Question 1

Which power quantity is zero for an ideal capacitor, indicating that it dissipates no real power?

Accepted Answer

Introduction: This question probes understanding of power in reactive components. An ideal capacitor stores and releases energy without dissipating it as heat. Distinguishing between true, reactive, and apparent power clarifies why real power dissipation is zero in an ideal capacitor. Given Data / Assumptions: Ideal capacitor with no equivalent series resistance (ESR). Sinusoidal steady state AC. Voltage and current are 90 degrees out of phase (current leads). Concept / Approach: Power types: true (real) power P, reactive power Q, and apparent power S. For a purely reactive element, average power over a cycle is zero because energy taken from the source during part of the cycle is returned during another part. Thus, true power equals zero. Reactive power is nonzero and measured in VAR, representing energy exchange with the field. Step-by-Step Solution: Instantaneous power p(t) = v(t) * i(t) oscillates positive and negative.Average over one full cycle: P = average[p(t)] = 0 for an ideal capacitor.Reactive power: Q = V * I * sin(phi); for capacitor, phi = −90°, so |Q| = V * I.Apparent power: S = V * I (VA) is nonzero but does not represent dissipation. Verification / Alternative check: Energy stored per cycle W = 0.5 * C * V^2 at peak and is returned each half cycle. No net energy accumulates as heat, confirming P = 0 W for the ideal case. Why Other Options Are Wrong: Instantaneous power: Not always zero; it alternates in sign. Reactive power: Not zero; it quantifies energy exchange (VAR), not dissipation. Resistive power: Not a standard term; dissipation requires resistance, which an ideal capacitor lacks. Common Pitfalls: Confusing reactive and true power, assuming zero current, or ignoring ESR in practical capacitors, which introduces small but nonzero losses. Final Answer: True power is zero for an ideal capacitor, meaning it dissipates no real power.

Question 2

Effect of decreasing dielectric constant (relative permittivity k) on a capacitor’s capacitance, with geometry unchanged

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Introduction: Capacitance depends on the physical geometry and the dielectric material between the plates. This question examines how changing the dielectric constant alone affects capacitance when plate area and separation remain fixed. Given Data / Assumptions: Parallel-plate model is applicable. Plate area A and separation d are unchanged. Dielectric constant (relative permittivity) k is decreased. Concept / Approach: The basic relation for a parallel-plate capacitor is C = k * epsilon_0 * A / d, where epsilon_0 is the permittivity of free space and k is the relative permittivity of the dielectric. Capacitance is directly proportional to k; therefore, reducing k reduces C linearly when A and d are fixed. Step-by-Step Solution: Start from C = k * epsilon_0 * A / d.Treat A, d, and epsilon_0 as constants.Decrease k → the numerator decreases → C decreases proportionally. Verification / Alternative check: Practical observation: inserting a material with higher k (e.g., ceramic with high-k dielectric) increases capacitance for the same size, while switching to a lower-k material decreases it. Measurements align with the formula above. Why Other Options Are Wrong: Increase: Opposite to the formula's direct proportionality to k. Remain the same: Would require k to be unchanged or A/d to counterbalance. Be destroyed: Changing k does not inherently damage the capacitor. Common Pitfalls: Confusing dielectric constant with dielectric strength, or assuming geometry changes when only material changes. Also, mixing AC breakdown behavior with static capacitance value. Final Answer: Decrease.

Question 3

A series capacitor–resistor network with the output taken across the resistor is driven by an AC signal. What filter action results?

