Question 1

Filter selectivity — “The less selective a filter is, the steeper the slope (sharper the roll-off) at its cutoff frequencies.” Is this statement accurate?

Accepted Answer

Introduction / Context: Selectivity describes how well a filter separates passband from stopband. It is tightly linked to roll-off steepness, transition bandwidth, and quality factor (Q). This question tests whether you can correctly relate selectivity and slope at the cutoff region. Given Data / Assumptions: Standard low-pass/high-pass/band-pass/band-stop filters. Slope discussed in dB/decade or dB/octave per order. Concept / Approach: Greater selectivity corresponds to sharper transition regions and steeper skirts. Higher-order filters (or higher Q for resonant responses) achieve steeper roll-off, which improves selectivity. Therefore, a filter that is “less selective” has gentler slopes and broader transition regions, not steeper ones. Step-by-Step Solution: 1) Define selectivity: ability to attenuate unwanted frequencies near the cutoff.2) Recognize slope per order: first order ~20 dB/decade, second order ~40 dB/decade, etc.3) Higher order → steeper slope → more selective, not less.4) Conclude the statement as written is false. Verification / Alternative check: Bode plots of Butterworth/Chebyshev/Elliptic filters show that tighter selectivity comes with steeper roll-off (or ripples/zeros) near cutoff. Why Other Options Are Wrong: “True/only for active/only below 100 Hz/only for high-Q”: selectivity principles are frequency-scale invariant and do not depend on “active vs passive” per se (though implementation choices affect practical limits). Common Pitfalls: Confusing passband flatness with selectivity; assuming “less selective” equals “more slope,” which reverses the real relationship. Final Answer: False

Question 2

Series RLC above resonance — “In a series RLC circuit, at frequencies above resonance the circuit behaves more capacitive than inductive.” Is this correct?

Accepted Answer

Introduction / Context: Knowing how a series RLC behaves relative to its resonant frequency helps in filter and matching design. The sign of net reactance determines whether current leads or lags voltage and which component dominates energy storage. Given Data / Assumptions: Series RLC (single current path). Sinusoidal steady state. Concept / Approach: Net series reactance is X = XL − XC, where XL = ωL and XC = 1/(ωC). At resonance, XL = XC and X = 0 (purely resistive). For frequencies above resonance, ω is larger, so XL grows while XC shrinks; therefore XL > XC and X is positive (inductive). Thus the circuit is more inductive above resonance and more capacitive below resonance. Step-by-Step Solution: 1) Identify resonance condition: XL = XC.2) Increase frequency above f0: XL increases, XC decreases.3) Compute X = XL − XC > 0 → inductive behavior.4) Conclude the statement is false. Verification / Alternative check: Phasor diagrams show current lagging voltage above resonance (inductive), and leading below resonance (capacitive). Why Other Options Are Wrong: Any “True” variants contradict the algebraic combination of reactances for series RLC. Common Pitfalls: Remembering the rule in reverse; mixing series with parallel behavior. Final Answer: False

Question 3

Filter bandwidth — by definition, the bandwidth (BW) of a band-pass (or band-stop) filter is the span between its two half-power (−3 dB) frequencies. Is this correct?

Accepted Answer

Introduction / Context: Bandwidth is a fundamental descriptor of how wide a filter’s pass (or stop) region extends around a center frequency. The industry-standard definition uses the half-power (−3 dB) points for linear time-invariant systems unless otherwise specified. Given Data / Assumptions: Band-pass or band-stop context. Power proportional to voltage squared in constant-impedance systems. Half-power corresponds to 1/√2 in voltage magnitude (−3 dB). Concept / Approach: The half-power frequencies f1 and f2 are where the output power is half the peak (or where |H(jω)| drops by 3 dB from the reference). The bandwidth BW is defined as f2 − f1. This convention is ubiquitous in communications and filter design, enabling consistent comparison of selectivity and Q (Q ≈ f0 / BW for narrowband band-pass filters). Step-by-Step Solution: 1) Identify f1 and f2 as the −3 dB points around the center.2) Compute BW = f2 − f1.3) For narrowband, Q ≈ f0 / BW; for broader responses, use exact definitions.4) Conclude that the statement is correct as a general definition. Verification / Alternative check: Standard texts and datasheets define bandwidth using −3 dB points unless a different metric (e.g., ripple limits) is explicitly stated. Why Other Options Are Wrong: “False/only Butterworth/only high Q/depends solely on ripple” are inconsistent with the general, technology-neutral definition. Common Pitfalls: Confusing −3 dB (voltage) with absolute voltage drop; mixing alternative definitions used in specific standards without noting the context. Final Answer: True

Question 4

Operating relative to resonance (repaired) — “With the stated frequency, the given circuit is operating above resonance.” Without a schematic and component values, can we determine whether operation is above resonance?

