Q: In a series RC circuit, both the source frequency f and resistance R are halved. Without numerical values of R, C, and f, how does the total impedance |Z| change?

Introduction / Context: The magnitude of impedance in a series RC network depends on both resistance and frequency-dependent reactance. When two parameters change simultaneously, the combined effect on |Z| cannot be predicted uniquely without specific values. Given Data / Assumptions: Series RC with R and C constant initially. Change: R → R/2 and f → f/2. Capacitance C unchanged. Concept / Approach: Impedance magnitude is |Z| = √(R^2 + Xc^2), where Xc = 1 / (2 * π * f * C). Halving f doubles Xc; halving R halves R. The new magnitude depends on the relative sizes of R and Xc before the change. Step-by-Step Solution: Original: |Z| = √(R^2 + Xc^2).After change: R′ = R/2, Xc′ = 2 * Xc.New |Z′| = √((R/2)^2 + (2Xc)^2) = √(R^2/4 + 4Xc^2).The ratio |Z′| / |Z| = √((R^2/4 + 4Xc^2) / (R^2 + Xc^2)) depends on the R/Xc balance and has no single fixed value. Verification / Alternative check: If R ≫ Xc, halving R may dominate and |Z′| decreases; if Xc ≫ R, doubling Xc may dominate and |Z′| increases. Different starting conditions give different outcomes. Why Other Options Are Wrong: 'Doubles', 'halved', and 'one-fourth' assert universal factors that ignore the relative sizes of R and Xc. Common Pitfalls: Assuming proportional changes carry over to the square-root sum; forgetting Xc depends on f. Final Answer: cannot be determined without values

Q: In a series RC circuit, as the frequency of the applied sinusoidal voltage increases, how does the circuit phase angle θ (of impedance) change?

Introduction / Context: The phase angle θ of a series RC circuit indicates how far the current leads the applied voltage. Understanding how θ varies with frequency is central to filter design and Bode-plot intuition. Given Data / Assumptions: Series RC with resistance R and capacitance C fixed. Frequency f increases. θ is defined from the impedance Z = R − jXc: θ = arctan(−Xc/R) (capacitive angle is negative). Concept / Approach: As frequency increases, capacitive reactance Xc = 1/(2 * π * f * C) decreases. With a smaller Xc relative to R, the magnitude |θ| shrinks toward 0°, meaning the circuit becomes more resistive and the phase lag (negative angle) reduces in magnitude. Step-by-Step Solution: Start: θ = arctan(−Xc/R).Increase f ⇒ Xc decreases ⇒ |−Xc/R| decreases.Therefore, |θ| decreases toward 0°; the phase angle (more precisely its magnitude) decreases. Verification / Alternative check: At very high f, Xc → 0 and θ → 0°. At very low f, Xc ≫ R and θ → −90°. This monotonic trend confirms decreasing |θ| with increasing f. Why Other Options Are Wrong: 'Increases' contradicts the trend. 'Remains the same' is not true because Xc changes with f. 'Becomes erratic' does not describe linear, time-invariant behavior. Common Pitfalls: Ignoring the negative sign (capacitive lag) and comparing signed angles instead of magnitude; mixing RC with RL behavior. Final Answer: decreases

Q: A 120 Ω resistor is in parallel with a capacitor having Xc = 40 Ω, both across a 12 V AC source. What is the magnitude of the total circuit impedance |Z|?

