Question 1

An RLC circuit resonates at f0 = 150 kHz with quality factor Q = 30. What is the half-power bandwidth range, that is, f1 to f2 enclosing the frequencies at which current (series) or impedance (parallel) reaches the 70.7% points?

Accepted Answer

Introduction: For narrowband resonant RLC networks, bandwidth and quality factor are tightly related. This problem tests your ability to convert a given resonant frequency and Q into the half-power (3 dB) frequencies f1 and f2 that bound the passband where current remains at least 70.7% of its peak (series) or where impedance criteria are met (parallel). Given Data / Assumptions: Resonant frequency f0 = 150 kHz. Quality factor Q = 30. Standard half-power definition: bandwidth BW = f2 - f1 = f0 / Q for high-Q circuits. Moderately high Q so the standard approximation applies well. Concept / Approach: The fundamental relation is BW = f0 / Q. Once BW is known, the half-power frequencies are centered about f0 so that f1 = f0 - BW/2 and f2 = f0 + BW/2. This symmetry is an accurate approximation for practical Q values of 10 or more. Step-by-Step Solution: 1) Compute bandwidth: BW = f0 / Q = 150 kHz / 30 = 5 kHz.2) Split BW evenly around f0: BW/2 = 2.5 kHz.3) Lower half-power corner: f1 = 150 kHz - 2.5 kHz = 147.5 kHz.4) Upper half-power corner: f2 = 150 kHz + 2.5 kHz = 152.5 kHz. Verification / Alternative check: For Q = 30, relative bandwidth BW/f0 = 1/30 ≈ 3.33%. This is small enough that the symmetric approximation about f0 is valid, reinforcing the computed f1 and f2 values. Why Other Options Are Wrong: 100.0 kHz to 155.0 kHz: Bandwidth is far larger than f0/Q; does not straddle f0 symmetrically. 4500 kHz to 295.5 kHz: Nonsensical range; not centered on 150 kHz. 149,970 Hz to 150,030 Hz: Implies BW = 60 Hz, which would correspond to Q ≈ 2500, not 30. 140 kHz to 160 kHz: BW = 20 kHz, four times the correct 5 kHz. Common Pitfalls: Forgetting BW = f0 / Q, mis-centering the range about f0, or confusing series versus parallel definitions. At moderate-to-high Q, the same BW formula applies and the 70.7% criterion leads to the quoted half-power points. Final Answer: 147.5 kHz to 152.5 kHz

Question 2

At resonance, a passive parallel RLC network is often described as exhibiting a "flywheel" effect. Which statement best characterizes its behavior at resonance?

Accepted Answer

Introduction: Parallel resonant (tank) circuits are widely used in tuning and filtering. The term "flywheel" refers to how energy shuttles between electric and magnetic fields, sustaining oscillatory behavior with minimal loss near resonance. This question probes conceptual understanding of energetic exchange and the practical meaning of resonance in passive networks. Given Data / Assumptions: Parallel RLC network, passive components. Sinusoidal steady state near resonance. Finite but small losses (nonzero R) so Q is finite. Concept / Approach: At resonance, reactive branch currents in L and C are large and largely equal and opposite, causing minimal source current. Energy alternately stores in C (electric field) and L (magnetic field), akin to a flywheel storing kinetic energy. With a sustaining source, this exchange produces a strong selective response that we describe as oscillatory behavior at or near the resonant frequency. Step-by-Step Solution: 1) Recognize resonance: net reactive susceptance ≈ 0 in parallel tanks.2) Energy transfer cycle: capacitor charges and discharges, inductor current builds and collapses.3) With minimal losses, stored energy persists over many cycles, giving a "flywheel" effect.4) The response is oscillatory in time when excited; with an active sustaining mechanism, the tank can form the core of an oscillator. Verification / Alternative check: Consider ring-down: excite the tank briefly and remove the source. The voltage/current decay exponentially at a rate set by Q, demonstrating free oscillation and the energy-storage "flywheel" effect until losses dissipate energy. Why Other Options Are Wrong: Flywheel only: captures energy storage but omits the inherently oscillatory exchange. Oscillate only: describes the motion but ignores the energy-storage metaphor. None of the above: contradicts standard textbook descriptions of resonant tanks. Ring only when driven by DC: DC does not sustain AC resonance; a transient can start ringing, but sustained oscillation needs AC drive or active feedback. Common Pitfalls: Thinking passive tanks self-oscillate without excitation (they require initial energy or active feedback) or that resonance means zero current everywhere. In parallel tanks, branch currents are large but cancel at the input. Final Answer: both of the above

Question 3

A series RLC circuit is driven by a sinusoidal source with VT = 100 V RMS. The reactances are XL = 160 Ω and XC = 80 Ω, and the resistance is R = 60 Ω. What is the RMS current in the circuit?