Accepted Answer

Introduction: Simple RC networks can implement first-order filters. Whether the network is low-pass or high-pass depends on which component's voltage is used as the output. This question checks understanding of a series C–R with the output across the resistor. Given Data / Assumptions: Input: AC signal applied in series to capacitor and resistor. Output: Voltage across the resistor (not across the capacitor). Linear, time-invariant components and small-signal conditions. Concept / Approach: In a series C–R network, the capacitor's impedance is Xc = 1 / (2 * pi * f * C). At low frequencies, Xc is large, most input drops across the capacitor, and the resistor sees little voltage. At high frequencies, Xc becomes small, more voltage appears across R, so the output increases with frequency. Therefore, the configuration acts as a high-pass filter with cutoff fc = 1 / (2 * pi * R * C). Step-by-Step Solution: Compute Xc = 1 / (2 * pi * f * C).At f → 0: Xc → ∞ → V_R ≈ 0 → attenuates low frequencies.At f → ∞: Xc → 0 → V_R ≈ V_in → passes high frequencies.Therefore, magnitude response rises with frequency and the circuit is high-pass. Verification / Alternative check: Bode magnitude plot of |V_R/V_in| shows a +20 dB/decade slope below cutoff and 0 dB above cutoff, confirming a first-order high-pass behavior. Why Other Options Are Wrong: Bandpass: Requires at least second-order behavior or RLC; not present here. Low-pass: Would occur if output were taken across the capacitor. Band-stop: Needs a topology that attenuates a band; not applicable. Common Pitfalls: Confusing which node is the output or mixing up capacitor and resistor positions; also assuming cutoff uses 0.707 amplitude without linking it to the correct fc expression. Final Answer: High-pass.

Question 4

In an RC AC circuit, what is the correct expression for true (real) power PR?

Accepted Answer

Introduction: True (real) power is the portion of power converted to heat or useful work. In an RC AC circuit, only the resistor dissipates real power; the capacitor exchanges energy with the source but does not dissipate it (ideal case). This item checks recognition of the correct algebraic form for PR. Given Data / Assumptions: Series or parallel RC under sinusoidal steady state. Ideal capacitor with zero loss. R is the only dissipative element. Concept / Approach: The resistor's real power is PR = I_rms^2 * R. Alternative forms include PR = V_R(rms)^2 / R. Apparent power S = V_rms * I_rms, while reactive power Q is associated with the capacitor. The product VS * I does not generally equal real power when there is phase shift between voltage and current. Step-by-Step Solution: Identify the dissipative element: the resistor R.Use PR = I^2 * R (rms quantities).Capacitor contributes Q, not P, because current and voltage are 90° out of phase for the reactive branch. Verification / Alternative check: Compute P = V * I * cos(phi). For a series RC, cos(phi) = R / Z. With I the same through both components, PR = I^2 * R, which matches P = V * I * cos(phi). Why Other Options Are Wrong: VS × I: This is apparent power when there is a phase shift; not necessarily real power. VT × RT: Not a standard power expression; has unit inconsistencies. Common Pitfalls: Equating V * I with real power regardless of phase, or neglecting that only resistors dissipate real power in the ideal model. Final Answer: I^2 * R.

Question 5

Diagnosing an open (failed) capacitor using an analog ohmmeter: what needle behavior should you expect?

Accepted Answer

Introduction: Using an analog ohmmeter to assess capacitors relies on observing transient charging behavior. A healthy capacitor briefly conducts as it charges from the meter's internal battery, then its indicated resistance rises. An open capacitor behaves differently. This question targets practical troubleshooting skills. Given Data / Assumptions: Analog ohmmeter with internal DC source. Capacitor disconnected from circuit and discharged before testing. Looking for qualitative needle movement. Concept / Approach: A good capacitor initially appears as a low resistance (charging current flows) and then the resistance increases as the capacitor voltage builds, approaching infinity when fully charged. An open capacitor has a broken internal connection, so no charging current flows and the meter reads open circuit constantly. Step-by-Step Solution: Connect leads with correct polarity (for electrolytics) or either way (for nonpolarized).Observe needle: a good capacitor moves from low resistance toward high resistance over a second or two.For an open capacitor, no movement occurs; the needle remains at infinity because current cannot flow. Verification / Alternative check: Cross-check with a capacitance meter or ESR meter. An open capacitor will show extremely low capacitance (near zero) and often infinite ESR. Why Other Options Are Wrong: Stay on zero: Indicates a shorted capacitor, not open. Move from zero to infinity: Describes a healthy capacitor's transient response. Move from infinity to zero: Suggests a short developing; not open. Common Pitfalls: Failing to discharge the capacitor before testing, misinterpreting the analog scale, or testing in-circuit where parallel paths can mask the result. Final Answer: Stay on infinity (open circuit).

