Q: RL pulse shaping — in a basic RL differentiator (step/pulse shaping network), is the output conventionally taken across the inductor to emphasize rapid changes (spikes at edges)?

Introduction / Context: Differentiator networks generate outputs that are large during rapid input changes and small during steady levels. An RL differentiator uses the inductor’s property v_L = L * di/dt, producing voltage spikes at rising and falling edges of a pulse. Knowing where the output is taken (across which element) is essential for correct classification and design. Given Data / Assumptions: Series RL excited by steps/pulses. Small-signal, linear behavior; ideal inductor. Load is high-impedance measurement so as not to disturb the network. Concept / Approach: For a series RL with an applied step, the current cannot change instantaneously, so the inductor initially takes nearly the entire input voltage and then its voltage decays exponentially as current builds. This “edge-peaked” v_L is proportional to di/dt and therefore implements differentiation. Hence, the conventional RL differentiator output is measured across the inductor. Step-by-Step Solution: Write inductor relation: v_L(t) = L * di/dt.Apply a step or pulse: di/dt is large at edges, small in between.Therefore v_L is large at edges (spikes) and small between edges → differentiation.Conclusion: take the output across L for RL differentiator behavior. Verification / Alternative check: Transient plots show v_L peaks with exponential decay after each transition, while the resistor’s voltage tends to follow the slowly changing current, not the derivative highlight. Why Other Options Are Wrong: Incorrect or “only for square waves”: differentiation property holds for any changing waveform; square waves just make spikes obvious. Constraints like R ≫ X_L or core material: affect shape and amplitude but not the fundamental placement of the output for differentiation. Common Pitfalls: Confusing RL differentiator with RC differentiator (where output is taken across the resistor). Final Answer: Correct — the output of an RL differentiator is taken across the inductor.

Q: Settling criterion — for a capacitor to “completely” (≈99%) charge during the on-time of a pulse in an RC network, the pulse width should be related to the time constant τ how?

Introduction / Context: In first-order systems, “complete” charging is a practical notion. Engineers often use 5τ as a rule of thumb for ≈99% settling of an exponential response. This item asks how the pulse width (the available charging time) must compare with τ to achieve near-full charging during a pulse interval. Given Data / Assumptions: RC network with time constant τ = R * C. Exponential approach to a new level during a pulse. “Completely” interpreted as ≈99% (5τ rule). Concept / Approach: The capacitor voltage follows v_C(t) = V_final + (V_initial − V_final) * exp(−t/τ). After t = 5τ, the residual error is exp(−5) ≈ 0.0067, or about 0.7%. Thus, a pulse that lasts at least 5τ allows the capacitor to get within about 1% of its asymptotic value for that interval. Shorter pulses yield proportionally less settling. Step-by-Step Solution: Define settling goal: ≈99% in practical design.Use the exponential error term e^(−t/τ).Solve e^(−t/τ) ≤ 0.01 → t ≥ 4.6τ; designers round to 5τ.Therefore, PW ≥ 5τ is a standard engineering guideline. Verification / Alternative check: Simulation or measurement of step response shows 63% at 1τ, 95% at 3τ, ≈99% at 5τ, reinforcing the rule. Why Other Options Are Wrong: “Less than 5τ” or “independent of τ”: would not reach ≈99% reliably. “Equal to exactly 5τ”: a useful benchmark, but “greater than or equal” is the safer, more general statement. Common Pitfalls: Treating “complete” as 100% in finite time; exponentials only approach asymptote, so 5τ is a practical design target. Final Answer: Greater than or equal to 5 time constants.

Q: Repair for missing schematic — “If the pulse width were cut in half in the given RC pulse circuit, the voltage across the resistor at the end of the pulse would be ______.” Without R, C, τ, and the original pulse width, can a unique numeric value be…

Introduction / Context: For an RC network excited by a pulse, the voltage across the resistor at the end of the pulse depends on the exponential evolution during the on-time. That evolution is set by the time constant τ = R * C and the actual on-time (pulse width). Halving the pulse width changes the settling time, but without the numeric τ and original timing, the final value cannot be pinned down to a single number. Given Data / Assumptions: No values for R, C, or τ are provided. No explicit pulse amplitude or original pulse width is given. Output node location is not fully specified in the missing figure. Concept / Approach: The resistor voltage in a simple RC driven by a step is v_R(t) = V_in * exp(−t/τ) (for a series RC with output across R after a step to V_in), or the complement thereof depending on the exact topology. At t = PW (end of pulse), v_R(PW) depends on exp(−PW/τ). If PW is halved, the new value depends on exp(−(PW/2)/τ). Without τ and PW, the numeric outcome is undetermined. Step-by-Step Solution: State general form: v_end ∝ exp(−PW/τ) (topology dependent scale).Halve PW → v_end,new depends on exp(−(PW/2)/τ).No τ or PW given → cannot compute a numeric result.Therefore, a single numeric choice is not justifiable. Verification / Alternative check: Try two examples: If PW = 5τ, exp(−PW/τ) ≈ 0.007; if PW = 0.5τ, exp(−PW/τ) ≈ 0.607. The outcomes are vastly different, proving dependence on missing data. Why Other Options Are Wrong: 0 V / equals source / fixed fractions: these assert specific conditions that may or may not hold; they require τ and PW. “One-half of the original”: exponential processes do not in general scale linearly with time halving. Common Pitfalls: Assuming a universal “0.37” or “0.5” result; those come from particular τ:PW ratios, not from the act of halving itself. Final Answer: Cannot be determined from the information provided.