Accepted Answer

Introduction: Current in a series RLC circuit is found from the magnitude of total impedance Z. Because elements are in series, the same current flows through R, L, and C. This problem reinforces phasor addition of reactances and the Pythagorean form of impedance magnitude. Given Data / Assumptions: R = 60 Ω, XL = 160 Ω, XC = 80 Ω. Source VT = 100 V RMS. Sinusoidal steady-state operation. Concept / Approach: For series RLC, net reactance X = XL - XC. The impedance magnitude is |Z| = sqrt(R^2 + X^2). Then I = VT / |Z|. Phase affects voltage division but not current magnitude once |Z| is known. Step-by-Step Solution: 1) Compute net reactance: X = 160 − 80 = 80 Ω.2) Compute impedance magnitude: |Z| = sqrt(60^2 + 80^2) = sqrt(3600 + 6400) = sqrt(10000) = 100 Ω.3) Compute current: I = VT / |Z| = 100 / 100 = 1 A RMS.4) Express answer in amperes: 1 A. Verification / Alternative check: Check reasonableness: With |Z| = 100 Ω and 100 V RMS, current near 1 A is expected. Also, XL larger than XC indicates net inductive behavior, so current should lag voltage by arctan(X/R) = arctan(80/60), consistent with the phasor picture. Why Other Options Are Wrong: 1 mA and 0.2 A: too small given 100 V across about 100 Ω. 6.28 A and 10 A: far too large; would require |Z| much smaller than 100 Ω. Common Pitfalls: Adding reactances arithmetically to resistance without squaring, or forgetting to subtract XC from XL for net X. Always use |Z| = sqrt(R^2 + (XL − XC)^2). Final Answer: 1 A

Question 4

In AC phasor analysis of a parallel RLC circuit, which quantity can always be used as the universal 0° vector reference because it is common to all branches?

Accepted Answer

Introduction: Phasor diagrams require a reference. Selecting the shared quantity simplifies analysis and reduces mistakes. In parallel networks, voltage across each branch is identical, making it the most logical and universal reference for angles and vector sums. Given Data / Assumptions: Parallel RLC topology with sinusoidal steady state. Ideal components and a single operating frequency. R, L, and C are connected in parallel across the same source. Concept / Approach: Because the node-to-node potential is the same for every branch, choosing voltage as the 0° phasor lets us express branch currents with simple phase relations: IR is in phase with V, IL lags V by 90° (for inductive susceptance, current lags voltage), and IC leads V by 90°. The total source current is the vector sum of these branch currents relative to the shared V reference. Step-by-Step Solution: 1) Identify the network as parallel: common voltage, different branch currents.2) Set V to 0° on the horizontal axis.3) Draw IR in phase with V, IL at −90°, IC at +90°.4) Add currents tip-to-tail to obtain the source current phasor. Verification / Alternative check: Attempting to use current as a universal reference fails because each branch has a different current phase. Voltage avoids this inconsistency and clarifies the diagram immediately. Why Other Options Are Wrong: Current and inductive current: not common across branches; phases differ. Reactance and resistance: component properties, not phasor references; they do not define a universal phase. Common Pitfalls: Using series-circuit intuition (where current is common) in a parallel context, or forgetting the 90° phase relationships of L and C currents relative to voltage. Final Answer: voltage

Question 5

At resonance in a simple RLC system, the bandwidth is defined between the half-power points. The corresponding current at these corner frequencies is what percentage of the maximum current?

Accepted Answer

Introduction: Bandwidth in resonant circuits is commonly specified by the half-power (3 dB) points. Recognizing the current or voltage ratios at these points is crucial for filter design, tuning, and quality factor calculations. Given Data / Assumptions: Linear, time-invariant RLC circuit. Half-power definition used for bandwidth. Sinusoidal steady-state operation. Concept / Approach: Half-power means the power is one-half of the peak value. Since power in a resistive element is proportional to the square of current (or voltage), halving power implies the amplitude of current (or voltage) equals 1/sqrt(2) of the maximum, which is 0.707 or 70.7%. Step-by-Step Solution: 1) Half-power condition: P = 0.5 * Pmax.2) Because P ∝ I^2, we have (I/Imax)^2 = 0.5.3) Take square root: I/Imax = 1 / sqrt(2) ≈ 0.707.4) Express as a percentage: 0.707 * 100% ≈ 70.7%. Verification / Alternative check: The 3 dB rule: a drop of 3 dB corresponds to a power ratio of 0.5 and an amplitude ratio of 0.707, confirming the same percentage for either current or voltage. Why Other Options Are Wrong: 50: This would be amplitude halved, but half power corresponds to amplitude reduced to 70.7%. 62.3 and 84.1: Do not correspond to standard half-power amplitude ratios. 95.3: Too close to the peak; not representative of half-power points. Common Pitfalls: Confusing power ratio with amplitude ratio, or assuming a 50% power reduction means 50% amplitude reduction. Always convert through the square relationship. Final Answer: 70.7

Question 6

In filter and resonant-circuit design, if the bandwidth (BW) of a band-pass response becomes larger while the center (resonant) frequency f0 is held fixed, how does the quality factor Q = f0 / BW change?

