Q: Rectangular form conversion: Express the polar-form complex number 40∠55° as its equivalent rectangular form (a + jb).

Introduction / Context: Converting between polar (magnitude ∠ angle) and rectangular (a + jb) forms is essential for phasor arithmetic in AC circuits. Rectangular form is convenient for addition and subtraction; polar form is handy for multiplication and division. Given Data / Assumptions: Magnitude M = 40. Angle θ = 55° (degrees). Use a = M * cosθ and b = M * sinθ for rectangular components. Concept / Approach: Apply trigonometric projections to map the polar vector onto the real and imaginary axes. Ensure the calculator is in degrees when using degree angles. Step-by-Step Solution: Compute real part: a = 40 * cos(55°) ≈ 40 * 0.5736 ≈ 22.94.Compute imaginary part: b = 40 * sin(55°) ≈ 40 * 0.8192 ≈ 32.77 ≈ 32.76 (rounded).Therefore, rectangular form ≈ 22.94 + j32.76. Verification / Alternative check: Back-convert: magnitude √(22.94^2 + 32.76^2) ≈ 40; angle arctan(32.76/22.94) ≈ 55°, confirming correctness. Why Other Options Are Wrong: 40 + j40 and 55 + j55 imply θ = 45° with different magnitudes. 45.88 + j65.52 corresponds to a magnitude much larger than 40. Common Pitfalls: Using radians inadvertently; swapping sin and cos; rounding too early leading to noticeable errors. Final Answer: 22.94 + j32.76

Q: Polar form conversion: Convert the rectangular complex number 6 + j6 into polar form (magnitude ∠ angle in degrees).

Introduction / Context: Converting between rectangular and polar forms lets us pick the most convenient representation for circuit computations. Equal real and imaginary parts indicate a 45° angle in the first quadrant for positive values. Given Data / Assumptions: Rectangular form: a = 6, b = 6 (first quadrant). Magnitude formula: M = √(a^2 + b^2). Angle formula: θ = arctan(b/a) in degrees for a > 0. Concept / Approach: Calculate magnitude and angle directly. Because a = b > 0, the angle should be 45°, serving as a quick sanity check. Step-by-Step Solution: Magnitude: M = √(6^2 + 6^2) = √(36 + 36) = √72 ≈ 8.485 ≈ 8.48.Angle: θ = arctan(6/6) = arctan(1) = 45°.Polar form: 8.48∠45°. Verification / Alternative check: Back-conversion: a = M cosθ ≈ 8.48 * 0.707 ≈ 6; b = M sinθ ≈ 8.48 * 0.707 ≈ 6. Why Other Options Are Wrong: 6∠45° has correct angle but wrong magnitude. 8.48∠90° has correct magnitude but wrong angle. 36∠45° has magnitude squared error (not how polar magnitude is defined). Common Pitfalls: Confusing RMS/peak ideas with complex magnitude; forgetting that equal components imply 45° in quadrant I. Final Answer: 8.48∠45°

Question 1

Series RC impedance in rectangular form: A 47 Ω resistor is in series with a capacitor whose reactance is 150 Ω. What is the total impedance Z expressed in rectangular form?

Accepted Answer

Introduction / Context: Representing impedances in rectangular form (R ± jX) is central to AC circuit analysis. For capacitors, the reactance is negative imaginary; for inductors, it is positive imaginary. This sign convention enables straightforward phasor addition of series elements. Given Data / Assumptions: Resistor R = 47 Ω (real, in-phase with voltage). Capacitive reactance magnitude X_C = 150 Ω. Series connection; sinusoidal steady-state. Concept / Approach: Impedances in series add directly as complex numbers. For a capacitor, X_C is represented as −jX_C. Therefore, Z_total = R + (−jX_C) = R − jX_C. Avoid converting to polar unless requested; the problem asks specifically for rectangular form. Step-by-Step Solution: Express capacitor: Z_C = −j 150 Ω.Add series resistor: Z = 47 + (−j150) Ω.Therefore, Z = 47 Ω − j150 Ω. Verification / Alternative check: Polar magnitude: |Z| = √(47^2 + 150^2) ≈ √(2209 + 22500) ≈ √24709 ≈ 157.2 Ω; angle ≈ arctan(−150/47) ≈ −72.7°. Converting back to rectangular reproduces 47 − j150 Ω, confirming correctness. Why Other Options Are Wrong: Z = 47 Ω + j150 Ω: Wrong sign for capacitive reactance (that would be inductive). Z = 197 Ω or Z = 103 Ω: These look like real-only magnitudes; impedance has both real and imaginary parts here. Common Pitfalls: Using +j for capacitors instead of −j. Treating reactance magnitudes as simple arithmetic with resistance without complex notation. Final Answer: Z = 47 Ω – j150 Ω

Question 2

Complex numbers and quadrants: In the complex plane, determine the quadrant in which the number 14 − j5 (real part 14, imaginary part −5) lies.

