Question 1

Repair for missing figure — “The capacitor voltage at the beginning of the second pulse is ______. Assume the capacitor is initially uncharged.” Without the component values, period, and τ relative to the pulse timing, can we determine the exact vol…

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Introduction / Context: The initial voltage of a capacitor at the start of a subsequent pulse depends on how much it charged during the previous high interval and how much it discharged during the low interval. Both are governed by τ = R * C and the specific high/low times (duty cycle). With no numeric values or figure, the exact voltage cannot be computed. Given Data / Assumptions: Capacitor starts uncharged at t = 0. Repetitive pulse train drives an RC network. Component values and period/duty are not provided. Concept / Approach: Over one period, the capacitor charges toward a level during “high” and discharges during “low,” each following v_C(t) = V_final + (V_initial − V_final) * exp(−t/τ). The steady-state beginning-of-pulse value is found by equating the end of one interval to the start of the next. This requires τ and the timing parameters; absent them, there is no unique voltage. Step-by-Step Solution: Let v1 be the start-of-high voltage and v2 the end-of-high.Compute v2 from v1 using the charge exponential over t_high.Compute the start of the next pulse from v2 using the discharge exponential over t_low.Solve for steady-state v1; without τ and timings, this is indeterminate. Verification / Alternative check: Examples with τ ≪ period produce near-zero start voltages; τ ≫ period produces significant memory (v1 near prior average), proving dependence on missing data. Why Other Options Are Wrong: Fixed percentages (36.8%, 63.2%) correspond to exactly 1τ intervals, not universal outcomes. “Always equals amplitude” or “always zero” are special cases only. Common Pitfalls: Assuming the 1τ percentages apply to any arbitrary pulse timing; they apply only when interval length equals τ. Final Answer: Cannot be determined from the information provided.

Question 2

Exponential decay reality check — after the rising edge in a first-order RC pulse network, how long does it take for the voltage across the resistor to “decrease to zero”? (Interpret “zero” as a practical threshold; without τ the value is not unique…

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Introduction / Context: In a first-order RC network, voltages across elements change exponentially. They never reach mathematical zero in finite time; design practice uses thresholds such as 1τ (~63% change) or 5τ (~99% change). If a problem asks for “time to zero” but does not specify τ or a practical threshold, no single numeric answer exists. Given Data / Assumptions: No R, C, or τ given. No threshold definition (e.g., 1% or 0.1%) provided. Standard first-order behavior assumed. Concept / Approach: The resistor voltage after a step is an exponential of the form v_R(t) = K * exp(−t/τ). For any finite t, exp(−t/τ) > 0. A practical engineering “zero” requires choosing a percentage (e.g., 5τ for ~99% decay). Without τ and a threshold, the requested time cannot be fixed. Step-by-Step Solution: Recognize exponential decay behavior.Note the absence of τ and threshold.Conclude that the time to “zero” is undefined unless a percentage is specified.Therefore, a numeric selection is impossible from the given information. Verification / Alternative check: Assuming τ = 100 µs gives “≈zero” near 500 µs; assuming τ = 1 ms shifts this to ≈5 ms. Different τ yield different times. Why Other Options Are Wrong: 0 s / 1τ / 5τ / half period: each presumes a specific definition or value of τ or period not provided. Common Pitfalls: Taking 5τ as a universal answer even when τ is unknown; ignoring that “zero” needs a tolerance. Final Answer: Cannot be determined from the information provided.

Question 3

Fault effect on time constant — in an RC integrator or differentiator, which component fault most directly decreases the effective time constant τ = R * C (thus collapsing the time-based behavior)?

