Q: Parallel-plate formula check: Capacitance C is directly proportional to the dielectric constant (relative permittivity) and the plate area A, and inversely proportional to the plate separation d (C = ε * A / d). Is the claim “directly proportional t…

Introduction / Context: The geometry of a capacitor determines its capacitance. For the common parallel-plate model, a simple formula relates material properties and dimensions. This question tests whether the stated proportionalities match the well-known relationship used in design and analysis. Given Data / Assumptions: Parallel-plate approximation with negligible fringing. Dielectric fully fills the region between plates. Relative permittivity ε_r multiplies ε_0 to yield ε. Concept / Approach: The canonical formula is C = ε * A / d, where ε = ε_r * ε_0. Capacitance increases with plate area A and dielectric constant ε, and decreases as the separation d increases. Therefore, any statement claiming direct proportionality to distance and inverse to area is the exact opposite and is incorrect for this standard geometry. Step-by-Step Solution: State the formula: C = ε * A / d. Identify proportionalities: C ∝ ε and C ∝ A; C ∝ 1/d. Compare with the claim: it reverses the roles of A and d. Conclude the claim is incorrect for parallel plates. Verification / Alternative check: Dimensional and physical intuition: larger area gives more field-coupled region (higher C); greater distance weakens the field coupling (lower C). Measurements of variable capacitors confirm these trends. Why Other Options Are Wrong: Correct only for cylindrical/fringing cases: the sign of dependence does not reverse for standard configurations; while exact formulas differ, C does not increase with distance in such ideal models. Common Pitfalls: Mixing up proportionalities; overlooking that dielectric constant boosts capacitance by concentrating the electric field (higher ε reduces effective field for the same charge, enabling more Q per volt). Final Answer: Incorrect

Q: Filter intuition check: Does a high-pass RC filter strongly attenuate the higher frequencies (rather than the lower ones) in its passband/stopband behavior?

Introduction / Context: Filters shape spectra by passing some frequency ranges while attenuating others. A high-pass filter is designed to pass high frequencies and reject low frequencies. This question probes whether that definition is being applied correctly in the context of basic RC networks. Given Data / Assumptions: First-order RC high-pass implementation (series C, shunt R, output across R). Ideal components; small-signal linear operation. Standard −3 dB cutoff at f_c = 1/(2πRC). Concept / Approach: In a high-pass filter, attenuation decreases (gain increases) with frequency above the cutoff; low frequencies are blocked because the capacitor's reactance is high (Xc is large). Thus the claim that a high-pass “heavily attenuates the higher frequencies” contradicts the definition and expected Bode magnitude plot. Step-by-Step Solution: Recall: high-pass ⇒ pass highs, reject lows. At low f: Xc is large ⇒ small output across R (attenuated). At high f: Xc is small ⇒ output approaches input (passed). Therefore, the statement is false. Verification / Alternative check: Plot |H(jω)| for the canonical high-pass transfer H(jω) = jωRC / (1 + jωRC). The magnitude tends to 1 as ω → ∞ and to 0 as ω → 0. Why Other Options Are Wrong: True only at cutoff / active filters only: the trend of passing highs and attenuating lows holds generally; implementation details do not reverse it. Common Pitfalls: Mixing up low-pass and high-pass; confusing amplitude with phase behavior around cutoff. Final Answer: Incorrect

Q: Frequency dependence of capacitive reactance: Is the statement “capacitive reactance (Xc) is inversely proportional to frequency” correct for linear capacitors, where Xc = 1 / (2π f C)?

Introduction / Context: Reactance quantifies opposition to AC current due to energy storage in reactive elements. For capacitors, reactance falls as frequency rises, which is why capacitors can pass high-frequency components more easily than low-frequency ones. This question confirms the textbook formula and its proportionality. Given Data / Assumptions: Ideal capacitor of capacitance C farads. Sinusoidal steady-state analysis. Xc defined as the magnitude of the reactive part of impedance. Concept / Approach: The standard relation is Xc = 1 / (2π f C). As f increases, the denominator grows, making Xc smaller. This inverse proportionality underpins the behavior of coupling capacitors, differentiators, and high-pass filters, where high-frequency signals encounter less opposition. Step-by-Step Solution: Write the formula: Xc = 1 / (2π f C). Identify proportionality: Xc ∝ 1/f for fixed C. Conclude that the statement is correct for linear capacitors. Verification / Alternative check: Measure current through a capacitor at two frequencies with the same applied voltage. The higher-frequency current increases in proportion to f, consistent with Xc declining as 1/f. Why Other Options Are Wrong: Only true above cutoff / only for electrolytics: the relation is universal for ideal capacitors across frequencies where the component remains linear and parasitics are negligible. Common Pitfalls: Confusing reactance with impedance magnitude when resistive elements are in play; ignoring ESR and ESL at very high frequencies where the simple model breaks down. Final Answer: Correct

Question 1

Capacitor opposition to AC: Is the correct term for the opposition that a capacitor presents to alternating current known as capacitive reactance rather than reluctance?