Accepted Answer

Introduction / Context: Quadrant identification for complex numbers is a foundational skill in AC circuit analysis and phasor algebra. A complex number a + jb corresponds to a point with horizontal coordinate a (real axis) and vertical coordinate b (imaginary axis). Knowing the sign of each part immediately tells you the quadrant. Given Data / Assumptions: Complex number: 14 − j5. Real part Re = 14 (positive). Imaginary part Im = −5 (negative). Standard quadrant convention: I (+,+), II (−,+), III (−,−), IV (+,−). Concept / Approach: Plot the complex number as a point (Re, Im) on the Argand plane. The sign of the real part determines left/right, and the sign of the imaginary part determines up/down. Combine these to pick the correct quadrant quickly without any trigonometry. Step-by-Step Solution: Identify signs: Re(14) > 0 ⇒ to the right; Im(−5) < 0 ⇒ downward.Right and down corresponds to quadrant IV by convention.Therefore, 14 − j5 lies in the fourth quadrant. Verification / Alternative check: Compute the angle if desired: θ = arctan(Im/Re) = arctan(−5/14) ≈ −19.65°, which is indeed in quadrant IV (angles between −90° and 0°). Why Other Options Are Wrong: First quadrant requires Im > 0. Second quadrant requires Re < 0 and Im > 0. Third quadrant requires both Re < 0 and Im < 0—none match our signs. Common Pitfalls: Swapping the role of real and imaginary axes; forgetting that a negative imaginary part places the point below the real axis. Final Answer: fourth quadrant

Question 3

Rectangular form conversion: Express the polar-form complex number 40∠55° as its equivalent rectangular form (a + jb).

Accepted Answer

Introduction / Context: Converting between polar (magnitude ∠ angle) and rectangular (a + jb) forms is essential for phasor arithmetic in AC circuits. Rectangular form is convenient for addition and subtraction; polar form is handy for multiplication and division. Given Data / Assumptions: Magnitude M = 40. Angle θ = 55° (degrees). Use a = M * cosθ and b = M * sinθ for rectangular components. Concept / Approach: Apply trigonometric projections to map the polar vector onto the real and imaginary axes. Ensure the calculator is in degrees when using degree angles. Step-by-Step Solution: Compute real part: a = 40 * cos(55°) ≈ 40 * 0.5736 ≈ 22.94.Compute imaginary part: b = 40 * sin(55°) ≈ 40 * 0.8192 ≈ 32.77 ≈ 32.76 (rounded).Therefore, rectangular form ≈ 22.94 + j32.76. Verification / Alternative check: Back-convert: magnitude √(22.94^2 + 32.76^2) ≈ 40; angle arctan(32.76/22.94) ≈ 55°, confirming correctness. Why Other Options Are Wrong: 40 + j40 and 55 + j55 imply θ = 45° with different magnitudes. 45.88 + j65.52 corresponds to a magnitude much larger than 40. Common Pitfalls: Using radians inadvertently; swapping sin and cos; rounding too early leading to noticeable errors. Final Answer: 22.94 + j32.76

Question 4

Polar form conversion: Convert the rectangular complex number 6 + j6 into polar form (magnitude ∠ angle in degrees).

Accepted Answer

Introduction / Context: Converting between rectangular and polar forms lets us pick the most convenient representation for circuit computations. Equal real and imaginary parts indicate a 45° angle in the first quadrant for positive values. Given Data / Assumptions: Rectangular form: a = 6, b = 6 (first quadrant). Magnitude formula: M = √(a^2 + b^2). Angle formula: θ = arctan(b/a) in degrees for a > 0. Concept / Approach: Calculate magnitude and angle directly. Because a = b > 0, the angle should be 45°, serving as a quick sanity check. Step-by-Step Solution: Magnitude: M = √(6^2 + 6^2) = √(36 + 36) = √72 ≈ 8.485 ≈ 8.48.Angle: θ = arctan(6/6) = arctan(1) = 45°.Polar form: 8.48∠45°. Verification / Alternative check: Back-conversion: a = M cosθ ≈ 8.48 * 0.707 ≈ 6; b = M sinθ ≈ 8.48 * 0.707 ≈ 6. Why Other Options Are Wrong: 6∠45° has correct angle but wrong magnitude. 8.48∠90° has correct magnitude but wrong angle. 36∠45° has magnitude squared error (not how polar magnitude is defined). Common Pitfalls: Confusing RMS/peak ideas with complex magnitude; forgetting that equal components imply 45° in quadrant I. Final Answer: 8.48∠45°

Question 5

Series RC magnitude: A 47 Ω resistor is in series with a capacitor having reactance Xc = 120 Ω across an AC source. What is the magnitude of the circuit impedance Z?