Introduction / Context: For parallel RC circuits, it is often easier to work in admittance (Y) because conductances and susceptances add directly. The magnitude of impedance is the reciprocal of the magnitude of admittance. Given Data / Assumptions: Resistive branch: R = 120 Ω ⇒ G = 1/R = 0.008333 S. Capacitive branch: Xc = 40 Ω ⇒ Bc = −1/Xc = −0.025 S (capacitive susceptance). Ideal sinusoidal conditions. Concept / Approach: Total admittance: Y = G + jBc. Magnitude |Y| = √(G^2 + Bc^2). Then |Z| = 1 / |Y|. Step-by-Step Solution: Compute G = 1/120 = 0.008333 S.Compute |Bc| = 1/40 = 0.025 S (sign negative for capacitor; magnitude used for |Y|).|Y| = √(0.008333^2 + 0.025^2) ≈ √(0.00006944 + 0.000625) ≈ √(0.00069444) ≈ 0.02638 S.Therefore, |Z| = 1 / 0.02638 ≈ 37.9 Ω. Verification / Alternative check: Sanity check: Parallel combinations lower impedance below the smallest branch magnitude (min of 120 Ω and 40 Ω is 40 Ω). The result, ~37.9 Ω, is slightly below 40 Ω, which is consistent. Why Other Options Are Wrong: 3.7 Ω is an order-of-magnitude error. 14,400 Ω and 4,800 Ω result from misapplied series reasoning or arithmetic inversions. Common Pitfalls: Adding impedances directly in parallel; forgetting to invert admittance to return to impedance. Final Answer: 37.9 Ω

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

Question 1

Right-triangle length: Determine the hypotenuse of a right triangle whose perpendicular sides measure 12 units and 18 units.

Accepted Answer

Introduction / Context: The Pythagorean theorem is widely used across electrical engineering and physics, particularly in phasor magnitude calculations (e.g., combining orthogonal voltage or current components). Here, it is applied directly to a geometric right triangle to find the hypotenuse from two perpendicular legs. Given Data / Assumptions: Leg a = 12 units. Leg b = 18 units. Right triangle; hypotenuse c is opposite the right angle. Concept / Approach: Use the Pythagorean relation c^2 = a^2 + b^2. Then compute c = √(a^2 + b^2). This is the same vector magnitude formula used in phasor addition and impedance magnitude calculations in AC circuits. Step-by-Step Solution: Compute squares: a^2 = 12^2 = 144, b^2 = 18^2 = 324.Sum: a^2 + b^2 = 144 + 324 = 468.Take square root: c = √468.Simplify numerically: √468 ≈ 21.633 (since 21.633^2 ≈ 468).Rounded to two decimals: c ≈ 21.63. Verification / Alternative check: Express √468 = √(4 * 117) = 2√117. With √117 ≈ 10.8167, c ≈ 2 * 10.8167 ≈ 21.633, which matches the decimal computation and ensures consistent rounding. Why Other Options Are Wrong: 4.24, 3.46, 2.16: These are much smaller than either leg and cannot be a hypotenuse (which must be the longest side). Common Pitfalls: Accidentally subtracting the squares (appropriate only for projecting one component from hypotenuse and other leg). Rounding too early and producing a noticeably inaccurate result. Final Answer: 21.63

Question 2

In AC circuit analysis, a pure capacitor with capacitive reactance Xc = 150 Ω is connected across a sinusoidal source. Express the capacitor’s impedance Z in polar form (magnitude ∠ angle, in ohms).

Accepted Answer

Introduction / Context: A capacitor in steady-state AC is represented by a purely imaginary impedance that is negative on the imaginary axis. Converting this to polar form reinforces phasor skills used in RC/RLC analysis and filter design. Given Data / Assumptions: Pure capacitor with capacitive reactance Xc = 150 Ω. No series resistance or parasitics; ideal sinusoidal steady state. Polar form is written as |Z| ∠ angle, in ohms. Concept / Approach: The impedance of a capacitor is Zc = −jXc. On the complex plane, −j corresponds to an angle of −90° (pointing straight down the negative imaginary axis). The magnitude is |Z| = Xc for a pure capacitor. Step-by-Step Solution: Write impedance: Zc = −j * 150 Ω.Magnitude: |Zc| = 150 Ω.Angle: ∠Zc = −90° (purely capacitive).Polar form: Zc = 150 ∠ −90° Ω. Verification / Alternative check: Rectangular-to-polar check: a point at (0, −150) has magnitude √(0^2 + 150^2) = 150 and angle arctan(−∞) = −90°, confirming the polar form. Why Other Options Are Wrong: 150∠90° Ω describes +j150 Ω (inductor). 'Z = 150' omits the essential angle and would imply a resistor. 150∠180° Ω would be a negative real resistor, which is not a capacitor. Common Pitfalls: Confusing sign of the imaginary part; swapping inductor and capacitor angles; leaving off the angle in polar form. Final Answer: Z = 150 ∠–90°Ω