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Introduction / Context: Terminology matters in circuit theory. The frequency-dependent opposition of capacitors and inductors in AC circuits is described by reactance, while 'reluctance' belongs to magnetic circuit analysis and has different units and meaning. Mixing these terms leads to conceptual errors. Given Data / Assumptions: Sinusoidal steady state analysis context. Ideal capacitor behavior in linear circuits. Standard definitions of impedance and reactance. Concept / Approach: Capacitive reactance is Xc = 1 / (2 * pi * f * C), expressed in ohms, and decreases with frequency. Together with resistance R, it forms complex impedance Z = R − j * Xc in an RC series. 'Reluctance' describes magnetic opposition to flux: ℛ = l / (μ * A), with units of A/Wb, not ohms. Therefore, calling a capacitor's opposition 'reluctance' is incorrect; the correct term is capacitive reactance. Step-by-Step Solution: Identify the phenomenon: AC opposition in electrical circuits.Recall the formula Xc = 1 / (2 * pi * f * C).Contrast with magnetic circuits: reluctance relates to flux, not current.Conclude that the proper term is capacitive reactance. Verification / Alternative check: Textbook impedance triangles and phasor diagrams use R and Xc to compute current and phase angle in RC networks, confirming terminology and equations. Why Other Options Are Wrong: Incorrect / DC / resonance qualifiers do not change terminology.Reluctance equals reactance: false; they belong to different physical models. Common Pitfalls: Confusing resistance, reactance, and impedance; importing magnetic-circuit terms into electric-circuit contexts. Final Answer: Correct

Question 2

Parallel connection rule: Do capacitors connected in parallel produce a total capacitance greater than any individual capacitor in the group (i.e., they add algebraically)?

Accepted Answer

Introduction / Context: Combining capacitors allows designers to achieve target values and performance. In parallel, plate areas effectively add, increasing the ability to store charge per volt. Understanding series versus parallel rules prevents design and troubleshooting mistakes in filters, timing networks, and decoupling arrays. Given Data / Assumptions: Ideal capacitors, negligible parasitics for first-order rules. Two or more capacitors connected in parallel across the same nodes. Same voltage across all elements. Concept / Approach: For parallel capacitors, the equivalent capacitance is C_eq = C1 + C2 + ... + Cn. Because the sum of positive values exceeds each individual value, C_eq is greater than any single capacitor. This contrasts with series connection, where charges equalize and 1 / C_eq = 1 / C1 + 1 / C2 + ... leading to a smaller equivalent than the smallest element. Step-by-Step Solution: Define Q_total = Q1 + Q2 + ... under the same voltage V.Since Qi = Ci * V, then Q_total = (C1 + C2 + ...) * V.Therefore C_eq = Q_total / V = C1 + C2 + ...Hence C_eq exceeds any single Ci. Verification / Alternative check: Bench measurements with an LCR meter confirm that paralleling parts sums their capacitances within tolerances; designers often parallel different dielectrics to broaden frequency performance. Why Other Options Are Wrong: Incorrect: contradicts the formula.Requirements for identical type, electrolytic only, or high frequency are unnecessary for the basic rule (parasitics aside). Common Pitfalls: Forgetting voltage rating limits when paralleling; neglecting ESR/ESL which influence high-frequency behavior though not the DC-value sum. Final Answer: Correct

Question 3

Series capacitors voltage division: In a purely series capacitive chain, does the smallest capacitance experience the largest voltage drop for a given charge?

Accepted Answer

Introduction / Context: Voltage division across series capacitors follows the inverse relation to capacitance values. Because series capacitors carry the same charge, the voltage across each is determined by V_i = Q / C_i. This principle is essential when designing capacitive dividers or placing series capacitors across high-voltage rails for rating purposes. Given Data / Assumptions: Ideal series connection, same current and hence same charge magnitude on each element. Negligible leakage and identical steady-state charge distribution. Sinusoidal or DC conditions where reactive steady state is reached. Concept / Approach: In series, Q is common to all capacitors. Therefore V_i = Q / C_i: a smaller C produces a larger V_i. The total applied voltage divides inversely with capacitance values. Designers often include voltage-balancing resistors in high-voltage stacks to mitigate imbalance due to leakage and tolerances. Step-by-Step Solution: State series rule: same charge on all capacitors.Compute V_i = Q / C_i for each element.Compare voltages: the smallest C has the largest V_i.Therefore, the statement is correct. Verification / Alternative check: AC phasor analysis also shows larger magnitude across the smaller capacitance because Xc = 1 / (2 * pi * f * C) is larger, yielding a larger share of the phasor drop. Why Other Options Are Wrong: Incorrect or conditional options add constraints that are not required by the ideal series rule. Common Pitfalls: Ignoring leakage and tolerance which can distort division in practice; forgetting to add balancing networks in HV applications. Final Answer: Correct

Question 4

Impedance of an RC series circuit: Is it correct that the magnitude of impedance is (R^2 + Xc^2)^(1/2), where Xc = 1 / (2 * pi * f * C), rather than (R^2 + C^2)^(1/2)?