Q: Repair for missing values — “If the pulse source has an internal resistance of 80 Ω in the given circuit, it will take ______ for the output voltage to decrease to zero.” Can a time-to-zero be specified for a first-order RC without knowing C (and to…

Introduction / Context: First-order RC voltages decay exponentially and never reach absolute zero in finite time; designers use practical thresholds (e.g., 1τ ≈ 63% settled, 5τ ≈ 99% settled). The question as written lacks the capacitance and full topology, yet asks for a specific time “to zero,” which is not physically precise and cannot be computed numerically without τ. Given Data / Assumptions: Only the source resistance Rs = 80 Ω is mentioned. Capacitance C and any additional series/parallel resistances are unknown. No threshold (e.g., to 1% or to 0.1%) is specified. Concept / Approach: Decay time constant is τ = R_eq * C, where R_eq is the effective resistance seen by the capacitor during discharge (often Rs plus any series elements). Without C and R_eq, τ is unknown. Moreover, an exponential strictly reaches zero only as t → ∞; practical “zero” must be defined via a percentage, such as 5τ for ≈99% decay. Step-by-Step Solution: Identify missing parameters: C (and full R_eq).Recognize exponential nature: v(t) = V0 * exp(−t/τ).Select a practical threshold (if provided) to compute t; absent this, no unique time exists.Conclude that a numeric answer cannot be given. Verification / Alternative check: Pick example C values: with C = 1 µF, τ = 80 µs; with C = 10 µF, τ = 800 µs. Times differ by 10×, showing sensitivity to the missing parameter. Why Other Options Are Wrong: “Exactly 5τ” or “1τ”: these are rules of thumb for percentages, not true zero. “Independent of C” or “equals the pulse period”: incorrect generalizations. Common Pitfalls: Forgetting that first-order exponentials only asymptotically reach zero; ignoring the need to specify a decay threshold. Final Answer: Cannot be determined from the information provided.

Question 1

In pulse and digital electronics, which type of circuit is specifically designed to produce very short-duration spikes (narrow pulse triggers) from an input signal or event?

Accepted Answer

Introduction: Short-duration spikes, often called trigger pulses or narrow pulses, are essential in timing, synchronization, counting, and digital control. This question checks recognition of the circuit class whose primary function is to create sharp, brief pulses on demand, rather than to average, integrate, or otherwise reshape signals into long-duty cycles. Given Data / Assumptions: The target pulse must be narrow (short time width) with a quick rise and fall. Applications include clocking flip-flops, triggering monostables, or providing synchronization edges. We are selecting among distinct functional blocks used in basic electronics. Concept / Approach: A trigger pulse generator is explicitly built to deliver narrow pulses. Implementations range from differentiator-based edge generators with diode shaping, to Schmitt-trigger RC networks, to monostable multivibrators (one-shots) set for small time constants. By contrast, integrators emphasize area/averaging, and DC converters emphasize producing a steady level rather than a spike. Step-by-Step Solution: 1) Identify the required output: a narrow pulse (spike) suitable for triggering.2) Map candidate circuits to function: integrators average, converters rectify/average, timing blocks may produce long intervals.3) Choose the class designed for short pulses: the trigger pulse generator.4) Conclude that the correct answer is the purpose-designed generator of narrow pulses. Verification / Alternative check: Consider a monostable multivibrator configured with a small RC constant; it outputs a short pulse when triggered. This is a canonical trigger pulse generator use-case and matches the requirement. Why Other Options Are Wrong: RL integrator: shapes/averages signals rather than producing sharp, adjustable trigger spikes. Timing circuit: generic term; often used for long intervals rather than intentionally minimal widths. Pulse waveform-to-dc converter: averages or rectifies a pulse train into a DC level, not spikes. Monolithic voltage regulator: provides regulated DC voltage, not pulses. Common Pitfalls: Equating any RC network with a pulse generator. Differentiators can produce edges, but a dedicated trigger pulse generator includes amplitude control, shaping, and consistent pulse width across conditions. Final Answer: A trigger pulse generator

Question 2

Which statements about an ideal capacitor's behavior are correct in circuit theory (choose the best overall statement)?

Accepted Answer

Introduction: Capacitors are energy-storage elements whose current-voltage relationship is i = C * dv/dt. This relationship yields several widely used rules of thumb about transient and steady-state behavior. The question asks you to identify the consolidated truth among the listed statements. Given Data / Assumptions: Ideal capacitor with capacitance C (no leakage, ESR, or ESL). Standard circuit-theory interpretations of "instantaneous" and "dc steady state". Small-signal, linear operating region. Concept / Approach: Because i = C * dv/dt, an instantaneous voltage step (large dv/dt) implies a very large current impulse; conversely, changing the capacitor's voltage requires time (non-instantaneous) unless infinite current is available. In dc steady state (dv/dt = 0), the current is zero, so the ideal capacitor behaves as an open circuit. Step-by-Step Solution: 1) Voltage cannot jump instantly: that would require infinite current since dv/dt would be infinite.2) For fast transients, the capacitor presents very low impedance, effectively shunting rapid changes and acting "like a short" for the highest-frequency components.3) At dc steady state, dv/dt = 0, so i = 0 and the capacitor behaves as an open circuit.4) Therefore, each individual statement is correct; the combined best choice is "All of the above". Verification / Alternative check: Frequency-domain view: Xc = 1 / (2 * pi * f * C). As f → ∞, Xc → 0 (short-like). As f → 0 (dc), Xc → ∞ (open-like). This confirms the time-domain reasoning. Why Other Options Are Wrong: Any single statement alone (a, b, or c) is incomplete; the question asks for the best comprehensive truth. None of the above: contradicts standard capacitor behavior taught in basic circuit theory. Common Pitfalls: Confusing "short to instantaneous changes" with "short to all AC". Real capacitors have finite impedance and parasitics; only very high-frequency components see near-short behavior. Final Answer: All of the above

Question 3

In a simple RC integrator, if the capacitor accidentally becomes shorted (zero impedance), what happens to the output measured across the capacitor node?