Question 3

A 2 kΩ resistor in series with a 0.015 µF capacitor is connected to a 15 kHz AC source. Determine the total impedance magnitude |Z| and the circuit phase angle θ (degrees).

Accepted Answer

Introduction / Context: Series RC impedance problems combine frequency-dependent reactance with resistance via vector (phasor) addition. You must compute capacitive reactance, then form the magnitude and angle of the series impedance to understand current and voltage phase relations. Given Data / Assumptions: R = 2 kΩ. C = 0.015 µF = 15 nF. f = 15 kHz (sinusoidal steady state). Ideal components; no stray inductance or ESR. Concept / Approach: Compute Xc = 1 / (2 * π * f * C). For a series RC, Z = R − jXc. The magnitude is |Z| = √(R^2 + Xc^2), and the phase angle is θ = arctan(−Xc / R) (negative for a capacitive circuit). Step-by-Step Solution: Xc = 1 / (2 * π * 15,000 * 15e−9) ≈ 707 Ω.Z = 2000 − j707 Ω.|Z| = √(2000^2 + 707^2) ≈ √(4,000,000 + 499,849) ≈ 2121 Ω.θ = arctan(−707 / 2000) ≈ −19.5°. Verification / Alternative check: Since Xc < R, |Z| should be slightly above R and angle modestly negative, matching 2121 Ω and −19.5°. Why Other Options Are Wrong: 707 Ω is Xc, not |Z|. 734 Ω and −38.9° are inconsistent with the series values. 73.4 Ω is off by a factor-of-10 slip. Common Pitfalls: Mistaking Xc for |Z|; forgetting the negative sign of the capacitive angle; calculator in radians instead of degrees. Final Answer: 2121 Ω and θ = –19.5°

Question 4

In a series RC circuit, both the source frequency f and resistance R are halved. Without numerical values of R, C, and f, how does the total impedance |Z| change?

Accepted Answer

Introduction / Context: The magnitude of impedance in a series RC network depends on both resistance and frequency-dependent reactance. When two parameters change simultaneously, the combined effect on |Z| cannot be predicted uniquely without specific values. Given Data / Assumptions: Series RC with R and C constant initially. Change: R → R/2 and f → f/2. Capacitance C unchanged. Concept / Approach: Impedance magnitude is |Z| = √(R^2 + Xc^2), where Xc = 1 / (2 * π * f * C). Halving f doubles Xc; halving R halves R. The new magnitude depends on the relative sizes of R and Xc before the change. Step-by-Step Solution: Original: |Z| = √(R^2 + Xc^2).After change: R′ = R/2, Xc′ = 2 * Xc.New |Z′| = √((R/2)^2 + (2Xc)^2) = √(R^2/4 + 4Xc^2).The ratio |Z′| / |Z| = √((R^2/4 + 4Xc^2) / (R^2 + Xc^2)) depends on the R/Xc balance and has no single fixed value. Verification / Alternative check: If R ≫ Xc, halving R may dominate and |Z′| decreases; if Xc ≫ R, doubling Xc may dominate and |Z′| increases. Different starting conditions give different outcomes. Why Other Options Are Wrong: 'Doubles', 'halved', and 'one-fourth' assert universal factors that ignore the relative sizes of R and Xc. Common Pitfalls: Assuming proportional changes carry over to the square-root sum; forgetting Xc depends on f. Final Answer: cannot be determined without values

Question 5

In a series RC circuit, as the frequency of the applied sinusoidal voltage increases, how does the circuit phase angle θ (of impedance) change?