Accepted Answer

Introduction / Context: Determining whether an RLC circuit operates above or below resonance requires knowing the actual resonant frequency f0 and the applied frequency f. Without the circuit values or topology, any claim about “above resonance” is unsupported. This repaired stem evaluates your reasoning about required data. Given Data / Assumptions: No values for L and C are given; R may be present but does not set f0. Topology (series vs parallel) is unspecified. A single operating frequency is referenced but not related to f0. Concept / Approach: Resonant frequency is f0 = 1/(2π√(L*C)) for ideal RLC (series or parallel). To decide if operation is “above resonance,” you must compare f to f0. Without L and C, f0 is unknown; thus the statement cannot be validated or refuted objectively. Step-by-Step Solution: 1) Recognize that deciding “above/below” requires both f and f0.2) Note that f0 depends on L and C only (ideal), not R.3) Since L and C are missing, f0 is unknown.4) Therefore, the claim cannot be determined from the given data. Verification / Alternative check: Example: if L and C produce f0 = 10 kHz, then f = 20 kHz is above resonance; if L and C produce f0 = 100 kHz, the same f = 20 kHz is below resonance. The conclusion flips with different L and C. Why Other Options Are Wrong: “Always above/below”: incorrect generalizations.Topology-based answers: both series and parallel use the same f0 relation.“Resistance sets resonance”: f0 is not set by R. Common Pitfalls: Assuming “high frequency” implies “above resonance” without checking component values. Final Answer: Cannot be determined from the information provided

Question 5

Output tapping and filter type (repaired) — for a series RLC network driven by a source, if the output is taken across the resistor (R), does the resulting two-port behave as a band-pass filter?

Accepted Answer

Introduction / Context: Filter behavior in simple RLC networks depends on where you probe or take the output. A single series RLC path can realize different transfer functions depending on whether the output is measured across L, C, or R. This question clarifies the case when output is taken across R. Given Data / Assumptions: Single series RLC driven by a voltage source. Output is the voltage across the resistor, VR. Linear time-invariant conditions; moderate Q. Concept / Approach: The magnitude of VR is low at both very low and very high frequencies: far below resonance, current is limited by large XC; far above resonance, current is limited by large XL. At resonance (where XL = XC), impedance is minimum and current peaks, so VR = I * R peaks. This “low at extremes, high at center” behavior is characteristic of a band-pass filter centered at f0. Step-by-Step Solution: 1) Write |I(jω)| = |V_source| / |Z_series|, with Z_series minimum at resonance.2) Compute VR = I * R; since I peaks at f0, so does VR.3) Observe that VR diminishes away from f0 in both directions.4) Conclude that the transfer |VR/Vs| exhibits a band-pass profile. Verification / Alternative check: Bode plots for |VR/Vs| in series RLC show a peak at f0 with attenuation on both sides, confirming band-pass behavior. Why Other Options Are Wrong: “False” or “always low-pass”: contradicts the current-peaking nature at resonance when measuring across R.“Only if Q>5/ideal components”: while Q affects sharpness and peak gain, the qualitative band-pass shape persists for finite Q with real components. Common Pitfalls: Assuming the network type (LPF/HPF/BPF) is intrinsic to the parts rather than the measurement node. Final Answer: True

Question 6

Impedance value claim (repaired) — “The impedance is 1177 Ω.” Without a schematic, frequency, and component values, can this numeric impedance be verified?