Accepted Answer

Introduction: An RC integrator typically consists of a series resistor feeding a capacitor to ground, with the output taken across the capacitor. This question explores the failure mode when the capacitor is shorted, a common troubleshooting scenario that produces a distinctive symptom at the output node. Given Data / Assumptions: Passive RC integrator: input → series R → node → capacitor to ground, output at the node. Steady operation with a bounded input signal. "Shorted capacitor" means the capacitor's impedance has collapsed to ~0 Ω. Concept / Approach: If the shunt element (capacitor) is a short, the output node is directly clamped to ground potential. Regardless of the input waveform, the node voltage cannot rise above approximately 0 V because any attempted change is immediately diverted to ground through the short. Step-by-Step Solution: 1) In a healthy RC integrator, the capacitor charges/discharges, creating an output proportional to the time-integral of the input.2) With a shorted capacitor, the output node is effectively a direct connection to ground.3) Therefore, the measured output voltage is ≈ 0 V for all practical purposes.4) The circuit no longer integrates; the input sees a series resistor into a ground short, which may stress the source. Verification / Alternative check: Measure with a multimeter or oscilloscope: the node reads near 0 V despite input changes. Upstream, current is limited only by the series resistor, confirming the fault signature. Why Other Options Are Wrong: Same as input: impossible; the node is clamped to ground. None of the above: incorrect because 0 V is the expected reading. Is at ground vs would measure zero volts: both descriptions point to 0 V, but the most precise measurement-oriented choice is "would measure zero volts". Common Pitfalls: Assuming the integrator will simply "stop integrating" but still pass some signal. A hard short forces the output to 0 V; any remaining signal appears only as current through the series resistor. Final Answer: would measure zero volts

Question 4

For an RC differentiating circuit driven by a periodic pulse waveform, which statement correctly captures the two possible regimes relating the pulse width (tw) to the circuit time constant (tau)?

Accepted Answer

Introduction: In RC differentiators, performance depends on comparing the input pulse width tw to the circuit time constant tau = R * C. Designers commonly distinguish between "long" pulses (tw much greater than tau) and "short" pulses (tw much less than tau) to anticipate output spike shapes and amplitudes. Given Data / Assumptions: First-order RC differentiator with time constant tau. Periodic pulse input with width tw and period Tperiod (not used directly here). Rule-of-thumb boundary at ~5 * tau separates long and short regimes. Concept / Approach: When tw ≥ 5 * tau, the capacitor has time to charge and discharge substantially within each pulse edge, yielding narrow, well-defined spikes primarily at transitions. When tw < 5 * tau, the circuit operates in a strongly differentiating regime where the output resembles brief spikes with limited decay during the pulse, emphasizing edges over the flat top. Step-by-Step Reasoning: 1) Define tau = R * C as the characteristic time scale.2) For tw ≥ 5 * tau, treat the pulse as "long"; the capacitor's exponential response largely settles between edges.3) For tw < 5 * tau, treat the pulse as "short"; output is dominated by edge spikes, with minimal settling between edges.4) Hence the two practical conditions are tw ≥ 5 * tau or tw < 5 * tau. Verification / Alternative check: Simulate with typical values (e.g., R = 1 kΩ, C = 1 nF → tau = 1 µs). Compare outputs for tw = 10 µs (≥ 5 * tau) versus tw = 0.2 µs (< 5 * tau). The first shows distinct charge/discharge, the second shows sharp, narrow spikes. Why Other Options Are Wrong: Options a and b omit the complementary short-pulse regime or duplicate the same inequality; option c covers only short cases; option e is a single value and not a regime. Common Pitfalls: Treating 5 * tau as an exact boundary; it is a heuristic. Real component tolerances and loading shift the practical split, but the two regimes remain conceptually valid. Final Answer: tw ≥ 5Τ or tw < 5Τ

Question 5

Filter behavior mapping: An RL integrator behaves as which kind of first-order filter, and an RC differentiator behaves as which kind, respectively?

Accepted Answer

Introduction: First-order RL and RC networks can be arranged to integrate or differentiate signals. Each arrangement corresponds to a classic filter response, useful for quick intuition about frequency-domain behavior and for selecting components in signal conditioning. Given Data / Assumptions: RL integrator topology: output taken so that low-frequency components are passed and fast changes are attenuated. RC differentiator topology: output taken so that high-frequency transitions are emphasized. Linear, time-invariant approximation with ideal components. Concept / Approach: Integrators pass low-frequency components (they "accumulate" slowly varying signals) and attenuate high-frequency content, which is the hallmark of a low-pass filter. Differentiators emphasize rapid changes—high-frequency components—while attenuating slow variations, the hallmark of a high-pass filter. Step-by-Step Solution: 1) RL integrator: with appropriate node selection, the inductor resists rapid current changes; the output follows the averaged (integrated) behavior, passing lows → low-pass.2) RC differentiator: the capacitor's impedance decreases with frequency, creating spikes at edges and passing highs → high-pass.3) Therefore, the correct mapping is low-pass for the RL integrator and high-pass for the RC differentiator.4) This mapping aligns directly with Bode magnitude slopes: ±20 dB/dec around the cutoff. Verification / Alternative check: Frequency-domain impedances: Xl = 2 * pi * f * L increases with f (blocking highs in the integrator path); Xc = 1 / (2 * pi * f * C) decreases with f (passing highs in the differentiator path). The resulting transfer functions match low-pass and high-pass behaviors, respectively. Why Other Options Are Wrong: Low-pass, low-pass / high-pass, high-pass: both entries cannot simultaneously describe integrator and differentiator behavior. High-pass, low-pass: reverses the correct mapping. Band-pass, band-stop: require at least second-order behavior or combined networks, not simple first-order integrator/differentiator forms. Common Pitfalls: Confusing where the output is taken in the network; swapping nodes can convert an apparent integrator into a different response. Always confirm topology and node definitions. Final Answer: low-pass, high-pass

Question 6

RC integrator — single narrow pulse behavior: “The output voltage of an RC integrator will eventually equal the amplitude of a single input pulse if the pulse width is less than 5 time constants.” Decide whether this statement is valid.