Accepted Answer

Introduction: Understanding the relationship between bandwidth (BW), center frequency (f0), and quality factor (Q) is essential in analog filter design and resonant circuits. Q is a measure of selectivity: high Q means a narrow passband around f0, while low Q means a wider passband. This question tests the direct inverse relationship between Q and BW when f0 is fixed. Given Data / Assumptions: Band-pass response centered at f0. Bandwidth BW = f2 - f1 between the two half-power (−3 dB) cutoff frequencies. Center frequency f0 is held constant while BW increases. Standard definition: Q = f0 / BW. Concept / Approach: By definition, Q is inversely proportional to BW for a given f0. If BW grows larger, the ratio f0 / BW becomes smaller. Therefore, the circuit becomes less selective and more broadbanded, which is interpreted as a decrease in Q. Step-by-Step Solution: Start with the identity: Q = f0 / BWHold f0 fixed (constant).Increase BW → denominator increases.Result: Q decreases because the ratio f0 / BW becomes smaller. Verification / Alternative check: Consider two designs at the same f0. If Design A has BW_A = 100 Hz and Design B has BW_B = 200 Hz, then Q_A = f0 / 100 and Q_B = f0 / 200 = Q_A / 2. The broader design (larger BW) indeed has half the Q, confirming the inverse relationship. Why Other Options Are Wrong: The roll-off rate increases: Roll-off mainly depends on filter order, not directly on BW. The half-power frequency decreases: f1 and f2 move apart as BW increases; this statement is incomplete/misleading. The center frequency decreases: f0 is specified constant in the scenario. Q increases: Opposite of the Q = f0 / BW definition when f0 is fixed. Common Pitfalls: Confusing Q with gain; Q characterizes selectivity, not passband gain directly. Assuming roll-off steepness changes with BW independent of order. Mixing arithmetic mean with geometric mean definitions of center frequency in wideband cases. Final Answer: Q decreases

Question 7

In an RLC circuit that behaves net inductively (current lags the applied voltage), what is the corresponding power-factor angle and qualitative power factor?

Accepted Answer

Introduction: Power factor describes how effectively alternating current power is converted to useful work. In reactive circuits, voltage and current are not in phase, leading or lagging by an angle φ. Inductive behavior means current lags the voltage (negative phase angle with the usual reference), impacting both the angle and the numerical power factor cos(φ). Given Data / Assumptions: Sinusoidal steady-state operation. RLC circuit with net inductive reactance (X_L > X_C). Idealized discussion of limiting behaviors (purely reactive vs. mixed R–X). Concept / Approach: For a purely inductive circuit, the current lags the voltage by 90 degrees. The power factor is cos(−90°) = 0, meaning no average real power is consumed (only reactive exchange). In a general inductive R–L–C with resistance, the lag angle would be between 0 and −90 degrees, and the power factor between 0 and 1 lagging. Among the offered choices, the statement that best represents an inductive power-factor condition is '−90 degrees lagging' (the purely inductive limit). Step-by-Step Solution: Identify inductive behavior: X_L > X_C ⇒ current lags voltage.Limit case for pure inductor: φ = −90°.Power factor PF = cos(φ) ⇒ PF = cos(−90°) = 0.Therefore, the representative angle/power-factor description is '−90 degrees lagging' (PF approaching zero). Verification / Alternative check: Phasor analysis: Z = R + j(X_L − X_C). If R → 0 and X_L > X_C, arg(Z) → +90°, so arg(I) = −90° relative to V. This confirms the lagging current and purely reactive behavior at the limit. Why Other Options Are Wrong: +90 degrees leading / about +45 degrees leading: These indicate capacitive behavior (current leads), not inductive. one: PF = 1 only for purely resistive circuits (φ = 0°), not inductive. zero: While the numeric PF is zero in the pure inductor limit, the option lacks the crucial lagging/angle qualifier; the more accurate descriptive choice is '−90 degrees lagging'. Common Pitfalls: Equating inductive behavior with a specific numeric PF without stating lagging angle. Assuming PF = 1 can occur in reactive circuits; it requires zero reactance. Confusing sign convention for leading vs. lagging angles. Final Answer: −90 degrees lagging

Question 8

An RLC circuit resonates at f0 = 2000 Hz and exhibits a bandwidth BW = 250 Hz. Using the narrowband approximation, what is the high cutoff (upper half-power) frequency f2?

Accepted Answer

Introduction: For band-pass responses around resonance, the two half-power (−3 dB) frequencies f1 and f2 bound the passband. The bandwidth BW is defined as f2 − f1. When the bandwidth is small compared with the center frequency (narrowband), simple arithmetic approximations around f0 are accurate and widely used in design and quick checks. Given Data / Assumptions: Resonance (center) frequency: f0 = 2000 Hz. Bandwidth: BW = 250 Hz. Narrowband approximation (BW ≪ f0), so f1 ≈ f0 − BW/2 and f2 ≈ f0 + BW/2. Concept / Approach: Use the identity BW = f2 − f1 and the narrowband symmetry around f0. Thus f2 ≈ f0 + (BW / 2). This method is accurate enough for filters with modest Q where BW/f0 is small. Step-by-Step Solution: Compute half-bandwidth: BW/2 = 250 / 2 = 125 HzUpper cutoff: f2 ≈ f0 + BW/2 = 2000 + 125 = 2125 HzLower cutoff (for reference): f1 ≈ 2000 − 125 = 1875 HzCheck: f2 − f1 = 2125 − 1875 = 250 Hz (matches given BW) Verification / Alternative check: The exact relationship for narrowband resonators also satisfies f0 ≈ sqrt(f1 * f2). With the approximated f1 and f2 above, sqrt(1875 * 2125) ≈ 1999.999 Hz, essentially 2000 Hz, validating the arithmetic method. Why Other Options Are Wrong: 2250 Hz: This would imply BW/2 = 250 Hz, doubling the bandwidth incorrectly. 1750 Hz: That is below f0 and cannot be the upper cutoff. 8.0 Hz: Irrelevant magnitude; not connected to the given data. 2000 Hz: The resonant (center) frequency, not the high cutoff. Common Pitfalls: Confusing center frequency f0 with cutoff frequencies f1 and f2. Forgetting that BW is the difference f2 − f1, not f0 − f1 or f2 − f0 alone. Misapplying the geometric-mean identity outside the narrowband regime. Final Answer: 2125 Hz