Accepted Answer

Introduction / Context: Impedance (Z) in AC circuits depends on frequency and component values. A bare numerical claim like “Z = 1177 Ω” is meaningless without context. This repaired item checks whether you recognize the missing data necessary to compute or confirm impedance. Given Data / Assumptions: No circuit diagram or element values (R, L, C) are supplied. No frequency is given; reactances depend on ω. We assume linear components; parasitics ignored. Concept / Approach: For a general network, Z(jω) = R + j(XL − XC) for simple series RLC, or more complex functions for other topologies. The magnitude |Z| and phase ∠Z vary with ω. Without ω and component values, any specific value such as 1177 Ω cannot be computed or validated. Therefore, the only defensible choice is that the claim cannot be determined from the given information. Step-by-Step Solution: 1) Recognize that impedance is frequency-dependent whenever L or C is present.2) Note that even for purely resistive networks, you still need resistor values and connections to compute Z.3) Since neither the schematic nor the values nor the frequency are given, Z is indeterminate.4) Conclude: the numeric claim cannot be verified. Verification / Alternative check: Provide any two different hypothetical sets of R, L, C and frequencies that yield different |Z| values; the same number 1177 Ω will not be universally correct. Why Other Options Are Wrong: “Always 1177 Ω / always resistive”: physically incorrect generalizations.“True only for DC”: still need resistor values and topology.“False — cannot be 1177 Ω”: it could be 1177 Ω for some specific case; we simply lack the data to assert it. Common Pitfalls: Accepting numeric statements without verifying the underlying parameters; forgetting the role of frequency in impedance. Final Answer: Cannot be determined from the information provided

Question 7

Series RLC — same current in all series elements: Assume a single-loop series RLC circuit in steady-state AC where a clamp-on ammeter at the source lead reads 5 A (rms). What is the inductor current (rms)?

Accepted Answer

Introduction / Context: In any single-loop series circuit (R, L, and/or C in series), the same instantaneous current flows through every element because there is only one current path. This question tests the core idea of series topology and how it applies to an inductor’s branch current reading. Given Data / Assumptions: Single-loop series RLC driven by a sinusoidal source. The measured circuit (source-lead) current is 5 A rms. Ideal lumped elements; parasitics and measurement burden are negligible. Concept / Approach: Kirchhoff’s Current Law applied to a series loop implies a single branch; therefore the current is identical through R, L, and C at every instant. Phasor relationships change how voltage distributes (phase shifts), but not the series current equality. Thus the inductor current must equal the source current. Step-by-Step Solution: Recognize a single series path → same current in all components.Given I_source(rms) = 5 A → I_L(rms) = 5 A by series rule.Voltage magnitudes will differ (V_R = IR, V_L = IX_L, V_C = I*X_C), but the current is common. Verification / Alternative check: Measure with an ammeter inserted in series before the inductor and again between L and C; both readings match the source current (within instrument tolerance). Why Other Options Are Wrong: “Depends only on L” or “greater than 5 A”: violates series current continuity.“Zero at any frequency”: only true if the loop is open; here the loop is closed.“Cannot be determined”: the series rule determines it directly. Common Pitfalls: Confusing series (same current) with parallel (same voltage). Also, mixing up current magnitude with reactive voltage magnitudes, which can be large at resonance without changing current equality. Final Answer: 5 A

Question 8

Resonance in a series RLC — total current behavior: If the given circuit operates exactly at series resonance, is the total current drawn from the source at its maximum value (for a fixed source voltage)?

Accepted Answer

Introduction / Context: Series resonance is a cornerstone concept in AC circuit theory and filter/tank design. It occurs when the inductive reactance equals the capacitive reactance in magnitude (X_L = X_C), causing reactances to cancel so the impedance is purely resistive and minimal. Given Data / Assumptions: Series RLC circuit driven by a fixed RMS voltage. Ideal L and C, with a finite series resistance R. Frequency is adjustable and can be set to the resonant frequency. Concept / Approach: Impedance of a series RLC is Z = R + j(X_L − X_C). At resonance, X_L = X_C → Z = R (minimum magnitude). Ohm’s law shows I = V/Z is largest when |Z| is smallest (for fixed V), hence current peaks at resonance. Large reactive voltages may appear across L and C individually, but the net reactive effect in the loop is zero at resonance. Step-by-Step Solution: Write Z(f) = √(R^2 + (X_L − X_C)^2).At f = f_0, X_L = X_C → |Z| = R (minimum possible for that circuit).Thus I(f_0) = V/R is the maximum attainable loop current for that applied V. Verification / Alternative check: Plot I versus frequency (a resonance curve). The peak occurs at f_0. Bandwidth relates to quality factor Q = X_L/R at resonance. Why Other Options Are Wrong: “Incorrect”: contradicts Z minimization at resonance.“Only if R = 0”: even with finite R, current is maximal at resonance for that V.“Parallel resonance”: refers to a different topology where current dips in the feed branch.“XL = 2XC”: not the resonance condition. Common Pitfalls: Confusing series and parallel resonance characteristics; focusing on large VL or VC and missing that total impedance is minimized, not maximized. Final Answer: Correct

Question 9

Resonance duplicate (clarified wording): For a series RLC circuit at its resonant frequency (where XL = XC), the total current from a fixed-voltage source reaches a maximum. Is this statement accurate?