Accepted Answer

Introduction / Context: An RC integrator is designed so that the capacitor voltage represents the time integral (area) of the input waveform, particularly when the input changes much faster than the circuit’s time constant τ = R * C. The prompt asks whether a single narrow pulse (width < 5τ) will make the output eventually equal the pulse amplitude. This clarifies a common misconception between an integrator and a follower. Given Data / Assumptions: Series RC integrator with output taken across the capacitor. Single rectangular input pulse of width T_p where T_p < 5τ. Initial capacitor voltage is zero; ideal components, linear operation. Concept / Approach: For a finite pulse width T_p, the capacitor charges only for time T_p, reaching V_out(T_p) = V_in * (1 − exp(−T_p / τ)). If T_p is small compared to τ, the exponential term exp(−T_p / τ) ≈ 1 − (T_p / τ), giving V_out ≈ V_in * (T_p / τ) which is much less than V_in. Thus, the output does not “equal the pulse amplitude.” Instead, it is a smaller value proportional to the pulse area V_in * T_p divided by τ. Step-by-Step Solution: Define τ = R * C and pulse width T_p < 5τ.Compute capacitor rise over T_p: V_out(T_p) = V_in * (1 − exp(−T_p / τ)).For T_p ≪ τ, approximate V_out ≈ V_in * (T_p / τ), clearly less than V_in.After the pulse ends, the capacitor discharges back toward zero with time constant τ; it never “catches up” to the input amplitude of that already-ended pulse. Verification / Alternative check: Simulate with τ = 10 ms and T_p = 2 ms: V_out(T_p) = V_in * (1 − e^(−0.2)) ≈ 0.181 * V_in, far below V_in. As T_p approaches several τ, V_out approaches V_in, but the statement restricts T_p to be less than 5τ and claims “equal,” which is not guaranteed nor typical. Why Other Options Are Wrong: Correct: Not correct; integrators respond to pulse area, not necessarily to amplitude. Valid only if the capacitor is ideal / pulse amplitude small: Ideal parts or small amplitude do not make V_out equal to V_in for short pulses. Common Pitfalls: Confusing RC charging to source (a follower with long pulses) with integrating behavior for short pulses; assuming the 5τ “near steady-state” rule for steps applies to narrow pulses. Final Answer: Incorrect.

Question 7

Differentiator function — DC output claim: “A differentiator circuit can convert a pulse input into a nearly constant DC output.” Assess the validity of this statement for a standard RC differentiator.

Accepted Answer

Introduction / Context: An RC differentiator emphasizes rapid changes in the input signal and suppresses slowly varying components. Designers often use it to detect edges of pulses and to create narrow trigger spikes. The claim that such a circuit produces a nearly constant DC output from a pulse input reverses its intended function and needs to be evaluated critically. Given Data / Assumptions: Simple RC differentiator: series capacitor followed by a resistor to ground with output taken across the resistor. Input: a rectangular pulse or a train of pulses. Small-signal, linear, time-invariant behavior; no active elements. Concept / Approach: Differentiation is mathematically v_out ∝ dv_in/dt. For a rectangular pulse, dv/dt is large at the rising and falling edges and nearly zero during the flat top. Consequently, the differentiator outputs a positive spike at the rising edge and a negative spike at the falling edge, with almost zero output between edges (apart from small exponential tails). There is no mechanism to produce a steady DC level from a constant input segment in a passive RC differentiator. Step-by-Step Solution: Model the input as step up, constant interval, step down.At step up: dv/dt is large positive ⇒ positive spike across the resistor.During the constant interval: dv/dt ≈ 0 ⇒ output decays toward 0 V.At step down: dv/dt is large negative ⇒ negative spike. Verification / Alternative check: Scope captures of an RC differentiator fed by a pulse train show symmetric positive/negative spikes bracketing each pulse, with the mean output near zero. This is standard edge-detection behavior used in timing and trigger circuits. Why Other Options Are Wrong: Correct: Contradicts the fundamental purpose of a differentiator. Valid only with a very large capacitor / very low frequency: Changing RC alters spike width, not the fact that steady DC is rejected. Common Pitfalls: Confusing differentiators with integrators or low-pass filters that can produce average/DC components; overlooking that the differentiator’s transfer magnitude increases with frequency and tends to zero at DC. Final Answer: Incorrect.

Question 8

Differentiator in steady state — average output: During steady-state operation with a symmetrical pulse train input, is the average output voltage of an RC differentiator approximately zero volts?

Accepted Answer

Introduction / Context: RC differentiators respond to changes, not steady levels. For periodic pulse inputs, the output consists of alternating spikes at each transition. Understanding the average (DC) component at the output is important for coupling considerations, measurement offsets, and avoiding unintended biasing of downstream stages. Given Data / Assumptions: Standard RC differentiator with output across the resistor. Input is a repetitive pulse train without DC offset. Linear, time-invariant operation; steady state reached. Concept / Approach: A differentiator ideally has zero gain at DC. For a symmetrical pulse train that rises and falls to the same baseline, the positive spike at each rising edge and the negative spike at each falling edge have equal and opposite areas (charge transfer through the capacitor). Over a complete period, the time-average (integral over one cycle divided by the period) is therefore approximately zero, ignoring minor leakage and parasitic effects. Step-by-Step Solution: Represent the pulse as a sum of a DC component and AC transitions.The differentiator blocks DC; the AC transitions produce spikes whose average over a full period cancels.Compute average over one period: ∫ v_out dt over T is ≈ 0 for equal positive/negative spike areas.Conclude the average (DC) output is approximately 0 V. Verification / Alternative check: On an oscilloscope with AC coupling disabled, the baseline of the differentiated waveform hovers near 0 V for symmetric pulses. Any small average offset typically points to leakage, unequal edges, or measurement bias, not to inherent DC generation by the ideal differentiator. Why Other Options Are Wrong: Incorrect: Conflicts with the differentiator’s DC blocking property. Applies only if duty cycle is 50%: Even with other duty cycles, equal and opposite edge contributions can keep the average near zero as long as the waveform returns to the same baseline each cycle. Applies only if there is DC offset: A DC offset at the input would shift the baseline but a pure differentiator does not pass steady DC. Common Pitfalls: Assuming unequal spikes due to saturation or slew limits; using a non-symmetric waveform with different high/low levels, which can introduce a small average. Final Answer: Correct.

Question 9

RC integrator capacitor voltage — how it changes: “The voltage across the capacitor in an RC integrator cannot change exponentially; it can change only instantaneously.” Judge this statement.