Accepted Answer

Introduction / Context: Impedance generalizes resistance to AC by incorporating phase and frequency dependence. For a series RC, the impedance vector is Z = R − j * Xc. Confusing the capacitance C (in farads) with the capacitive reactance Xc (in ohms) leads to dimensional errors and wrong results. Given Data / Assumptions: Sinusoidal steady state at frequency f. Ideal R and C elements. Magnitude operation on complex impedance. Concept / Approach: Capacitive reactance is Xc = 1 / (2 * pi * f * C). The series impedance is Z = R − j * Xc. The magnitude is |Z| = sqrt(R^2 + Xc^2). Substituting C directly into sqrt(R^2 + C^2) is incorrect because C has units of farads, not ohms. Proper dimensional analysis reinforces the correct formula. Step-by-Step Solution: Write Z = R − j * (1 / (2 * pi * f * C)).Compute |Z| = sqrt(R^2 + (1 / (2 * pi * f * C))^2).Note the unit consistency: both terms under the square root are in ohms squared.Conclude that (R^2 + C^2)^(1/2) is dimensionally and numerically wrong. Verification / Alternative check: Phasor diagrams and impedance triangles used in laboratory measurements match the sqrt(R^2 + Xc^2) expression and produce correct current and phase predictions. Why Other Options Are Wrong: Conditional statements (high frequency, R = 0, specific capacitor types) do not change the fundamental formula. Common Pitfalls: Plugging capacitance directly where reactance belongs; forgetting the frequency dependence of Xc. Final Answer: Correct

Question 5

True power in an RC series circuit: Is the real power dissipated given by P_R = I^2 * R, with the capacitor contributing zero average real power?

Accepted Answer

Introduction / Context: In AC circuits, resistors dissipate real power as heat, while ideal capacitors and inductors exchange reactive power with the source, averaging to zero over a complete cycle. Understanding this distinction is essential for power factor correction and thermal design. Given Data / Assumptions: Sinusoidal steady state with RMS current I. Ideal capacitor with no dielectric loss (no ESR for first-order analysis). Series RC topology. Concept / Approach: Real or true power is P = V_RMS * I_RMS * cos(phi). For an RC series, the phase angle phi satisfies cos(phi) = R / |Z|. Equivalently, the real power equals I^2 * R because only the resistor dissipates energy. The capacitor's voltage and current are 90 degrees out of phase, so its average power over a cycle is zero, though it stores and returns energy within each cycle. Step-by-Step Solution: Write Z = R − j * Xc and compute I = V / |Z|.Real power: P = I^2 * R (since the resistive part alone dissipates).Reactive exchange in C: Q_C = V * I * sin(phi) with zero average real power in an ideal capacitor.Therefore, P_R = I^2 * R is the correct expression. Verification / Alternative check: Wattmeter readings in lab confirm that power scales with the resistive element; adding series capacitance at fixed current does not increase true power, but changes phase and current at fixed voltage. Why Other Options Are Wrong: Incorrect: contradicts basic AC power relations.DC-only or cutoff-only qualifiers are unnecessary; the expression holds across frequencies for ideal components. Common Pitfalls: Confusing apparent power S = V * I with real power P; ignoring capacitor ESR which introduces small real-loss in practice. Final Answer: Correct

Question 6

Fundamental definition of capacitance: Is it correct to say that “the unit of capacitance is the farad; one farad stores one coulomb of charge,” without stating the voltage reference (i.e., per volt), given that 1 coulomb ≈ 6.24 × 10^18 electrons?

Accepted Answer

Introduction / Context: Capacitance quantifies how much electric charge a capacitor stores per unit voltage across its plates. The SI unit is the farad (F). A common misconception is to define a farad simply as “one coulomb of charge,” which omits the critical dependence on voltage. This question tests whether the definition is understood precisely in terms of charge-per-volt and clarifies the role of the familiar conversion 1 coulomb ≈ 6.24 × 10^18 electrons. Given Data / Assumptions: Capacitance C relates to charge Q and voltage V via C = Q / V. 1 coulomb ≈ 6.24 × 10^18 elementary charges (electrons). No leakage, ideal dielectric, and small-signal conditions are assumed for the definition. Concept / Approach: The correct definition is: a capacitor has capacitance C if it stores charge Q when a voltage V is applied, with C = Q / V. Therefore, 1 farad means the device stores 1 coulomb when 1 volt is applied (i.e., 1 F = 1 C per V). Saying “one farad stores one coulomb of charge” is incomplete and misleading because it ignores the required voltage reference of 1 volt. The electron count merely restates what 1 coulomb represents and does not alter the per-volt relationship. Step-by-Step Solution: Define capacitance: C = Q / V. Interpret 1 F: Q = 1 C when V = 1 V (thus 1 F = 1 C/V). Evaluate the statement: it omits “per volt,” so it is not a correct definition. Conclude: the statement is incorrect as written. Verification / Alternative check: Dimensional analysis confirms farad units are coulombs per volt (C/V). Any valid definition must include the voltage term. Manufacturer datasheets and textbooks consistently state 1 F = 1 C per V, reinforcing the necessity of the per-volt reference. Why Other Options Are Wrong: Correct: would only be true if it explicitly read “one coulomb per volt.” Depends on the capacitor's ESR only: ESR affects loss, not the definition of units. Cannot be determined: the SI definition is precise; no extra data are required. Common Pitfalls: Forgetting that stored charge scales with voltage; assuming “one coulomb” is absolute rather than per volt; confusing unit definitions with practical limitations like leakage or ESR. Final Answer: Incorrect

