Q: In a 20 Vac series RC circuit, the measured voltage across the resistor is 20 V and the measured voltage across the capacitor is 40 V. What is the magnitude of the applied source voltage (phasor sum)?

Introduction / Context: In series RC circuits, the resistor and capacitor voltages are out of phase by 90 degrees (resistive drop in phase with current; capacitive drop lags the current by 90 degrees). Therefore, the source voltage is not the arithmetic sum of the magnitudes but the vector (phasor) sum. This question checks correct phasor addition of Vr and Vc. Given Data / Assumptions: Series RC circuit with sinusoidal steady state. Vr = 20 V (in phase with current). Vc = 40 V (lags current by 90 degrees). We seek |Vs|, the magnitude of the applied source voltage. Concept / Approach: Because Vr and Vc are orthogonal in phase, |Vs| is found using the Pythagorean relationship: |Vs| = sqrt(Vr^2 + Vc^2). Numerical substitution yields the correct applied voltage magnitude. Step-by-Step Solution: Given Vr = 20 V and Vc = 40 V.Compute Vs magnitude: |Vs| = sqrt(Vr^2 + Vc^2).|Vs| = sqrt(20^2 + 40^2) = sqrt(400 + 1600) = sqrt(2000) ≈ 44.72 V.Rounded to the nearest whole-volt choice: 45 Vac. Verification / Alternative check: Phasor diagram: draw Vr on the horizontal axis and Vc downward by 90 degrees. The source vector is the hypotenuse of the right triangle formed, clearly less than Vr + Vc (60 V) and greater than max(Vr, Vc), which fits ≈ 44.7 V. Why Other Options Are Wrong: 50 Vac, 60 Vac, 65 Vac: These imply arithmetic addition or incorrect geometry; the correct operation is vector addition with orthogonal components. Common Pitfalls: Adding voltage magnitudes directly in series reactive circuits; phase must be considered. Confusing capacitor voltage phase (90 degrees out of phase with Vr) with in-phase addition. Final Answer: 45 Vac

Q: In AC power analysis, the quantity measured in volt–amperes (VA) is referred to as which type of power?

Introduction / Context: AC circuits with reactance exhibit a distinction between power that performs real work and power that circulates between source and reactive elements. Proper terminology—true, reactive, and apparent power—is foundational for sizing equipment and understanding power factor. Given Data / Assumptions: Sinusoidal steady-state assumptions. Voltage and current may be out of phase by angle phi due to reactance. Standard definitions of power components. Concept / Approach: Apparent power S is defined as S = V_rms * I_rms and is measured in volt–amperes (VA). True (real) power P = V_rms * I_rms * cos(phi) is measured in watts (W). Reactive power Q = V_rms * I_rms * sin(phi) is measured in volt–ampere reactive (VAR). Thus, when an instrument reports results in VA, it is indicating apparent power. Step-by-Step Solution: Identify units: VA corresponds to S = V_rms * I_rms.Relate to P and Q: P in watts, Q in VAR, S in VA.Therefore, the power measured in volt–amperes is apparent power. Verification / Alternative check: Power triangle relationships: S^2 = P^2 + Q^2. Instruments like apparent power meters or nameplate ratings for transformers/UPS specify VA because they must supply the vector sum capacity regardless of power factor. Why Other Options Are Wrong: Impedance power: Not a standard term in AC power theory. Reactive power: Measured in VAR, not VA. True power: Measured in watts. Common Pitfalls: Confusing VA with W; at unity power factor, VA equals W, but the units and concept remain distinct. Using VA ratings to assume real work capability without considering power factor. Final Answer: apparent power

Q: Series RC phasor sum — a resistor shows 6 V (rms) and a capacitor shows 8 V (rms) in the same series RC branch. Using vector (phasor) addition, what is the magnitude of the source voltage?

Introduction / Context: In AC circuits containing reactance, voltage drops do not add arithmetically. Instead, they combine as vectors (phasors) because they are out of phase. This question tests your ability to use the right kind of addition for a series RC circuit where resistor and capacitor voltages are in quadrature (90 degrees apart). Given Data / Assumptions: Series RC circuit under sinusoidal steady state. Measured magnitudes: V_R = 6 V (in phase with current), V_C = 8 V (lags current by 90 degrees). The source voltage equals the phasor sum of V_R and V_C. Concept / Approach: In a series RC circuit, current is common. The resistor voltage is V_R = I * R and is in phase with the current; the capacitor voltage is V_C = I * Xc, but since Xc introduces -90 degrees, V_C is 90 degrees behind the current. Therefore, V_R and V_C are orthogonal, making the source magnitude Vs the Pythagorean sum: Vs = sqrt(V_R^2 + V_C^2). Step-by-Step Solution: Identify orthogonal components: V_R along the real axis, V_C along the negative imaginary axis.Compute magnitude: Vs = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 V.Conclude the statement is correct: the source voltage magnitude is 10 V. Verification / Alternative check: If you convert to impedance and current, the same result emerges. For some I, V_R = I*R and V_C = I*Xc; geometry dictates a right triangle, yielding Vs = 10 V when the legs are 6 V and 8 V. Why Other Options Are Wrong: Arithmetic sum (14 V) ignores the 90-degree phase difference. “2 V” is non-physical for these magnitudes; phasor subtraction is not appropriate here. Frequency is not needed once V_R and V_C are given; their vector relationship is fixed. Common Pitfalls: Adding AC drops like DC series voltages or forgetting that resistor and capacitor voltages are orthogonal. Final Answer: Correct — Vs = 10 V.

Q: Resistance versus impedance — in AC analysis of an RC network, is resistance the same as impedance? Choose the most accurate statement.

Introduction / Context: In AC circuits, impedance generalizes resistance to include the effects of reactance from capacitors and inductors. Confusing resistance with impedance leads to incorrect magnitudes and phase angles. This question checks basic vocabulary and the mathematical form of impedance for an RC network. Given Data / Assumptions: Sinusoidal steady-state analysis. Components are ideal, with a resistor R and capacitor C. Capacitive reactance Xc = 1 / (omega * C). Concept / Approach: Resistance is a real scalar measured in ohms and relates voltage and current without phase shift: V = I * R. Impedance is a complex quantity, Z = R + jX, that relates phasor voltage and current: V = I * Z. For a capacitor, the reactance is negative imaginary (Xc is treated as -|Xc|), so in series RC we write Z = R - jXc. The magnitude is |Z| = sqrt(R^2 + Xc^2), and the phase angle is phi = arctan(Xc / R), with current leading voltage in a capacitive circuit. Step-by-Step Solution: Write Z for series RC: Z = R - jXc.Compute magnitude and angle: |Z| = sqrt(R^2 + Xc^2); phi = arctan(Xc / R).Recognize resistance equals impedance only if Xc = 0 (not true for a real capacitor at finite frequency). Verification / Alternative check: At extremely high frequencies Xc approaches 0, so Z approaches R (still not literally equal except in the ideal limit). At DC, Xc tends to infinity, the circuit blocks current, and Z is not equal to R. Why Other Options Are Wrong: “Correct — interchangeable” ignores the imaginary component from reactance. “Valid only at resonance” does not apply to a lone RC branch; resonance requires both L and C. “Valid only at high frequency” is an approximation, not an identity. Common Pitfalls: Using R instead of |Z| in current calculations or neglecting phase when combining AC voltages. Final Answer: Incorrect — impedance includes both R and Xc and is not the same as resistance.