Question 6

Phase relationship in a purely capacitive AC circuit: how does current relate to voltage?

Accepted Answer

Introduction: Understanding phase relationships is fundamental to AC circuit analysis. In purely reactive elements, voltage and current are out of phase. This question asks for the specific relationship in a purely capacitive circuit. Given Data / Assumptions: Ideal capacitor (no resistance). Sinusoidal steady state. Single-frequency excitation. Concept / Approach: The impedance of a capacitor is Zc = 1 / (j * 2 * pi * f * C) = −j * Xc. The negative imaginary sign indicates current leads voltage by 90 degrees. Equivalently, i(t) = C * dv/dt; the derivative emphasizes that current peaks when voltage crosses zero and vice versa, producing a +90° current lead. Step-by-Step Solution: Write Zc = 1 / (j * ω * C) = −j / (ω * C).Phasor current I = V / Zc = V * (j * ω * C) → +90° relative to V.Hence current leads voltage by a quarter cycle. Verification / Alternative check: Time-domain relation i(t) = C * dv/dt shows current is maximum at voltage zero crossings and zero at voltage peaks, consistent with a 90° lead. Why Other Options Are Wrong: Voltage leads by 90° or current lags by 90°: These describe inductive, not capacitive, circuits. In phase: Requires resistive behavior; not applicable to ideal capacitors. Common Pitfalls: Confusing the sign convention of j or mixing inductive and capacitive phase relations. Remember ‘‘ICE’’: in a Capacitor, Current leads Voltage. Final Answer: Current leads voltage by 90°.

Question 7

What is the name of the power that is alternately stored in a capacitor’s electric field and then returned to the source?

Accepted Answer

Introduction: AC circuits feature different kinds of power: true (real), reactive, and apparent. This item asks you to identify the kind associated with energy shuttling in and out of reactive fields without net dissipation over a cycle. Given Data / Assumptions: Ideal capacitor with negligible losses. Sinusoidal AC excitation. Periodic steady-state conditions. Concept / Approach: Reactive power Q describes energy exchange between the source and the reactive element's field. For capacitors, energy is stored as W = 0.5 * C * V^2 and then returned to the source later in the cycle. Average real power P over a full cycle is zero for the ideal component, but Q is nonzero and measured in VAR (volt-ampere reactive). Step-by-Step Solution: Recognize that no heat is produced in the ideal capacitor → P = 0 W.Energy sloshes between source and field → quantified by Q, not P.Therefore, the correct term is reactive power. Verification / Alternative check: Phasor analysis: For a capacitor, current leads voltage by 90°, so cos(phi) = 0 and sin(phi) = −1. Real power P = V * I * cos(phi) = 0; reactive power magnitude |Q| = V * I is nonzero. Why Other Options Are Wrong: Apparent power: S = V * I (VA) includes both P and Q; not specifically the stored-and-returned component. Impedance power: Not a standard term in AC power theory. True power: Represents average dissipation; zero in an ideal capacitor. Common Pitfalls: Equating apparent with reactive power, or thinking reactive power is ‘‘waste’’; actually it represents necessary energy exchange to sustain fields in reactive elements. Final Answer: Reactive power.

Question 8

A 100 V DC source charges a capacitor. What is the capacitor voltage after one time constant τ from the start of charging?