Question 9

In a general RLC circuit, when the capacitive reactance equals the inductive reactance (X_C = X_L), what fundamental operating condition is satisfied?

Accepted Answer

Introduction: Resonance in RLC circuits is a cornerstone concept: it occurs when the inductive and capacitive reactances cancel each other in magnitude (X_L = X_C). Recognizing this condition helps predict current, voltage distribution, and power factor behavior in both series and parallel topologies. Given Data / Assumptions: Sinusoidal steady state, linear and ideal components. RLC network in either series or parallel configuration. Condition under study: X_L = X_C. Concept / Approach: At resonance, the net reactive component of impedance is zero in series circuits (Z is purely resistive), or the net reactive component of admittance is zero in parallel circuits (Y is purely conductive). This is the unifying definition of resonance and is directly tied to X_L = X_C. Step-by-Step Solution: Write X_L = ωL and X_C = 1 / (ωC)Set X_L = X_C ⇒ ωL = 1 / (ωC)Solve for resonance radian frequency: ω_0 = 1 / sqrt(L * C)Thus f_0 = ω_0 / (2 * π) = 1 / (2 * π * sqrt(L * C))At f_0, the reactive effects cancel, defining resonance. Verification / Alternative check: Series RLC: Z = R + j(X_L − X_C) ⇒ at resonance Z = R (minimum magnitude), current is maximized for a given applied voltage. Parallel RLC: input admittance Y has zero imaginary part at resonance, leading to minimum source current. In both cases, the defining condition is X_L = X_C. Why Other Options Are Wrong: The circuit draws maximum current in every topology: True for series only, false for parallel (it is minimum at resonance). The applied voltage is zero: Not implied by resonance. The circuit draws minimum current in every topology: True for parallel only, false for series. The power factor is always zero: At resonance in series, PF = 1; not zero. Common Pitfalls: Overgeneralizing series behavior to parallel circuits (or vice versa). Confusing the condition X_L = X_C with purely reactive behavior; in series it yields purely resistive input. Final Answer: The circuit is at resonance

Question 10

A band-pass RLC circuit resonates at f0 = 150 kHz and the inductor has a coil quality factor Q = 30. Using the narrowband approximation, what are the −3 dB cutoff frequencies (F1 and F2) that define the passband?

Accepted Answer

Introduction: Band-pass filters around a resonant frequency f0 are characterized by a bandwidth BW and a quality factor Q = f0 / BW. Determining the lower and upper cutoff frequencies (F1 and F2) from Q and f0 is a routine but vital task in RF design, selectivity analysis, and tuning stages. Given Data / Assumptions: Resonant (center) frequency: f0 = 150 kHz. Coil quality factor: Q = 30 (assumed representative of the filter Q). Narrowband assumption applies (Q ≫ 1). Concept / Approach: For narrowband resonators, BW = f0 / Q. The cutoff frequencies are approximately symmetric around f0: F1 ≈ f0 − BW/2 and F2 ≈ f0 + BW/2. This symmetry gives a quick, accurate estimate when BW is small compared with f0. Step-by-Step Solution: Compute bandwidth: BW = f0 / Q = 150 kHz / 30 = 5 kHzHalf-bandwidth: BW/2 = 2.5 kHzLower cutoff: F1 ≈ 150 kHz − 2.5 kHz = 147.5 kHzUpper cutoff: F2 ≈ 150 kHz + 2.5 kHz = 152.5 kHzCheck: F2 − F1 = 5.0 kHz = BW (consistent) Verification / Alternative check: Using the geometric-mean identity for narrowband filters, f0 ≈ sqrt(F1 * F2). With F1 = 147.5 kHz and F2 = 152.5 kHz, sqrt(147.5 * 152.5) kHz ≈ 150.0 kHz, validating the approximation. Why Other Options Are Wrong: 100.0–155.0 kHz: Far too wide; BW would be 55 kHz, not 5 kHz. 295.5 kHz–4500 kHz: Nonsensical relative to f0. 149,970–150,030 Hz: Extremely narrow (BW 60 Hz) inconsistent with Q = 30. 130–170 kHz: BW 40 kHz; contradicts Q = 30. Common Pitfalls: Using BW itself as the offset from f0 instead of BW/2. Confusing coil Q with loaded Q; here we assume they are approximately equal. Final Answer: 147.5 kHz to 152.5 kHz

Question 11

In a series RLC circuit, the branch magnitudes are V_R = I_T * R, V_L across the inductor, and V_C across the capacitor. Given I_T = 3 mA, V_L = 30 V, V_C = 18 V, and R = 1000 Ω, what is the magnitude of the applied source voltage?