Accepted Answer

Introduction / Context: The phase angle θ of a series RC circuit indicates how far the current leads the applied voltage. Understanding how θ varies with frequency is central to filter design and Bode-plot intuition. Given Data / Assumptions: Series RC with resistance R and capacitance C fixed. Frequency f increases. θ is defined from the impedance Z = R − jXc: θ = arctan(−Xc/R) (capacitive angle is negative). Concept / Approach: As frequency increases, capacitive reactance Xc = 1/(2 * π * f * C) decreases. With a smaller Xc relative to R, the magnitude |θ| shrinks toward 0°, meaning the circuit becomes more resistive and the phase lag (negative angle) reduces in magnitude. Step-by-Step Solution: Start: θ = arctan(−Xc/R).Increase f ⇒ Xc decreases ⇒ |−Xc/R| decreases.Therefore, |θ| decreases toward 0°; the phase angle (more precisely its magnitude) decreases. Verification / Alternative check: At very high f, Xc → 0 and θ → 0°. At very low f, Xc ≫ R and θ → −90°. This monotonic trend confirms decreasing |θ| with increasing f. Why Other Options Are Wrong: 'Increases' contradicts the trend. 'Remains the same' is not true because Xc changes with f. 'Becomes erratic' does not describe linear, time-invariant behavior. Common Pitfalls: Ignoring the negative sign (capacitive lag) and comparing signed angles instead of magnitude; mixing RC with RL behavior. Final Answer: decreases

Question 6

A 120 Ω resistor is in parallel with a capacitor having Xc = 40 Ω, both across a 12 V AC source. What is the magnitude of the total circuit impedance |Z|?

Accepted Answer

Introduction / Context: For parallel RC circuits, it is often easier to work in admittance (Y) because conductances and susceptances add directly. The magnitude of impedance is the reciprocal of the magnitude of admittance. Given Data / Assumptions: Resistive branch: R = 120 Ω ⇒ G = 1/R = 0.008333 S. Capacitive branch: Xc = 40 Ω ⇒ Bc = −1/Xc = −0.025 S (capacitive susceptance). Ideal sinusoidal conditions. Concept / Approach: Total admittance: Y = G + jBc. Magnitude |Y| = √(G^2 + Bc^2). Then |Z| = 1 / |Y|. Step-by-Step Solution: Compute G = 1/120 = 0.008333 S.Compute |Bc| = 1/40 = 0.025 S (sign negative for capacitor; magnitude used for |Y|).|Y| = √(0.008333^2 + 0.025^2) ≈ √(0.00006944 + 0.000625) ≈ √(0.00069444) ≈ 0.02638 S.Therefore, |Z| = 1 / 0.02638 ≈ 37.9 Ω. Verification / Alternative check: Sanity check: Parallel combinations lower impedance below the smallest branch magnitude (min of 120 Ω and 40 Ω is 40 Ω). The result, ~37.9 Ω, is slightly below 40 Ω, which is consistent. Why Other Options Are Wrong: 3.7 Ω is an order-of-magnitude error. 14,400 Ω and 4,800 Ω result from misapplied series reasoning or arithmetic inversions. Common Pitfalls: Adding impedances directly in parallel; forgetting to invert admittance to return to impedance. Final Answer: 37.9 Ω

Question 7

For the series RC voltages considered at some frequency, what change in frequency will make the capacitor’s voltage exceed the resistor’s voltage (i.e., |Vc| > |Vr|)?