Accepted Answer

Introduction: First-order RC charging follows a well-known exponential. This question checks your ability to apply the standard time-constant result to find the capacitor voltage after one τ when charging from a step DC source. Given Data / Assumptions: Step input Vs = 100 V applied at t = 0. Capacitor initially uncharged. Time of interest t = τ = R * C. Concept / Approach: The capacitor voltage during charging is Vc(t) = Vs * (1 − e^(−t/τ)). Substituting t = τ yields Vc(τ) = Vs * (1 − e^(−1)). The numerical factor 1 − e^(−1) ≈ 0.632 applies to any RC with a step input. Step-by-Step Solution: Write Vc(t) = Vs * (1 − e^(−t/τ)).Set t = τ → Vc(τ) = Vs * (1 − e^(−1)).Compute 1 − e^(−1) ≈ 1 − 0.3679 ≈ 0.6321.With Vs = 100 V → Vc(τ) ≈ 100 * 0.6321 ≈ 63.21 V.Closest given option is 63 V. Verification / Alternative check: The complementary current at t = τ is i(τ) = (Vs/R) * e^(−1) ≈ 0.368 * (Vs/R); this pairs with Vc(τ) ≈ 0.632 * Vs, confirming standard RC behavior. Why Other Options Are Wrong: 37.8 V or 38 V: These correspond to the remaining fraction e^(−1) of the initial step on the resistor, not the capacitor voltage at τ. 101 V: Exceeds the source; impossible without an inductor or overshoot mechanism. Common Pitfalls: Confusing the capacitor's voltage with the resistor's drop at t = τ, or rounding errors that swap 0.632 and 0.368 fractions. Final Answer: 63 V.

Question 9

In basic electronics and circuit theory, the material inserted between the conducting plates of a capacitor plays a critical role in how the device stores electric energy. What is the technical name for this material placed between the plates of a c…

Accepted Answer

Introduction: Capacitors store electric energy by maintaining an electric field between two conductive plates separated by a non-conductive medium. This question checks your understanding of the correct terminology for that medium and why it determines key properties of a capacitor. Given Data / Assumptions: Two conductive plates form a capacitor. A non-conductive medium fills the space between the plates. We are naming that medium and relating it to capacitor behavior. Concept / Approach: The non-conductive medium between plates is called the dielectric. Its dielectric constant (relative permittivity) and breakdown strength strongly affect capacitance, leakage, and maximum working voltage. Step-by-Step Solution: 1) Identify the role of the inter-plate medium: prevent direct conduction while supporting an electric field.2) Recognize the standard term: dielectric.3) Connect term to effects: larger dielectric constant increases capacitance for given plate area and separation.4) Note practical implications: dielectric breakdown sets safe voltage limits; dielectric losses affect AC performance. Verification / Alternative check: Reference the parallel-plate formula: C = (epsilon_r * epsilon_0 * A) / d. The term epsilon_r is the dielectric constant, confirming that the material is specifically called a dielectric. Why Other Options Are Wrong: Insulation material: generic term; correct technical term here is dielectric.Separation material: imprecise; does not capture electrical properties.Plate-divider material: non-standard wording. Common Pitfalls: Confusing general insulation with dielectric specificity, and ignoring that dielectric properties directly set capacitance and maximum voltage rating. Final Answer: dielectric material

Question 10

In RC charging theory for first-order circuits, engineers use the time constant τ = R * C to estimate charging progress. After how many time constants can a capacitor be considered 'practically' fully charged (about 99% of the supply voltage) in typ…

Accepted Answer

Introduction: Time constant analysis provides a quick way to estimate how fast a capacitor charges in a first-order RC circuit. This question targets the rule of thumb engineers use to declare a capacitor essentially fully charged without waiting for true mathematical completion. Given Data / Assumptions: Series RC charging from a DC source. Time constant τ defined as R * C. Fully charged in practice means near the final value. Concept / Approach: Capacitor voltage during charging is Vc(t) = Vs * (1 - exp(-t / τ)). Exact final value is approached asymptotically, but design convention treats about 99% as practically full. Step-by-Step Solution: 1) At t = 1τ, Vc ≈ 0.632 * Vs.2) At t = 3τ, Vc ≈ 0.950 * Vs.3) At t = 5τ, Vc ≈ 0.993 * Vs, commonly accepted as practically full charge.4) Beyond 5τ the change becomes negligible for most applications. Verification / Alternative check: Compute exp(-5) ≈ 0.0067, so Vc(5τ) ≈ (1 - 0.0067) * Vs ≈ 0.993 * Vs, which is near 99.3% of final value. Why Other Options Are Wrong: First: only ~63% charged.Third: about 95%, not typically called fully charged.Seventh: unnecessary waiting; 5τ already meets practical criteria. Common Pitfalls: Assuming 100% occurs at a finite time; in first-order systems, the exponential never truly reaches the final value, so engineers use 5τ as a practical benchmark. Final Answer: fifth

