Question 1

Low-pass filter behavior — as the input frequency increases beyond the cutoff, the output (across the load) decreases in magnitude. Is this statement correct?

Accepted Answer

Introduction / Context: Frequency response curves characterize how filters pass or attenuate signals versus frequency. A low-pass filter (LPF) passes low-frequency components while attenuating higher frequencies above its cutoff (fc). Given Data / Assumptions: Simple first-order or higher-order LPF topology. Linear, time-invariant behavior around operating point. Concept / Approach: The magnitude response |H(jω)| of an LPF decreases with ω beyond fc. For a first-order RC LPF, |H(jω)| = 1 / sqrt(1 + (ωRC)^2), clearly decreasing as ω increases. Step-by-Step Solution: 1) Identify the filter as low-pass.2) Recognize cutoff frequency fc = 1/(2πRC) for RC case.3) For f ≫ fc, output magnitude falls approximately as 1/(ωRC) per pole.4) Therefore, as input frequency increases, output decreases. Verification / Alternative check: Bode plots show −20 dB/decade per pole beyond cutoff for first-order LPFs. Why Other Options Are Wrong: Statements tying behavior to resonance, DC, or ideal inductors confuse LPFs with band-pass or RLC resonant circuits. Common Pitfalls: Assuming “low-pass” means no attenuation at high frequencies; real filters have finite roll-off and phase shift. Final Answer: Correct

Question 2

Capacitance rule (repaired stem) — in a parallel combination that includes capacitor C4 along with other capacitors, the total capacitance is larger than the value of C4 alone. Is this statement correct?

Accepted Answer

Introduction / Context: Knowing how total capacitance compares to individual branch values helps validate designs quickly. In parallel, capacitances add directly, so the total must exceed each individual branch value. Given Data / Assumptions: Capacitors are connected in parallel, including C4 among them. Capacitors behave linearly in the operating region. Concept / Approach: Parallel rule: CT = C1 + C2 + … + Cn. Since each Ci ≥ 0, CT ≥ any Ci, with strict inequality if at least one additional capacitor is present besides C4. Step-by-Step Solution: 1) Write CT = C4 + Σ(other Ci).2) If any other Ci > 0, then CT > C4.3) Conclude the statement is correct for a genuine parallel group.4) Use this as a quick plausibility check after calculations. Verification / Alternative check: Example: C4 = 10 µF in parallel with 2.2 µF → CT = 12.2 µF > 10 µF. Why Other Options Are Wrong: Identical/AC/ESR caveats are unnecessary for the ideal addition rule; ESR affects losses, not the ideal capacitance sum. Common Pitfalls: Confusing series vs parallel behavior; in series, the total is less than the smallest element, which is the opposite trend. Final Answer: Correct

Question 3

RC series circuit (time-domain KVL): In a single-loop resistor–capacitor series circuit, is the instantaneous source voltage equal to the sum of the instantaneous drops across the capacitor and the resistor, i.e., v_s(t) = v_C(t) + v_R(t)?

Accepted Answer

Introduction / Context: Kirchhoff's Voltage Law (KVL) states that the algebraic sum of voltages around any closed loop is zero at every instant. In a simple RC series loop, this means the applied source voltage must equal the sum of the instantaneous voltage drops across the resistor and the capacitor. The statement v_s(t) = v_C(t) + v_R(t) is therefore a direct application of KVL in the time domain. Given Data / Assumptions: Single source, single-loop series circuit with components R and C. Ideal lumped components, no parasitic elements or mutual coupling. Voltages are instantaneous time-domain quantities (not merely magnitudes). Concept / Approach: KVL is a fundamental conservation law resulting from the conservative nature of electrostatic fields in lumped circuits. For a series RC, the same current i(t) flows through both elements. The resistor drop is v_R(t) = i(t) * R, while the capacitor drop is v_C(t) related to charge q(t) by v_C(t) = q(t) / C and i(t) = dq/dt. Regardless of the excitation (step, sinusoid, pulse), KVL holds instant by instant: v_s(t) = v_R(t) + v_C(t). Step-by-Step Solution: Write KVL: v_s(t) − v_R(t) − v_C(t) = 0.Substitute element laws: v_R(t) = i(t) * R and i(t) = C * dv_C/dt.Therefore v_s(t) = R * C * dv_C/dt + v_C(t).This identity holds for any excitation waveform when lumped-circuit assumptions are valid. Verification / Alternative check: For a DC step, v_C(t) rises from 0 toward V, v_R(t) decays from V toward 0, and their instantaneous sum remains V at all times, confirming KVL. For AC sinusoidal excitation, the phasor relation is Ṽs = ṼR + ṼC; note that magnitudes do not add arithmetically, but the complex phasors do. Why Other Options Are Wrong: Incorrect / Only DC steady state: KVL applies at all times, not only at steady state.Only at resonance: RC circuits have no resonance; KVL is not conditional.Cannot be determined: Directly determined by KVL. Common Pitfalls: Confusing instantaneous addition with RMS magnitude addition; in AC, phasor vector addition is required. Saying magnitudes add is wrong, but the time-domain sum is always valid. Final Answer: Correct