Q: Parallel RC phase behavior — as the capacitance C increases in a parallel RC circuit excited by a sinusoid, how does the overall phase angle between the source voltage and total current change?

Introduction / Context: Phase angle in parallel circuits depends on the relative contributions of conductance and susceptance. In a parallel RC network, changing the capacitor value alters the reactive branch current dramatically, which changes the total current phasor and the angle it makes relative to the impressed voltage. Given Data / Assumptions: Sinusoidal steady-state excitation. Parallel RC: one branch is a resistor R (conductance G = 1/R), the other is a capacitor C (susceptance Bc = omega * C). Phase angle theta is the angle of total current relative to voltage (voltage is the reference in admittance analysis). Concept / Approach: Total admittance is Y = G + jBc, where Bc = omega * C. As C increases, Bc increases proportionally, so the reactive component of current grows relative to the resistive component. The phase angle of the total current is theta = arctan(Bc / G). Therefore, increasing C increases Bc, which increases |theta| toward +90 degrees (capacitive behavior). Step-by-Step Solution: Write Y: Y = G + jBc with Bc = omega * C.Compute angle: theta = arctan(Bc / G).Increase C → increase Bc → larger theta magnitude → more capacitive phase. Verification / Alternative check: Consider extreme cases. If C is very small, Bc ≈ 0 and the circuit is nearly resistive (theta ≈ 0). If C is very large, Bc dominates and theta approaches +90 degrees, indicating strongly capacitive current. Why Other Options Are Wrong: “More inductive” contradicts the nature of a capacitor. “No change” ignores the dependence of Bc on C. “Reverses sign / pure resistor” has no basis; adding capacitance increases capacitive character. Common Pitfalls: Confusing series and parallel behavior or using reactance Xc = 1/(omega*C) directly in parallel when admittance is more convenient. Final Answer: Correct — increasing C increases the capacitive phase angle magnitude.

Q: RC low-pass filter terminology — for a single-pole RC low-pass network, is the bandwidth equal to the -3 dB cutoff frequency f_c?

Introduction / Context: Bandwidth is a fundamental specification in filter design and signal processing. For simple, first-order filters, the definition is straightforward and often used in practical datasheets. This item ensures you understand how bandwidth relates to the -3 dB cutoff frequency in a single-pole RC low-pass network. Given Data / Assumptions: First-order RC low-pass filter (one resistor, one capacitor). Cutoff frequency f_c defined at the magnitude response where output power is half the midband (voltage down by 3 dB): |H(f_c)| = 1/sqrt(2). “Bandwidth” refers to the range of frequencies for which the filter passes signals with minimal attenuation. Concept / Approach: By convention, for a first-order low-pass, the passband is taken from DC (0 Hz) up to the -3 dB point f_c. Since that interval length is simply f_c − 0, the bandwidth equals f_c. This contrasts with band-pass filters, where bandwidth is f_2 − f_1 between the two half-power frequencies. Step-by-Step Solution: Identify filter type: first-order low-pass (transfer function magnitude decreases beyond f_c).Define bandwidth: from 0 Hz to f_c.Compute bandwidth: BW = f_c − 0 = f_c. Verification / Alternative check: The -3 dB definition implies that below f_c the attenuation is less than or equal to 3 dB; since there is only one cutoff, this region’s width equals f_c. Many textbooks and manufacturer notes explicitly state BW = f_c for first-order low-pass sections. Why Other Options Are Wrong: “Twice f_c” has no basis for a single-pole low-pass. “Defined only for high-pass” is incorrect; bandwidth is a universal metric. “Requires Q” applies to resonant second-order sections, not a first-order RC. Common Pitfalls: Confusing band-pass bandwidth (difference between two cutoffs) with low-pass bandwidth (single cutoff from DC). Final Answer: Correct — bandwidth equals f_c for a first-order RC low-pass filter.

Q: Reactance and susceptance equivalence — a capacitive reactance of 10 kΩ corresponds to what capacitive susceptance in siemens or microsiemens?

Introduction / Context: Reactance and susceptance are reciprocal concepts used in impedance and admittance analysis. For capacitors, engineers often flip between series (reactance) and parallel (susceptance) viewpoints. This question checks your fluency with the reciprocal relationship for a specific numeric value. Given Data / Assumptions: Capacitive reactance magnitude Xc = 10 kΩ. We are converting to capacitive susceptance Bc in siemens (S). Ideal sinusoidal steady-state conditions. Concept / Approach: Admittance Y is the reciprocal of impedance Z. For a purely capacitive element, the magnitude relationships simplify: |Z| = Xc and |Y| = Bc, and therefore Bc = 1 / Xc. Once Xc is given, its reciprocal yields susceptance in siemens. Converting to microsiemens helps match typical instrumentation scales. Step-by-Step Solution: Start from Bc = 1 / Xc.Insert Xc = 10 kΩ = 10,000 Ω.Compute: Bc = 1 / 10,000 = 0.0001 S = 100 × 10^-6 S = 100 µS. Verification / Alternative check: Cross-check by dimensional reasoning: ohms and siemens are reciprocals. If Xc is large (10 kΩ), its admittance must be small (100 µS), which is consistent. Why Other Options Are Wrong: 10 µS is off by a factor of 10; there is no extra factor of 10 in the reciprocal. 1 mS equals 1000 µS, which contradicts the reciprocal of 10 kΩ. Frequency is not needed here; Xc has already been provided, and Bc = 1/Xc. Common Pitfalls: Forgetting to convert kilo-ohms to ohms before taking the reciprocal or mixing up mS and µS. Final Answer: Correct — 100 µS.

Q: RC high-pass behavior — as input frequency decreases, does the output voltage magnitude of a simple RC high-pass filter increase or decrease?