Question 11

An AC signal conditioning stage uses a series RC network driven by a square wave, with the output taken across the resistor. In basic signal processing terms, this series capacitor followed by a resistor (output across R) behaves as which type of ci…

Accepted Answer

Introduction: RC networks are widely used to shape waveforms. A classic configuration places a capacitor in series with the input and a resistor to ground, with the output taken across the resistor. This question identifies the functional behavior of that network when excited by a square wave. Given Data / Assumptions: Series capacitor, then resistor to reference. Output node measured across the resistor. Input excitation is a square wave. Concept / Approach: A series C with output across R forms a high-pass RC. For fast edges (square wave), the output approximates the time derivative of the input, producing narrow positive and negative spikes at transitions. Hence it acts as a differentiator when τ is small compared to the input period. Step-by-Step Solution: 1) Identify topology: capacitor first, resistor second, output across R.2) Recognize high-pass behavior: low frequencies blocked, rapid changes passed.3) For a square wave, edges are rapid changes; output shows spikes proportional to dVin/dt.4) Practical design sets RC such that τ << waveform period for good differentiation. Verification / Alternative check: Observe oscilloscope response: spikes at rising and falling edges confirm differentiator action; steady-state between edges decays toward zero due to capacitor AC coupling. Why Other Options Are Wrong: Integrator: requires output across C (low-pass), not R in a high-pass path.Multiplier: nonlinear function; RC linear and time-invariant.Divider: passive division is frequency-dependent, but the key behavior here is differentiation. Common Pitfalls: Confusing placement of the output node; swapping which element you measure across inverts the function (integrator vs differentiator). Final Answer: a differentiator

Question 12

For a parallel-plate capacitor, which set of physical factors primarily determines its capacitance value in practice, assuming ideal geometry?

Accepted Answer

Introduction: Capacitance is a geometric and material property that dictates how much charge a capacitor stores per volt. This question asks you to identify the parameters that directly appear in the capacitance expression for an ideal parallel-plate structure. Given Data / Assumptions: Parallel-plate approximation. Uniform dielectric fills the gap. Fringing and losses neglected. Concept / Approach: The ideal formula is C = (epsilon_r * epsilon_0 * A) / d, where A is plate area, d is separation, epsilon_r is dielectric constant, and epsilon_0 is vacuum permittivity. These factors directly control capacitance. Step-by-Step Solution: 1) Start from C = (epsilon_r * epsilon_0 * A) / d.2) Identify variables: A increases C linearly; larger epsilon_r increases C; larger d reduces C.3) Recognize that dielectric strength and temperature coefficient affect ratings and stability, not the first-order ideal C value. Verification / Alternative check: Compare two capacitors with same A and d but different materials; the one with larger epsilon_r always has larger capacitance, confirming the direct dependence on dielectric constant. Why Other Options Are Wrong: Temperature coefficient: influences change with temperature, not base value.Voltage rating: mostly tied to dielectric strength and construction safety, not C directly.Dielectric strength: a breakdown property; does not set C in the ideal formula. Common Pitfalls: Confusing dielectric strength (breakdown) with dielectric constant (permittivity). They are different properties with different roles. Final Answer: Plate area, dielectric constant, and plate separation

Question 13

A capacitor is charged from an 8 V DC source through a resistor (first-order RC). What voltage will appear across the capacitor exactly at t = 2τ (two time constants) during the charging process?