Accepted Answer

Introduction / Context: The time constant of a simple RC stage is τ = R * C. Faults that effectively reduce R or C will reduce τ and undermine the intended timing/shape of the response. Identifying which failures drive τ downward helps with troubleshooting pulse-shaping networks. Given Data / Assumptions: Simple first-order RC network used as an integrator or differentiator. Time constant τ = R * C for the effective series R and capacitance. We consider the dominant effect of common faults. Concept / Approach: A shorted capacitor effectively makes C → 0 in the time-constant product, collapsing τ toward zero and destroying the intended time-based behavior (the node becomes a near-short for AC transients in many topologies). By contrast, an open resistor removes the current path entirely (no defined exponential); an open capacitor removes C from the circuit (again no defined RC exponential). Increased leakage (finite parallel resistance) modifies the effective network but does not predictably “decrease τ” in the intended series path sense; instead it distorts the response and DC bias. Step-by-Step Solution: Start from τ = R * C.Consider the shorted capacitor fault → C ≈ 0 → τ ≈ 0.Other faults either invalidate the RC model (open) or add leakage paths without a clean τ ↓ conclusion.Therefore, the shorted capacitor most directly decreases τ. Verification / Alternative check: Lab observation: with a shorted capacitor, edge responses vanish into immediate jumps limited only by wiring; with an open capacitor, the node stops integrating/differentiating entirely rather than becoming “faster.” Why Other Options Are Wrong: Open resistor / open capacitor: break the RC function entirely; τ is not meaningfully reduced but rather undefined. Leaky capacitor: adds a parallel resistance that skews behavior, not a simple τ reduction. Shorted resistor bypassing the capacitor: not listed in original options, but such a bypass also collapses behavior; the classic textbook answer is the shorted capacitor. Common Pitfalls: Assuming any fault that “speeds up” a response means τ simply decreased; some faults destroy the RC mechanism instead of altering τ cleanly. Final Answer: Shorted capacitor.

Question 4

Topology swap thought experiment — starting with a classic RC integrator (series resistor feeding a capacitor to ground, output across the capacitor), if the positions of the components are swapped (series capacitor feeding a resistor to ground, out…

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Introduction / Context: First-order RC networks can be arranged to emphasize integration (slow change) or differentiation (edge emphasis). An “RC integrator” typically has a series resistor and a shunt capacitor with the output taken across the capacitor. Swapping the positions of R and C and taking the output across the resistor yields the complementary high-pass behavior associated with an RC differentiator. Given Data / Assumptions: Original circuit: series R, capacitor to ground, output across C. Modified circuit: series C, resistor to ground, output across R. High-impedance measurement so loading is negligible; small-signal linear behavior. Concept / Approach: In the differentiator arrangement (series C, shunt R, output across R), the capacitor passes rapid changes (high-pass action) while the resistor converts the changing current to a voltage. The output magnitude is therefore proportional to the rate of change of the input within the useful bandwidth, producing edge spikes to steps/pulses. Conversely, the integrator (series R, shunt C, output across C) accumulates charge and emphasizes low-frequency content (low-pass action). Step-by-Step Solution: Identify original: low-pass RC with V_out at the capacitor → integrator behavior for τ ≫ signal width.Swap series/shunt positions: now a high-pass RC with V_out at the resistor.Recognize differentiator behavior: output highlights edges and fast changes.Conclude the new circuit is an RC differentiator. Verification / Alternative check: Frequency response: integrator has gain ∝ 1/ω at high frequencies (low-pass), while differentiator has gain ∝ ω at low frequencies (high-pass within limits). Swapping elements flips the response class. Why Other Options Are Wrong: RL differentiator / RL integrator: incorrect component family. RC integrator: describes the original, not the swapped configuration. “Pure attenuator only”: while both topologies can attenuate, the key function here is differentiation. Common Pitfalls: Confusing where the output is taken; with RC differentiator it must be across the resistor, not the capacitor. Final Answer: RC differentiator.

Question 5

RC charging (repaired for solvability): An RC low-pass (first-order) is driven by two identical rectangular pulses from 0 V to 12 V, applied back-to-back with no gap. Each pulse width equals 5 time constants (5τ), so the node sees a total “high” dur…

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Introduction / Context: Time-response questions with “the given circuit” often reference an RC network responding to pulses. To make the problem solvable without the missing diagram, we specify a standard first-order RC low-pass driven by two consecutive 0→12 V pulses, each lasting 5τ. This lets us compute the capacitor voltage using the classic exponential charging law and determine the voltage at the precise moment the second pulse ends. Given Data / Assumptions: First-order RC, initially uncharged (V_C(0) = 0 V). Input = 12 V high for 5τ, immediately followed by another 12 V high for 5τ (total 10τ high), then low. Ideal buffer/op-amp if used; no loading on the capacitor; τ = R * C is constant. Concept / Approach: For a step from 0 to V_S, the capacitor charges as V_C(t) = V_S * (1 − e^(−t/τ)). After a duration of nτ, the residual error to the final value is e^(−n). At 5τ, the charge has reached ≈ 99.3% of its final value; at 10τ, ≈ 99.995% of final. With V_S = 12 V, the end-of-second-pulse voltage is essentially the final value, i.e., approximately 12 V. Step-by-Step Solution: Model each pulse as a high level of 12 V.Use V_C(10τ) = 12 * (1 − e^(−10)).Compute e^(−10) ≈ 0.000045; thus V_C ≈ 12 * (0.999955) ≈ 11.9995 V.Round sensibly: ≈ 12 V at the end of the second pulse. Verification / Alternative check: If each pulse were only 3τ wide, the end of the second pulse would be V_C(6τ) = 12 * (1 − e^(−6)) ≈ 12 * 0.9975 ≈ 11.97 V—still very close to 12 V. Increasing to 10τ makes it indistinguishable from 12 V at normal resolution. Why Other Options Are Wrong: 3.32 V / 1.06 V / 7.56 V: These correspond to much shorter charge intervals (small multiples of τ) or to intermediate times, not to 10τ of continuous charging. Common Pitfalls: Forgetting that the RC never reaches the final value mathematically in finite time; confusing a single 5τ pulse with two back-to-back 5τ pulses; neglecting that “virtual ground” buffers prevent loading of the RC node. Final Answer: 12 V

