Q: 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 …

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

Q: 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?

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

Q: 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)?

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)

Q: 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.

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

Q: 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|)?

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

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

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

Q: 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?

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

Q: Series R–L–C notch output: In a series-tuned notch configuration where the output is taken across the series L–C branch, what is the ideal output voltage at exact resonance (assume a stiff voltage source and negligible component loss)?

Introduction / Context: A common way to realize a band-stop (notch) response is to use a series R–L–C path and take the output across the reactive branch (L–C). At the notch (resonant) frequency, the inductive and capacitive reactances cancel in magnitude, dramatically changing the voltage division. This question asks for the ideal output at that frequency. Given Data / Assumptions: Series R–L–C driven by an ideal voltage source. Output measured across the L–C series sub-branch. Ideal components (no resistance in L or leakage in C) to illustrate the principle. Concept / Approach: In a series R–L–C, the total impedance is Z = R + j(XL − XC). At resonance, XL = XC so the net reactive term is zero and Z ≈ R. Importantly, the inductor and capacitor voltages are equal and opposite at resonance, so the net voltage across the L–C series pair is ideally zero, even though the individual element voltages can be large and opposite in sign. Therefore, a detector connected across the L–C pair reads a deep minimum (the notch). Step-by-Step Solution: Set resonance: XL = XC. Write v_L = i * XL and v_C = i * (−XC). At resonance: v_L + v_C = i * (XL − XC) = 0. Hence the measured output across the L–C pair is ideally 0 V. Verification / Alternative check: Phasor diagram shows v_L and v_C 180 degrees apart with equal magnitude at f0, summing to zero. Practical parts have finite Q, so a small but nonzero residual appears; the ideal target remains 0 V. Why Other Options Are Wrong: 557 mV, 6.08 V, 6.5 V: these imply a nonzero residual inconsistent with the ideal cancellation at resonance in a pure series L–C branch. Common Pitfalls: Forgetting that while the L and C each can have large voltages at f0, their sum across the series pair is zero. Confusing this notch with the band-pass response seen when measuring across the resistor. Final Answer: 0 V

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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 11

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

Question 12

Series R–L–C notch output: In a series-tuned notch configuration where the output is taken across the series L–C branch, what is the ideal output voltage at exact resonance (assume a stiff voltage source and negligible component loss)?

Accepted Answer

Introduction / Context: A common way to realize a band-stop (notch) response is to use a series R–L–C path and take the output across the reactive branch (L–C). At the notch (resonant) frequency, the inductive and capacitive reactances cancel in magnitude, dramatically changing the voltage division. This question asks for the ideal output at that frequency. Given Data / Assumptions: Series R–L–C driven by an ideal voltage source. Output measured across the L–C series sub-branch. Ideal components (no resistance in L or leakage in C) to illustrate the principle. Concept / Approach: In a series R–L–C, the total impedance is Z = R + j(XL − XC). At resonance, XL = XC so the net reactive term is zero and Z ≈ R. Importantly, the inductor and capacitor voltages are equal and opposite at resonance, so the net voltage across the L–C series pair is ideally zero, even though the individual element voltages can be large and opposite in sign. Therefore, a detector connected across the L–C pair reads a deep minimum (the notch). Step-by-Step Solution: Set resonance: XL = XC. Write v_L = i * XL and v_C = i * (−XC). At resonance: v_L + v_C = i * (XL − XC) = 0. Hence the measured output across the L–C pair is ideally 0 V. Verification / Alternative check: Phasor diagram shows v_L and v_C 180 degrees apart with equal magnitude at f0, summing to zero. Practical parts have finite Q, so a small but nonzero residual appears; the ideal target remains 0 V. Why Other Options Are Wrong: 557 mV, 6.08 V, 6.5 V: these imply a nonzero residual inconsistent with the ideal cancellation at resonance in a pure series L–C branch. Common Pitfalls: Forgetting that while the L and C each can have large voltages at f0, their sum across the series pair is zero. Confusing this notch with the band-pass response seen when measuring across the resistor. Final Answer: 0 V

Question 13

Definition check for series resonance: In a series R–L–C circuit, which condition defines resonance?