Introduction / Context: High-pass filters are ubiquitous in coupling networks, audio systems, and sensor interfaces. Their defining characteristic is that they pass high-frequency content while attenuating low-frequency components and DC. This question probes your qualitative understanding of how output changes with frequency. Given Data / Assumptions: Single-pole RC high-pass filter (series capacitor, shunt resistor or equivalent topology). Sinusoidal steady-state behavior. No significant load interaction beyond the intended resistor. Concept / Approach: In a high-pass RC network, the capacitor’s reactance Xc = 1 / (omega * C) is large at low frequencies and small at high frequencies. The transfer function magnitude |H(jomega)| increases with frequency, approaching unity at very high frequency and approaching zero at very low frequency. Therefore, as frequency decreases, output voltage magnitude falls, not rises. Step-by-Step Solution: Recognize the high-pass role: blocks DC and very slow variations.At low frequency: Xc is large → series impedance high → output across the resistor is small.At high frequency: Xc is small → signal passes → output increases toward input magnitude. Verification / Alternative check: Check the -3 dB cutoff f_c = 1 / (2 * pi * R * C). Below f_c, output rolls off at about -20 dB/decade (first order). Above f_c, output asymptotically approaches full scale. Why Other Options Are Wrong: “Output increases as frequency decreases” contradicts high-pass behavior. “Only when load is inductive” is irrelevant to the basic trend; standard topology already determines the response. “Undetermined” is not correct; the qualitative direction is known without exact values. Common Pitfalls: Mixing up high-pass with low-pass or overlooking the role of the series capacitor as a frequency-dependent coupling element. Final Answer: Incorrect — the output of an RC high-pass decreases as frequency decreases.

Question 1

For a parallel RC network fed by a 5 Vac (RMS) source at 500 kHz, with R = 4.7 kΩ and C = 47 pF connected in parallel, what is the overall phase angle of the total current relative to the applied voltage (use the passive sign convention)?

Accepted Answer

Introduction / Context: In a parallel RC circuit, the total current equals the vector (phasor) sum of the resistive branch current (in phase with voltage) and the capacitive branch current (leading the voltage by 90 degrees). The phase angle of the total current indicates how “capacitive” the overall admittance is. Given Data / Assumptions: Frequency f = 500 kHz. Source voltage Vrms = 5 V (magnitude not needed for angle). R = 4.7 kΩ, C = 47 pF in parallel. Ideal components; small-signal, linear operation. Concept / Approach: Work with admittance: Y = G + jB. For a resistor, G = 1/R. For a capacitor, susceptance B_C = 2 * π * f * C (positive for capacitive). The phase angle of total current relative to voltage is θ = arctan(B / G). Because capacitive current leads voltage, θ is positive (leading) when B > 0. Step-by-Step Solution: Compute G: G = 1 / 4700 ≈ 0.0002128 S.Compute B: B = 2 * π * 500000 * 47e-12 ≈ 0.0001477 S.Find angle: θ = arctan(B / G) ≈ arctan(0.0001477 / 0.0002128) ≈ arctan(0.6939) ≈ 34.8°.Interpretation: total current leads voltage by about 34.8°, indicating a moderately capacitive net admittance. Verification / Alternative check: Power factor pf = cos(θ) ≈ cos(34.8°) ≈ 0.82 (leading). Simulation or an LCR meter in parallel mode would show similar angle at this frequency. Why Other Options Are Wrong: ±55.3° imply a much stronger susceptance relative to conductance, not supported by B/G ≈ 0.694. Negative angles denote lagging (inductive) behavior, which is not present here. 0° would mean purely resistive with B = 0, not the case with a capacitor in parallel. Common Pitfalls: Mixing series and parallel formulas; for parallel circuits, always compute with admittance (G and B). Also note sign convention: capacitive susceptance is positive, giving a leading current. Final Answer: 34.8°

Question 2

For phasor diagrams of parallel RC circuits, which quantity is the most convenient and commonly used reference vector for measuring angles of branch and total currents?

Accepted Answer

Introduction / Context: Phasor diagrams in AC analysis require a reference vector from which other quantities are measured. In parallel circuits, all branches share the same voltage. Choosing the applied voltage as the reference simplifies the diagram and makes angle relationships intuitive. Given Data / Assumptions: Parallel RC network (and, by extension, any parallel AC network). Phasor method for sinusoidal steady-state analysis. Voltage is common across all parallel branches. Concept / Approach: Because voltage is identical on every branch in a parallel circuit, setting the voltage phasor V along the real axis (0°) lets each branch current be drawn relative to V: resistive branch current in phase with V, capacitive branch current leading V by 90°, and inductive branch current lagging V by 90°. The total current I is the vector sum of branch currents referenced to V. Step-by-Step Solution: Pick V as 0° reference (horizontal axis).Draw I_R in phase with V; draw I_C leading V by 90°.Vector-sum to obtain total I_T; its angle θ is measured from V.Hence, V is the standard reference vector for parallel RC circuits. Verification / Alternative check: Textbook conventions, lab exercises, and simulation tools default to voltage as reference in parallel networks. This yields immediate insight into power factor and reactive current components. Why Other Options Are Wrong: R and X_C are scalar component values, not phasors. I can be used as a reference in series circuits (current common), but in parallel circuits currents differ by branch. Z is overall impedance; using it as a reference is less straightforward than choosing the shared voltage. Common Pitfalls: Using current as reference for a parallel network complicates angle interpretation because each branch current has a different phase relative to V. Always align reference choice with the quantity common to all branches. Final Answer: V

Question 3

In a series RC (resistor–capacitor) AC circuit, what qualitative effect does increasing the resistance R have on the circuit's behavior, particularly on the phase shift between the source voltage and the current?

Accepted Answer

Introduction / Context: This question examines qualitative behavior in a series RC AC circuit when the resistance R is increased while frequency and capacitance remain fixed. In such circuits, the capacitor's reactance and the resistor determine both the current magnitude and the phase angle between the total applied voltage and the current. Given Data / Assumptions: Series RC circuit excited by a sinusoidal source of fixed frequency. Capacitance C is constant; only R is increased. Standard phasor sign convention: in a capacitive circuit, current leads voltage by angle theta. Concept / Approach: For a series RC, the impedance is Z = R - jXc, where Xc = 1 / (2 * pi * f * C). The phase angle of the current relative to the applied voltage is theta = arctan(-Xc / R). As R increases with Xc held constant, the ratio Xc / R gets smaller, so the magnitude of the phase angle decreases in absolute value; the circuit becomes less capacitive and more resistive in behavior. Step-by-Step Solution: Define Xc = 1 / (2 * pi * f * C) (fixed here).Impedance: Z = R - jXc; phase angle (voltage vs current) satisfies tan(theta) = -Xc / R.Increase R → Xc / R decreases → |theta| decreases (phase shift reduces toward 0 degrees).Current magnitude I = V / |Z| also decreases because |Z| increases with larger R. Verification / Alternative check: Consider two cases with the same Xc: R1 and R2 where R2 > R1. Then tan(|theta2|) = Xc / R2 < Xc / R1 = tan(|theta1|), so |theta2| < |theta1|. This confirms that increasing R reduces the phase shift magnitude. Why Other Options Are Wrong: No effect: False; phase and current both change with R. The current will increase: False; increasing R increases |Z|, so current decreases. The input voltage will increase: Source voltage is independent of R in this context. Common Pitfalls: Confusing capacitive phase lead (current leads) with inductive lag; in RC the current leads but the lead shrinks as R grows. Assuming Xc changes when R changes; Xc depends only on f and C. Final Answer: The phase shift will decrease.

