Question 1

RC filter cutoff frequency from given R and C  A first-order RC filter (single pole) is built with R = 1.00 kΩ and C = 11.0 nF. Assuming an ideal op-amp buffer or negligible loading, what is the cutoff frequency (defined as the –3 dB frequency)?

Accepted Answer

Introduction / Context: First-order RC filters (high-pass or low-pass) have a single time constant τ = R * C and a cutoff frequency where the magnitude response is 1/√2 of the passband, commonly called the –3 dB point. This frequency sets bandwidth, rise time, and phase behavior for countless analog front ends. Given Data / Assumptions: R = 1.00 kΩ. C = 11.0 nF. Single-pole RC, negligible loading (e.g., buffered output). Cutoff definition: fc = 1 / (2πRC). Concept / Approach: The pole frequency for a single-pole RC is fc = 1 / (2πRC). With R and C provided, plug in the values directly. Ensure consistent SI units to avoid order-of-magnitude mistakes. Step-by-Step Solution: Compute τ = R * C = 1.00 kΩ * 11.0 nF = (1000 Ω) * (11 × 10^–9 F) = 11 × 10^–6 s.Apply fc formula: fc = 1 / (2π * 11 × 10^–6).Evaluate: 2π * 11 × 10^–6 ≈ 69.115 × 10^–6 s.Invert: fc ≈ 1 / (69.115 × 10^–6) ≈ 14,470 Hz = 14.47 kHz. Verification / Alternative check: Use the rule-of-thumb fc ≈ 0.159 / (R[kΩ] * C[µF]). Here R = 1 kΩ and C = 0.011 µF, so fc ≈ 0.159 / (1 * 0.011) ≈ 14.45 kHz, which agrees with the precise calculation. Why Other Options Are Wrong: 5 kHz / less than 5 kHz: imply a larger time constant; inconsistent with R and C given.more than 20 kHz: would require a smaller RC product.9.00 kHz: not supported by 1/(2πRC) with the stated values. Common Pitfalls: Confusing rad/s with Hz; forgetting to convert nF to F; neglecting load resistance which can shift fc if not buffered. Final Answer: 14.47 kHz

Question 2

Series RC phase dependence  In a series RC circuit driven by a sinusoidal source, the phase difference between the total source voltage and the current is determined by which factor?

Accepted Answer

Introduction / Context: The phase relationship between voltage and current in reactive circuits shapes filter characteristics, stability margins, and timing. In a series RC, the resistor and capacitor share the same current but their voltage drops are out of phase. The net phase between the source voltage and current is not constant; it varies with component values and frequency. Given Data / Assumptions: Topology: series RC. Sinusoidal steady-state excitation. Ideal components for clarity. Phase defined between total source voltage and circuit current. Concept / Approach: The series RC impedance is Z = R − j|XC| with |XC| = 1/(ωC). The phase angle φ between total voltage and current satisfies tan(φ) = −|XC|/R. Therefore the magnitude |φ| = arctan(|XC|/R). Changing R, C, or the frequency (which changes |XC|) alters |XC|/R and hence the phase. Step-by-Step Solution: Write Z = R − j|XC|.Compute φ = arctan(Im(Z)/Re(Z)) = arctan(−|XC|/R).Observe dependence: |XC| = 1/(ωC), so φ changes with R, C, and ω.Conclude: phase depends on component values (and frequency), not on voltage amplitude. Verification / Alternative check: Limit cases support the conclusion: if R ≫ |XC|, φ ≈ 0° (almost resistive). If |XC| ≫ R, |φ| approaches 90° (dominantly capacitive). Continuous variation between these limits confirms dependence on R and C (via |XC|). Why Other Options Are Wrong: equals 0° / 90°: only true in limiting cases, not generally.depends on amplitude: linear passive networks have phase independent of amplitude.fixed at 45°: occurs only when |XC| = R. Common Pitfalls: Confusing individual component voltage phases with the overall source-to-current phase; assuming linear circuits exhibit amplitude-dependent phase shift (they do not). Final Answer: depends on the values of R and C

Question 3

Impedance triangle for series RC circuits  In the right-triangle (phasor) representation of a series RC circuit with sides R and |XC|, what does the hypotenuse represent?