Accepted Answer

Introduction / Context: This item reaffirms the behavior of a series RLC at resonance, emphasizing the equality of reactances and the resulting purely resistive impedance that dictates the current peak for a given source voltage. Given Data / Assumptions: Series RLC network, sinusoidal steady state. Source RMS voltage held constant. Losses modeled by a series resistance R. Concept / Approach: At resonance, |Z| = R (minimum), so I = V/|Z| is largest. Although VL and VC can be large and cancel phasorially, they do not increase |Z| at resonance; instead, they produce internal voltage magnification characterized by Q, while the source sees a small resistive |Z|. Step-by-Step Solution: Impose XL = XC → Z = R.Compare current off-resonance: |Z| grows as |XL − XC| increases → current falls.Therefore, the maximum current occurs at the resonant frequency. Verification / Alternative check: Measure current while sweeping frequency; the plot yields a bell-shaped resonance curve with peak at f0 and −3 dB bandwidth linked to Q. Why Other Options Are Wrong: Requiring L = C numerically is irrelevant; resonance depends on ωL = 1/(ωC).Parallel resonance property is different (feed current minimum).Q affects sharpness of the peak, not whether a peak exists. Common Pitfalls: Confusing internal reactive voltages with source current; overlooking that the source only “sees” R at resonance. Final Answer: Accurate

Question 10

Coil Q at resonance — effect of series resistance: Does increasing a coil’s series resistance increase its quality factor Q at resonance?

Accepted Answer

Introduction / Context: The quality factor Q expresses how under-damped or selective a resonant circuit is. For a series RLC at resonance, Q relates the reactive energy exchange to real power dissipation. Coil resistance is a major contributor to series loss, so its effect on Q is fundamental to RF design and filters. Given Data / Assumptions: Series resonant circuit model with inductor series resistance R. Sinusoidal steady state at resonant frequency f0. Ideal L and C except for the noted resistance. Concept / Approach: For series resonance, Q = X_L / R at f0 (equivalently Q = ω0L / R). Increasing R in the denominator reduces Q. Lower Q widens the bandwidth (BW ≈ f0/Q) and reduces peak voltage magnification across L and C at resonance. Step-by-Step Solution: Start with Q = ω0L / R.Increase R → denominator increases → Q decreases.Result: broader bandwidth, lower selectivity, lower tank magnification. Verification / Alternative check: Network analyzer measurements show increased 3 dB bandwidth and lower peak response as series resistance rises. Simulation likewise confirms Q ∝ 1/R for fixed L, C at f0. Why Other Options Are Wrong: Claims that higher R raises Q contradict the formula.Frequency or ESR caveats do not reverse the basic series relation Q = X_L/R at resonance. Common Pitfalls: Mixing up series and parallel definitions of Q; overlooking that parallel-resonant “Rp” enters Q_p = Rp / X_L, a different configuration. Final Answer: Incorrect — increasing resistance lowers Q

Question 11

Inductive behavior — phase relation: In a predominantly inductive series R–L circuit driven by a sinusoid (|X_L| > R), does the current lag the applied voltage?

Accepted Answer

Introduction / Context: Knowing whether current leads or lags voltage is central to AC analysis and power factor correction. Inductors store energy in magnetic fields and impose a positive reactive component, shifting current in time relative to voltage. Given Data / Assumptions: Series R–L circuit, sinusoidal source. |X_L| > R indicates inductive dominance. No capacitor is present in this basic scenario. Concept / Approach: In phasor form, Z = R + jX_L. The current phasor I = V/Z lags V by angle φ where tan φ = X_L / R > 0. A positive φ means current lags voltage. As X_L increases or R decreases, φ approaches 90 degrees. Step-by-Step Solution: Write Z = R + jX_L.Angle of Z is φ = arctan(X_L / R) > 0.Current angle is −φ relative to voltage → current lags. Verification / Alternative check: Oscilloscope with dual channels (voltage and shunt-resistor current sense) shows current zero crossings occurring after voltage zero crossings in inductive circuits. Why Other Options Are Wrong: “Leads” describes capacitive behavior (X_C < 0).“In phase” only when X_L ≈ 0 or reactive parts cancel.Capacitor value is irrelevant here; none is present. Common Pitfalls: Confusing mnemonic “ELI the ICE man”: voltage (E) leads current (I) in an inductor (L); current leads voltage in a capacitor (C). Final Answer: Yes, current lags the voltage