Accepted Answer

Introduction / Context: Capacitor behavior in first-order RC networks is a cornerstone of electronics. A common misunderstanding is that capacitor voltage can jump instantly. In reality, the defining property of a capacitor is that current is proportional to the rate of change of voltage, which leads to an exponential change in voltage for step inputs—not an instantaneous jump—when driven through a resistor. Given Data / Assumptions: RC integrator topology with capacitor connected to ground and driven through a resistor. Ideal components for the conceptual model. Input waveforms include steps and pulses. Concept / Approach: In an RC network, i_C = C * dv_C/dt. For a finite series resistance, dv_C/dt is finite for finite current; therefore, v_C cannot change instantaneously. For a step input, the solution is v_C(t) = V_final * (1 − exp(−t/τ)) + V_initial * exp(−t/τ), which is exponential in time. Only an ideal zero-impedance source directly across a capacitor (or infinite current) could enforce an instantaneous change—neither occurs in a standard RC integrator. Step-by-Step Solution: Identify that the current into the capacitor is limited by the series resistor.Relate current and voltage rate: i_C = C * dv_C/dt ⇒ finite i_C gives finite dv_C/dt.For a step input, write the first-order solution showing exponential charging/discharging with τ = R * C.Conclude that capacitor voltage changes exponentially, not instantaneously, in an RC integrator. Verification / Alternative check: Oscilloscope measurements of an RC charge curve show the well-known exponential approach to a new level, reaching ~63% in 1τ, ~86% in 2τ, ~95% in 3τ, ~98% in 4τ, and ~99% in 5τ. None of these represent an instantaneous jump. Why Other Options Are Wrong: Correct: Opposite of the physical behavior in an RC network. Valid only for ideal zero-ESR capacitors / very high frequency: ESR or frequency does not permit instantaneous voltage steps without infinite current. Common Pitfalls: Confusing an observed fast but finite rise with an instantaneous jump; ignoring series resistance or source limitations that necessarily enforce exponential transitions. Final Answer: Incorrect.

Question 10

Differentiator output polarity — falling edge behavior: In a standard RC differentiator (output taken across the resistor), does the output go negative on the falling edge of a positive input pulse?

Accepted Answer

Introduction / Context: Polarity of the spikes produced by differentiators matters when designing edge-triggered circuits and pulse shapers. A standard passive RC differentiator places a capacitor in series with the input and a resistor to ground, taking the output across the resistor. Understanding the sign of the output on rising and falling edges is essential for trigger polarity selection. Given Data / Assumptions: Series capacitor, shunt resistor, output across the resistor. Input: a positive rectangular pulse returning to zero baseline. Ideal, linear behavior; no saturation. Concept / Approach: At a rising edge (positive step), current flows through the capacitor into the resistor, dropping voltage across the resistor that is positive at the top node, so the output shows a positive spike. At a falling edge (negative step), current reverses direction; the resistor voltage reverses its polarity, producing a negative spike. Between edges, current decays toward zero and the output returns toward zero as the capacitor charges/discharges. Step-by-Step Solution: Model the falling edge as a negative step: ΔV = −V_pulse.The capacitor current is i_C = C * dV/dt, which is negative at the instant of the falling edge.Negative current through the resistor produces a negative voltage across it (top node goes negative relative to ground).Therefore, the output spike at the falling edge is negative. Verification / Alternative check: Oscilloscope traces of a differentiator driven by a pulse train show a positive spike at each rising edge and a corresponding negative spike at each falling edge, matching the sign convention presented here. Why Other Options Are Wrong: Incorrect: Ignores the direction reversal of current at the falling edge. Conditional options: Duty cycle or exact R/Xc ratio shape spike width/amplitude but not the sign at an ideal instant. Common Pitfalls: Taking the output across the capacitor instead of the resistor (which changes interpretation); losing the sign convention on current direction at edges. Final Answer: Correct.

Question 11

Differentiator with very long pulses — spike approximation: When the input pulse width is much longer than 5 time constants (T_p ≫ 5τ), can the output of an RC differentiator be considered a pair of narrow spikes at the transitions?

Accepted Answer

Introduction / Context: Edge detection is a primary use case for RC differentiators. When a rectangular pulse is very wide compared with the circuit time constant, the differentiator s output between edges is negligible. Recognizing when the spike approximation is valid helps simplify timing analysis in digital interfacing and trigger circuitry. Given Data / Assumptions: RC differentiator: series C, shunt R, output across R. Pulse width T_p much larger than 5τ = 5 * R * C. Symmetric high and low levels with a return to baseline. Concept / Approach: At each transition, dv/dt is large, producing a current impulse through the capacitor and a correspondingly sharp voltage across the resistor. After a few time constants, the capacitor current decays toward zero and the output settles very close to 0 V for the remainder of the pulse plateau. With T_p ≫ 5τ, most of the period is flat at zero output, and the response looks like two spikes per pulse—one positive at the rising edge and one negative at the falling edge. Step-by-Step Solution: Identify condition: T_p ≫ 5τ ensures complete decay between edges.At edges: i_C = C * dv/dt is large → narrow output spikes.Between edges: i_C ≈ 0 → v_out ≈ 0.Thus the waveform is well-approximated by impulse-like spikes at transitions. Verification / Alternative check: Simulations with τ = 1 ms and pulses of width 50 ms show spikes of a few milliseconds width and nearly zero output elsewhere. Scope captures in labs confirm this classical behavior used for trigger generation. Why Other Options Are Wrong: Incorrect: Contradicts standard differentiator operation for T_p ≫ 5τ. True only if the capacitor is leaky / only for certain duty cycles: Leakage and duty cycle adjust amplitudes and baselines but do not negate the spike approximation under the stated condition. Common Pitfalls: Using T_p only slightly greater than τ and expecting perfect spikes; forgetting that finite source and load impedances shape spike width and amplitude. Final Answer: Correct.

Question 12

Integrator settling — “5τ regardless of pulses” claim: “The steady-state condition of an RC integrator is reached after 5 time constants regardless of how many input pulses occur in that interval.” Evaluate this statement.