Question 4

As the excitation frequency increases, how do the impedances of series RC and parallel RC networks qualitatively change?

Accepted Answer

Introduction / Context: RC networks are fundamental frequency-selective elements. Understanding how their impedances vary with frequency is essential for filter design, coupling/decoupling, and timing applications. This question contrasts series and parallel RC behaviors as frequency rises. Given Data / Assumptions: Capacitive reactance Xc = 1 / (2 * pi * f * C). Resistor R is frequency-independent (ideal). Comparison is qualitative: increase frequency f while R and C are fixed. Concept / Approach: As frequency increases, Xc decreases. In a series RC, total impedance magnitude |Z_series| = sqrt(R^2 + Xc^2) falls because the reactive term shrinks. In a parallel RC, the total admittance increases because the capacitive susceptance (1 / Xc) grows with frequency, so the equivalent impedance |Z_parallel| = 1 / |Y| decreases as well. Step-by-Step Solution: Compute Xc = 1 / (2 * pi * f * C); increasing f → decreasing Xc.Series RC: |Z_series| = sqrt(R^2 + Xc^2) → decreases as Xc decreases.Parallel RC: Y_total = 1 / R + 1 / (jXc); |Y| increases with frequency, so |Z_parallel| = 1 / |Y| decreases.Therefore, both series and parallel RC impedances decrease with increasing frequency. Verification / Alternative check: Limiting cases: as f → 0, Xc → infinity; series RC behaves like largely resistive with large |Z|, parallel RC tends toward R (capacitor open). As f → infinity, Xc → 0; series RC tends to R (smaller than at low f), parallel RC tends to near 0 (capacitor acts short), confirming decreasing trends. Why Other Options Are Wrong: Series decreases, parallel increases: Parallel does not increase; added capacitive path lowers impedance. Series increases, parallel decreases: Opposite of Xc behavior trend. Both increase: Contradicts the fact that Xc decreases as frequency rises. Common Pitfalls: Confusing impedance with reactance; while Xc alone decreases, total Z must be evaluated for the configuration. Assuming R changes with frequency; ideal R is constant. Final Answer: both series and parallel RC impedance decrease

Question 5

In a 20 Vac series RC circuit, the measured voltage across the resistor is 20 V and the measured voltage across the capacitor is 40 V. What is the magnitude of the applied source voltage (phasor sum)?

Accepted Answer

Introduction / Context: In series RC circuits, the resistor and capacitor voltages are out of phase by 90 degrees (resistive drop in phase with current; capacitive drop lags the current by 90 degrees). Therefore, the source voltage is not the arithmetic sum of the magnitudes but the vector (phasor) sum. This question checks correct phasor addition of Vr and Vc. Given Data / Assumptions: Series RC circuit with sinusoidal steady state. Vr = 20 V (in phase with current). Vc = 40 V (lags current by 90 degrees). We seek |Vs|, the magnitude of the applied source voltage. Concept / Approach: Because Vr and Vc are orthogonal in phase, |Vs| is found using the Pythagorean relationship: |Vs| = sqrt(Vr^2 + Vc^2). Numerical substitution yields the correct applied voltage magnitude. Step-by-Step Solution: Given Vr = 20 V and Vc = 40 V.Compute Vs magnitude: |Vs| = sqrt(Vr^2 + Vc^2).|Vs| = sqrt(20^2 + 40^2) = sqrt(400 + 1600) = sqrt(2000) ≈ 44.72 V.Rounded to the nearest whole-volt choice: 45 Vac. Verification / Alternative check: Phasor diagram: draw Vr on the horizontal axis and Vc downward by 90 degrees. The source vector is the hypotenuse of the right triangle formed, clearly less than Vr + Vc (60 V) and greater than max(Vr, Vc), which fits ≈ 44.7 V. Why Other Options Are Wrong: 50 Vac, 60 Vac, 65 Vac: These imply arithmetic addition or incorrect geometry; the correct operation is vector addition with orthogonal components. Common Pitfalls: Adding voltage magnitudes directly in series reactive circuits; phase must be considered. Confusing capacitor voltage phase (90 degrees out of phase with Vr) with in-phase addition. Final Answer: 45 Vac

Question 6

In AC power analysis, the quantity measured in volt–amperes (VA) is referred to as which type of power?

Accepted Answer

Introduction / Context: AC circuits with reactance exhibit a distinction between power that performs real work and power that circulates between source and reactive elements. Proper terminology—true, reactive, and apparent power—is foundational for sizing equipment and understanding power factor. Given Data / Assumptions: Sinusoidal steady-state assumptions. Voltage and current may be out of phase by angle phi due to reactance. Standard definitions of power components. Concept / Approach: Apparent power S is defined as S = V_rms * I_rms and is measured in volt–amperes (VA). True (real) power P = V_rms * I_rms * cos(phi) is measured in watts (W). Reactive power Q = V_rms * I_rms * sin(phi) is measured in volt–ampere reactive (VAR). Thus, when an instrument reports results in VA, it is indicating apparent power. Step-by-Step Solution: Identify units: VA corresponds to S = V_rms * I_rms.Relate to P and Q: P in watts, Q in VAR, S in VA.Therefore, the power measured in volt–amperes is apparent power. Verification / Alternative check: Power triangle relationships: S^2 = P^2 + Q^2. Instruments like apparent power meters or nameplate ratings for transformers/UPS specify VA because they must supply the vector sum capacity regardless of power factor. Why Other Options Are Wrong: Impedance power: Not a standard term in AC power theory. Reactive power: Measured in VAR, not VA. True power: Measured in watts. Common Pitfalls: Confusing VA with W; at unity power factor, VA equals W, but the units and concept remain distinct. Using VA ratings to assume real work capability without considering power factor. Final Answer: apparent power

Question 7

Series RC circuits — phase relationship clarification: Which statement about a series RC (resistor–capacitor) AC circuit is true regarding the phase of current and the source voltage?