Accepted Answer

Introduction: First-order RC charging follows an exponential law. Knowing how to compute the capacitor voltage at a specified multiple of the time constant is essential for timing, settling, and design margin decisions. Given Data / Assumptions: Supply voltage Vs = 8 V. Time instant t = 2τ. Standard RC charging starting from 0 V initial condition. Concept / Approach: The charging law is Vc(t) = Vs * (1 - exp(-t / τ)). Substitute t = 2τ to compute the fraction of the final value reached at two time constants. Step-by-Step Solution: 1) Start with Vc(t) = 8 * (1 - exp(-t / τ)).2) At t = 2τ, Vc = 8 * (1 - exp(-2)).3) exp(-2) ≈ 0.1353.4) 1 - 0.1353 = 0.8647.5) Vc ≈ 8 * 0.8647 ≈ 6.9176 V, which rounds to about 6.9 V, commonly reported as 7 V. Verification / Alternative check: Check limits: at t = 0, Vc = 0; as t increases, Vc approaches 8 V. Two time constants giving roughly 86% aligns with standard RC charging tables. Why Other Options Are Wrong: 5 volts: corresponds to about 62.5% of 8 V, closer to t ≈ 1τ.3 volts: too low; would be early in the transient.1 volt: far too early; not consistent with 2τ. Common Pitfalls: Forgetting the exponential form and attempting linear interpolation; RC charging is not linear with time. Final Answer: 7 volts

Question 14

During the charging of a capacitor through a resistor from a DC source, how does the current flowing into the capacitor change with time?

Accepted Answer

Introduction: The transient behavior of current in an RC circuit reveals how quickly a capacitor charges and how the system settles. This question probes your understanding of the current waveform during the charging phase. Given Data / Assumptions: Series RC charging from a constant DC source. Capacitor initially uncharged. Ideal first-order model. Concept / Approach: In RC charging, i(t) = (Vs / R) * exp(-t / τ). The current is maximal at t = 0 and decays exponentially to zero as the capacitor approaches the supply voltage and the net voltage across the resistor falls. Step-by-Step Solution: 1) At t = 0, capacitor voltage is 0, so i(0) = Vs / R (maximum).2) As Vc rises, the voltage across R drops, reducing current.3) Exponential decay continues until steady state, where current ideally becomes zero. Verification / Alternative check: Differentiate Vc(t) = Vs * (1 - exp(-t / τ)) to see i(t) = C * dVc/dt = (Vs / R) * exp(-t / τ), confirming exponential decrease. Why Other Options Are Wrong: Increase: opposite of actual behavior; current is highest at the start.Remain the same: contradicts exponential law.Cannot tell: the first-order model clearly predicts the trend. Common Pitfalls: Assuming constant current as with an ideal current source; here the source is a constant voltage and the resistor limits current dynamically. Final Answer: decrease

Question 15

Capacitor code reading: Many small capacitors use a three-digit marking. For a device labeled 102, what is its capacitance value expressed in pF?

Accepted Answer

Introduction: Component codes compress values onto tiny packages. Understanding the capacitor three-digit code is essential for quick identification during prototyping and repair. Given Data / Assumptions: Code format: XY Z, where XY are significant digits and Z is the multiplier in pF. Marking is 102. Units are interpreted in picofarads unless otherwise stated. Concept / Approach: Use the rule: capacitance (pF) = (first two digits) * 10^(third digit). Then, convert to nF or µF only if asked. Here we keep pF directly. Step-by-Step Solution: 1) Significant digits: 10.2) Multiplier: 10^2 = 100.3) Capacitance: 10 * 100 = 1000 pF.4) Optional conversions: 1000 pF = 1 nF = 0.001 µF. Verification / Alternative check: Cross-check with standard EIA code tables for ceramic capacitors: 102 always maps to 1000 pF. Why Other Options Are Wrong: 10,000 pF: corresponds to 103, not 102.1000 µF and 10,000 µF: unrealistic for small coded ceramics; these are electrolytic-scale values. Common Pitfalls: Misreading the third digit as a literal digit rather than a power-of-ten multiplier; forgetting default units are pF for this code system. Final Answer: 1000 pF

Question 16

Real-world modeling: A practical capacitor is not ideal. Which simple equivalent model best represents its behavior for DC leakage and AC response in many applications?