Question 7

RC series phase relationship (magnitude vs. sign): For a series R–C circuit, is the phase angle magnitude given by |φ| = arctan(Xc / R) = arctan(Vc / Vr), with the understanding that the capacitor causes a negative (lagging) angle for the total volt…

Accepted Answer

Introduction / Context: Phase angle in AC circuits relates voltages and currents through impedance. For a series R–C circuit, engineers often use impedance triangles and phasors to express the relationship between resistive and reactive drops. This question checks whether the common magnitude formula for the phase angle in a series R–C circuit is correctly stated, and clarifies the sign convention for capacitive behavior. Given Data / Assumptions: Series R (ohms) and C (farads) driven by a sinusoidal source. Capacitive reactance Xc = 1 / (2π f C). Phasor notation with current as the reference; ideal components; steady state. Concept / Approach: In series R–C, the impedance is Z = R − jXc. The phase angle φ between total voltage and current satisfies tan(φ) = −Xc / R. The magnitude is |φ| = arctan(Xc/R). Because the drop across the capacitor lags the current by 90°, the overall angle is negative (capacitive). The ratio of drops also reflects this: Vc / Vr = Xc / R, consistent with the triangle of voltages in a series path. Step-by-Step Solution: Write Z = R − jXc and tan(φ) = Im(Z)/Re(Z) with sign = −Xc/R. Take magnitude: |φ| = arctan(Xc/R). Use divider: Vc = I*Xc and Vr = I*R ⇒ Vc/Vr = Xc/R. Conclude equality of |φ| and arctan(Vc/Vr), with φ negative for capacitive circuits. Verification / Alternative check: Construct the right triangle with sides R (adjacent) and Xc (opposite). The angle at the series node has tangent Xc/R (magnitude). Sign is assigned by the reactive sign (capacitive negative). Measurements with a scope confirm Vc leads Vr in quadrature, matching the geometry. Why Other Options Are Wrong: Incorrect: conflicts with standard phasor analysis. Only true at resonance: R–C series has no resonance without an L; the relation holds at all f. Cannot be used unless R >> Xc: no such restriction exists for the identity. Common Pitfalls: Forgetting the negative sign for capacitive circuits; mixing up magnitude and signed angle; confusing Vc/Vr with Vr/Vc. Final Answer: Correct

Question 8

Basic maintenance check: Can an ohmmeter (resistance meter) be used for a simple go/no-go test of a capacitor (e.g., to detect a shorted or open device) even if it cannot measure true capacitance accurately?

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Introduction / Context: Technicians often need a quick field test to identify gross capacitor faults without a dedicated LCR meter. An ohmmeter cannot report capacitance value directly, but it can still reveal shorts, opens, or severe leakage by observing the time-varying resistance as the capacitor charges from the meter's internal source. Given Data / Assumptions: Handheld ohmmeter provides a small DC voltage for measurement. We look for qualitative behavior, not precise capacitance. Capacitor is discharged before testing; polarity respected for electrolytics. Concept / Approach: When connected, a healthy capacitor initially appears as a low resistance (inrush as it charges) and then the indicated resistance rises toward a high value as the charge current decays. A shorted capacitor reads near-zero resistance and stays there; an open capacitor shows no charging action (reading may remain infinite). Excessive leakage yields a steady mid-range resistance. Thus an ohmmeter provides a practical go/no-go assessment. Step-by-Step Solution: Discharge the capacitor safely. Connect the ohmmeter and watch the resistance climb in time for a good part. Interpret steady zero as short, infinite as open, mid-range as leaky. Use a proper LCR meter for quantitative capacitance/ESR if needed. Verification / Alternative check: Technician guides describe this classic method; bench LCR readings can confirm findings. Oscilloscope with a known resistor can also time the charge curve (V(t) = V_s * (1 − e^(−t/RC))). Why Other Options Are Wrong: Incorrect: discards a widely used diagnostic technique. Only true for electrolytics / above 1 μF: the behavior occurs for any capacitor; larger C simply makes the effect more visible on slow meters. Common Pitfalls: Failing to discharge first; reversing polarity on electrolytics; concluding “bad” solely from a fast change on a digital meter with autoranging. Final Answer: Correct

Question 9

Terminology origin: The term “integrator” used for certain RC/operational-amplifier circuits derives from calculus (integration), not from trigonometry. Is this statement accurate?