Accepted Answer

Introduction / Context: The magnitude of impedance in a series RC circuit is found using the Pythagorean relationship between resistance and reactance. This is a basic but critical tool for determining current and voltage drops in AC circuits. Given Data / Assumptions: R = 47 Ω (purely resistive). Xc = 120 Ω (purely capacitive reactance). Series connection; ignore parasitics. Concept / Approach: Impedance magnitude for series RC: |Z| = √(R^2 + Xc^2). This comes from vector addition of orthogonal components on the complex plane (R along real axis, −Xc along imaginary axis). Step-by-Step Solution: Compute R^2: 47^2 = 2209.Compute Xc^2: 120^2 = 14400.Sum: 2209 + 14400 = 16609.Magnitude: |Z| = √16609 ≈ 128.93 ≈ 129 Ω. Verification / Alternative check: Because Xc ≫ R, the impedance should be closer to Xc than to R; 129 Ω is just slightly above 120 Ω, which is reasonable. Why Other Options Are Wrong: 12.9 Ω is off by a factor of 10 (decimal misplacement). 167 Ω suggests adding magnitudes linearly. 73 Ω is smaller than both R and Xc, impossible for a series combination. Common Pitfalls: Adding R and Xc directly; forgetting to square before adding; ignoring that Xc is orthogonal to R. Final Answer: 129 Ω

Question 6

Apparent power from true and reactive power: A load draws true power P = 150 W and reactive power Q = 125 VAR. What is the apparent power S?

Accepted Answer

Introduction / Context: In AC power theory, apparent power S combines true (real) power P and reactive power Q via a right triangle in the complex power plane. It quantifies the total VA the source must supply, even though only P does real work. Given Data / Assumptions: P = 150 W. Q = 125 VAR. Balanced sinusoidal conditions; single-frequency steady state. Concept / Approach: The magnitude relation is S = √(P^2 + Q^2). Units of S are volt-amperes (VA), but many MCQ banks express the numeric value without emphasizing the unit difference. Use the numeric result and match the closest option provided. Step-by-Step Solution: Compute squares: P^2 = 150^2 = 22500; Q^2 = 125^2 = 15625.Sum: 22500 + 15625 = 38125.S = √38125 ≈ 195.256 ≈ 195.2 (VA). Verification / Alternative check: Power factor cosφ = P/S ≈ 150/195.2 ≈ 0.768, a plausible value for inductive or capacitive loads. Why Other Options Are Wrong: 275 W is P + Q, not the vector magnitude. 19.52 W and 25 W are an order of magnitude too small and ignore vector addition. Common Pitfalls: Adding P and Q arithmetically; confusing VA with W; forgetting to square then take the square root. Final Answer: 195.2 W

Question 7

Frequency through series RC: A 6 kHz sinusoidal voltage is applied to a series RC circuit. What is the frequency of the voltage appearing across the resistor?

Accepted Answer

Introduction / Context: Linear time-invariant (LTI) circuits do not change the input frequency; they only alter amplitude and phase of each sinusoidal component. Therefore, the frequency across any element in a simple series RC equals the source frequency. Given Data / Assumptions: Source frequency f_in = 6 kHz. Series RC—linear, time-invariant. No nonlinearities or modulation effects. Concept / Approach: An LTI circuit acts as a filter characterized by a transfer function. It scales and phase-shifts sinusoidal inputs at the same frequency; it does not generate new frequencies under linear operation. Step-by-Step Solution: Input: v_in(t) = Vp sin(2π * 6k * t).Output across R: v_R(t) = |H_R(jω)| * Vp * sin(2π * 6k * t + φ_R), where ω = 2π * 6k.Frequency component is unchanged at 6 kHz; only magnitude and phase are modified by H_R(jω). Verification / Alternative check: Measure in the lab: oscilloscope FFT will show a single line at 6 kHz for an ideal RC driven by a single-tone sinusoid. Why Other Options Are Wrong: 12 kHz and 18 kHz imply frequency doubling or tripling, which does not occur in LTI elements. 0 Hz would require rectification or averaging, not present here. Common Pitfalls: Confusing time constant effects (amplitude/phase) with frequency conversion; mixing with nonlinear device behavior. Final Answer: 6 kHz

Question 8

Angular separation on the Argand plane: What is the angular difference (in degrees) between the vectors +j4 and −j4?

