negligible series resistance for this conceptual analysis. System voltage magnitude is considered fixed while frequency is varied. Charging MVAR is the reactive power associated with the cable's shunt capacitance. Concept / Approach: The fundamental relation for capacitive reactive current is I_c = V * ω * C, where ω = 2 * π * f. The corresponding reactive power is Q_c = V^2 * ω * C when line-to-neutral quantities are used. Therefore, Q_c is directly proportional to frequency f for a given voltage and capacitance. Step-by-Step Solution: Start from I_c = ω * C * V.Reactive power for a capacitor: Q_c = V * I_c = V * (ω * C * V) = V^2 * ω * C.Since C and V are held constant, Q_c ∝ ω ∝ f.Thus, as frequency increases, charging MVAR increases linearly. Verification / Alternative check: Utility practice shows that raising system frequency (or long EHV cable lengths) noticeably increases reactive charging demand, often requiring shunt reactors for compensation. Why Other Options Are Wrong: 'decreases' and 'remain the same': contradict Q_c ∝ f. 'decreases or remains the same': not supported by the capacitive model. 'first increases then decreases': no such behavior without additional frequency-dependent mechanisms (e.g., resonance), which are not implied here. Common Pitfalls: Confusing copper losses (which depend on current and skin effect) with reactive charging; assuming inductive line behavior dominates in short overhead lines—cables are different due to higher capacitance. Final Answer: increases

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