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Introduction / Context: Signal-processing terms often come directly from mathematical operations. An “integrator” produces an output proportional to the time integral of the input. In electronics, RC and op-amp circuits implement this operation over a bandwidth, hence the name. The question probes whether the term is rooted in calculus or trigonometry. Given Data / Assumptions: Linear time-invariant system behavior within a frequency range. Integrator approximates y(t) = ∫ x(t) dt scaled by a constant. We refer to continuous-time integration from calculus. Concept / Approach: Integration is a calculus operation (antiderivative/area under a curve). While sinusoids are convenient test signals and appear in trigonometry, the integrator's definition does not rely on trigonometric identities; it is defined by the operator s^−1 in Laplace terms. RC and active integrators realize transfer functions proportional to 1/(RC s) in the intended region. Step-by-Step Solution: Define integrator: output proportional to the integral of the input. Connect to math: integral is a calculus construct. Relate to circuits: ideal transfer H(s) ∝ 1/s; practical RC/op-amp approximations implement this over a band. Conclude: the term originates from calculus, not trigonometry. Verification / Alternative check: Textbooks present the integrator via differential equations and Laplace transforms. Trigonometric test signals aid analysis but do not define the operator itself. Why Other Options Are Wrong: Incorrect: contradicts standard mathematical origin. Depends on sinusoidal waveforms: the operator applies to any input, not just sinusoids. Only true for digital integrators: both analog and digital integrators stem from the same calculus concept. Common Pitfalls: Confusing the prevalence of sine waves in testing with the mathematical source of the term; equating “integrator” with “low-pass filter” without specifying operating region. Final Answer: Correct

Question 10

RC integrator behavior: A simple series R–C network with the output taken across the capacitor acts as an integrator for a sufficiently high input frequency (f >> 1 / (2π R C)), producing an output proportional to the time integral of the input. Is …

Accepted Answer

Introduction / Context: An “RC integrator” is a foundational analog building block used for pulse shaping, ramp generation, and as part of active filter designs. The classic passive form is a series R followed by a capacitor to ground, with output across the capacitor. Under the right frequency conditions, its output approximates the integral of the input waveform. Given Data / Assumptions: Series resistor R, shunt capacitor C, output across C. Linear components, small-signal regime. Input frequency sufficiently high relative to the corner f_c = 1/(2π R C). Concept / Approach: The transfer function of the passive network is H(s) = 1 / (1 + s R C). For frequencies well above f_c, the magnitude of H(s) becomes small and the capacitor voltage approximates the time integral of the input scaled by 1/(R C). Step inputs yield exponential ramps initially; narrow pulses produce proportional area at the output, which is the essence of integration over the pulse width. Step-by-Step Solution: Write H(s) = 1 / (1 + sRC). For s large (f >> f_c), 1 + sRC ≈ sRC, so H(s) ≈ 1/(sRC). The 1/s factor corresponds to integration; the scaling is 1/(RC). Therefore, the circuit behaves as an integrator in the stated high-frequency limit. Verification / Alternative check: Time-domain check: for a narrow input pulse of area A (volt-seconds), the change in capacitor voltage ΔV ≈ A / (R C), consistent with integration of the input. Why Other Options Are Wrong: Incorrect: contradicts the standard asymptotic behavior. “Only true if R << Xc at all frequencies”: condition is frequency-dependent; at high f, |Xc| becomes small and the approximation holds. “Only true when the input is DC”: integration of DC produces a ramp until limited; the integrator definition is broader and frequency-domain based. Common Pitfalls: Assuming perfect integration at any frequency; ignoring output amplitude droop and finite RC leakage; forgetting loading effects that spoil the ideal 1/s response. Final Answer: Correct

Question 11

Parallel-plate formula check: Capacitance C is directly proportional to the dielectric constant (relative permittivity) and the plate area A, and inversely proportional to the plate separation d (C = ε * A / d). Is the claim “directly proportional t…

Accepted Answer

Introduction / Context: The geometry of a capacitor determines its capacitance. For the common parallel-plate model, a simple formula relates material properties and dimensions. This question tests whether the stated proportionalities match the well-known relationship used in design and analysis. Given Data / Assumptions: Parallel-plate approximation with negligible fringing. Dielectric fully fills the region between plates. Relative permittivity ε_r multiplies ε_0 to yield ε. Concept / Approach: The canonical formula is C = ε * A / d, where ε = ε_r * ε_0. Capacitance increases with plate area A and dielectric constant ε, and decreases as the separation d increases. Therefore, any statement claiming direct proportionality to distance and inverse to area is the exact opposite and is incorrect for this standard geometry. Step-by-Step Solution: State the formula: C = ε * A / d. Identify proportionalities: C ∝ ε and C ∝ A; C ∝ 1/d. Compare with the claim: it reverses the roles of A and d. Conclude the claim is incorrect for parallel plates. Verification / Alternative check: Dimensional and physical intuition: larger area gives more field-coupled region (higher C); greater distance weakens the field coupling (lower C). Measurements of variable capacitors confirm these trends. Why Other Options Are Wrong: Correct only for cylindrical/fringing cases: the sign of dependence does not reverse for standard configurations; while exact formulas differ, C does not increase with distance in such ideal models. Common Pitfalls: Mixing up proportionalities; overlooking that dielectric constant boosts capacitance by concentrating the electric field (higher ε reduces effective field for the same charge, enabling more Q per volt). Final Answer: Incorrect

Question 12

Filter intuition check: Does a high-pass RC filter strongly attenuate the higher frequencies (rather than the lower ones) in its passband/stopband behavior?