Question 12

Series RLC numeric check — inductor voltage from current and reactance: Assume the same series RLC as in the earlier item with loop current 5 A (rms). If the inductor’s reactance is X_L = 15.56 Ω at the operating frequency, what is the magnitude of …

Accepted Answer

Introduction / Context: Voltage drops across reactive elements in a series AC circuit follow the same IX rule as resistive drops follow IR, but with a 90-degree phase shift. This question links a previously established series current to the inductor’s voltage magnitude using its reactance. Given Data / Assumptions: Series loop current I = 5 A (rms). Inductive reactance X_L = 15.56 Ω at the test frequency. Ideal inductor (ignore winding resistance for this calculation). Concept / Approach: The magnitude of an inductor’s RMS voltage is V_L = I * X_L. Phase considerations matter for vector addition in KVL, but magnitude computation is straightforward with the given current and reactance. Step-by-Step Solution: Use V_L = I * X_L.Substitute I = 5 A and X_L = 15.56 Ω → V_L = 5 * 15.56 = 77.8 V.Therefore the correct magnitude is 77.8 V (rms). Verification / Alternative check: Phasor diagram: V_L is orthogonal to V_R and opposite in direction to V_C. Its magnitude still equals I * X_L, independent of R and C values. Why Other Options Are Wrong: 31.1 V and 15.56 V: correspond to using a wrong current or confusing R with X_L.3.11 V: off by an order of magnitude.“Cannot be determined”: incorrect; I and X_L suffice for |V_L|. Common Pitfalls: Adding magnitudes instead of phasors when forming the source voltage; here we only compute the single-element magnitude. Final Answer: 77.8 V

Question 13

RC network identification — high-pass or low-pass?: Consider a simple RC with a series capacitor feeding a resistor to ground; the output is taken across the resistor. Is this a first-order high-pass filter?

Accepted Answer

Introduction / Context: Recognizing canonical RC topologies is essential for quick design judgements. Two complementary first-order forms exist: RC low-pass (R in series, C to ground, output across C) and RC high-pass (C in series, R to ground, output across R). Given Data / Assumptions: Series capacitor C from input node to output node. Resistor R from output node to ground. Output measured across R; linear, time-invariant operation. Concept / Approach: For the series-C / shunt-R network, the transfer magnitude |H(jω)| = |V_out/V_in| increases with frequency: |H| = ωRC / √(1 + (ωRC)^2). At low frequency (ω → 0), the capacitor blocks, |H| → 0. At high frequency (ω → ∞), the capacitor is a short, |H| → 1. This is precisely the behavior of a first-order high-pass filter, with cutoff f_c = 1/(2πRC). Step-by-Step Solution: Identify topology: series C, shunt R, output across R.Recall high-pass transfer |H| = ωRC / √(1 + (ωRC)^2).Low frequency → |H| ≈ 0; high frequency → |H| ≈ 1.Therefore classify as first-order high-pass. Verification / Alternative check: Bode plot shows +20 dB/decade slope below f_c and flat passband above f_c; phase approaches +90 degrees at high frequency. Why Other Options Are Wrong: Low-pass: opposite component placement (R series, C shunt, output across C).Band-pass and all-pass require different structures.“Cannot be determined” is incorrect; topology uniquely defines it. Common Pitfalls: Confusing where the output is taken; swapping where you probe flips high-pass vs low-pass. Final Answer: Yes, it is a first-order high-pass filter

Question 14

Series resonance quality factor — voltage ratios: At resonance in a series RLC, is the quality factor given by Q = V_R / V_L, or is it correctly Q = V_L / V_R (which equals X_L / R)?