Accepted Answer

Introduction: In series RLC circuits, branch voltages are not added arithmetically because they are out of phase. The resistor voltage V_R is in phase with the current, while V_L leads and V_C lags the current by 90 degrees, making V_L and V_C 180 degrees apart from each other. The source voltage is the phasor (vector) sum of these components. Given Data / Assumptions: Total current: I_T = 3 mA. Inductor voltage: V_L = 30 V (leads I by +90°). Capacitor voltage: V_C = 18 V (lags I by −90°). Resistance: R = 1000 Ω ⇒ V_R = I_T * R. Sinusoidal steady state, ideal elements. Concept / Approach: Compute V_R = I_T * R. Reactive voltages along j-axis combine algebraically: V_X = V_L − V_C. The applied voltage magnitude is |V_S| = sqrt(V_R^2 + V_X^2). This uses orthogonal decomposition of phasors (real axis for V_R, imaginary axis for net reactance voltage). Step-by-Step Solution: V_R = I_T * R = 0.003 A * 1000 Ω = 3.0 VNet reactive voltage: V_X = V_L − V_C = 30 − 18 = 12 VApplied voltage magnitude: |V_S| = sqrt(V_R^2 + V_X^2)|V_S| = sqrt(3.0^2 + 12^2) = sqrt(9 + 144) = sqrt(153) ≈ 12.369… VRounded to two decimals: 12.37 V Verification / Alternative check: Phasor diagram: V_R on real axis = 3 V; V_L and V_C on imaginary axis, opposing. Resultant vector length matches the computed magnitude. The value is sensible because the reactive voltages largely cancel, leaving a moderate resultant dominated by the 12 V net reactance and a small 3 V resistive component. Why Other Options Are Wrong: 3.00 V: Ignores the net reactive contribution (12 V). 34.98 V / 48.00 V: Resemble arithmetic sums or partial sums of branch magnitudes; phasors must be combined vectorially. 9.00 V: Equal to V_R + (V_L − V_C) linearly; again, incorrect because of orthogonal components. Common Pitfalls: Adding V_R, V_L, and V_C directly without phasor resolution. Forgetting that V_L and V_C oppose each other by 180 degrees along the reactive axis. Rounding too early; keep sufficient precision until the final step. Final Answer: 12.37 V

Question 12

When computing totals for series RLC circuits (voltages or impedances), which method correctly accounts for the phase relationships among resistive and reactive components?

Accepted Answer

Introduction: Series RLC analysis hinges on the fact that resistive and reactive elements introduce phase shifts between voltage and current. Therefore, totals cannot be computed with simple arithmetic sums of magnitudes; one must respect phase using vector (phasor) methods. Given Data / Assumptions: Sinusoidal steady state. Series connection of R, L, and C components. Voltages across elements can be larger than the source due to phase relations. Concept / Approach: Represent voltages and impedances as complex numbers: R on the real axis, X_L = +jωL and X_C = −j/(ωC) on the imaginary axis. Resultants are obtained by vector (phasor) addition, i.e., adding real and imaginary components separately to form a complex total, then taking magnitude/angle as needed. Step-by-Step Solution: Write impedances: Z_R = R, Z_L = jX_L, Z_C = −jX_CTotal impedance: Z_total = R + j(X_L − X_C)Voltage totals: V_S = V_R + V_L + V_C as phasors, not scalar magnitudesMagnitude: |Z_total| = sqrt(R^2 + (X_L − X_C)^2)Angle: φ = arctan((X_L − X_C) / R) Verification / Alternative check: Phasor diagrams graphically confirm that orthogonal components must be combined vectorially. Numerical examples (e.g., V_R = 3 V, V_L = 30 V, V_C = 18 V) produce |V_S| = sqrt(3^2 + (30 − 18)^2) = sqrt(153) ≈ 12.37 V, which differs fundamentally from any simple sum/difference of magnitudes. Why Other Options Are Wrong: Subtracting / multiplying / averaging magnitudes: Disregard phase, leading to large errors. Graphing the angles: A visualization aid, not the computation method; one must still perform vector addition. Common Pitfalls: Assuming the source voltage equals the arithmetic sum of branch voltages. Ignoring that V_L and V_C oppose along the imaginary axis. Confusing rms with peak values; consistency is required for phasor math. Final Answer: Adding vectors (phasor addition)

Question 13

Parallel RLC phasor analysis: Given branch currents IL = 15.3 A (inductive branch, current lags voltage by 90°), IC = 0.43 A (capacitive branch, current leads voltage by 90°), and IR = 3.5 A (resistive branch, in phase with voltage), determine the c…