Accepted Answer

Introduction / Context: In AC network analysis, understanding the phase relationship of current and voltage in a series RC circuit is fundamental. Because capacitors store and release energy in an electric field, they cause current to lead voltage. Recognizing which quantity leads or lags helps with phasor diagrams, impedance calculations, and power factor correction in electronics and electrical engineering. Given Data / Assumptions: Series circuit containing a single resistor R and capacitor C. Sinusoidal source voltage applied (steady-state AC). Ideal components (no parasitic inductance or resistance in C beyond R). Concept / Approach: For a capacitor, current i leads capacitor voltage v_C by 90 degrees because i = C * dv/dt. In a series RC, the same current flows through both R and C. The resistor voltage v_R is in phase with current, while the capacitor voltage v_C lags current by 90 degrees. The source voltage is the phasor sum v_S = v_R + v_C, which therefore lags the current by an angle 0 < phi < 90 degrees. Thus, current leads the source voltage by phi in a series RC circuit. Step-by-Step Solution: Represent impedances: Z_R = R (angle 0), Z_C = 1 / (j * ω * C) (angle −90°).Total impedance: Z = R − j * X_C; the impedance angle is negative.Current phasor I = V / Z thus has a positive angle relative to V, meaning current leads voltage.Component voltages: v_R in phase with I; v_C lags I by 90°; v_S is vector sum and lags I by phi. Verification / Alternative check: Create a phasor diagram: draw I on the positive real axis; v_R along I; v_C straight downward (−j axis). The source voltage is the diagonal from the origin to the tip of the v_R + v_C vector, clearly lagging I. Therefore, I leads v_S. Why Other Options Are Wrong: The capacitor's voltage and the resistor's voltage are not in phase; they differ by 90°. The current lags the source voltage: incorrect for RC (that would be RL where current lags). The resistor voltage lags the current: v_R is in phase with current, not lagging. Source leading resistor voltage by 90°: the phase shift is less than 90° and depends on R and X_C. Common Pitfalls: Confusing RC and RL behavior; in RC, current leads, in RL, current lags. Assuming the phase shift is always 90°; it is between 0° and 90° for series RC. Final Answer: The current leads the source voltage.

Question 8

Ideal capacitor power consideration in AC circuits: Evaluate the statement: “In a purely capacitive AC circuit (ideal capacitor, no resistive losses), the energy converted to heat is proportional to the circuit current.” Choose the most accurate ass…

Accepted Answer

Introduction / Context: Reactive components (ideal capacitors and inductors) in AC circuits exchange energy with the source but, ideally, do not dissipate real power as heat. This distinction between reactive power and real power is foundational for power factor, filter design, and AC analysis. Given Data / Assumptions: Purely capacitive AC circuit with an ideal capacitor. No series resistance or dielectric loss is considered (lossless assumption). Steady sinusoidal operation. Concept / Approach: Instantaneous power in a pure capacitor alternates between positive and negative as the capacitor charges and discharges. The average over a full cycle is zero because the current leads the voltage by 90 degrees. Real power P_avg = V_rms * I_rms * cos(phi). For a pure capacitor, phi = +90 degrees, so cos(phi) = 0 and P_avg = 0 W. Therefore, no average energy is converted to heat in the ideal case, even though current can be large. Any heating in real components comes from non-ideal losses (equivalent series resistance or dielectric losses), not from the ideal reactive behavior itself. Step-by-Step Solution: Recall that for a pure capacitor, current leads voltage by 90 degrees.Real power formula: P_avg = V_rms * I_rms * cos(phi).Set phi = 90 degrees ⇒ cos(phi) = 0 ⇒ P_avg = 0 W.Therefore, the statement about heat proportional to current is incorrect in the ideal case. Verification / Alternative check: Compute instantaneous power p(t) = v(t) * i(t) for v = Vp sin(ωt) and i = C dv/dt = ωC Vp cos(ωt). Over one period, the average of sin(ωt)cos(ωt) is zero, confirming zero average real power. Why Other Options Are Wrong: Valid / applies only at resonance / true for low frequencies only: These either contradict the zero real power result or introduce conditions (like resonance) that still do not create real power in an ideal, lossless capacitor. Common Pitfalls: Equating high reactive current with heating; ignoring that only resistive components dissipate real power; forgetting the role of phase angle in P_avg calculation. Final Answer: Invalid

Question 9

In alternating-current (AC) analysis of a series RC circuit driven by a sinusoidal source, does the current lag or lead the applied source voltage? Choose the most accurate statement.

Accepted Answer

Introduction / Context: The phase relationship between current and voltage is a core concept in AC circuit analysis. In reactive networks, capacitors and inductors shift phase, causing current to lead or lag voltage depending on the component. This question focuses on the classic series RC circuit to test whether you recall which quantity leads and by how much. Given Data / Assumptions: A sinusoidal steady-state source is applied. Components are ideal: one resistor R in series with one capacitor C. Angular frequency is omega, capacitive reactance is Xc = 1 / (omega * C). No measurement loading effects are considered. Concept / Approach: The impedance of a series RC circuit is Z = R - jXc. The current I is the same through both elements and equals V / Z. Because the impedance has a negative imaginary part (capacitive), the current phasor leads the voltage phasor by an angle phi where tan(phi) = Xc / R. A helpful memory aid is “ICE” (current leads voltage in a Capacitive circuit; current lags in an Inductive circuit). Step-by-Step Solution: Write the impedance: Z = R - jXc.Determine the phase: phi = arctan(Xc / R) (positive for capacitive, indicating current leads).Conclude the relationship: current leads source voltage by phi, not lags. Verification / Alternative check: At very high frequency, Xc approaches 0, so Z approaches R and the phase approaches 0 degrees, consistent with small lead. At very low frequency, Xc is very large, the current is small, and the capacitor dominates, with current attempting to lead by up to 90 degrees in the ideal limit. Why Other Options Are Wrong: “Current lags” is characteristic of inductive circuits, not capacitive ones. “Only at low frequencies” is misleading; the lead exists at all frequencies, though the angle varies. DC with ripple is irrelevant; pure DC has no steady AC phase, and ripple still follows the AC rule: capacitive currents lead. Common Pitfalls: Confusing inductive and capacitive behavior (“ELI the ICE man” mnemonic helps). Another mistake is assuming the phase is fixed; it depends on R and Xc. Final Answer: Incorrect — in a series RC circuit, the current leads the source voltage by arctan(Xc/R).

Question 10

Series RC phasor sum — a resistor shows 6 V (rms) and a capacitor shows 8 V (rms) in the same series RC branch. Using vector (phasor) addition, what is the magnitude of the source voltage?