Accepted Answer

Introduction / Context: Series resonance is a key operating point where a series R–L–C presents minimum impedance and maximum current. Recognizing the proper equality that marks this condition is essential for setting center frequency and predicting bandwidth and Q. Given Data / Assumptions: Series connection of R, L, and C driven by a sinusoidal source. Inductive reactance XL = 2 * pi * f * L; capacitive reactance XC = 1 / (2 * pi * f * C). No exotic elements; standard notation (XT sometimes denotes total reactance). Concept / Approach: In a series path, the total reactance is XT = XL − XC. Resonance occurs when the reactive terms cancel, i.e., XL = XC, making XT = 0. The loop impedance collapses to approximately R, maximizing current and producing a band-pass profile when the output is taken across R. Alternatives like XC = XC (tautology), XL = XB (undefined symbol), or XT = R (mixing reactance with resistance) do not define resonance. Step-by-Step Solution: Write total reactance: XT = XL − XC. Set XT = 0 for resonance ⇒ XL = XC. Solve f0 = 1 / (2 * pi * sqrt(L * C)) if needed for design. Use Q = (1/R) * sqrt(L/C) to relate selectivity and bandwidth. Verification / Alternative check: At resonance, current is maximal and the phase angle between source voltage and current is zero (power factor unity), confirming XT = 0 and XL = XC. Why Other Options Are Wrong: XC = XC: identity, gives no frequency condition. XL = XB: symbol XB not defined; meaningless here. XT = R: mixing reactive and resistive quantities; resonance requires XT = 0, not equality with R. Common Pitfalls: Forgetting that series and parallel resonance share the XL = XC magnitude equality, but differ in whether impedance is minimized or maximized. Final Answer: XL = XC

Question 14

Tank-in-series filter behavior: When a parallel L–C tank is placed in series with an output load resistor and driven by a voltage source, what kind of filter response results at the load?

Accepted Answer

Introduction / Context: Placing a parallel resonant (tank) circuit in series with a load is a classic way to create a notch or band-stop response. At its resonant frequency, the tank's impedance becomes large, impeding current to the load and thus attenuating the output. Away from resonance, the tank's impedance is lower and the signal passes more freely. Given Data / Assumptions: Ideal parallel L–C tank with finite but high Q. Tank is in series between source and load resistor. We read the output across the load resistor. Concept / Approach: For a parallel L–C at resonance, the inductor and capacitor currents circulate within the tank, making its input impedance very high. Because the tank is in series with the load, this high impedance throttles current, yielding a deep attenuation (notch) at f0. At frequencies well below or above f0, the tank's impedance decreases, so more current flows and the load voltage recovers, which is characteristic of a band-stop filter centered at f0. Step-by-Step Solution: Tune L and C so XL = XC at f0 ⇒ tank input impedance is large. Place the tank in series with the load: series path current minimizes at f0. Load voltage v_out = i * R_load therefore dips at f0 (notch). Away from f0 the tank impedance falls ⇒ v_out increases. Verification / Alternative check: Frequency sweeps of such a network show a distinct dip at f0 in |V_out/V_in|. Moving the tank to a shunt branch produces complementary band-pass behavior across the load. Why Other Options Are Wrong: Low-pass/high-pass: these have monotonic pass/stop behavior, not a notch centered at f0. Band-pass: would require a topology that maximizes output near f0, not minimizes it (e.g., series RLC with output across R). Common Pitfalls: Confusing series vs shunt placement of the tank; overlooking how loading alters Q and notch depth. Final Answer: band-stop

Question 15

RLC filter terminology: which property of an electrical circuit best describes its ability to respond strongly to selected (desired) frequencies while rejecting or greatly attenuating all other frequencies?

Accepted Answer

Introduction / Context: In RLC circuits and filters, engineers often need a concise term for how well a network responds to certain frequencies and rejects others. This capability is crucial in radio receivers, instrumentation, and audio crossovers. The question asks for the correct property name that captures this frequency discrimination ability. Given Data / Assumptions: Circuit context: linear time-invariant RLC or active filter networks. Desired behavior: pass (or amplify) a target band and suppress out-of-band content. No specific numeric values are required; this is a definition/terminology check. Concept / Approach: “Selectivity” describes how selectively a circuit accepts desired frequencies while rejecting undesired ones. It is a qualitative attribute, often quantified by shape factor, out-of-band attenuation, and skirt steepness around the passband or notch. While bandwidth and quality factor are related metrics (e.g., Q = f0 / BW for narrowband resonators), they do not by themselves state the overall ability to reject out-of-band signals; sensitivity refers to signal magnitude response, not frequency discrimination. Step-by-Step Solution: 1) Identify the capability being described: strong response at desired frequencies, weak response elsewhere. 2) Map terms: “selectivity” is the accepted descriptor of frequency discrimination in filters and tuned circuits. 3) Recognize supporting metrics: bandwidth and quality factor relate to the sharpness near resonance but are not the general descriptor. 4) Conclude that “Selectivity” is the correct term. Verification / Alternative check: Receiver specifications list “selectivity” as the measure of adjacent-channel rejection, often using dB attenuation at offsets from the center frequency. This aligns precisely with the verbal definition in the stem. Why Other Options Are Wrong: Bandwidth: measures the width of the passband but not overall rejection behavior. Sensitivity: indicates required input level for a given output; not frequency discrimination. Quality factor: relates to resonance sharpness; a high Q suggests narrow bandwidth but is not the general property name. Dynamic range: span between minimum detectable and maximum allowable signal; not about frequency selectivity. Common Pitfalls: Equating narrow bandwidth or high Q with complete selectivity; ignoring skirt steepness and stopband attenuation which are also part of real-world selectivity performance. Final Answer: Selectivity

Question 16

Series vs. parallel RLC far from resonance: at frequencies well above and well below the resonant frequency, the series RLC network behaves like ______ while the parallel RLC network behaves like ______.