Accepted Answer

Introduction: Ideal capacitors block DC and pass AC according to reactance, but practical devices also exhibit leakage and losses. This question targets the most common first-order model used in circuit analysis and datasheets. Given Data / Assumptions: Focus on a simple, low-order equivalent. Consider DC leakage and basic AC behavior. Ignore series inductance and ESR for this introductory model. Concept / Approach: A practical capacitor can be modeled as an ideal capacitor in parallel with a large resistor (leakage path). More detailed models add ESR in series and ESL, but the parallel resistor captures finite insulation resistance and DC leakage. Step-by-Step Solution: 1) Identify leakage: real dielectrics allow tiny DC current.2) Represent leakage with a high-value resistor in parallel.3) Retain ideal C element to represent energy storage in the electric field.4) Recognize that ESR and ESL are refinements for high-frequency work. Verification / Alternative check: Datasheets specify insulation resistance or leakage current, which maps directly to a parallel resistance model at rated conditions. Why Other Options Are Wrong: Is a short to DC: false; capacitors block DC ideally and leak only slightly.Is an open to AC: false; capacitive reactance allows AC current.Stores energy in a magnetic field: that describes inductors; capacitors store energy in an electric field. Common Pitfalls: Assuming ideal behavior and ignoring leakage can cause bias drift and timing errors in precision circuits. Final Answer: Looks like a capacitor in parallel with a resistor

Question 17

A capacitor is connected directly across a 20 V DC source through a series resistor and allowed to charge. After five time constants (5τ), what is the approximate voltage across the capacitor?

Accepted Answer

Introduction: Engineers often estimate settling times using the 5τ rule for first-order systems. This question asks for the capacitor voltage after five time constants during charging from a fixed DC source. Given Data / Assumptions: Supply voltage Vs = 20 V. Charging duration t = 5τ. Standard RC charging starting from 0 V. Concept / Approach: Use Vc(t) = Vs * (1 - exp(-t / τ)). At t = 5τ, exp(-5) is very small, so the capacitor is essentially at the supply voltage within about 0.7% error. Step-by-Step Solution: 1) Vc(5τ) = 20 * (1 - exp(-5)).2) exp(-5) ≈ 0.0067.3) 1 - 0.0067 = 0.9933.4) Vc ≈ 20 * 0.9933 ≈ 19.866 V.5) Rounded to typical engineering precision, Vc ≈ 20 V. Verification / Alternative check: Compare to the 5τ rule of thumb: about 99.3% settled, which is close enough to treat as the final value in most applications. Why Other Options Are Wrong: 13.5 V and 12.8 V: correspond to earlier times (between 1τ and 2τ), not 5τ.5.0 V: far too low for 5τ; would be near t much less than τ. Common Pitfalls: Reporting exact infinite-time behavior; first-order exponentials approach but do not instantaneously reach the final value. Engineers round at 5τ for practicality. Final Answer: 20 V

Question 18

In AC phasor analysis of a parallel RC circuit, which quantity is typically chosen as the 0° reference vector for drawing the phasor diagram and measuring all other angles?

Accepted Answer

Introduction: Understanding which phasor to choose as the reference is essential for clean and correct AC analysis. In a parallel RC circuit, all branches share the same node-to-node potential. This question checks whether you know that the common quantity, voltage, is the natural 0° reference for the phasor diagram. Given Data / Assumptions: Parallel RC network with sinusoidal steady-state excitation. Ideal components (linear R and ideal capacitor C). All branches experience the same voltage because nodes are common. Phasors represent RMS quantities and steady frequency. Concept / Approach: In a parallel circuit, the voltage across each branch is identical. Choosing that shared voltage as the 0° reference simplifies analysis: resistive branch current aligns with V, while capacitive branch current leads V by 90°. The total current is the vector sum of branch currents relative to the same V reference. Step-by-Step Solution: 1) Identify the common quantity: in parallel, V is common to every branch.2) Set V at 0°, i.e., V is the horizontal reference axis.3) Draw IR (through R) in phase with V.4) Draw IC (through C) leading V by +90°.5) Sum IR and IC vectorially to obtain the total source current. Verification / Alternative check: If you were to choose current as the reference in a parallel network, each branch would require a different phase relation (IC leads V, IR is in phase with V), resulting in extra steps. Using V as reference keeps the diagram consistent and minimal. Why Other Options Are Wrong: R and C: These are components, not phasor references; their values do not define a universal phase. I: In parallel circuits, currents differ by branch; no single current phasor is common to all branches. Phase angle of capacitor current: That angle is branch-specific and not shared by the resistor branch. Common Pitfalls: Confusing series and parallel conventions; in series, current is common so I is often a convenient reference, but in parallel circuits V is common and thus the correct reference baseline. Final Answer: V