Question 4

Applications of capacitors in electronics: Are capacitors commonly employed as filters for smoothing, bypassing, and decoupling unwanted signal components?

Accepted Answer

Introduction / Context: Capacitors are foundational components in analog, digital, and power electronics. One of their most frequent roles is filtering: shaping frequency content by shunting or blocking undesirable components of a signal or supply. This includes power-supply smoothing, AC coupling, RF bypassing, and decoupling at integrated-circuit pins. Given Data / Assumptions: Typical electronic systems with DC rails overlaid by ripple/noise. Signal paths where certain frequency bands must be reduced or blocked. Linear time-invariant behavior of ideal capacitors in the relevant bands. Concept / Approach: Capacitive reactance Xc = 1 / (2 * pi * f * C) decreases as frequency increases. Thus, at high frequencies a capacitor presents a low impedance path to ground, bypassing noise (decoupling). In RC low-pass filters, the capacitor shunts high-frequency components, smoothing rectifier ripple. In coupling networks, a series capacitor blocks DC while passing AC above a chosen cutoff frequency f_c = 1 / (2 * pi * R * C). Step-by-Step Solution: Identify the filtering need: suppress ripple/noise or set a bandwidth.Choose RC topology: low-pass, high-pass, or with active elements for precise shaping.Select C to set cutoff with surrounding R (f_c formula above).Place decoupling capacitors close to IC power pins to reduce supply impedance at high frequencies. Verification / Alternative check: Power integrity measurements show reduced rail ripple and faster transient response with proper bulk and high-frequency capacitors (e.g., electrolytic plus ceramic near loads). Bode plots of RC networks confirm attenuation above or below f_c as required. Why Other Options Are Wrong: Incorrect: contradicts widespread practice.Only DC or only HV: capacitors are used in DC, AC, RF, and mixed-signal domains.Only with inductors: LC filters exist, but RC filters and bypass caps operate effectively without inductors. Common Pitfalls: Ignoring ESR/ESL at high frequency; not using multiple values in parallel to cover wide bands; poor layout that adds inductance. Final Answer: Correct

Question 5

Capacitor charging mechanism: During charging of an ideal capacitor, does conduction current flow through the dielectric material itself, or is the observed current due to displacement current and external conduction?

Accepted Answer

Introduction / Context: A common misconception is that when a capacitor charges, electrons literally traverse the dielectric. In an ideal capacitor, the dielectric is an insulator; no conduction current passes through it. Yet a measurable current exists in the external circuit during charging and discharging, which is explained by time-varying electric fields and displacement current concepts. Given Data / Assumptions: Ideal capacitor with insulating dielectric between plates. Charging from a source through some series resistance or network. Lumped-circuit, quasi-static operation. Concept / Approach: The conduction path is external: electrons accumulate on one plate while leaving the other, establishing an electric field in the dielectric. Maxwell's displacement current term accounts for the continuity of current: as the electric field changes, a displacement current density exists in the dielectric without charge carriers crossing it. In real dielectrics, tiny leakage currents may exist, but they are not the primary charging mechanism and are ideally negligible. Step-by-Step Solution: Apply KCL at a plate: the line current equals the rate of change of stored charge, i = C * dv/dt.Recognize that within the dielectric, displacement current density equals epsilon * dE/dt.Note that no electron drift through the dielectric is required for charge to build on plates.Conclude that saying current flows through the dielectric is incorrect for ideal capacitors. Verification / Alternative check: Open-circuit steady state shows zero conduction current through the dielectric while a voltage remains across the plates, confirming the insulating behavior. Leakage specs (insulation resistance) quantify non-ideal conduction, typically very small. Why Other Options Are Wrong: Correct: contradicts ideal behavior.Only at high frequency or specific types: even then, the principal mechanism is displacement current; conduction implies loss. Common Pitfalls: Confusing external line current with through-the-dielectric conduction; ignoring leakage versus ideal behavior; misinterpreting electrolytic construction. Final Answer: Incorrect