Accepted Answer

Introduction / Context: In a series RC circuit, the ratio of capacitor to resistor voltage depends on the ratio of reactance to resistance. Tuning frequency can shift which element has the larger drop, a key idea in filters and frequency-selective dividers. Given Data / Assumptions: Series RC, fixed R and C. We compare |Vc| and |Vr| at different frequencies. Ideal, linear, sinusoidal operation. Concept / Approach: Voltage division in series gives |Vc| : |Vr| ≈ Xc : R. If Xc > R, the capacitor drop dominates; if Xc < R, the resistor drop dominates. Since Xc = 1 / (2 * π * f * C), lowering frequency increases Xc. Step-by-Step Solution: Goal: |Vc| > |Vr| ⇒ Xc > R.Adjust frequency: decrease f so that Xc = 1/(2πfC) grows and exceeds R.Therefore, to make Vc larger than Vr, the frequency must be decreased. Verification / Alternative check: At very low f, Xc ≫ R and almost all source voltage appears across the capacitor; at very high f, Xc → 0 and the resistor drop dominates. This confirms the frequency trend. Why Other Options Are Wrong: Increasing f shrinks Xc, favoring the resistor. Holding constant or 'no effect' contradicts the strong frequency dependence of Xc. Common Pitfalls: Confusing RC with RL behavior; ignoring that voltages in series divide in proportion to impedance magnitudes. Final Answer: must be decreased

Question 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

Operations with Complex Numbers in AC/Signal Analysis  Assess the statement: “Complex numbers can be added and subtracted but not multiplied or divided.”

Accepted Answer

Introduction / Context: Complex arithmetic underpins phasor analysis, impedance calculations, and control systems. Engineers routinely add, subtract, multiply, and divide complex quantities representing amplitudes and phase angles. The statement challenges whether all four operations are valid. Given Data / Assumptions: Complex numbers expressed in rectangular (a + j b) or polar (r ∠ θ) form. Nonzero denominator when performing division. Standard algebraic rules for complex arithmetic apply. Concept / Approach: The field of complex numbers supports addition, subtraction, multiplication, and division (except division by zero). In rectangular form, multiplication and division use distributive/associative properties and complex conjugates. In polar form, multiplication multiplies magnitudes and adds angles; division divides magnitudes and subtracts angles. Step-by-Step Solution: Multiplication (rectangular): (a + j b)(c + j d) = (ac − bd) + j(ad + bc).Division (rectangular): (a + j b)/(c + j d) = [(a + j b)(c − j d)]/(c^2 + d^2) = [(ac + bd) + j(bc − ad)]/(c^2 + d^2).Multiplication (polar): (r1 ∠ θ1)(r2 ∠ θ2) = (r1 r2) ∠ (θ1 + θ2).Division (polar): (r1 ∠ θ1)/(r2 ∠ θ2) = (r1/r2) ∠ (θ1 − θ2). Verification / Alternative check: Phasor operations in AC circuits rely on these rules: multiplying by j rotates a vector by +90 degrees; impedance division calculates current from voltage and impedance; transfer functions use complex rational expressions routinely. Why Other Options Are Wrong: Claims restricting multiplication or division contradict the algebraic structure of complex numbers. Both rectangular and polar forms fully support all operations (with the usual caveat of nonzero divisors). Common Pitfalls: Confusing angle units (degrees vs. radians) in polar form; forgetting to rationalize the denominator in rectangular form; or mishandling signs when using the conjugate. Final Answer: False

Question 16

Number Line Convention: Location of Positive Numbers on the Cartesian Axes  Evaluate the statement: “Positive numbers are represented by points to the right of the origin on the horizontal (x) axis.”