Accepted Answer

Introduction / Context: Understanding the limiting behavior of RLC circuits far from resonance allows quick mental checks when sketching Bode plots or predicting filter action. This question contrasts the series and parallel forms when the operating frequency is much lower or much higher than the resonant frequency f0. Given Data / Assumptions: Linear, ideal R, L, and C elements. Frequency well below and well above resonance (|ω − ω0| large). Qualitative “looks like” refers to net impedance magnitude relative to load: open (very large impedance) vs short (very low impedance). Concept / Approach: For a series RLC, the net impedance is Z = R + j(ωL − 1/(ωC)). Far below resonance (ω → 0), 1/(ωC) is very large in magnitude, so |Z| is large; far above resonance (ω → ∞), ωL is very large; either way the series path is “large impedance,” akin to an open. For a parallel RLC, the net admittance is the sum of branch admittances; away from resonance either the capacitor (at high ω) or the inductor (at low ω) provides a very low-impedance path, so the network tends toward a “short.” Step-by-Step Solution: 1) Series RLC: consider limits ω → 0 and ω → ∞; the reactive magnitude becomes very large in either limit. 2) Conclude series RLC appears open-like (very high |Z|) far from resonance. 3) Parallel RLC: consider branch impedances; at low ω the inductor is near short, at high ω the capacitor is near short. 4) Conclude parallel RLC appears short-like (very low |Z|) far from resonance. Verification / Alternative check: Impedance loci or Bode plots corroborate: series resonance gives a deep impedance minimum at ω0 and rises sharply away; parallel resonance gives a large impedance maximum at ω0 and falls off away, approaching a short at extremes. Why Other Options Are Wrong: Like a short, like an open: reverses the true limiting behavior. Inductive, capacitive / Capacitive, inductive: describe phase, not the open/short magnitude analogy across both extremes. Resistive, reactive: too vague and incorrect for the limits described. Common Pitfalls: Mixing series vs parallel intuition; focusing on phase sign rather than overall magnitude; forgetting that “open/short” here is a qualitative magnitude analogy, not an exact boundary condition. Final Answer: Like an open, like a short

Question 17

Filter behavior using a tank: when a parallel resonant (tank) circuit is inserted in series with an output load resistor, what overall filter type does the network implement around the tank’s resonant frequency?

Accepted Answer

Introduction / Context: Resonant tanks (LC networks) are the building blocks of many passive filters. A “parallel tank” has a very high impedance at its resonant frequency and a much lower impedance away from resonance. The way it is placed with respect to the signal path determines whether it passes or rejects a band of frequencies. Given Data / Assumptions: The tank is a parallel LC tuned to a resonant frequency f0. The tank is placed in series with the output load resistor. Components are ideal; we reason qualitatively around f0. Concept / Approach: A parallel LC exhibits maximum impedance at f0. If this high-impedance element is in series with the load, the signal path “blocks” at f0, producing a deep attenuation (a notch). Frequencies far from f0 see a comparatively low impedance through the series path and are less attenuated. This is the defining behavior of a band-stop (notch) filter with the stopband centered on f0. Step-by-Step Solution: 1) Recognize the series placement of a high-Z element at f0. 2) At f0: Z_tank → large; series path attenuation is high; output drops. 3) Away from f0: Z_tank → small; series path allows signal to reach the load. 4) Therefore the network rejects a narrow band around f0 and passes frequencies on either side. Verification / Alternative check: Plotting the transfer function magnitude shows a minimum (notch) at f0. Swapping the topology (placing the tank in shunt across the load) would produce a bandpass behavior, confirming the role of placement. Why Other Options Are Wrong: Low-pass / High-pass: these pass one side of the spectrum and reject the other, not a narrow band around f0. Bandpass: would require low impedance at f0 in the series path (a series-resonant element), not a high impedance. All-pass: preserves magnitude while shifting phase; not applicable here. Common Pitfalls: Confusing parallel vs series resonance; assuming any tank in the path yields bandpass; forgetting that parallel tanks have high impedance at resonance while series tanks have low impedance at resonance. Final Answer: Band-stop

RLC Circuits and Resonance Questions