Accepted Answer

Introduction / Context: Engineers often visualize the relationship between resistance and reactance using a right triangle known as the impedance triangle. This geometric view clarifies how magnitudes combine and how phase angle arises in AC circuits. Given Data / Assumptions: Series RC circuit with resistance R and capacitive reactance magnitude |XC|. Right triangle drawn with adjacent side R and opposite side |XC|. Phasor convention: current is the reference axis for series analysis. Concept / Approach: For series RC, the complex impedance is Z = R − j|XC|. The magnitude is |Z| = sqrt(R^2 + |XC|^2). When represented as a right triangle, the legs are R and |XC|, and the hypotenuse is |Z|, the magnitude of the impedance. The angle between the hypotenuse and the R-axis equals the phase angle. Step-by-Step Solution: Assign adjacent leg to R (in-phase component).Assign opposite leg to |XC| (quadrature component).Compute hypotenuse: |Z| = sqrt(R^2 + |XC|^2).Interpret hypotenuse: total opposition to AC = impedance magnitude. Verification / Alternative check: Ohm’s law for phasors states V = I * Z. If the current magnitude is 1 A, the voltage magnitude equals |Z| volts. That ties the hypotenuse length directly to the observed voltage magnitude for unit current, reinforcing that it represents |Z|. Why Other Options Are Wrong: current: in series analysis, current is the reference phasor, not a triangle side.voltage: voltage would be I * Z; the hypotenuse is Z, not V unless I = 1 A.power: power depends on both magnitude and phase (P = VI cos φ), not a triangle side.admittance: reciprocal of impedance; not represented by this triangle. Common Pitfalls: Swapping series and parallel triangles; forgetting that capacitive reactance is plotted on the negative imaginary axis, though magnitude is used for the triangle. Final Answer: impedance

Question 4

Power components in AC circuits  Which quantity is alternately stored and returned (shuttled back and forth) between the source and reactive elements in an AC circuit?

Accepted Answer

Introduction / Context: Power in AC systems can be decomposed into real (true), reactive, and apparent components. Understanding the physical meaning of each is essential for power factor correction, transformer sizing, and minimizing energy costs in industrial systems. Given Data / Assumptions: Sinusoidal steady-state operation. Reactive elements (capacitors or inductors) are present. Losses in ideal reactive components are negligible (they do not dissipate average power). Concept / Approach: True power (P, measured in watts) represents net energy converted to heat, work, or light per unit time. Reactive power (Q, measured in var) represents energy that oscillates between the source and the electric or magnetic fields of capacitors and inductors each cycle. Apparent power (S, in VA) is the product of RMS voltage and current and is the vector sum S^2 = P^2 + Q^2. Step-by-Step Solution: Identify that only inductors and capacitors can store and release energy cyclically.Recognize that the mean energy dissipation of ideal L and C over a full cycle is zero.Therefore, the back-and-forth energy exchange is quantified by reactive power Q.Conclude: “Reactive power” is the correct answer. Verification / Alternative check: Check limiting cases: a purely resistive circuit has Q = 0 and only P flows; a purely reactive circuit has P = 0 and Q ≠ 0 (no net energy transfer over a cycle, only exchange). Why Other Options Are Wrong: True power: represents irreversible energy conversion, not energy exchange.Apparent power: magnitude combination of P and Q; not specifically the shuttled component.Impedance power / Harmonic power: not standard power terms in this context. Common Pitfalls: Equating high apparent power with high losses; forgetting that poor power factor increases current without increasing useful work. Final Answer: Reactive power

Question 5

Magnitude of impedance in a specified series RC case  A series RC circuit has R = 50 Ω and capacitive reactance magnitude |XC| = 50 Ω at the operating frequency. What is the magnitude of the total impedance?

Accepted Answer

Introduction / Context: Impedance magnitude in a series RC circuit combines resistance and reactance vectorially, not arithmetically. This magnitude determines current for a given applied RMS voltage and therefore influences power and filter behavior. Given Data / Assumptions: R = 50 Ω. |XC| = 50 Ω (capacitive). Series connection, sinusoidal steady-state. Seek |Z|, the magnitude of total impedance. Concept / Approach: For series RC, Z = R − j|XC|. The magnitude is |Z| = sqrt(R^2 + |XC|^2). Equal magnitudes of resistance and reactance form a 45° impedance triangle (since opposite and adjacent legs are equal). Step-by-Step Solution: Compute squared terms: R^2 = 50^2 = 2500; |XC|^2 = 50^2 = 2500.Sum: 2500 + 2500 = 5000.Square root: |Z| = sqrt(5000) ≈ 70.71 Ω.Round to the nearest option: approximately 70 Ω. Verification / Alternative check: Use the 1–1–√2 triangle analogy: equal legs imply the hypotenuse is leg × √2 → 50 × 1.414 ≈ 70.7 Ω, matching the calculation and the chosen option. Why Other Options Are Wrong: 17.14 Ω / 24 Ω: far below the vector sum of equal 50 Ω legs.More information is needed: not true; R and |XC| suffice.100 Ω: that would be a simple arithmetic sum, which is incorrect for orthogonal components. Common Pitfalls: Adding R and |XC| directly; forgetting to take the square root; mixing magnitudes with signed reactance values. Final Answer: 70 Ω

Question 6

Relative phase between resistor and capacitor voltages in series RC  In any series RC circuit at a single frequency, what is the phase difference between the voltage across the capacitor and the voltage across the resistor?