Accepted Answer

Introduction / Context: Filters shape spectra by passing some frequency ranges while attenuating others. A high-pass filter is designed to pass high frequencies and reject low frequencies. This question probes whether that definition is being applied correctly in the context of basic RC networks. Given Data / Assumptions: First-order RC high-pass implementation (series C, shunt R, output across R). Ideal components; small-signal linear operation. Standard −3 dB cutoff at f_c = 1/(2πRC). Concept / Approach: In a high-pass filter, attenuation decreases (gain increases) with frequency above the cutoff; low frequencies are blocked because the capacitor's reactance is high (Xc is large). Thus the claim that a high-pass “heavily attenuates the higher frequencies” contradicts the definition and expected Bode magnitude plot. Step-by-Step Solution: Recall: high-pass ⇒ pass highs, reject lows. At low f: Xc is large ⇒ small output across R (attenuated). At high f: Xc is small ⇒ output approaches input (passed). Therefore, the statement is false. Verification / Alternative check: Plot |H(jω)| for the canonical high-pass transfer H(jω) = jωRC / (1 + jωRC). The magnitude tends to 1 as ω → ∞ and to 0 as ω → 0. Why Other Options Are Wrong: True only at cutoff / active filters only: the trend of passing highs and attenuating lows holds generally; implementation details do not reverse it. Common Pitfalls: Mixing up low-pass and high-pass; confusing amplitude with phase behavior around cutoff. Final Answer: Incorrect

Question 13

Frequency dependence of capacitive reactance: Is the statement “capacitive reactance (Xc) is inversely proportional to frequency” correct for linear capacitors, where Xc = 1 / (2π f C)?

Accepted Answer

Introduction / Context: Reactance quantifies opposition to AC current due to energy storage in reactive elements. For capacitors, reactance falls as frequency rises, which is why capacitors can pass high-frequency components more easily than low-frequency ones. This question confirms the textbook formula and its proportionality. Given Data / Assumptions: Ideal capacitor of capacitance C farads. Sinusoidal steady-state analysis. Xc defined as the magnitude of the reactive part of impedance. Concept / Approach: The standard relation is Xc = 1 / (2π f C). As f increases, the denominator grows, making Xc smaller. This inverse proportionality underpins the behavior of coupling capacitors, differentiators, and high-pass filters, where high-frequency signals encounter less opposition. Step-by-Step Solution: Write the formula: Xc = 1 / (2π f C). Identify proportionality: Xc ∝ 1/f for fixed C. Conclude that the statement is correct for linear capacitors. Verification / Alternative check: Measure current through a capacitor at two frequencies with the same applied voltage. The higher-frequency current increases in proportion to f, consistent with Xc declining as 1/f. Why Other Options Are Wrong: Only true above cutoff / only for electrolytics: the relation is universal for ideal capacitors across frequencies where the component remains linear and parasitics are negligible. Common Pitfalls: Confusing reactance with impedance magnitude when resistive elements are in play; ignoring ESR and ESL at very high frequencies where the simple model breaks down. Final Answer: Correct

Question 14

RC parallel impedance check: For a resistor R in parallel with a capacitor C at angular frequency ω, is the magnitude of the equivalent impedance given by |Z_p| = 1 / sqrt( (1/R)^2 + (ω C)^2 )?

Accepted Answer

Introduction / Context: Computing the equivalent impedance of parallel networks is essential for filter and bias network design. This question examines whether a commonly used closed-form expression for the magnitude of the impedance of an R || C network is correctly stated. Given Data / Assumptions: Linear components; sinusoidal steady state. Admittance addition for parallels: Y_total = Y_R + Y_C. Angular frequency ω = 2π f. Concept / Approach: For the parallel combination, Y_R = 1/R (conductance) and Y_C = jωC (susceptance). The magnitude of total admittance is |Y| = sqrt( (1/R)^2 + (ωC)^2 ). The magnitude of impedance is |Z| = 1 / |Y|, giving |Z| = 1 / sqrt( (1/R)^2 + (ωC)^2 ). This result is general for any ω and does not require small- or large-signal approximations beyond linearity. Step-by-Step Solution: Form admittance: Y = 1/R + jωC. Compute magnitude: |Y| = sqrt( (1/R)^2 + (ωC)^2 ). Invert to get impedance magnitude: |Z| = 1 / |Y|. Conclude the given expression is correct. Verification / Alternative check: Check limiting cases: as ω → 0, |Z| → R; as ω → ∞, |Z| → 0, consistent with a capacitor short at high frequency. Why Other Options Are Wrong: Incorrect: contradicts basic admittance arithmetic. Valid only when ωRC << 1 or >> 1: the expression holds for all ω under linear assumptions. Common Pitfalls: Mixing series and parallel formulas; forgetting to invert admittance magnitude to get impedance magnitude. Final Answer: Correct

Question 15

RC low-pass intuition: For a first-order RC low-pass filter where the output is taken across the capacitor, does the output amplitude decrease as the input frequency increases (i.e., higher f ⇒ lower |Vout|/|Vin|)?