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Introduction / Context: In AC circuits containing reactance, voltage drops do not add arithmetically. Instead, they combine as vectors (phasors) because they are out of phase. This question tests your ability to use the right kind of addition for a series RC circuit where resistor and capacitor voltages are in quadrature (90 degrees apart). Given Data / Assumptions: Series RC circuit under sinusoidal steady state. Measured magnitudes: V_R = 6 V (in phase with current), V_C = 8 V (lags current by 90 degrees). The source voltage equals the phasor sum of V_R and V_C. Concept / Approach: In a series RC circuit, current is common. The resistor voltage is V_R = I * R and is in phase with the current; the capacitor voltage is V_C = I * Xc, but since Xc introduces -90 degrees, V_C is 90 degrees behind the current. Therefore, V_R and V_C are orthogonal, making the source magnitude Vs the Pythagorean sum: Vs = sqrt(V_R^2 + V_C^2). Step-by-Step Solution: Identify orthogonal components: V_R along the real axis, V_C along the negative imaginary axis.Compute magnitude: Vs = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 V.Conclude the statement is correct: the source voltage magnitude is 10 V. Verification / Alternative check: If you convert to impedance and current, the same result emerges. For some I, V_R = I*R and V_C = I*Xc; geometry dictates a right triangle, yielding Vs = 10 V when the legs are 6 V and 8 V. Why Other Options Are Wrong: Arithmetic sum (14 V) ignores the 90-degree phase difference. “2 V” is non-physical for these magnitudes; phasor subtraction is not appropriate here. Frequency is not needed once V_R and V_C are given; their vector relationship is fixed. Common Pitfalls: Adding AC drops like DC series voltages or forgetting that resistor and capacitor voltages are orthogonal. Final Answer: Correct — Vs = 10 V.

Question 11

Resistance versus impedance — in AC analysis of an RC network, is resistance the same as impedance? Choose the most accurate statement.

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Introduction / Context: In AC circuits, impedance generalizes resistance to include the effects of reactance from capacitors and inductors. Confusing resistance with impedance leads to incorrect magnitudes and phase angles. This question checks basic vocabulary and the mathematical form of impedance for an RC network. Given Data / Assumptions: Sinusoidal steady-state analysis. Components are ideal, with a resistor R and capacitor C. Capacitive reactance Xc = 1 / (omega * C). Concept / Approach: Resistance is a real scalar measured in ohms and relates voltage and current without phase shift: V = I * R. Impedance is a complex quantity, Z = R + jX, that relates phasor voltage and current: V = I * Z. For a capacitor, the reactance is negative imaginary (Xc is treated as -|Xc|), so in series RC we write Z = R - jXc. The magnitude is |Z| = sqrt(R^2 + Xc^2), and the phase angle is phi = arctan(Xc / R), with current leading voltage in a capacitive circuit. Step-by-Step Solution: Write Z for series RC: Z = R - jXc.Compute magnitude and angle: |Z| = sqrt(R^2 + Xc^2); phi = arctan(Xc / R).Recognize resistance equals impedance only if Xc = 0 (not true for a real capacitor at finite frequency). Verification / Alternative check: At extremely high frequencies Xc approaches 0, so Z approaches R (still not literally equal except in the ideal limit). At DC, Xc tends to infinity, the circuit blocks current, and Z is not equal to R. Why Other Options Are Wrong: “Correct — interchangeable” ignores the imaginary component from reactance. “Valid only at resonance” does not apply to a lone RC branch; resonance requires both L and C. “Valid only at high frequency” is an approximation, not an identity. Common Pitfalls: Using R instead of |Z| in current calculations or neglecting phase when combining AC voltages. Final Answer: Incorrect — impedance includes both R and Xc and is not the same as resistance.

Question 12

Parallel RC phase behavior — as the capacitance C increases in a parallel RC circuit excited by a sinusoid, how does the overall phase angle between the source voltage and total current change?

Accepted Answer

Introduction / Context: Phase angle in parallel circuits depends on the relative contributions of conductance and susceptance. In a parallel RC network, changing the capacitor value alters the reactive branch current dramatically, which changes the total current phasor and the angle it makes relative to the impressed voltage. Given Data / Assumptions: Sinusoidal steady-state excitation. Parallel RC: one branch is a resistor R (conductance G = 1/R), the other is a capacitor C (susceptance Bc = omega * C). Phase angle theta is the angle of total current relative to voltage (voltage is the reference in admittance analysis). Concept / Approach: Total admittance is Y = G + jBc, where Bc = omega * C. As C increases, Bc increases proportionally, so the reactive component of current grows relative to the resistive component. The phase angle of the total current is theta = arctan(Bc / G). Therefore, increasing C increases Bc, which increases |theta| toward +90 degrees (capacitive behavior). Step-by-Step Solution: Write Y: Y = G + jBc with Bc = omega * C.Compute angle: theta = arctan(Bc / G).Increase C → increase Bc → larger theta magnitude → more capacitive phase. Verification / Alternative check: Consider extreme cases. If C is very small, Bc ≈ 0 and the circuit is nearly resistive (theta ≈ 0). If C is very large, Bc dominates and theta approaches +90 degrees, indicating strongly capacitive current. Why Other Options Are Wrong: “More inductive” contradicts the nature of a capacitor. “No change” ignores the dependence of Bc on C. “Reverses sign / pure resistor” has no basis; adding capacitance increases capacitive character. Common Pitfalls: Confusing series and parallel behavior or using reactance Xc = 1/(omega*C) directly in parallel when admittance is more convenient. Final Answer: Correct — increasing C increases the capacitive phase angle magnitude.

Question 13

RC low-pass filter terminology — for a single-pole RC low-pass network, is the bandwidth equal to the -3 dB cutoff frequency f_c?

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Introduction / Context: Bandwidth is a fundamental specification in filter design and signal processing. For simple, first-order filters, the definition is straightforward and often used in practical datasheets. This item ensures you understand how bandwidth relates to the -3 dB cutoff frequency in a single-pole RC low-pass network. Given Data / Assumptions: First-order RC low-pass filter (one resistor, one capacitor). Cutoff frequency f_c defined at the magnitude response where output power is half the midband (voltage down by 3 dB): |H(f_c)| = 1/sqrt(2). “Bandwidth” refers to the range of frequencies for which the filter passes signals with minimal attenuation. Concept / Approach: By convention, for a first-order low-pass, the passband is taken from DC (0 Hz) up to the -3 dB point f_c. Since that interval length is simply f_c − 0, the bandwidth equals f_c. This contrasts with band-pass filters, where bandwidth is f_2 − f_1 between the two half-power frequencies. Step-by-Step Solution: Identify filter type: first-order low-pass (transfer function magnitude decreases beyond f_c).Define bandwidth: from 0 Hz to f_c.Compute bandwidth: BW = f_c − 0 = f_c. Verification / Alternative check: The -3 dB definition implies that below f_c the attenuation is less than or equal to 3 dB; since there is only one cutoff, this region’s width equals f_c. Many textbooks and manufacturer notes explicitly state BW = f_c for first-order low-pass sections. Why Other Options Are Wrong: “Twice f_c” has no basis for a single-pole low-pass. “Defined only for high-pass” is incorrect; bandwidth is a universal metric. “Requires Q” applies to resonant second-order sections, not a first-order RC. Common Pitfalls: Confusing band-pass bandwidth (difference between two cutoffs) with low-pass bandwidth (single cutoff from DC). Final Answer: Correct — bandwidth equals f_c for a first-order RC low-pass filter.