Accepted Answer

Introduction / Context: Although the overall source-to-current phase in a series RC depends on component values and frequency, the voltages across the individual elements have a fixed quadrature relationship. Recognizing this helps in phasor diagram construction and in diagnosing measurement results on oscilloscopes. Given Data / Assumptions: Topology: series RC. Sinusoidal steady-state. Ideal components. We compare only VR (across R) with VC (across C). Concept / Approach: In series RC, the current I is common to both elements. The resistor voltage is in phase with I: VR = I * R (0° relative to I). The capacitor voltage lags the current by 90°: VC = I * (−j|XC|). Therefore, VC lags VR by 90° regardless of magnitude, making their phase difference fixed at 90°. Step-by-Step Solution: Reference current I at 0°.VR is aligned with I (0°).VC is −90° relative to I.Hence, phase(VC) − phase(VR) = −90°, magnitude of difference = 90°. Verification / Alternative check: Draw the phasor diagram: place VR along the positive real axis and VC along the negative imaginary axis. The right angle between them is evident. This geometric proof does not depend on values of R or C. Why Other Options Are Wrong: depends on R and C: incorrect for the inter-element voltage relationship.equals 0° / 45°: contradicts capacitor voltage’s inherent 90° lag to current.depends on input voltage amplitude: phase in linear passive networks is amplitude independent. Common Pitfalls: Confusing the fixed VR–VC phase with the variable source-to-current phase; mis-assigning lead/lag directions. Final Answer: equals 90°

Question 7

Impedance of a series RC from given R and XC magnitudes  A series RC circuit has resistance R = 120 kΩ and capacitive reactance magnitude |XC| = 160 kΩ at the operating frequency. What is the magnitude of the total impedance?

Accepted Answer

Introduction / Context: Combining resistance and reactance magnitudes correctly is crucial to predict current and voltage division in AC networks. For series RC, the total impedance magnitude follows a Pythagorean relationship due to the orthogonality of resistive and reactive components in the complex plane. Given Data / Assumptions: R = 120 kΩ. |XC| = 160 kΩ. Series connection under sinusoidal excitation. We seek |Z|, not the phase angle (though it can be found as arctan(|XC|/R)). Concept / Approach: The complex impedance is Z = R − j|XC|. The magnitude is |Z| = sqrt(R^2 + |XC|^2). The numbers 120 and 160 form a 3–4–5 multiple (since 12–16–20 scales to 3–4–5), which simplifies the square root. Step-by-Step Solution: Compute squares: 120^2 = 14,400; 160^2 = 25,600 (units kΩ^2).Sum: 14,400 + 25,600 = 40,000.Square root: sqrt(40,000) = 200 (kΩ preserved from magnitudes).Therefore, |Z| = 200 kΩ. Verification / Alternative check: Recognize 12–16–20 as the 3–4–5 family scaled by 4; multiply by 10 to reach 120–160–200. The Pythagorean identity confirms the result immediately without calculator use. Why Other Options Are Wrong: 280 kΩ: arithmetic (R + |XC|) sum, not vector sum.120 kΩ / 160 kΩ: represent only one leg, not the hypotenuse.40 kΩ: unrelated; perhaps a misread of the squared sum. Common Pitfalls: Adding impedances as scalars; forgetting unit consistency (kΩ vs. Ω); ignoring that reactance contributes orthogonally to resistance. Final Answer: 200 kΩ

Question 8

RC circuit — evaluating real (true) power from incomplete data: In the referenced RC network, the “true power” consumed equals what value? Use the definition P_true = I_rms^2 * R or P_true = V_rms^2 / R (for the resistive branch only). If the source…