Accepted Answer

Introduction / Context: First-order RC low-pass filters are fundamental in smoothing, anti-aliasing, and bandwidth limiting. The hallmark of a low-pass is that it passes low frequencies with little attenuation and attenuates high frequencies. This question verifies that intuition for the common configuration with output across the capacitor. Given Data / Assumptions: Series R, shunt C, output measured across C. Ideal linear components, sinusoidal steady state. Cutoff frequency f_c = 1 / (2π R C). Concept / Approach: The transfer function is H(jω) = 1 / (1 + jωRC). Its magnitude |H(jω)| = 1 / sqrt(1 + (ωRC)^2). As frequency increases, the denominator grows, so the magnitude decreases. Therefore, the output amplitude drops with higher input frequency, which is exactly what a low-pass filter is designed to do. Step-by-Step Solution: Write H(jω) = 1 / (1 + jωRC). Compute magnitude: |H| = 1 / sqrt(1 + (ωRC)^2). Observe monotonic decrease with increasing ω. Conclude the statement is correct for this configuration. Verification / Alternative check: At ω = 0, |H| = 1 (no attenuation). At ω = ω_c, |H| = 1/√2 (−3 dB). As ω → ∞, |H| → 0. Scope measurements on a simple RC confirm the roll-off behavior. Why Other Options Are Wrong: Incorrect: contradicts the low-pass definition. Inductive loads / frequency constraint options do not alter the basic monotonic trend for the ideal RC low-pass with output across C. Common Pitfalls: Confusing the low-pass with the high-pass (output taken across R); forgetting that loading by the next stage can shift the effective corner frequency. Final Answer: Correct

Question 16

Capacitor technology comparison: considering typical constructions and applications, is the blanket claim that “ceramic capacitors have higher voltage ratings than electrolytic capacitors” correct?

Accepted Answer

Introduction / Context: Designers choose capacitor types based on voltage rating, capacitance per volume, ESR, polarity, and reliability. A common misconception is that one class (ceramic) universally supports higher voltages than another (electrolytic). This question asks whether that blanket statement is correct in practical electronics contexts. Given Data / Assumptions: Comparing mainstream multilayer ceramic capacitors (MLCCs) and aluminum electrolytics. Voltage rating refers to the specified maximum working voltage under standard conditions. No special-purpose niche parts implied (e.g., exotic HV ceramics vs ultra-HV electrolytics). Concept / Approach: Voltage rating is a function of dielectric strength, layer thickness, construction, and package clearance. Aluminum electrolytics commonly span 6.3 V to 450 V (and beyond in some series). MLCCs are widely available from 6.3 V to a few hundred volts; specialty ceramic discs/stacks can reach kV, but so can certain film or electrolytic designs. Therefore, there is no universal ordering in which “ceramic > electrolytic” for voltage rating. The claim, as an absolute rule, is incorrect. Step-by-Step Solution: 1) Identify that both technologies cover overlapping voltage ranges. 2) Note common catalog ranges: electrolytics up to hundreds of volts; MLCCs typically lower, with special series higher. 3) Conclude that a blanket superiority statement is false. Verification / Alternative check: Surveying supplier parametric tables shows electrolytic and ceramic parts offered across broad, overlapping ranges; selection depends on series and manufacturer, not solely on dielectric type. Why Other Options Are Wrong: “True only below 10 V / above 1 kV / depends only on capacitance” each oversimplifies; voltage rating depends on construction and series, not just C value or a fixed threshold. Common Pitfalls: Generalizing from a limited lab kit; ignoring derating and package size effects on rating. Final Answer: Incorrect

Question 17

RC time constant behavior: for a first-order RC response to a step, does the capacitor voltage (or charge) move about 63% of the remaining difference toward its final value over each successive interval of one time constant?

Accepted Answer

Introduction / Context: The first-order RC time constant (tau = R * C) governs how quickly a capacitor charges or discharges toward a new steady state after a step. Understanding the “63% rule” helps engineers estimate settling time without a calculator. Given Data / Assumptions: Single-pole RC circuit driven by a step. Ideal linear components. We consider changes over equal intervals of length tau. Concept / Approach: For a step from V_initial to V_final, the solution is V(t) = V_final + (V_initial - V_final) * exp(-t/tau). Over any interval of duration tau starting at t0, the change toward the final value equals (1 - e^-1) ≈ 0.632 of the remaining difference at t0. This is true for charge and discharge; only the sign of the change differs. Step-by-Step Solution: 1) Write V(t0 + tau) - V_final = (V(t0) - V_final) * e^-1. 2) Therefore, reduction in the remaining gap over that tau is (1 - e^-1) ≈ 0.632 (≈ 63%). 3) This holds for each successive tau, not just the first. Verification / Alternative check: At t = tau from the initial step, the state is ~63% closer to final than at t = 0. Reapplying the same interval later yields the same fractional progress on the then-remaining difference. Why Other Options Are Wrong: “Incorrect” contradicts the exponential law; discharge vs charge and temperature qualifiers do not alter the first-order mathematics. Common Pitfalls: Confusing “63% of total change” with “63% of remaining change” when looking at later intervals; mixing series vs parallel RC forms. Final Answer: Correct