Accepted Answer

Introduction: In a parallel RLC circuit, the total current is the phasor (vector) sum of branch currents: resistive current IR is in phase with the supply voltage, inductive current IL lags the voltage by 90 degrees, and capacitive current IC leads the voltage by 90 degrees. The net phase angle of the total current with respect to the voltage reveals whether the circuit is overall inductive (negative angle, current lags) or capacitive (positive angle, current leads). Given Data / Assumptions: Inductive branch current IL = 15.3 A (lags V by 90 degrees) Capacitive branch current IC = 0.43 A (leads V by 90 degrees) Resistive branch current IR = 3.5 A (in phase with V) Sinusoidal steady state; ideal components Concept / Approach: Let the in-phase axis be along the voltage (and IR). The reactive axis is orthogonal. Net reactive current is Ireact = IC - IL (capacitive positive, inductive negative). The phase angle of total current is θ = arctan(Ireact / IR). A negative θ means the total current lags the voltage (inductive dominance). Step-by-Step Solution: Compute net reactive current: Ireact = IC - IL = 0.43 - 15.3 = -14.87 AUse θ = arctan(Ireact / IR)θ = arctan((-14.87) / 3.5) ≈ arctan(-4.2486)θ ≈ -76.7 degrees (inductive overall) Verification / Alternative check: The magnitude of the total current is |I| = sqrt(IR^2 + Ireact^2) ≈ sqrt(3.5^2 + 14.87^2) ≈ 15.3 A. Since |Ireact| > IR, the angle must be near ±90 degrees; negative sign confirms an inductive net response. Why Other Options Are Wrong: 76.7 degrees: Wrong sign; would imply strong capacitive lead. -4.25 degrees: Magnitude far too small given |Ireact| ≫ IR. 88.8 degrees: Too large; |Ireact|/IR ≈ 4.25, not near infinity. -15.0 degrees: Underestimates inductive dominance. Common Pitfalls: Adding magnitudes arithmetically instead of using phasors. Forgetting that Ireact = IC - IL with sign convention (capacitive positive). Final Answer: -76.7 degrees

Question 14

Series RLC phasor convention: In a series RLC circuit, what quantity is most appropriately chosen as the universal phasor reference for analyzing voltages and angles across the elements?

Accepted Answer

Introduction: Phasor analysis requires choosing a consistent reference phasor so that all other phasors can be measured in magnitude and angle. In series RLC circuits, the same current flows through R, L, and C, making current an ideal reference for comparing element voltages and determining impedance angles. Given Data / Assumptions: Series RLC topology (single loop) Sinusoidal steady state Goal: choose a stable, universal reference for phasor diagrams Concept / Approach: Because series elements share the same current, setting the current phasor at 0 degrees (as the reference) simplifies analysis: VR is in phase with I, VL leads I by 90 degrees, and VC lags I by 90 degrees. Impedance angle then directly equals the angle of the voltage relative to the current, and power factor is cos(angle). Step-by-Step Solution: Choose I as the 0-degree phasor (the reference).Express VR = I * R (0-degree), VL = I * XL (+90 degrees), VC = I * XC (-90 degrees).Sum voltages vectorially to get V and the total impedance angle.Interpret power factor directly from the angle between V and I. Verification / Alternative check: Textbook convention universally sets current as the reference for series circuits, minimizing ambiguity and keeping sign conventions consistent across R, L, and C voltage drops. Why Other Options Are Wrong: a leading vector / a lagging vector: Describe relative angles, not a universal reference choice. an angle: An angle is a measurement, not a phasor. a phasor of voltage: Possible, but it complicates series analysis; using I is standard and simpler. Common Pitfalls: Picking different references mid-solution, causing sign mistakes. Confusing the power factor angle (between V and I) with individual element voltage angles. Final Answer: a reference

Question 15

Bandwidth from quality factor: A tuned RLC circuit resonates at f0 = 150 kHz and has quality factor Q = 30. Determine the -3 dB bandwidth BW and the approximate half-power frequencies f1 and f2 (centered around resonance).

Accepted Answer

Introduction: Quality factor (Q) characterizes the sharpness of resonance in RLC circuits. For narrowband resonant circuits, Q links the center frequency f0, the -3 dB bandwidth BW, and the half-power (cutoff) frequencies f1 and f2 that bound the passband of a bandpass behavior. Given Data / Assumptions: Resonant frequency f0 = 150 kHz Quality factor Q = 30 Narrowband assumption valid (Q ≫ 1) Concept / Approach: For a bandpass resonant network, BW = f2 - f1 and Q = f0 / BW. Hence BW = f0 / Q. For high Q, the half-power frequencies are approximately symmetric about f0, so f1 ≈ f0 - BW/2 and f2 ≈ f0 + BW/2. Step-by-Step Solution: Compute bandwidth: BW = f0 / Q = 150 kHz / 30 = 5 kHzCompute f1: f1 ≈ 150 kHz - 2.5 kHz = 147.5 kHzCompute f2: f2 ≈ 150 kHz + 2.5 kHz = 152.5 kHzHence the passband is approximately 147.5 kHz to 152.5 kHz Verification / Alternative check: Since BW is much less than f0 (5 kHz ≪ 150 kHz), the symmetry assumption holds well. Also, Q = f0 / BW = 150/5 = 30, matching the given Q. Why Other Options Are Wrong: 100.0 kHz to 155.0 kHz: BW = 55 kHz ⇒ Q ≈ 2.7, not 30. 4500 kHz to 295.5 kHz: Nonsensical ordering and values. 149,970 Hz to 150,030 Hz: BW = 60 Hz ⇒ Q = 2500, not 30. 140.0 kHz to 160.0 kHz: BW = 20 kHz ⇒ Q = 7.5, not 30. Common Pitfalls: Confusing BW = f2 - f1 with fractional bandwidth (BW / f0). Forgetting that higher Q means narrower bandwidth. Final Answer: 147.5 kHz to 152.5 kHz