Accepted Answer

Introduction / Context: In AC analysis of an RC circuit, “true power” (also called real or active power) is the portion of power that is actually dissipated as heat in the resistive element. Unlike reactive power, which alternately stores and releases energy in the capacitor, true power depends only on the resistive component of the impedance and on the portion of current through it. Correctly identifying whether enough information has been provided is a vital engineering skill that prevents calculation errors and misinterpretation of measurements. Given Data / Assumptions: The prompt references a specific circuit but does not specify R, C, source voltage, or the measured current. We are asked for a single numeric value of true power. Acceptable formulas include P_true = I_rms^2 * R (branch current through R) or P_true = V_R_rms^2 / R (voltage across R). For total-circuit values, power factor may be needed. Concept / Approach: True power for RC circuits is determined by resistive loss only. If only total current or total impedance is known, we still require the power factor cos(phi) to map apparent power S = V_rms * I_rms into real power P = S * cos(phi). Alternatively, we must know the exact voltage across the resistor or the current through the resistor. Without any of these, a unique numeric result cannot be produced responsibly. Step-by-Step Solution: Check for R, C, and source V or I: none provided explicitly.Assess whether power factor or phase angle is given: not provided.Attempt alternative: P_true = I_rms^2 * R requires I_R (resistive-branch current), which is unknown.Conclusion: a unique numeric value cannot be determined from the information supplied. Verification / Alternative check: If a magnitude of impedance Z and the total current I were available, we would still need the power factor to separate true and reactive components. If V_R or I_R were known, P_true could be computed directly. The absence of these confirms indeterminacy. Why Other Options Are Wrong: The listed milliwatt values presume specific circuit parameters or a known power factor, which are not given. Selecting any fixed number would be speculative rather than analytical. Common Pitfalls: Using apparent power V_rms * I_rms as if it were real power, forgetting that a capacitor shifts the phase and changes power factor; assuming Z alone determines P_true without R or cos(phi). Final Answer: More information is required; the given data are insufficient to compute true power reliably.

Question 9

Parallel RC fundamentals — convert impedance to admittance: If the equivalent impedance magnitude of a parallel RC circuit is given as 2 kΩ, what is the corresponding admittance magnitude?

Accepted Answer

Introduction / Context: Admittance is the reciprocal of impedance and is a convenient way to analyze parallel AC networks. While impedance Z is measured in ohms, admittance Y is measured in siemens. For magnitude-only questions, the conversion is straightforward when a single impedance magnitude is specified. Given Data / Assumptions: Equivalent impedance magnitude Z = 2 kΩ (i.e., 2000 Ω). We need the admittance magnitude Y = 1 / |Z|. The question asks only for magnitude, not phase or rectangular components. Concept / Approach: For magnitude-only conversion, use Y = 1 / Z. Units invert from ohms to siemens. For example, 1 / 1000 Ω = 1 mS. Keeping track of prefixes avoids rounding errors and reinforces comfort with engineering notation. Step-by-Step Solution: Compute reciprocal: Y = 1 / 2000 Ω = 0.0005 S.Express using engineering prefix: 0.0005 S = 500 µS.State final value clearly: 500 µS. Verification / Alternative check: Use kΩ to mS reciprocal pairing: 1 kΩ ↔ 1 mS. Therefore, 2 kΩ ↔ 0.5 mS = 500 µS. Both methods agree. Why Other Options Are Wrong: 2 kS and 4 kS are far too large (they would correspond to 0.5 mΩ and 0.25 mΩ, respectively). “More information is required” is not true when only magnitude is requested. 50 µS corresponds to 20 kΩ, not 2 kΩ. Common Pitfalls: Confusing siemens with microsiemens, or reciprocating the numeric value without converting units properly (kΩ ↔ mS). Final Answer: 500 µS

Question 10

Series RC phase angle intuition — comparing resistance and capacitive reactance: When the resistance R is greater than the capacitive reactance Xc in a series RC circuit, the circuit phase angle (voltage leading current) is what range?

Accepted Answer

Introduction / Context: The phase angle in a series RC circuit indicates how far the source voltage leads the circuit current. Understanding the relationship between R and Xc helps predict phase behavior without detailed computation, which is valuable for quick checks in design and troubleshooting. Given Data / Assumptions: Series RC network with resistance R and capacitive reactance Xc. Condition: R > Xc. Phase angle φ for series RC is defined by tan(φ) = Xc / R (voltage leads current). Concept / Approach: Because tan(φ) = Xc / R, if R is larger than Xc, the ratio Xc/R is less than 1. The arctangent of a value less than 1 yields an angle less than 45 degrees but greater than 0 degrees, reflecting modest leading behavior. Extreme cases: if Xc ≫ R, φ approaches 90°; if Xc = R, φ = 45°; if Xc ≪ R, φ approaches 0°. Step-by-Step Solution: Start with tan(φ) = Xc / R.Given R > Xc ⇒ Xc / R < 1.Therefore 0° < φ < 45°.Hence the correct descriptive range is “between 0° and 45°.” Verification / Alternative check: Choose a numeric example: let R = 1000 Ω and Xc = 500 Ω. Then tan(φ) = 0.5 ⇒ φ ≈ 26.6°, which lies squarely in the stated range. Why Other Options Are Wrong: 0° and exactly 45° correspond to limiting or equality conditions not satisfied here. “Between 45° and 90°” would require Xc > R, which contradicts the premise. 90° is unattainable with finite R unless R approaches zero. Common Pitfalls: Confusing series RC with series RL sign conventions, or misremembering whether voltage leads or lags in capacitive circuits. Final Answer: between 0° and 45°

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