Question 19

In basic capacitor theory, select the correct relationship between charge (Q), capacitance (C), and voltage (V).

Accepted Answer

Introduction: The most fundamental relation for a capacitor ties stored charge to the applied voltage through the capacitance. Mastering this identity is essential for voltage, energy, and transient calculations in analog and power electronics. Given Data / Assumptions: Ideal capacitor with constant capacitance C over the operating range. Steady-state or instantaneous relation (no explicit time dependency needed). SI units: Q in coulombs, C in farads, V in volts. Concept / Approach: By definition, capacitance is the ratio of stored charge to the voltage across the device: C = Q / V. Rearranging yields Q = C * V and V = Q / C. These forms are interchangeable and are the starting point for many capacitor problems, including energy E = 0.5 * C * V^2 and dynamic current i(t) = C * dv/dt in time-varying signals. Step-by-Step Solution: 1) Start from the definition: C = Q / V.2) Multiply both sides by V to isolate Q: Q = C * V.3) Recognize the inverse form for voltage: V = Q / C (useful when Q and C are known).4) Apply as needed in design or analysis to compute stored charge or voltage. Verification / Alternative check: Dimensional analysis: farad equals coulomb per volt. Therefore C * V has units (C/V) * V = C, matching charge in coulombs. This confirms the correctness of Q = C * V. Why Other Options Are Wrong: C = Q * V: Incorrect algebra; this would give units of C = C*V, dimensionally inconsistent. V = I * R: Ohm law for resistors, unrelated to capacitors. Q = V / C: Inverted; correct is V = Q / C, not Q = V / C. Common Pitfalls: Mixing up Q = C * V and V = Q / C, or trying to apply Ohm law to capacitive behavior in place of the capacitor definition. Always start from C = Q / V to avoid errors. Final Answer: Q = C * V

Question 20

Capacitors in parallel — to compute total capacitance of capacitors connected in parallel, sum the individual capacitances (CT = C1 + C2 + …). Is this statement correct?

Accepted Answer

Introduction / Context: Capacitor combining rules mirror resistor combining rules but with roles swapped: capacitances add in parallel and combine reciprocally in series. This property is widely used to realize non-standard values and increase energy storage. Given Data / Assumptions: Capacitors are connected node-to-node (parallel). Ideal components (no leakage or parasitics) for the rule statement. Concept / Approach: In parallel, plates effectively increase total plate area while maintaining the same voltage across each capacitor, so charge adds linearly: Q_total = ΣCi * V. Since CT = Q_total / V, we get CT = ΣCi. Step-by-Step Solution: 1) Write Qi = Ci * V for each branch (same V in parallel).2) Sum charges: Q_total = ΣCi * V.3) Define CT by Q_total = CT * V → CT = ΣCi.4) Conclude the statement is correct. Verification / Alternative check: Measure with an LCR meter: two capacitors 10 µF // 22 µF ≈ 32 µF (ignoring tolerances). Why Other Options Are Wrong: Only identical / only AC / depends on dielectric: the addition rule holds for any values and dielectrics under linear conditions. Common Pitfalls: Confusing series and parallel rules; forgetting voltage ratings still apply and should not be exceeded. Final Answer: Correct

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