Question 6

Foundations — what do we call the characteristic time measure for first-order charging/discharging (e.g., in RC circuits) that sets the exponential response speed?

Accepted Answer

Introduction / Context: In first-order systems such as RC and RL circuits, the output follows an exponential trajectory toward a new value after a step input. The pace of this response is governed by a single parameter that engineers use constantly for design, quick estimates, and troubleshooting. Recognizing and naming this parameter correctly is essential for time-domain intuition and for converting between time and frequency viewpoints. Given Data / Assumptions: We consider linear first-order systems (e.g., RC, RL). The response to a step is exponential with a single scale factor. Terminology must be standard and unambiguous. Concept / Approach: The exponential law for an RC step is V(t) = V_final + (V_initial − V_final) * e^(−t/τ). The parameter τ (tau) is called the time constant. It equals R * C for RC circuits and L / R for RL circuits. After 1τ, the response has moved about 63.2% toward its final value; after 5τ, it is within about 0.7% of final, which is often treated as practically settled or “steady state” for many applications. Step-by-Step Solution: Identify the first-order exponential form.Note that τ scales the time axis of the exponential.Recall τ = RC for RC, τ = L/R for RL.Therefore, the name of the characteristic time is “time constant.” Verification / Alternative check: Frequency domain linkage: the cutoff frequency f_c of a first-order filter is 1 / (2π τ). Thus time constant also determines bandwidth, confirming its centrality. Why Other Options Are Wrong: Steady state: refers to the final condition, not the time scale.Transient time / pulse response: general descriptors, not the specific parameter name. Common Pitfalls: Equating “steady state” with a fixed time like 5τ; in precision work, required settling might be 7–10τ depending on tolerance. Final Answer: time constant

Question 7

RC settling time (repaired for solvability): A first-order RC output is considered “at maximum value” for practical purposes when it has reached about 99.3% of its final level (i.e., after 5 time constants). If the circuit’s time constant is 243.9 µ…

Accepted Answer

Introduction / Context: Because an exponential response never reaches its final value in strictly finite time, engineers adopt practical settling criteria such as 63.2% (1τ), 95% (≈3τ), 99.3% (≈5τ), or tighter depending on accuracy needs. This question uses the widely accepted 5τ rule of thumb to define when the output is “at maximum value” for typical design work. Given Data / Assumptions: First-order RC circuit with time constant τ = 243.9 µs. “Maximum value” is defined here as 99.3% of final (≈ 5τ). Start from 0 V step to a new DC level; linear components. Concept / Approach: The exponential approach to the final value follows V(t) = V_final * (1 − e^(−t/τ)). At t = 5τ, e^(−5) ≈ 0.0067, so V(5τ) ≈ 0.993 * V_final. Multiply the given τ by 5 to find the practical settling time. Step-by-Step Solution: Compute 5τ: 5 * 243.9 µs = 1219.5 µs.Convert to ms: 1219.5 µs ≈ 1.22 ms.Therefore, the output is essentially at its maximum after ≈ 1.22 ms.This meets the 99.3% practical criterion. Verification / Alternative check: For a tighter 7τ criterion (≈ 99.9%), time would be ≈ 1.71 ms; for a looser 3τ criterion (≈ 95%), time would be ≈ 0.732 ms. These bracket the 1.22 ms result logically. Why Other Options Are Wrong: 0 s: impossible for a causal system.243.9 µs: that is only 1τ, i.e., 63.2% of final.1 ms: shy of 5τ for the given τ; ≈ 4.10τ. Common Pitfalls: Confusing “time constant” with “settling time”; assuming “steady state” occurs at a fixed calendar time regardless of τ; mixing µs and ms units. Final Answer: 1.22 ms

Time Response of Reactive Circuits Questions