Accepted Answer

Introduction / Context: Graphical representation of quantities is fundamental to mathematics and engineering. By convention, the Cartesian x-axis increases to the right and the y-axis increases upward. This convention underlies plotting signals, Bode diagrams, IV curves, and more. Given Data / Assumptions: Standard right-handed Cartesian coordinate system. Origin at (0, 0), horizontal axis labeled x, vertical axis labeled y. Uniform linear scale along axes. Concept / Approach: On the x-axis, positive values are located to the right of the origin, and negative values to the left. This orientation is part of the conventional right-handed system: the positive x direction to the right, positive y upwards, and when including z, the positive z comes out of the plane following the right-hand rule. Step-by-Step Solution: Mark the origin at x = 0.Place tick marks at x = +1, +2, +3 to the right of 0 and at x = −1, −2 to the left.By definition of the coordinate system, points to the right correspond to positive numbers on the x-axis.Therefore the statement is correct under standard plotting conventions. Verification / Alternative check: Any standard math or physics text uses this orientation. Software plotting tools (MATLAB, Python, spreadsheets) default to the same layout unless explicitly redefined. Why Other Options Are Wrong: “False” contradicts the universal convention. “True only in polar/logarithmic” is irrelevant; the statement refers to the Cartesian x-axis. While nonstandard axes could be relabeled or reversed, the default assumption is the standard convention unless stated otherwise. Common Pitfalls: Confusing screen pixel coordinates (some graphics APIs have y increasing downward) with mathematical axes; always rely on the plotted axis labels and arrows when in doubt. Final Answer: True

Question 17

Complex Numbers as Phasors in AC Analysis  Evaluate the statement: “A complex number represents a phasor quantity.”

Accepted Answer

Introduction / Context: Phasors offer a powerful way to analyze linear AC circuits by representing sinusoidal signals as rotating vectors (complex numbers) at a fixed angular frequency. This abstraction converts differential equations into algebraic ones in the frequency domain, enabling straightforward calculations of voltages, currents, and impedances. Given Data / Assumptions: Steady-state sinusoidal signals at angular frequency ω. Linear time-invariant circuits where phasor representation is valid. Complex numbers used in rectangular a + j b or polar r ∠ θ form. Concept / Approach: A phasor is the complex representation of a sinusoid’s amplitude and phase: x(t) = Re{X e^{j ω t}}, where the complex constant X encodes magnitude and phase. As such, every phasor corresponds to a complex number, and manipulations of phasors follow complex arithmetic rules. Impedances of L and C become j ω L and 1/(j ω C), respectively, fitting naturally into the complex framework. Step-by-Step Solution: Represent a sinusoid x(t) = X_m cos(ω t + φ).Define the phasor X = X_m / √2 ∠ φ (for RMS) or X = X_m ∠ φ (for peak).In rectangular form, X = a + j b with a = X cos φ and b = X sin φ.Circuit laws (KCL/KVL) apply to phasors using complex addition and multiplication with impedances. Verification / Alternative check: Solving an RLC series circuit via phasors yields I = V / (R + j ω L + 1/(j ω C)). Time-domain solution reconstructed by i(t) = Re{I e^{j ω t}} matches the sinusoidal steady-state solution, confirming the equivalence. Why Other Options Are Wrong: A magnitude-1 restriction is unnecessary; phasors carry arbitrary magnitude. Rectangular or polar are both valid; the form does not limit phasor validity. Phasors are inherently complex; insisting on real-only would lose phase information. Common Pitfalls: Mixing instantaneous values with phasors (phasors omit the explicit time factor e^{j ω t}); forgetting RMS vs. peak conventions when comparing magnitudes. Final Answer: True

Question 18

Series RC circuits — does the total impedance vary directly with frequency?