Accepted Answer

Introduction / Context: Quality factor Q is a key figure of merit for resonant selectivity and voltage magnification. For series resonance, several equivalent expressions exist, including reactance-to-resistance ratios and voltage ratios at resonance. Given Data / Assumptions: Series RLC circuit at resonance (X_L = X_C). Finite series resistance R representing all losses. Voltages and currents are RMS quantities. Concept / Approach: At series resonance, Q = X_L / R = V_L / V_R = V_C / V_R. The resistor drop V_R = IR and the inductor drop V_L = IX_L occur at the same current I, so their ratio equals X_L/R. In a high-Q circuit, V_L and V_C can greatly exceed the source voltage, even though the source sees only R at resonance. Step-by-Step Solution: Write Q = X_L / R at resonance.Using V = I*Z per element: V_L/V_R = (I X_L) / (I R) = X_L / R = Q.Therefore the correct voltage ratio is V_L / V_R, not V_R / V_L. Verification / Alternative check: Measure V_R (across R) and V_L (across L) at the resonant peak; their ratio equals the independently measured Q from bandwidth (Q = f0/BW) for linear circuits. Why Other Options Are Wrong: Q = V_R / V_L is 1/Q, the inverse of the correct ratio.“Depends only on C” or “cannot be defined with voltages” contradict standard series-resonance relations.V_source/V_R equals 1 at resonance (since Z = R), so it is not Q. Common Pitfalls: Forgetting that large reactive voltages are internal and do not mean large source impedance; interchanging numerator and denominator when forming the ratio. Final Answer: Q = V_L / V_R (correct for series resonance)

Question 15

Impedance claim sanity check — “The impedance is 9684 Ω”: Given a series RLC with R = 3.0 kΩ, X_L = 7.5 kΩ, and X_C = 5.5 kΩ at the operating frequency, evaluate the claim that the circuit impedance magnitude is 9684 Ω.

Accepted Answer

Introduction / Context: Numerical checks prevent order-of-magnitude mistakes in AC design. Series RLC impedance magnitude combines resistance and net reactance vectorially, not arithmetically. This item asks you to confirm or refute a specific impedance claim using given R, X_L, and X_C values. Given Data / Assumptions: Series RLC with R = 3.0 kΩ. Inductive reactance X_L = 7.5 kΩ. Capacitive reactance X_C = 5.5 kΩ. Linear, sinusoidal steady state. Concept / Approach: For series RLC, Z = R + j(X_L − X_C). The magnitude is |Z| = √(R^2 + (X_L − X_C)^2). Reactances must be netted before combining with R. Plug in values carefully and keep units consistent (ohms). Step-by-Step Solution: Compute net reactance: X = X_L − X_C = 7.5 kΩ − 5.5 kΩ = 2.0 kΩ.Form magnitude: |Z| = √(R^2 + X^2) = √((3.0 kΩ)^2 + (2.0 kΩ)^2).Square terms: 9.0 + 4.0 = 13.0 (in (kΩ)^2).Take square root: √13.0 ≈ 3.606 kΩ → |Z| ≈ 3606 Ω. Verification / Alternative check: Phasor diagram shows a modest inductive reactance remaining; the impedance should be a few kilohms, not almost 10 kΩ. A quick calculator or spreadsheet confirms 3.606 kΩ. Why Other Options Are Wrong: “Correct — 9684 Ω” is far off the computed value.“Only at resonance” is irrelevant; the given X_L and X_C are not equal.“Cannot be determined” is false; R, X_L, X_C suffice.“R much larger than reactances” contradicts the provided numbers. Common Pitfalls: Adding R and reactances directly; forgetting to difference X_L and X_C before forming the magnitude; dropping kilo- prefixes during arithmetic. Final Answer: Incorrect — |Z| ≈ 3606 Ω, not 9684 Ω

Question 16

Tuned amplifiers — series vs parallel LC usage: In RF/mid-frequency tuned amplifier stages, are parallel (tank) LC resonators used more commonly for selectivity and impedance transformation than series LC combinations?