Accepted Answer

Introduction / Context: The “5τ rule” is a convenient guideline for first-order responses to a step input: after about five time constants, a system is essentially settled near its final value. However, applying this rule blindly to RC integrators driven by pulse trains can be misleading. The steady-state level depends on waveform shape, duty cycle, and repetition rate, not just elapsed time. Given Data / Assumptions: RC integrator with output across the capacitor. Input: not a single step but pulses that may repeat within the 5τ window. No saturation; linear behavior assumed. Concept / Approach: For a single step, the capacitor voltage approaches its final value as v_C(t) = V_final − (V_final − V_initial) * exp(−t/τ). After ~5τ, the error is under 1%. For a pulse train, the output is the convolution of the input with the RC exponential; the final steady-state depends on the periodic input’s average and duty cycle. Multiple pulses within 5τ continually perturb the capacitor, preventing it from reaching the same level as a single uninterrupted step would. Therefore, “steady-state after 5τ regardless of pulses” is incorrect. Step-by-Step Solution: Recognize the 5τ guideline applies to step responses, not arbitrary repeated pulses.For pulse trains, compute or reason from average value and time constants; repeated charging/discharging leads to a periodic steady-state that depends on duty cycle and period.If pulses repeat faster than the RC can settle, the capacitor never reaches the single-step asymptote.Hence the blanket statement is false. Verification / Alternative check: Simulate with τ = 10 ms and a 2 ms wide pulse repeating every 4 ms. Even after many multiples of 5τ, the waveform remains a periodic ripple around a duty-cycle-dependent mean, not a step-like flat level. This contradicts the claim. Why Other Options Are Wrong: Correct: Misapplies the 5τ rule beyond its domain. True only for single-step inputs / Undetermined without duty cycle: These clarifications highlight the real dependency; the original statement’s universality is incorrect. Common Pitfalls: Using “5τ” as a magic number for every situation; forgetting that different inputs produce different steady-state behaviors even in the same RC network. Final Answer: Incorrect.

Question 13

Where to measure — RC differentiator output node: In a standard passive RC differentiator, is the output correctly taken across the capacitor, or across the resistor?

Accepted Answer

Introduction / Context: Knowing where to take the output in basic RC networks avoids functional mistakes. In the differentiator configuration, the location of the output determines whether the circuit emphasizes edges (high-pass behavior) or averages signals (low-pass behavior). This item checks whether the output of an RC differentiator is taken across the capacitor or across the resistor. Given Data / Assumptions: Passive differentiator: series capacitor, resistor to ground. Linear small-signal analysis; no loading that would fundamentally alter behavior. Objective: produce spikes at edges (differentiate). Concept / Approach: For differentiation, the output is taken across the resistor. A series capacitor blocks DC and passes fast-changing components as current pulses into the resistor; the resistor then converts these current pulses into voltage spikes (v_out = i * R). If, instead, one measures across the capacitor, the circuit behaves as a high-pass coupling network across the dielectric but does not provide the classic resistor-referenced spike output of a differentiator. Therefore, the statement “output of an RC differentiator is taken across the capacitor” is incorrect for the standard configuration. Step-by-Step Solution: Identify topology: input → C_series → node → R_to_ground.Take output at the node across R: v_out = i_C * R = C * dv_in/dt * R → spikes at edges.If output were across C, the behavior would not match the standard differentiator definition.Hence, the correct output node is across the resistor, not the capacitor. Verification / Alternative check: Textbook diagrams and lab demonstrations consistently show v_out measured across the resistor in the basic RC differentiator. Oscilloscope waveforms confirm positive and negative spikes corresponding to input edges. Why Other Options Are Wrong: Correct: Misidentifies the output node. Frequency- or polarity-conditional options: The topology determines the correct node regardless of frequency or capacitor polarity (for non-polarized parts). Common Pitfalls: Confusing the differentiator with the RC integrator (output across the capacitor) due to their dual nature; mixing schematic conventions that place probes differently. Final Answer: Incorrect.

Question 14

RC integrator fault diagnosis — zero output reading: “If the output of an RC integrator is zero volts, the capacitor might be open.” Decide whether this is a reliable diagnostic conclusion for the standard topology (output across the capacitor).

Accepted Answer

Introduction / Context: Fault-finding in simple RC networks often relies on qualitative symptoms. In an RC integrator (input → R_series → node → C_to_ground, with output measured at the node across C), a “zero volts” reading at the output suggests a specific type of failure. This question checks whether “capacitor open” is the likely cause or if another fault better explains the symptom. Given Data / Assumptions: Standard RC integrator topology with output across the capacitor. Assume the input provides a nonzero signal. Measurement made under load that does not by itself short the output to ground. Concept / Approach: In the integrator, the capacitor must charge and discharge to develop output voltage. If the capacitor is shorted, the output node is effectively clamped to ground potential, resulting in ~0 V regardless of input (apart from possible small source resistance drops). If the capacitor is open, the node becomes floating; depending on stray/leak paths, the output may drift, show spikes via parasitics, or even mirror portions of the input through stray capacitances—but it is not automatically a rock-solid 0 V. Thus, “0 V ⇒ capacitor open” is not the best conclusion; “0 V ⇒ capacitor short” is more consistent with the observed symptom. Step-by-Step Solution: Identify symptom: V_out ≈ 0 V while a signal is applied.Analyze shorted C case: node hard-tied to ground → output forced to ~0 V.Analyze open C case: node floating; without a discharge path, DC may float and AC coupling may produce some nonzero behavior.Conclude that “capacitor open” is not the most plausible cause for a strict zero reading; a shorted capacitor is. Verification / Alternative check: Practical troubleshooting guides list “shorted shunt capacitor” as a cause of no output in RC coupling/integrator stages. Replacing the capacitor or lifting one lead to test continuity confirms the diagnosis empirically. Why Other Options Are Wrong: Correct: Misidentifies the most likely fault given a zero-voltage output. Applies only if series resistor is shorted / Undetermined…: While other faults can also kill the output, the specific claim about an open capacitor being the cause is not reliable. Common Pitfalls: Equating “no signal” with any component being open; overlooking that an open component often yields a floating node (undefined), not a hard zero. Final Answer: Incorrect.

