Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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

Question 8

Discharged state identification: immediately after being fully discharged and left at equilibrium, what charge condition best describes the plates of an ideal capacitor?

Accepted Answer

Introduction / Context: Understanding the “discharged” condition helps with safe handling, measurement, and modeling initial conditions in transient analysis. This item asks what the plate charges are when an ideal capacitor has been fully discharged and allowed to settle. Given Data / Assumptions: Ideal capacitor model without dielectric absorption (DA) or leakage artifacts. Fully discharged via a low-impedance path to equalize potentials. Measurement after equilibrium is reached. Concept / Approach: In the ideal model, discharging removes the separated charge that created the electric field. With both plates at the same potential, there is no electric field in the dielectric; hence no stored energy and no separated plate charge. The device’s net charge is effectively zero, and so is the charge on each plate in this simplified view. Step-by-Step Solution: 1) Short the plates to the same potential to remove separated charge. 2) Wait for currents to cease; with ΔV = 0, E = 0 in the dielectric. 3) Conclude that separated plate charge is zero in the ideal case. Verification / Alternative check: Energy check: U = 0.5 * C * V^2; with V = 0, stored energy is zero, consistent with no separated charge. Why Other Options Are Wrong: Unequal/equal-opposite charges imply a nonzero field; “only negative charge” is nonsensical; “always a small DC charge” confuses real-world dielectric absorption with the ideal model used here. Common Pitfalls: Assuming DA “memory” is always present; ideal textbook questions typically ignore DA unless stated. Final Answer: No charge on either plate (zero net and zero separated charge)

Question 9

RC dynamics shape: the time evolution (rate of charge or discharge) of an ideal first-order capacitor circuit with a step input follows which mathematical function class?

Accepted Answer

Introduction / Context: First-order circuits with one energy-storage element (RC) respond to steps in a characteristic way. Identifying the function type is foundational for predicting settling time, bandwidth, and filter behavior. Given Data / Assumptions: Single-pole RC circuit, step input. Ideal linear components. No additional feedback or higher-order dynamics. Concept / Approach: The governing differential equation is dV_c/dt = (1/(R*C)) * (V_in - V_c). Its solution is exponential in time: V_c(t) = V_final + (V_initial - V_final) * exp(-t/(R*C)). The rate (slope) is initially largest and decays as the state approaches the final value, defining the exponential approach to equilibrium. Step-by-Step Solution: 1) Write the first-order ODE for the RC network. 2) Solve with constant coefficients to obtain an exponential term exp(-t/tau). 3) Interpret tau = R * C as the time constant setting the curve’s speed. Verification / Alternative check: Curve-fitting lab data for RC steps yields straight lines on a semi-log plot (log of the remaining error vs time), confirming exponential behavior. Why Other Options Are Wrong: Linear/parabolic/trigonometric/log-only forms do not satisfy the standard first-order homogeneous solution structure for RC steps. Common Pitfalls: Confusing the exponential state trajectory with “linear charging” due to constant current (which occurs only with a current source into C). Final Answer: An exponential

Question 10

RC charging — time to reach 95%: In a first-order RC circuit, how many time constants (tau = RC) are required for a capacitor’s voltage to charge to approximately 95% of its final value?

Accepted Answer

Introduction / Context: Charging in a first-order RC circuit follows an exponential law. Designers and technicians often use simple “time-constant rules of thumb” to estimate how fast a capacitor reaches a certain percentage of its final (steady-state) voltage. This question asks for the number of time constants needed to reach about 95% of the final value. Given Data / Assumptions: First-order RC network with time constant tau = R * C. Ideal source and lumped components. Target: capacitor voltage Vc reaching 95% of its final value Vf. Concept / Approach: The charging law is Vc(t) = Vf * (1 - exp(-t / tau)). To find when Vc/Vf = 0.95, solve 0.95 = 1 - exp(-t / tau), which implies exp(-t / tau) = 0.05 and t / tau = -ln(0.05) ≈ 2.996. Thus, it takes about 3 time constants to hit 95% in an ideal RC charge. Step-by-Step Solution: 1) Write the charging equation: Vc/Vf = 1 - exp(-t/tau).2) Set Vc/Vf = 0.95 → exp(-t/tau) = 0.05.3) Take natural log: t/tau = -ln(0.05) ≈ 2.996.4) Round to a rule of thumb: ≈ 3 time constants. Verification / Alternative check: Rule-of-thumb milestones: 63.2% at 1 tau, 86.5% at 2 tau, 95% at ~3 tau, 98.2% at 4 tau, and 99.3% at 5 tau. These standard checkpoints confirm the 95% ≈ 3 tau result. Why Other Options Are Wrong: 2: Only reaches ~86.5%, not enough for 95%. 4: Reaches ~98.2%, higher than 95% (not the first time constant count meeting 95%). 5: Reaches ~99.3%, more conservative than necessary. Common Pitfalls: Confusing the 5-tau “essentially fully charged” rule with the 95% mark; mixing discharge and charge curves; forgetting that the time constant is RC and not dependent on source magnitude for the percentage-of-final calculation. Final Answer: 3 time constants are required to reach approximately 95%.