Accepted Answer

Introduction / Context: Tuned amplifiers rely on resonant networks to select narrow frequency bands and to provide impedance matching. Two fundamental LC forms exist: series resonators (low impedance at resonance) and parallel resonators (high impedance at resonance). Understanding which is more common in amplifier input/output stages is important for practical RF design. Given Data / Assumptions: Discrete or integrated tuned stages in radio/IF/RF amplifiers. Desire for band-selective loading and impedance transformation. Use of passive, linear LC networks. Concept / Approach: A parallel LC at resonance presents a high impedance (tank), favoring voltage gain, selectivity, and isolation between stages. It can be tapped or transformer-coupled for impedance matching. Series LC resonates to a low impedance and is often used for coupling, notch/series traps, or as part of impedance-matching networks, but not as commonly as the main tuned load in amplifiers. Step-by-Step Solution: Recognize that amplifier loads often prefer high impedance at resonance to maximize voltage swing and selectivity.Parallel tanks provide this high impedance and can be lightly coupled for narrow bandwidth (high Q).Series LC elements appear, but typically in coupling/matching roles, not as the dominant tuned load. Verification / Alternative check: Examine classic tuned-collector or tuned-drain stages and double-tuned IF transformers; these use parallel LC tanks as resonant loads with transformer coupling for bandwidth shaping. Why Other Options Are Wrong: Claiming series LC is more common ignores widespread tank-load topologies.“Equal roles” oversimplifies practical stage design.RC-only or “active-only” claims contradict conventional RF practice where LC is standard for narrowband selectivity. Common Pitfalls: Confusing impedance behavior of series versus parallel resonance; overlooking that the choice depends on whether a high- or low-impedance condition at resonance is desired. Final Answer: Yes — parallel LC tanks are more commonly used

Question 17

Filter identification in a series R–L–C network: If the output voltage is taken across the series resistor (and the input drives the same series R–L–C loop from a voltage source), does the circuit behave as a band-stop (notch) filter or not?

Accepted Answer

Introduction / Context: Series R–L–C circuits are used to build simple frequency-selective filters. Whether the network is band-pass or band-stop depends on where the output is measured. This item asks you to judge the claim that taking the output across the series resistor makes the circuit a band-stop (notch) filter. Given Data / Assumptions: Single loop driven by an ideal voltage source. Elements R, L, and C in series; output measured across R only. Linear, time-invariant components; standard small-signal operation. Concept / Approach: In a series R–L–C, the current peaks at resonance because the net reactance cancels (XL = XC), leaving the loop impedance approximately R. The resistor's voltage is v_R = i * R, so when current peaks at resonance, the resistor's voltage also peaks. A response that peaks at a center frequency and falls off on both sides is the hallmark of a band-pass, not a band-stop. By contrast, taking the output across L or across C in the same series loop yields a deep minimum at resonance (because the reactive element's voltage tends toward zero there), which is a notch or band-stop behavior. Step-by-Step Solution: Write impedance: Z = R + j(XL − XC). At resonance: XL = XC ⇒ Z ≈ R ⇒ i = V_in / R (maximum). Output across R: v_out = i * R ≈ V_in at resonance (peak output). Hence the R output is band-pass, not band-stop. Verification / Alternative check: Plotting |v_R/V_in| vs frequency shows a peak near f_0; plotting |v_L/V_in| or |v_C/V_in| shows a dip at f_0. This matches standard lab measurements and textbook Bode plots. Why Other Options Are Wrong: Correct: contradicts series-R output behavior (band-pass). Applies only with an ideal current source: the statement concerns a voltage source; with a current source the mapping of outputs changes but does not make the R output a notch under a voltage-source assumption. Cannot be determined: with the stated series topology and output node, it is determinable. Common Pitfalls: Confusing the R output (band-pass) with L or C outputs (notch). Mixing up series vs parallel tuned circuits. Final Answer: Incorrect

Question 18

Series/parallel resonance criterion: In a linear R–L–C circuit, does resonance occur when the magnitudes of capacitive and inductive reactances are equal (|XC| = |XL|)?

Accepted Answer

Introduction / Context: Resonance is a foundational concept in filter design, oscillators, and matching networks. At resonance, reactive effects cancel in a way that produces either a minimum or maximum in impedance, depending on whether the tuned network is series or parallel. The simplest recognition rule uses the equality of reactance magnitudes. Given Data / Assumptions: Sine-wave steady state; linear components. Inductive reactance XL = 2 * pi * f * L. Capacitive reactance XC = 1 / (2 * pi * f * C). Concept / Approach: Setting XL = XC yields 2 * pi * f * L = 1 / (2 * pi * f * C), which solves to f0 = 1 / (2 * pi * sqrt(L * C)). In a series RLC, this makes the net reactance zero so impedance is near R (minimum), maximizing current (band-pass via the resistor). In a parallel RLC, the branch susceptances cancel, producing a large input impedance (a notch when used in series with a load). The criterion |XC| = |XL| is the unifying test for both cases. Step-by-Step Solution: Write equality: XL = XC. Solve for frequency: f0 = 1 / (2 * pi * sqrt(L * C)). Interpretation: series ⇒ Z_min; parallel ⇒ Z_max. Conclude the statement is correct for linear RLC circuits. Verification / Alternative check: Impedance/admittance plots show phase crossing zero at f0, with magnitude extremum consistent with the topology. Why Other Options Are Wrong: Incorrect: contradicts standard resonance math. Series-only / parallel-only: the equality holds for both; only the impedance extremum type differs. Common Pitfalls: Confusing equality of reactances with equality of impedances; forgetting that resistance alters Q and bandwidth but not the resonance condition itself. Final Answer: Correct