Question 15

RC pulse response with duty cycle — a repetitive rectangular pulse (50% duty) is applied to the input of an RC “integrator.” If one time constant τ is less than one-fifth of the pulse width (τ < PW/5), will the capacitor be able to essentially charg…

Accepted Answer

Introduction / Context: An RC “integrator” is simply an RC network driven by a pulse or step where the output of interest is usually taken across the capacitor. Whether it truly integrates or instead reaches near steady states during each pulse depends on the relationship between the time constant τ = R * C and the pulse width (PW) and duty cycle. This question asks if a capacitor can effectively charge and discharge to its final values within each half-cycle when τ is less than one-fifth of the pulse width in a 50% duty train. Given Data / Assumptions: Input: repetitive pulse, 50% duty (high time = low time = PW/2 if period = PW). RC network with time constant τ = R * C. “Fully” or “completely” is interpreted in practical design sense as ≈99% settling by about 5τ. Ideal source and components for first-order analysis; no loading beyond the RC itself. Concept / Approach: First-order exponentials settle to about 63% in 1τ, 86% in 2τ, 95% in 3τ, 98% in 4τ, and 99.3% in 5τ. Thus, when a pulse high (or low) interval exceeds roughly 5τ, the capacitor will reach within about 1% of its final value for that interval. With a 50% duty waveform, there is a charging interval and a discharging interval of the same length; the same 5τ rule applies to both halves. Step-by-Step Solution: Relate settling to time constant: ≈5τ → 99% of final value.Given τ < PW/5, the pulse high time (PW/2) is > 2.5τ, and if PW is defined as the high duration itself, then τ < PW/5 means 5τ < PW, satisfying the 99% criterion.Apply the same logic to the low interval (50% duty → equal time for discharge).Conclude the capacitor essentially charges and discharges each half-cycle under the stated inequality. Verification / Alternative check: Use v_C(t) = V_final + (V_initial − V_final) * exp(−t/τ). For t = 5τ, |v_C − V_final| ≈ 0.7% of the step. With τ < PW/5, t ≥ 5τ within each interval, yielding ≈99% settling. Why Other Options Are Wrong: Incorrect / depends only on duty cycle: settling depends primarily on τ relative to the interval length, not duty alone. Only if AC-coupled or R = C: irrelevant to the exponential time constant rule. Common Pitfalls: Confusing “integrator behavior” (which prefers τ ≫ PW) with “full charging” (which needs τ ≪ PW). The same RC can show either behavior depending on τ vs. PW. Final Answer: Correct — with τ < PW/5, the capacitor essentially reaches its final value each half-cycle.

Question 16

RL pulse shaping — in a basic RL differentiator (step/pulse shaping network), is the output conventionally taken across the inductor to emphasize rapid changes (spikes at edges)?

Accepted Answer

Introduction / Context: Differentiator networks generate outputs that are large during rapid input changes and small during steady levels. An RL differentiator uses the inductor’s property v_L = L * di/dt, producing voltage spikes at rising and falling edges of a pulse. Knowing where the output is taken (across which element) is essential for correct classification and design. Given Data / Assumptions: Series RL excited by steps/pulses. Small-signal, linear behavior; ideal inductor. Load is high-impedance measurement so as not to disturb the network. Concept / Approach: For a series RL with an applied step, the current cannot change instantaneously, so the inductor initially takes nearly the entire input voltage and then its voltage decays exponentially as current builds. This “edge-peaked” v_L is proportional to di/dt and therefore implements differentiation. Hence, the conventional RL differentiator output is measured across the inductor. Step-by-Step Solution: Write inductor relation: v_L(t) = L * di/dt.Apply a step or pulse: di/dt is large at edges, small in between.Therefore v_L is large at edges (spikes) and small between edges → differentiation.Conclusion: take the output across L for RL differentiator behavior. Verification / Alternative check: Transient plots show v_L peaks with exponential decay after each transition, while the resistor’s voltage tends to follow the slowly changing current, not the derivative highlight. Why Other Options Are Wrong: Incorrect or “only for square waves”: differentiation property holds for any changing waveform; square waves just make spikes obvious. Constraints like R ≫ X_L or core material: affect shape and amplitude but not the fundamental placement of the output for differentiation. Common Pitfalls: Confusing RL differentiator with RC differentiator (where output is taken across the resistor). Final Answer: Correct — the output of an RL differentiator is taken across the inductor.

Question 17

RC integrator under pulses — when a repetitive-pulse waveform drives an RC integrator, does the output waveshape depend on the relationship between the time constant τ and the pulse duty cycle/width?

Accepted Answer

Introduction / Context: RC “integrators” and “differentiators” are first-order networks whose actual output shape depends on how the input timing compares with the natural time constant τ = R * C. With repetitive pulses, designers can choose τ to emphasize integration (slow change) or near-square tracking (full settling). Duty cycle also matters because it changes the high/low durations of each cycle. Given Data / Assumptions: Input: repetitive pulses with a defined duty cycle D. Circuit: RC with output observed across the capacitor (typical integrator view). Lumped linear components and negligible loading. Concept / Approach: If τ ≫ pulse width, the capacitor voltage changes only slightly during each pulse, approximating a scaled integral of the waveform (ramp-like). If τ ≪ pulse width, the capacitor reaches near its final value during the high interval and returns near the low level afterward, producing a waveform closer to the rectangular input. Changing duty cycle alters the relative lengths of charge and discharge intervals, shifting the DC level and the waveform shape in steady state. Step-by-Step Solution: Relate τ to pulse high duration: charge extent depends on t_high/τ.Relate τ to pulse low duration: discharge extent depends on t_low/τ.Vary duty cycle D = t_high / period → moves the average and ripple shape.Conclude that both τ and duty cycle jointly determine the output waveshape. Verification / Alternative check: Use v_C(t) = V_final + (V_initial − V_final) * exp(−t/τ) on high and low intervals; steady-state levels result from equalizing end-of-interval conditions over one period. Why Other Options Are Wrong: “Only amplitude” or “only frequency”: incomplete; τ and the exact time lengths control exponential settling. “Depends solely on ESR”: ESR slightly perturbs τ and damping but is not the dominant factor. Common Pitfalls: Calling any RC with a pulse an “integrator” regardless of τ; neglecting the effect of non-50% duty on the average output level. Final Answer: Correct — output shape depends on τ and the duty-cycle timing.