Question 16

Dominance of reactance in a series RLC: Across operating frequencies, which quantity determines whether the net reactive behavior is inductive or capacitive in a series RLC circuit?

Accepted Answer

Introduction: Series RLC circuits exhibit frequency-dependent reactance. Whether the circuit looks inductive or capacitive depends on the relative magnitudes of the inductive reactance XL and the capacitive reactance XC at the operating frequency. This underlies resonance and filter behavior. Given Data / Assumptions: Series connection of R, L, and C Sinusoidal steady-state analysis Net reactance X = XL - XC (sign indicates inductive or capacitive) Concept / Approach: In a series RLC, the total impedance is Z = R + j(XL - XC). If XL > XC, X is positive (inductive). If XC > XL, X is negative (capacitive). At resonance, XL = XC and the net reactance is zero. Thus, the larger magnitude of the two reactances determines the reactive character away from resonance. Step-by-Step Solution: Compute XL = 2 * π * f * LCompute XC = 1 / (2 * π * f * C)Evaluate X = XL - XCInterpret sign: X > 0 → inductive; X < 0 → capacitive; X = 0 → resonance Verification / Alternative check: Plot XL and XC versus frequency: XL rises linearly with f, while XC falls hyperbolically. The crossover (XL = XC) marks resonance and confirms dominance flips around f0. Why Other Options Are Wrong: Capacitive reactance is always dominant.: False; dominance depends on frequency. Inductive reactance is always dominant.: False; at low f, capacitive dominates. Resistance is always dominant.: Resistance affects damping, not reactive sign. Both reactances cancel at all frequencies.: Cancellation happens only at resonance. Common Pitfalls: Thinking R determines inductive vs capacitive behavior; it does not. Forgetting sign conventions when computing X = XL - XC. Final Answer: The larger of the two reactances is dominant.

Question 17

Filter classification: When a continuous band of frequencies around a center frequency is allowed to pass to the output while others are attenuated, what type of resonant filter is this?

Accepted Answer

Introduction: Filters are categorized by how they treat frequency components. A band of frequencies centered around a resonant frequency is characteristic of a bandpass response, widely used in communication receivers, instrumentation, and audio crossovers to select desired signals while rejecting out-of-band noise. Given Data / Assumptions: We want a range (band) of frequencies to pass Out-of-band frequencies are attenuated Realizable with RLC resonators or active filter topologies Concept / Approach: A bandpass filter exhibits low attenuation within its passband (between f1 and f2) and increased attenuation below f1 and above f2. It can be implemented using series-parallel RLC combinations or active op-amp stages (e.g., multiple-feedback or state-variable filters). The resonator sets the center frequency f0, and Q controls bandwidth. Step-by-Step Solution: Identify requirement: Allow a contiguous frequency range.Map to filter type: This is the definition of a bandpass response.Note: Low-pass passes only below a cutoff; high-pass passes only above; band-stop (notch) rejects a narrow band. Verification / Alternative check: Sketching the ideal magnitude response confirms a pass window between f1 and f2 with roll-offs on either side, matching a bandpass filter’s behavior. Why Other Options Are Wrong: low-pass filter: Passes only low frequencies below fc. high-pass filter: Passes only high frequencies above fc. band-stop filter: Rejects a band around f0 instead of passing it. notch-pass filter: Not a standard category; usually “notch” means a narrow band-stop. Common Pitfalls: Confusing bandpass with band-stop because both involve two corner frequencies. Ignoring Q, which sets bandwidth and selectivity. Final Answer: bandpass filter

Question 18

Effect of a parallel (tank) resonant circuit placed in a filter path: At resonance, what impact does a parallel LC tank have on the downstream or final filter current?