Question 14

Reactance and susceptance equivalence — a capacitive reactance of 10 kΩ corresponds to what capacitive susceptance in siemens or microsiemens?

Accepted Answer

Introduction / Context: Reactance and susceptance are reciprocal concepts used in impedance and admittance analysis. For capacitors, engineers often flip between series (reactance) and parallel (susceptance) viewpoints. This question checks your fluency with the reciprocal relationship for a specific numeric value. Given Data / Assumptions: Capacitive reactance magnitude Xc = 10 kΩ. We are converting to capacitive susceptance Bc in siemens (S). Ideal sinusoidal steady-state conditions. Concept / Approach: Admittance Y is the reciprocal of impedance Z. For a purely capacitive element, the magnitude relationships simplify: |Z| = Xc and |Y| = Bc, and therefore Bc = 1 / Xc. Once Xc is given, its reciprocal yields susceptance in siemens. Converting to microsiemens helps match typical instrumentation scales. Step-by-Step Solution: Start from Bc = 1 / Xc.Insert Xc = 10 kΩ = 10,000 Ω.Compute: Bc = 1 / 10,000 = 0.0001 S = 100 × 10^-6 S = 100 µS. Verification / Alternative check: Cross-check by dimensional reasoning: ohms and siemens are reciprocals. If Xc is large (10 kΩ), its admittance must be small (100 µS), which is consistent. Why Other Options Are Wrong: 10 µS is off by a factor of 10; there is no extra factor of 10 in the reciprocal. 1 mS equals 1000 µS, which contradicts the reciprocal of 10 kΩ. Frequency is not needed here; Xc has already been provided, and Bc = 1/Xc. Common Pitfalls: Forgetting to convert kilo-ohms to ohms before taking the reciprocal or mixing up mS and µS. Final Answer: Correct — 100 µS.

Question 15

RC high-pass behavior — as input frequency decreases, does the output voltage magnitude of a simple RC high-pass filter increase or decrease?

Accepted Answer

Introduction / Context: High-pass filters are ubiquitous in coupling networks, audio systems, and sensor interfaces. Their defining characteristic is that they pass high-frequency content while attenuating low-frequency components and DC. This question probes your qualitative understanding of how output changes with frequency. Given Data / Assumptions: Single-pole RC high-pass filter (series capacitor, shunt resistor or equivalent topology). Sinusoidal steady-state behavior. No significant load interaction beyond the intended resistor. Concept / Approach: In a high-pass RC network, the capacitor’s reactance Xc = 1 / (omega * C) is large at low frequencies and small at high frequencies. The transfer function magnitude |H(jomega)| increases with frequency, approaching unity at very high frequency and approaching zero at very low frequency. Therefore, as frequency decreases, output voltage magnitude falls, not rises. Step-by-Step Solution: Recognize the high-pass role: blocks DC and very slow variations.At low frequency: Xc is large → series impedance high → output across the resistor is small.At high frequency: Xc is small → signal passes → output increases toward input magnitude. Verification / Alternative check: Check the -3 dB cutoff f_c = 1 / (2 * pi * R * C). Below f_c, output rolls off at about -20 dB/decade (first order). Above f_c, output asymptotically approaches full scale. Why Other Options Are Wrong: “Output increases as frequency decreases” contradicts high-pass behavior. “Only when load is inductive” is irrelevant to the basic trend; standard topology already determines the response. “Undetermined” is not correct; the qualitative direction is known without exact values. Common Pitfalls: Mixing up high-pass with low-pass or overlooking the role of the series capacitor as a frequency-dependent coupling element. Final Answer: Incorrect — the output of an RC high-pass decreases as frequency decreases.

Question 16

Choosing the output node — when a series RC network is used as a low-pass filter, from which component should the output be taken to obtain low-pass behavior?

Accepted Answer

Introduction / Context: First-order RC filters are formed by selecting which element develops the output. The same two components can implement either a high-pass or a low-pass, depending on where the output is taken. This question checks if you can identify the correct node for a low-pass configuration in a series RC network. Given Data / Assumptions: Series RC branch driven by a source. Sinusoidal steady-state analysis. Ideal components and light loading (or buffered output). Concept / Approach: In a series RC, the resistor and capacitor form a voltage divider whose division ratio varies with frequency. The capacitor’s reactance is high at low frequency and low at high frequency. Taking the output across the capacitor provides more voltage at low frequencies and less at high frequencies, which is the definition of low-pass behavior. Conversely, taking the output across the resistor yields a high-pass response. Step-by-Step Solution: Form the divider: Vout = Vin * Z_out / (R + Z_C).If Z_out = Z_C, then Vout = Vin * (Z_C / (R + Z_C)).As frequency increases, |Z_C| decreases, so Vout decreases → low-pass response. Verification / Alternative check: At DC (omega = 0), |Z_C| is very large, so almost all the source voltage appears across the capacitor (Vout near Vin). At very high frequency, |Z_C| is very small, so most voltage is across the resistor and the capacitor output tends toward zero. This exactly matches a low-pass characteristic. Why Other Options Are Wrong: Across the resistor is a high-pass configuration. “Either node is identical” is incorrect; the transfer function depends on the node. A parallel RC is not required; a series RC suffices for first-order low-pass. Common Pitfalls: Forgetting that “output across C = low-pass” and “output across R = high-pass” in the series RC topology. Final Answer: Correct — take the output across the capacitor.

Question 17

Combining R and Xc — in a series RC with R = 100 Ω and capacitive reactance Xc = 100 Ω, what is the magnitude of the total impedance Z?

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Introduction / Context: Calculating impedance magnitude in series RC and RL circuits requires vector addition, not simple arithmetic. This question verifies that you can combine resistance and reactance correctly using the Pythagorean relationship that arises from orthogonal phasor components. Given Data / Assumptions: Series RC circuit, sinusoidal steady state. R = 100 Ω, Xc = 100 Ω. Ideal components and no additional series elements. Concept / Approach: Impedance is complex: Z = R - jXc for series RC. The real part is R and the imaginary part is -Xc. The magnitude of a complex number a + jb is sqrt(a^2 + b^2). Therefore, the impedance magnitude is |Z| = sqrt(R^2 + Xc^2). This is analogous to the right triangle formed by resistive and reactive voltage drops or currents in phasor diagrams. Step-by-Step Solution: Write Z = 100 - j100.Compute |Z| = sqrt(100^2 + 100^2) = sqrt(10,000 + 10,000) = sqrt(20,000).Evaluate sqrt(20,000) ≈ 141.4 Ω (since 141.4^2 ≈ 20,000). Verification / Alternative check: If R equals Xc in magnitude, the impedance magnitude is R * sqrt(2). Here, 100 * 1.414 ≈ 141.4 Ω, matching the calculation. The phase angle is phi = arctan(Xc / R) = arctan(1) = 45 degrees (current leads), which is consistent but not required to get |Z|. Why Other Options Are Wrong: Adding magnitudes (200 Ω) ignores orthogonality; reactance is imaginary and must be combined vectorially. “100 Ω because reactance cancels” is false; cancellation does not occur between real and imaginary parts. “Undetermined” is incorrect; |Z| is computable without knowing the angle. Common Pitfalls: Adding R and Xc directly or trying to subtract them numerically, both of which disregard the complex nature of impedance. Final Answer: Correct — the impedance magnitude is approximately 141.4 Ω.