Question 18

Settling criterion — for a capacitor to “completely” (≈99%) charge during the on-time of a pulse in an RC network, the pulse width should be related to the time constant τ how?

Accepted Answer

Introduction / Context: In first-order systems, “complete” charging is a practical notion. Engineers often use 5τ as a rule of thumb for ≈99% settling of an exponential response. This item asks how the pulse width (the available charging time) must compare with τ to achieve near-full charging during a pulse interval. Given Data / Assumptions: RC network with time constant τ = R * C. Exponential approach to a new level during a pulse. “Completely” interpreted as ≈99% (5τ rule). Concept / Approach: The capacitor voltage follows v_C(t) = V_final + (V_initial − V_final) * exp(−t/τ). After t = 5τ, the residual error is exp(−5) ≈ 0.0067, or about 0.7%. Thus, a pulse that lasts at least 5τ allows the capacitor to get within about 1% of its asymptotic value for that interval. Shorter pulses yield proportionally less settling. Step-by-Step Solution: Define settling goal: ≈99% in practical design.Use the exponential error term e^(−t/τ).Solve e^(−t/τ) ≤ 0.01 → t ≥ 4.6τ; designers round to 5τ.Therefore, PW ≥ 5τ is a standard engineering guideline. Verification / Alternative check: Simulation or measurement of step response shows 63% at 1τ, 95% at 3τ, ≈99% at 5τ, reinforcing the rule. Why Other Options Are Wrong: “Less than 5τ” or “independent of τ”: would not reach ≈99% reliably. “Equal to exactly 5τ”: a useful benchmark, but “greater than or equal” is the safer, more general statement. Common Pitfalls: Treating “complete” as 100% in finite time; exponentials only approach asymptote, so 5τ is a practical design target. Final Answer: Greater than or equal to 5 time constants.

Question 19

Repair for missing schematic — “If the pulse width were cut in half in the given RC pulse circuit, the voltage across the resistor at the end of the pulse would be ______.” Without R, C, τ, and the original pulse width, can a unique numeric value be…

Accepted Answer

Introduction / Context: For an RC network excited by a pulse, the voltage across the resistor at the end of the pulse depends on the exponential evolution during the on-time. That evolution is set by the time constant τ = R * C and the actual on-time (pulse width). Halving the pulse width changes the settling time, but without the numeric τ and original timing, the final value cannot be pinned down to a single number. Given Data / Assumptions: No values for R, C, or τ are provided. No explicit pulse amplitude or original pulse width is given. Output node location is not fully specified in the missing figure. Concept / Approach: The resistor voltage in a simple RC driven by a step is v_R(t) = V_in * exp(−t/τ) (for a series RC with output across R after a step to V_in), or the complement thereof depending on the exact topology. At t = PW (end of pulse), v_R(PW) depends on exp(−PW/τ). If PW is halved, the new value depends on exp(−(PW/2)/τ). Without τ and PW, the numeric outcome is undetermined. Step-by-Step Solution: State general form: v_end ∝ exp(−PW/τ) (topology dependent scale).Halve PW → v_end,new depends on exp(−(PW/2)/τ).No τ or PW given → cannot compute a numeric result.Therefore, a single numeric choice is not justifiable. Verification / Alternative check: Try two examples: If PW = 5τ, exp(−PW/τ) ≈ 0.007; if PW = 0.5τ, exp(−PW/τ) ≈ 0.607. The outcomes are vastly different, proving dependence on missing data. Why Other Options Are Wrong: 0 V / equals source / fixed fractions: these assert specific conditions that may or may not hold; they require τ and PW. “One-half of the original”: exponential processes do not in general scale linearly with time halving. Common Pitfalls: Assuming a universal “0.37” or “0.5” result; those come from particular τ:PW ratios, not from the act of halving itself. Final Answer: Cannot be determined from the information provided.

Question 20

Repair for missing values — “If the pulse source has an internal resistance of 80 Ω in the given circuit, it will take ______ for the output voltage to decrease to zero.” Can a time-to-zero be specified for a first-order RC without knowing C (and to…

Accepted Answer

Introduction / Context: First-order RC voltages decay exponentially and never reach absolute zero in finite time; designers use practical thresholds (e.g., 1τ ≈ 63% settled, 5τ ≈ 99% settled). The question as written lacks the capacitance and full topology, yet asks for a specific time “to zero,” which is not physically precise and cannot be computed numerically without τ. Given Data / Assumptions: Only the source resistance Rs = 80 Ω is mentioned. Capacitance C and any additional series/parallel resistances are unknown. No threshold (e.g., to 1% or to 0.1%) is specified. Concept / Approach: Decay time constant is τ = R_eq * C, where R_eq is the effective resistance seen by the capacitor during discharge (often Rs plus any series elements). Without C and R_eq, τ is unknown. Moreover, an exponential strictly reaches zero only as t → ∞; practical “zero” must be defined via a percentage, such as 5τ for ≈99% decay. Step-by-Step Solution: Identify missing parameters: C (and full R_eq).Recognize exponential nature: v(t) = V0 * exp(−t/τ).Select a practical threshold (if provided) to compute t; absent this, no unique time exists.Conclude that a numeric answer cannot be given. Verification / Alternative check: Pick example C values: with C = 1 µF, τ = 80 µs; with C = 10 µF, τ = 800 µs. Times differ by 10×, showing sensitivity to the missing parameter. Why Other Options Are Wrong: “Exactly 5τ” or “1τ”: these are rules of thumb for percentages, not true zero. “Independent of C” or “equals the pulse period”: incorrect generalizations. Common Pitfalls: Forgetting that first-order exponentials only asymptotically reach zero; ignoring the need to specify a decay threshold. Final Answer: Cannot be determined from the information provided.

Time Response of Reactive Circuits Questions