Question 11

Terminology repaired — which term is directly named after the calculus operation that accumulates area under a curve (i.e., produces an output proportional to the integral of the input)?

Accepted Answer

Introduction / Context: Analog signal-processing blocks often take their names from calculus: circuits that compute time derivatives are called differentiators, while those that accumulate area under an input curve are called integrators. This question focuses on the term whose meaning comes directly from the integral operation in calculus. Given Data / Assumptions: We seek the term explicitly tied to accumulation (area under curve). Idealized op-amp models are assumed when thinking of canonical circuits. Basic continuous-time linear circuits context. Concept / Approach: An integrator produces an output proportional to the time integral of its input: v_out(t) = (1/RC) * ∫ v_in(t) dt for the canonical RC op-amp integrator (constant factors depend on configuration). Conversely, a differentiator outputs a scaled time derivative of the input. Therefore, “integrator” is the term directly derived from the calculus integral. Step-by-Step Solution: 1) Identify the calculus operation associated with accumulation: integration.2) Map operations to circuit names: “integrator” ↔ integral, “differentiator” ↔ derivative.3) Select the term that denotes integration: integrator. Verification / Alternative check: Standard op-amp texts show the inverting integrator with a capacitor in the feedback path and a resistor at the input, yielding v_out proportional to the integral of v_in. Why Other Options Are Wrong: Differentiator: corresponds to the derivative, not the integral. Filter: generic term; can be integrative in effect but is not a calculus operation name. Voltage-divider: a passive pair of resistors that scales voltage; unrelated to calculus operations. Common Pitfalls: Assuming both “differentiator” and “integrator” fit because both come from calculus. The stem was repaired to specify the integral (accumulation) concept to ensure a unique correct choice. Final Answer: Integrator.

Question 12

Capacitor plate area and stored charge — choose the correct completion: The ______ the area of the plates, the greater the number of electrons and ______ that can be stored (all else equal).

Accepted Answer

Introduction / Context: Capacitance measures a component’s ability to store electric charge per unit voltage. Two key geometric factors influence capacitance in the parallel-plate model: the plate area and the separation distance. This item asks you to complete a sentence about how plate area affects stored charge. Given Data / Assumptions: Parallel-plate intuition: C ≈ epsilon * A / d. Voltage across the capacitor is fixed for comparison. Material permittivity (epsilon) and plate separation (d) unchanged. Concept / Approach: From C = epsilon * A / d, increasing the plate area A increases capacitance C linearly if other variables are constant. For a given applied voltage V, stored charge is Q = C * V, so larger C means proportionally larger Q. Therefore, a larger plate area allows more charge to be stored. Step-by-Step Solution: 1) Start with the constitutive relation C ∝ A/d.2) Hold V, epsilon, and d constant to isolate the effect of A.3) Increasing A → increases C.4) Since Q = C * V, a larger C yields greater Q (stored charge). Verification / Alternative check: Practical capacitors that target higher capacitance within the same volume often increase effective surface area (e.g., etched/roughened foils, multi-layer structures) to realize greater charge storage per volt. Why Other Options Are Wrong: Smaller, charge: Reducing area reduces capacitance and stored charge. “Polarity” options: Polarity refers to terminals’ sign, not quantity of stored charge; plate area does not change “polarity.” Common Pitfalls: Confusing energy with charge; forgetting that charge scales with capacitance at fixed voltage; ignoring that dielectric properties and distance also affect C. Final Answer: larger, charge.

Question 13

Rate-of-change circuit — which block produces an output proportional to the time derivative (rate of change) of the input voltage?

Accepted Answer

Introduction / Context: Classic analog building blocks include differentiators and integrators. A differentiator accentuates rapid changes and edges, while an integrator accumulates the signal over time. This question checks recognition of the block that outputs a signal proportional to the input’s rate of change. Given Data / Assumptions: Continuous-time, small-signal linear analysis. Ideal op-amp representations where relevant. Focus on functional behavior, not implementation details. Concept / Approach: The ideal differentiator realizes v_out(t) ∝ dv_in(t)/dt. In frequency terms, it multiplies by jomega, boosting high-frequency components. In contrast, an integrator scales by 1/(jomega), emphasizing low-frequency content. Mixers and “averagers” are different functions altogether. Step-by-Step Solution: 1) Match “rate of change” to the mathematical derivative.2) Identify the circuit named for that operation: differentiator.3) Exclude integrator (opposite operation) and unrelated blocks.4) Conclude the correct choice is differentiator. Verification / Alternative check: Standard inverting differentiator: a series capacitor at the input and a feedback resistor produce v_out = -RC * dv_in/dt (sign and scale depend on topology). Why Other Options Are Wrong: Integrator: outputs an integral, not a derivative. Curve averager: not a standard calculus block; averaging is not differentiation. Mixer: multiplies/combines signals (e.g., RF), not a time derivative. Common Pitfalls: Assuming “averaging fast changes” is differentiation; conflating noise amplification (a practical differentiator issue) with the ideal mathematical operation. Final Answer: Differentiator.