Question 18

Geometry–capacitance relation: with all else equal (dielectric and plate area), does increasing the plate separation reduce the capacitance value of a parallel-plate capacitor?

Accepted Answer

Introduction / Context: Capacitance depends on physical geometry and dielectric properties. For parallel-plate capacitors, plate area, separation, and dielectric constant determine the value. This item checks the inverse relationship between separation and capacitance. Given Data / Assumptions: Parallel-plate model with uniform dielectric. Plate area and dielectric constant fixed. Fringing effects neglected for the basic relation. Concept / Approach: The ideal formula is C = ε * A / d, where ε is permittivity, A is plate area, and d is plate spacing. As d increases, A and ε fixed, C decreases proportionally. This is independent of frequency in the ideal electrostatic model (though real parts exhibit parasitics at high frequency). Step-by-Step Solution: 1) Start from C = ε * A / d. 2) Treat ε and A as constants; vary only d. 3) Increasing d increases the denominator, reducing C. Verification / Alternative check: Doubling spacing halves capacitance in the first-order model; measurements on adjustable capacitors show inverse proportionality for modest changes. Why Other Options Are Wrong: “Incorrect / depends only on frequency / only for electrolytics” each conflicts with the geometric dependence; the relation is technology-agnostic. Common Pitfalls: Forgetting that dielectric constant changes also affect C; over-interpreting fringing at extreme geometries. Final Answer: Correct

Question 19

Capacitive reactance comparison at 60 Hz: given two capacitors, 10 µF and 20 µF, which one has the larger capacitive reactance magnitude Xc in a 60 Hz AC circuit?

Accepted Answer

Introduction / Context: Capacitive reactance determines how strongly a capacitor opposes AC. Understanding how Xc depends on C and frequency helps in filter design and impedance matching. Given Data / Assumptions: Frequency f = 60 Hz. Capacitances: 10 µF and 20 µF. Ideal components; ESR and ESL neglected. Concept / Approach: Capacitive reactance is Xc = 1 / (2 * pi * f * C). For fixed f, Xc is inversely proportional to C. Therefore, the smaller capacitor (10 µF) exhibits the larger reactance magnitude; the larger capacitor (20 µF) has half that reactance. Step-by-Step Solution: 1) Xc(10 µF) = 1 / (2 * pi * 60 * 10e-6) ≈ 265.3 Ω. 2) Xc(20 µF) = 1 / (2 * pi * 60 * 20e-6) ≈ 132.7 Ω. 3) Compare: 265.3 Ω > 132.7 Ω, so 10 µF has larger reactance. Verification / Alternative check: Doubling C halves Xc at fixed frequency, consistent with the formula. Why Other Options Are Wrong: “20 µF larger / equal / independent of C / infinite at 60 Hz” each contradicts Xc = 1/(2*pi*f*C). Common Pitfalls: Mixing resistance with reactance; forgetting to convert microfarads to farads. Final Answer: 10 µF has the larger reactance

Question 20

Charge storage concept: is a capacitor considered “charged” when one plate holds an excess of electrons while the opposite plate holds an equal deficit, producing an electric field between them?

Accepted Answer

Introduction / Context: Capacitors store energy via separated charge on two conductors. Clarifying what “charged” means helps avoid confusion between net device charge and equal/opposite charges on the plates that create voltage. Given Data / Assumptions: Two conductive plates separated by a dielectric. External source applies a potential difference V. Ideal leakage (very high resistance) during the charging interval. Concept / Approach: When a voltage is applied, electrons accumulate on one plate (negative), while electrons are removed from the opposite plate (positive). The charges are equal in magnitude and opposite in sign, producing an electric field in the dielectric. This condition is precisely what we call a “charged capacitor,” whether the source is DC or an instantaneous AC value at a moment in time. Step-by-Step Solution: 1) Apply a voltage V; charges redistribute via the source. 2) One plate gains electrons; the other loses the same amount. 3) The resulting field stores energy U = 0.5 * C * V^2. Verification / Alternative check: Disconnecting the source leaves the charge separation and voltage (subject to leakage), confirming stored energy without continuous current. Why Other Options Are Wrong: Electrolytic/DC/air qualifiers are unnecessary: the definition applies to all capacitors, all dielectrics, and both DC and instantaneous AC states. Common Pitfalls: Thinking a “charged capacitor” must have net device charge; in reality, the net is near zero while plates have equal and opposite charges. Final Answer: Correct

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