Question 18

Parallel RC circuits and phase behavior with frequency  Evaluate the statement: “If the excitation frequency increases in a parallel RC circuit, the phase angle between the source voltage and the total current decreases.” State whether this claim is…

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Introduction / Context: In alternating-current (AC) circuit analysis, understanding how phase angle varies with frequency is fundamental for filters, sensor interfaces, and power systems. A parallel RC network exhibits a frequency-dependent current lead relative to the applied voltage. This question tests whether you know the direction of that change as frequency increases. Given Data / Assumptions: Topology: parallel RC circuit (a resistor R in parallel with a capacitor C). Source: sinusoidal voltage excitation. Ideal components (ignore capacitor ESR and wiring parasitics unless noted). Phase angle referenced between source voltage and total line current. Concept / Approach: The parallel RC is best analyzed using admittance Y. The total admittance is Y = G + jB where G = 1/R and B = ωC. The total current phasor leads the voltage by angle θ such that tan(θ) = B / G = (ωC) / (1/R) = ωCR. Because B increases linearly with frequency, the current’s lead angle increases with frequency. Step-by-Step Solution: Write Y = G + jB with G = 1/R and B = ωC.Compute the phase: θ = arctan(B/G) = arctan(ωCR).As frequency f (and ω = 2πf) increases, ωCR increases monotonically.Therefore θ increases (the total current leads the voltage by a larger angle). Verification / Alternative check: At very low frequency (ω → 0), B → 0 and θ → 0°, so the circuit behaves resistively. At very high frequency, B ≫ G and θ approaches 90°, dominated by capacitive current. This confirms that θ increases with frequency, not decreases. Why Other Options Are Wrong: Correct: contradicts admittance-phase behavior; θ does not decrease with frequency.Valid only at resonance: RC networks do not resonate without L; no resonance applies.Applies only when R = XC: that is a series condition for a specific phase in series RC; not relevant here.Zero ESR clause: ESR slightly perturbs G, but the trend (θ increasing with f) remains. Common Pitfalls: Confusing series vs. parallel behavior; mixing voltage-lead with current-lead language; assuming capacitor current behavior from series circuits applies identically to parallel circuits. Final Answer: Incorrect

Question 19

Phase angle of a series RC network with specified reactance ratio  A sinusoid drives a series RC circuit where the capacitive reactance magnitude is four-thirds of the resistance: |XC| : R = 4 : 3. What is the phase angle between the source voltage …

Accepted Answer

Introduction / Context: For a series RC circuit, the current leads the total voltage by an angle whose magnitude depends on the ratio of capacitive reactance to resistance. Designers use this relationship to set filter corner slopes, timing responses, and phase-shift networks. Here we compute the phase given a specific |XC| to R ratio. Given Data / Assumptions: Topology: series RC. Ratio: |XC| : R = 4 : 3 (so |XC| = 4k, R = 3k for some common factor k). Steady-state sinusoidal operation with ideal components. Phase angle φ defined as the angle between total voltage and current; for RC, current leads voltage by |φ|. We report the magnitude. Concept / Approach: In a series RC, the impedance is Z = R − j|XC|. The phase angle between V and I is φ = arctan(−|XC|/R); current leads by |φ|. The magnitude is |φ| = arctan(|XC|/R). With |XC|/R known, we can directly compute the arctangent. Step-by-Step Solution: Compute the ratio: |XC|/R = 4/3.Find the angle: |φ| = arctan(4/3).Use the 3–4–5 triangle identity: arctan(4/3) ≈ 53.13°.Therefore, the current leads the voltage by 53.13° (magnitude reported). Verification / Alternative check: Construct a right triangle with adjacent = R = 3 and opposite = |XC| = 4. The hypotenuse is 5. The angle opposite the “4” side is arcsin(4/5) = 53.13°, matching the arctan result. Impedance magnitude is 5k (arbitrary k), consistent with the triangle. Why Other Options Are Wrong: 90°: would require R → 0 (purely capacitive), not true here.36.87°: that angle corresponds to arctan(3/4), i.e., |XC| < R, which is not our case.0°: purely resistive (|XC| = 0), not true.18.43°: unrelated to the given 4:3 ratio. Common Pitfalls: Mixing sign and magnitude conventions; forgetting that in series RC, current leads voltage; confusing |XC|/R with R/|XC|. Final Answer: 53.13°

Question 20

Oscilloscope safety in RC networks  A single-ended oscilloscope (earth-referenced ground clip) is connected exactly as shown in a typical RC ladder where the probe’s ground clip is tied to a non-ground node that also connects to C1, R2, and C2. Pred…

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Introduction / Context: Oscilloscopes with single-ended inputs use a ground-referenced clip that is bonded to earth ground. If you attach this clip to a node that is not at ground potential, you create a low-impedance path to earth. In multi-node RC networks, this mistake can unintentionally short portions of the circuit and alter or damage the system under test. Given Data / Assumptions: Instrument: earth-referenced, single-ended oscilloscope channel. Connection: the probe’s ground clip is placed on a non-ground node shared by C1, R2, and C2. Assume typical bench wiring: scope chassis and ground clip are connected to mains earth. Concept / Approach: Earth-referenced ground clips clamp the chosen node to earth potential. Any network node tied by that clip becomes effectively shorted to ground. Components connected from that node to ground line are bypassed, altering impedances and measured voltages, and may cause high currents if a source drives across the unintended short. Step-by-Step Solution: Identify the node where the ground clip is attached (shared by C1, R2, and C2).Recognize that the clip forces this node to 0 V (earth).All elements from this node to ground are effectively shorted or bypassed.Hence C1, R2, and C2 are shorted to ground in the measurement setup. Verification / Alternative check: Compare with a differential probe or an isolated/battery-powered scope. Those instruments do not reference either lead to earth, preventing the short; the network behavior remains intact and the voltage is measured correctly. Why Other Options Are Wrong: Measure the source voltage: not if the connection shorts intermediate nodes.Measure across C1, R2, and C2: shorting them collapses their voltages to near zero.Short R1 only: the ground clip location dictates which parts are shorted; not necessarily R1 alone.Increase impedance: the action decreases impedance by creating a low-impedance path. Common Pitfalls: Using ground clips on floating or active nodes; forgetting that the probe shell is grounded; not using differential probes when required. Final Answer: short C1, R2, and C2

RC Circuits Questions