Question 14

What sets RC timing — which physical attribute of a capacitor directly affects charge/discharge time via its influence on capacitance (tau = RC)?

Accepted Answer

Introduction / Context: Timing in first-order networks depends on the time constant tau = R * C. Since tau scales with C, any physical attribute that increases capacitance increases the time constant and slows charging/discharging (for a given R). This item asks which attribute directly influences C through the parallel-plate model. Given Data / Assumptions: First-order resistor-capacitor timing, tau = RC. Parallel-plate intuition: C ≈ epsilon * A / d. All other variables held constant for comparison. Concept / Approach: Capacitance rises with plate area A and dielectric permittivity epsilon, and falls with greater plate separation d. Therefore, increasing plate area raises C and increases tau; decreasing area lowers C and shortens tau. Packaging, lead arrangement, or nominal voltage rating do not directly determine the capacitance value in the basic model. Step-by-Step Solution: 1) Use C ∝ A/d to identify geometry-dependent variables.2) Plate area A is a direct multiplier of C.3) Larger C → larger tau (for fixed R).4) Therefore, plate area affects charge/discharge time. Verification / Alternative check: Manufacturing techniques (e.g., multi-layer ceramic stacks, etched foils) increase effective area to realize higher capacitance in small volumes, which directly lengthens RC time constants in timing applications. Why Other Options Are Wrong: Package style / lead arrangement: may affect parasitics/ESR/ESL, but do not directly set C in the simple model. Voltage rating: specifies safe operating voltage; it does not by itself determine C. Common Pitfalls: Confusing voltage rating or package with capacitance value; overlooking that dielectric constant and geometry dominate C. Final Answer: Plate area.

Question 15

Capacitor structure — identify the non-conductive medium: The basic capacitor consists of two conductive plates separated by what kind of material?

Accepted Answer

Introduction / Context: All capacitors share a core structure: two conductors separated by a non-conductive medium that stores energy in an electric field. Recognizing this medium’s role is essential to understanding capacitance, leakage, breakdown, and frequency behavior. Given Data / Assumptions: Parallel-plate conceptual model for clarity. Two conductive electrodes (plates). Non-conductive separator between plates. Concept / Approach: The separator is the dielectric: an insulating material characterized by permittivity (epsilon) and breakdown strength. Its permittivity influences capacitance (C ∝ epsilon), and its loss tangent and stability affect AC behavior and temperature/frequency characteristics. Common dielectrics include air, paper, ceramics, mica, polymers, and oxide layers (e.g., in electrolytics). Step-by-Step Solution: 1) Identify the structural pieces: two conductors + one insulator.2) Name the insulator: dielectric.3) Link properties: higher epsilon → higher C for the same geometry.4) Note reliability limits: breakdown voltage and leakage are dielectric-dependent. Verification / Alternative check: Textbook definitions of capacitance explicitly assume an insulating dielectric layer between plates; replacing it with a conductor collapses the plates into a short circuit (no energy storage). Why Other Options Are Wrong: A conductor / an electrode: those are the plates themselves, not the separator. A spacer: a generic mechanical term; in capacitors the proper term is dielectric, which provides electrical insulation and field storage. Common Pitfalls: Using “spacer” or “insulator” loosely; “dielectric” is the precise term in capacitor theory. Final Answer: A dielectric.

Question 16

Signal coupling function — which capacitor role passes AC while blocking DC between two stages?