Question 6

Energy storage principle: Does a capacitor store electric energy primarily in the electric field established between its plates?

Accepted Answer

Introduction / Context: Understanding where and how energy is stored in passive components is fundamental to circuit design. Capacitors and inductors are complementary: capacitors store energy in electric fields, while inductors store energy in magnetic fields. Recognizing this helps explain transient behavior, resonance, and filter dynamics. Given Data / Assumptions: Ideal parallel-plate capacitor viewpoint. Uniform field approximation within the dielectric. Negligible leakage and series resistance for energy accounting. Concept / Approach: The energy stored in a capacitor is E = 0.5 * C * v^2. This energy physically resides in the electric field in the dielectric volume between the plates. Increasing voltage strengthens the field (E-field magnitude), increasing stored energy. Geometry and dielectric constant determine C, hence energy density for a given voltage. Step-by-Step Solution: Charge the plates to voltage v; equal/opposite charges reside on facing surfaces.An electric field forms across the dielectric, with energy density proportional to epsilon * E^2 / 2.Total energy integrates to 0.5 * C * v^2 across the field volume.Removing the source leaves the energy stored until discharged through a load or loss. Verification / Alternative check: Transient RC behavior releases the stored energy into the resistor as heat when discharging, consistent with energy conservation. Field simulations show energy concentrated within the dielectric region. Why Other Options Are Wrong: Incorrect: contradicts basic electromagnetics.Only DC / only with inductors / only at resonance: capacitor field energy exists whenever a voltage is present, regardless of AC or DC, and does not require inductors or resonance. Common Pitfalls: Assuming energy is stored on the plates rather than in the field; conflating charge quantity with stored energy magnitude. Final Answer: Correct

Question 7

Time-constant insight: Is it accurate to say a capacitor is fully charged in 4 time constants (4τ) in a first-order RC charging circuit?

Accepted Answer

Introduction / Context: First-order RC charging follows an exponential approach to the final value. Because exponentials are asymptotic, the capacitor voltage gets closer and closer to the source voltage but mathematically reaches it only as time approaches infinity. Practical rules of thumb use percentages at specific multiples of the time constant τ = R * C. Given Data / Assumptions: Standard step response v_C(t) = V * (1 − e^(−t/τ)). Definition of τ = R * C. Engineering tolerance for 'fully charged' is approximate, not exact. Concept / Approach: At t = 1τ, v_C ≈ 63%; 2τ ≈ 86%; 3τ ≈ 95%; 4τ ≈ 98%; 5τ ≈ 99.3%. Many practitioners call 5τ effectively full for practical purposes. Therefore, the claim that the capacitor is fully charged in 4τ is inaccurate; at 4τ it is close (about 98%), but not fully, and 5τ is the more common benchmark for 'essentially charged.' Step-by-Step Solution: Compute v_C(4τ) = V * (1 − e^(−4)) ≈ 98% of V.Compute v_C(5τ) = V * (1 − e^(−5)) ≈ 99.3% of V.Note asymptotic approach: exact equality requires infinite time.Conclude the statement is incorrect as written. Verification / Alternative check: Oscilloscope measurements of step response match the exponential curve and common 5τ guideline used in design texts. Why Other Options Are Wrong: Correct: contradicts the asymptotic nature of the response.Other qualifiers (sinusoidal, R = 0) do not fix the misstatement; the 4τ figure is still not 'fully charged.' Common Pitfalls: Equating 'nearly full' with 'fully' without specifying tolerance; ignoring that many specifications require explicit percentage thresholds. Final Answer: Incorrect

Question 8

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?

Accepted Answer

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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?

Accepted Answer

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 16

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

Accepted Answer

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 17

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 18

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 19

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

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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 20

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

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