Question 19

Selectivity versus bandwidth: In tuned circuits and filters, does greater selectivity imply a wider bandwidth, or the opposite?

Accepted Answer

Introduction / Context: Selectivity describes how well a circuit discriminates between a wanted frequency and nearby unwanted frequencies. Bandwidth is the range of frequencies passed (for band-pass) or rejected (for band-stop) within a defined amplitude criterion (often −3 dB points). Understanding their inverse relationship is vital when specifying filters or RF stages. Given Data / Assumptions: Standard definitions: bandwidth BW = f_H − f_L at the chosen reference (e.g., −3 dB), center frequency f0 = sqrt(f_H * f_L). Quality factor Q = f0 / BW for band-pass networks. Linear, time-invariant response; symmetrical response for simplicity. Concept / Approach: Higher selectivity means the response is more sharply peaked (band-pass) or deeper/narrower (notch), which mathematically corresponds to a higher Q and therefore a smaller bandwidth. Stated differently, for a fixed center frequency f0, increasing Q compresses BW = f0 / Q. Hence the claim that greater selectivity implies wider bandwidth is backward and must be rejected. Step-by-Step Solution: Define Q = f0 / BW for a band-pass filter. Greater selectivity ⇒ higher Q. Higher Q ⇒ BW = f0 / Q becomes smaller. Therefore, greater selectivity implies narrower (not wider) bandwidth. Verification / Alternative check: Plot responses for two band-pass filters with the same f0 but different Q. The higher-Q curve is visibly narrower, confirming the inverse relation. Why Other Options Are Wrong: Correct: reverses the true relationship. Q < 1 and active-only qualifiers: the selectivity/BW relationship stems from definitions, not from active vs passive implementation or Q below/above 1. Common Pitfalls: Confusing peak gain with bandwidth; assuming added stages always broaden the passband (they typically narrow it when tuned identically). Final Answer: Incorrect

Question 20

Energy transfer and tuning: Can a parallel tuned (tank) circuit be used to couple energy from one circuit to another at or near its resonant frequency?

Accepted Answer

Introduction / Context: Coupling stages in radios, filters, and oscillators often relies on tuned circuits to pass desired frequencies while rejecting others. A parallel resonant (tank) circuit presents a high impedance at its resonant frequency and can be arranged to transfer energy efficiently between circuits with appropriate coupling (inductive or capacitive). Given Data / Assumptions: Parallel L–C network with finite Q. Loose to moderate coupling to source and load (transformer or capacitive coupling). Operation near the tank's resonant frequency f0. Concept / Approach: At resonance, the tank's reactive currents circulate between L and C while the input/output see a transformed impedance that can pass energy selectively at f0. With inductive coupling (mutual inductance between coils) or capacitive tapping, the tank can be used as a frequency-selective interstage coupler, maximizing transfer at f0 and attenuating off-frequency signals. This is the basis of many RF front-ends and IF stages. Step-by-Step Solution: Tune L and C so XL = XC at f0. Arrange coupling (e.g., transformer secondary or capacitive tap) to feed the next stage. Exploit high impedance at f0 to favor power transfer at the selected frequency while rejecting others. Adjust coupling coefficient/Q to trade bandwidth vs selectivity. Verification / Alternative check: Classic AM radio IF transformers are parallel tanks with adjustable coupling; sweep responses show a peaked passband centered at f0 with controllable bandwidth. Why Other Options Are Wrong: Incorrect: ignores widespread use of parallel tanks in coupling networks. Series-only / DC-only: coupling by tuned circuits is inherently AC and frequency-selective; DC is blocked or bypassed as designed. Common Pitfalls: Over-coupling (excessive bandwidth or double-hump response); under-coupling (weak transfer). Forgetting loading reduces effective Q and shifts tuning. Final Answer: Correct

RLC Circuits and Resonance Questions