Accepted Answer

Introduction / Context: Capacitors in analog circuits serve different roles depending on where they are placed. A common use is to transfer the AC component of a signal from one stage to another while preventing any DC bias from shifting the next stage’s operating point. This use case has a specific name. Given Data / Assumptions: Two amplifier/processing stages with different DC bias points. A capacitor placed in series between stages. Linear small-signal operation around bias. Concept / Approach: A coupling capacitor is placed in series with the signal path. Its reactance is low at signal frequencies (so AC passes) and effectively infinite at DC (so DC is blocked). Bypass/decoupling capacitors are typically shunt elements to ground that stabilize supply rails or provide AC ground at certain nodes; they do not sit in series as a conduit for AC between stages. Step-by-Step Solution: 1) Identify the desired behavior: pass AC, block DC.2) Map to component placement: series capacitor between stages.3) Recognize the standard term: coupling capacitor.4) Eliminate shunt-type roles (bypass/decoupling) and generic “filter.” Verification / Alternative check: Basic small-signal models show that the series capacitor and the input resistance of the next stage form a high-pass network that sets the lower cutoff frequency f_c = 1/(2piR_in*C). Why Other Options Are Wrong: Filter: Non-specific; many filters exist and may not block DC by default. Bypass / Decoupling: Typically shunt to ground for supply or local AC stabilization, not series AC transfer. Common Pitfalls: Confusing “bypass” (to ground) with “coupling” (in series path); overlooking that decoupling ≈ bypass on power rails. Final Answer: Coupling capacitor.

Question 17

Capacitive reactance calculation — with a 5 V, 60 Hz source applied to a 4.7 µF capacitor, what is the circuit current (assume ideal components and RMS values)?

Accepted Answer

Introduction / Context: For a pure capacitor driven by a sinusoidal voltage source, the magnitude of the current is determined by the capacitive reactance Xc. This problem tests the ability to compute Xc and then use Ohm’s law in the phasor (RMS) sense to find current. Given Data / Assumptions: V_rms = 5 V (sinusoidal, 60 Hz). Capacitance C = 4.7 µF. Ideal capacitor, no series resistance; RMS quantities used. Concept / Approach: Capacitive reactance magnitude is Xc = 1 / (2 * pi * f * C). The current magnitude then is I_rms = V_rms / Xc for a purely reactive element. After computing Xc, divide 5 V by Xc to get the current in amperes and convert to milliamperes. Step-by-Step Solution: 1) Compute Xc: Xc = 1 / (2 * pi * 60 * 4.7e-6) ≈ 564.38 Ω.2) Compute I_rms: I = 5 / 564.38 ≈ 0.00886 A.3) Convert to mA: 0.00886 A ≈ 8.86 mA.4) Match to the closest option: 8.86 mA. Verification / Alternative check: A quick sanity check: increasing C lowers Xc and raises current; for a few microfarads at 60 Hz, a few milliamps is typical at low voltages, which agrees with 8.86 mA at 5 V. Why Other Options Are Wrong: 0.886 mA / 88.6 µA / 8.86 µA: Each is off by one or two decimal places relative to the computed value. Common Pitfalls: Mixing peak and RMS values; misplacing decimal points when converting microfarads to farads; forgetting that capacitive current leads voltage by 90 degrees (phase does not change magnitude here). Final Answer: 8.86 mA.

Question 18

Capacitor technology comparison — among common capacitor types used in electronics (electrolytic, ceramic, plastic-film, and mica), which type typically offers the highest capacitance values in practical, commercially available parts?

Accepted Answer

Introduction / Context: Selecting a capacitor type often begins with the required capacitance value. Different dielectric technologies (electrolytic, ceramic, plastic-film, mica) trade off size, voltage rating, tolerance, equivalent series resistance (ESR), and stability. This question asks which of the listed mainstream types typically provides the highest capacitance values in standard electronic components. Given Data / Assumptions: The comparison is among electrolytic, ceramic, plastic-film, and mica capacitors commonly used on PCBs. We consider typical catalog parts and practical size/voltage ranges. We exclude specialty devices such as supercapacitors from being the intended correct choice in this context (those are distinct technology classes used for energy storage rather than conventional decoupling or timing). Concept / Approach: Aluminum electrolytic (and tantalum electrolytic) capacitors employ a very thin dielectric oxide layer and an electrolyte, enabling large capacitance per volume. Ceramic MLCCs achieve high values with high-k dielectrics but practically top out well below large electrolytics for the same package and voltage. Plastic-film and mica prioritize stability and low loss, trading away volumetric capacitance density, thus offering lower maximum values. Step-by-Step Solution: 1) List maximum practical values by family: electrolytics >> ceramics > film > mica (typical ordering).2) Note that electrolytics commonly span microfarads to thousands of microfarads (and beyond).3) Conclude that the highest available capacitance among the listed mainstream types is electrolytic. Verification / Alternative check: Vendor catalogs show electrolytics in 10 µF to several thousand µF ranges in modest sizes, while film parts are usually nF to low µF, and mica far smaller. Why Other Options Are Wrong: Ceramic: high for their size but generally far below electrolytics at similar voltages.Plastic-film: prized for stability/low ESR, not extreme capacitance.Mica: precision/temperature stability, very low capacitance values.Supercapacitor: different device class; not the intended mainstream PCB capacitor type in this comparison. Common Pitfalls: Confusing volumetric efficiency with absolute maximum value; overlooking voltage derating and tolerance differences. Final Answer: Electrolytic

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