Accepted Answer

Introduction / Context: This statement probes how the magnitude of impedance in a series RC circuit changes with frequency. Understanding this relationship is fundamental for filter design, Bode plots, and time-constant intuition in electronics. Given Data / Assumptions: Series RC network driven by a sinusoidal source of angular frequency ω = 2 * π * f. Impedance of resistor R is R (independent of frequency). Impedance of capacitor C is Zc = 1 / (j * ω * C). Concept / Approach: The total series impedance is the vector sum of R and the capacitive reactance magnitude Xc = 1 / (ω * C). As frequency increases, Xc decreases. The impedance magnitude therefore decreases toward R, not increases “directly with frequency.” Step-by-Step Solution: Write total impedance: Z = R − j * (1 / (ω * C)).Compute magnitude: |Z| = sqrt( R^2 + (1 / (ω * C))^2 ).Observe dependence: as ω increases, (1 / (ω * C)) decreases, so |Z| monotonically decreases toward R.At low ω → 0, |Z| → very large (dominated by capacitor); at high ω → ∞, |Z| → R. Verification / Alternative check: Sketch a Bode magnitude plot for |Z|. The curve slopes downward with a corner near ω = 1 / (R * C), confirming inverse dependence on frequency due to the 1 / (ω * C) term. Why Other Options Are Wrong: Choosing “True” implies |Z| grows with f, contradicting |Z| = sqrt( R^2 + (1 / (ω * C))^2 ). Common Pitfalls: Confusing impedance of a capacitor (inversely proportional to frequency) with that of an inductor (directly proportional to frequency). Only inductive reactance Xl = ω * L increases with frequency. Final Answer: False

Question 19

Lag network (RC low-pass) — does the output voltage lag the input in phase?

Accepted Answer

Introduction / Context: Lag networks (often realized as RC low-pass filters with output taken across the capacitor) are used for phase lag and amplitude attenuation at higher frequencies. This question checks phase intuition for such networks. Given Data / Assumptions: A passive RC network configured so that output is across the capacitor (classic lag configuration). Sinusoidal steady-state operation at frequency ω. Concept / Approach: For an RC low-pass, the capacitor current leads the capacitor voltage by 90 degrees, but the capacitor voltage lags the input because the resistor introduces a voltage division with frequency-dependent phase. Thus, Vout (across C) lags Vin for ω > 0. Step-by-Step Solution: Transfer function (magnitude-phase): H(jω) = Vout/Vin = 1 / (1 + j * ω * R * C).Phase of H(jω) = −arctan(ω * R * C).Since −arctan(ωRC) ≤ 0 for ω ≥ 0, output angle is negative → output lags input.At ω = 0, phase = 0 degrees (no lag). As ω increases, lag increases toward −90 degrees. Verification / Alternative check: Phasor diagram: input phasor equals the vector sum of resistor and capacitor voltage phasors; the capacitor voltage trails the current by 90 degrees, placing it behind the input phasor in angle. Why Other Options Are Wrong: “False” would imply the output does not lag; that contradicts both H(jω) angle and basic RC phasor relations when output is taken across C. Common Pitfalls: Confusing lag vs lead networks. If output is taken across the resistor (RC high-pass), the output leads the input at many frequencies (lead network). The placement of the output node determines phase behavior. Final Answer: True

Question 20

Pure resistors — is the resistor’s voltage always out of phase with the current?

Accepted Answer

Introduction / Context: Phase relationships between voltage and current define how elements store or dissipate energy. This item checks the unique, purely dissipative behavior of an ideal resistor. Given Data / Assumptions: Ideal resistor R only; no reactive components. Sinusoidal steady-state excitation. Concept / Approach: In a pure resistor, voltage and current are in phase. Inductors and capacitors cause phase shifts of +90 degrees (current lags) and −90 degrees (current leads), respectively; resistors cause none. Step-by-Step Solution: Ohm’s law in phasor form: V = I * R (R is real).Phasor angle of R is 0 degrees, so angle(V) − angle(I) = 0.Therefore, v(t) and i(t) reach maxima and zeros simultaneously—no phase difference. Verification / Alternative check: Instantaneous power p(t) = v(t) * i(t) is always non-negative for a resistor, consistent with no reactive energy exchange and reinforcing zero phase shift. Why Other Options Are Wrong: Selecting “True” confuses resistors with reactive elements that store energy and introduce phase shifts. Common Pitfalls: Assuming some phase difference must exist because AC is involved. AC does not guarantee phase shifts; it depends on element type. Final Answer: False

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