Accepted Answer

Introduction: A parallel (tank) resonant circuit exhibits very high impedance at its resonant frequency. When inserted in the signal path or as a shunt in certain filter topologies, this high impedance behavior changes how much current can flow at resonance, which is critical for shaping filter responses and isolating stages. Given Data / Assumptions: Parallel LC tank operated at its resonant frequency Ideal components; losses only small unless specified Placed so that its impedance influences line current to the next stage Concept / Approach: At resonance, a parallel LC has large impedance because the equal and opposite reactive currents in L and C circulate internally, minimizing net current drawn from the line. Consequently, the branch presents a high barrier to current flow, effectively blocking current to the load in that path or sharply reducing it, depending on the exact placement in the filter. Step-by-Step Solution: Recognize that Z_parallel_LC → high at resonance.In a series path, a high Z element reduces line current significantly.In a shunt configuration, the tank draws minimal current at resonance, increasing the line impedance seen by the load.Net effect: Final (downstream) current is blocked or greatly reduced at resonance. Verification / Alternative check: Compare with a series resonant circuit, which has minimum impedance at resonance and thus passes maximum current. The parallel case is the complement, confirming the blocking behavior at resonance. Why Other Options Are Wrong: very little: Incorrect; effect is large at resonance due to very high impedance. The bandpass frequencies will change.: Placement may affect tuning slightly, but the principal effect at resonance is the high impedance (blocking), not retuning by itself. The frequency cutoff will change.: Cutoffs relate to design Q and component values; the primary immediate effect at resonance is current suppression. It will automatically double the Q.: Q depends on L, C, and losses; not automatically doubled by topology. Common Pitfalls: Confusing series versus parallel resonance effects on impedance. Assuming resonance always means maximum current; that is only for series resonance. Final Answer: The impedance will block output.

Question 19

Series RLC totals: When combining impedances or voltage phasors in a series RLC circuit, which method must always be used to obtain correct totals?

Accepted Answer

Introduction: Series RLC circuits contain resistive and reactive elements whose voltages and impedances have phase differences. To compute accurate totals, we must account for both magnitudes and angles, which requires vector (phasor) addition rather than simple arithmetic addition or subtraction. Given Data / Assumptions: Series connection: the same current flows through R, L, and C Voltages across R, L, and C differ in phase: 0°, +90°, and -90° with respect to current Goal: total voltage or total impedance calculation Concept / Approach: Represent voltages and impedances as complex numbers (phasors). The total series impedance is Z = R + j(XL - XC). The total supply voltage is V = VR + VL + VC, but these are vector sums because of their phase offsets. Therefore, the correct method is vector (phasor) addition in the complex plane. Step-by-Step Solution: Compute individual phasors: VR = I * R (0°), VL = I * XL (+90°), VC = I * XC (-90°).Sum phasors: V = VR + VL + VC (complex addition).Magnitude of total: |V| = sqrt((∑ real)^2 + (∑ imag)^2).Impedance: Z = V / I = R + j(XL - XC) (vector form). Verification / Alternative check: Graphical phasor diagrams or complex-number calculations both yield identical results. Arithmetic addition of magnitudes would overestimate totals, especially when VL and VC partially cancel. Why Other Options Are Wrong: graphing the angles: A visualization tool, not a calculation method. multiplying the values: No basis in phasor summation for series totals. subtracting the values: Partial cancellation occurs vectorially, not by blind subtraction of magnitudes. adding magnitudes arithmetically: Ignores phase; produces incorrect totals. Common Pitfalls: Summing magnitudes of VL and VC without recognizing their opposite phases. Forgetting to convert degrees to radians when using complex arithmetic programmatically. Final Answer: adding values vectorially

Question 20

In an AC series RLC circuit, under which operating condition does the circuit's voltage lag the circuit's current (i.e., current leads and the net reactance is negative/capacitive)?

Accepted Answer

Introduction: Understanding phase relationships between voltage and current is fundamental in AC circuit analysis. In a series RLC circuit, whether voltage leads or lags current depends on the net reactance (inductive or capacitive). This question asks you to identify the operating condition under which the voltage lags the current, meaning the current leads by a positive phase angle. Given Data / Assumptions: Series RLC circuit driven by a sinusoidal source. Ideal components used for conceptual clarity. We are comparing qualitative operating conditions: inductive, capacitive, resistive, and resonant. Concept / Approach: The total reactance of a series RLC circuit is X = XL − XC, where XL = 2 * pi * f * L and XC = 1 / (2 * pi * f * C). The phase angle phi between source voltage and current is given by tan(phi) = X / R. If X is positive (inductive), voltage leads current; if X is negative (capacitive), voltage lags current (equivalently, current leads). At resonance, X = 0 and voltage and current are in phase. Step-by-Step Solution: Write X = XL − XC.If XL > XC → X > 0 (inductive) → voltage leads current (current lags).If XL < XC → X < 0 (capacitive) → voltage lags current (current leads).If XL = XC → X = 0 (resonant) → voltage and current are in phase.Therefore, “voltage lags current” describes the capacitive operating condition. Verification / Alternative check: Using i(t) and v(t) phasors: for capacitive behavior, i(t) = C * dv/dt implies current peaks occur before voltage peaks (lead), confirming that the circuit voltage lags the current when the circuit acts capacitively. Why Other Options Are Wrong: Resistively: With purely resistive behavior, voltage and current are in phase; neither leads nor lags. Inductively: Inductive dominance makes voltage lead current, the opposite of what is asked. Resonantly: At resonance, phase angle is approximately 0 degrees; there is no lead/lag. Common Pitfalls: Students often reverse the sign of X or confuse “current leads” with “voltage leads.” A quick rule is “ICE”: in a Capacitor, current (I) leads voltage (V); in an Inductor (L), voltage leads current. Always check XL − XC to determine the sign of X. Final Answer: Capacitively.

RLC Circuits and Resonance Questions