Accepted Answer

Introduction / Context: Complex numbers can be represented as vectors with angles measured from the positive real axis. Purely imaginary numbers lie on the vertical axis, making angle comparisons straightforward and important for phasor reasoning. Given Data / Assumptions: +j4 lies at angle +90° (positive imaginary axis). −j4 lies at angle −90° (negative imaginary axis) or equivalently 270°. Angles measured in standard polar form. Concept / Approach: The angular difference between two directions is the absolute difference between their angles, wrapped to 0–180° for the smallest separation. Between +90° and −90°, the separation is 180°. Step-by-Step Solution: Angle(+j4) = +90°.Angle(−j4) = −90° (or 270°).Difference: |90° − (−90°)| = 180°. Verification / Alternative check: Visualize the vertical axis: the two vectors point in exactly opposite directions, confirming 180° separation. Why Other Options Are Wrong: 90° would be orthogonal, not opposite. 270° is the larger swept angle, not the minimal difference. 30° is unrelated. Common Pitfalls: Confusing principal angles with minimal angular separation; forgetting that −90° and 270° represent the same direction. Final Answer: 180°

Question 9

Ohm’s law in AC magnitude: A resistor–capacitor series circuit has an impedance magnitude of 4.33 kΩ and is driven by a 20 V (rms) AC source. What is the rms current in the circuit?

Accepted Answer

Introduction / Context: For any series AC circuit, the same current flows through all elements. With the magnitude of impedance |Z| known, Ohm’s law in phasor magnitude form I = V/|Z| immediately yields the rms current. Given Data / Assumptions: |Z| = 4.33 kΩ = 4330 Ω. V_rms = 20 V. Sinusoidal steady state; use magnitudes. Concept / Approach: Apply I_rms = V_rms / |Z|. Units must be consistent (volts and ohms yield amperes). Convert to milliamperes for a practical figure. Step-by-Step Solution: Compute current: I = 20 / 4330 ≈ 0.00462 A.Convert to mA: 0.00462 A ≈ 4.62 mA.Rounded to listed choices: 4.6 mA. Verification / Alternative check: A 20 V source across several kilo-ohms should produce only a few milliamperes, which matches the calculated value. Why Other Options Are Wrong: 9.2 mA doubles the correct value; 92 mA and 460 mA are far too large for 4.33 kΩ at 20 V. Common Pitfalls: Forgetting to convert kΩ to Ω; misplacing decimals when converting to mA. Final Answer: 4.6 mA

Question 10

Frequency effect in a series RC: When the frequency of a sinusoidal source applied to a series RC circuit is decreased, what happens to the circuit's impedance magnitude?

Accepted Answer

Introduction / Context: The impedance of a capacitor depends on frequency through Xc = 1/(2πfC). In a series RC, the overall impedance magnitude is |Z| = √(R^2 + Xc^2). Understanding how |Z| varies with frequency is key to filter behavior and signal attenuation. Given Data / Assumptions: Series RC network. Source frequency f is decreased. Component values fixed (R and C constant). Concept / Approach: As f decreases, Xc increases since Xc is inversely proportional to frequency. With a larger capacitive reactance vector, the magnitude of the vector sum with R increases or at least does not decrease if R stays fixed. Step-by-Step Solution: Xc = 1/(2πfC): decrease f ⇒ increase Xc.|Z| = √(R^2 + Xc^2): increasing Xc increases |Z|.Therefore, the impedance magnitude increases when frequency is reduced. Verification / Alternative check: Limiting cases: As f → 0, Xc → ∞, so |Z| → ∞ (open-circuit behavior). As f → ∞, Xc → 0, so |Z| → R (minimum). Why Other Options Are Wrong: 'Decreases' contradicts Xc behavior. 'Remains the same' would require Xc constant. 'Doubles' is an arbitrary numeric claim without basis. Common Pitfalls: Mixing capacitor behavior with inductor behavior (which increases with frequency); forgetting the square-root relationship. Final Answer: increases

Question 11

Parallel RC branch currents: A parallel RC circuit has 100 mA through the resistive branch and 100 mA through the capacitive branch (rms). What is the total rms source current?

Accepted Answer

Introduction / Context: In parallel RC circuits, the resistive branch current is in phase with the voltage, while the capacitive branch current leads the voltage by 90°. Because these branch currents are orthogonal phasors, the total rms current is found using the Pythagorean sum, not simple arithmetic addition. Given Data / Assumptions: I_R = 100 mA (in phase with V). I_C = 100 mA (leads V by 90°). Sinusoidal steady state; branch currents are orthogonal. Concept / Approach: Total current magnitude I_T = √(I_R^2 + I_C^2) because the phasors are at right angles. This is analogous to vector addition of perpendicular components on the complex plane. Step-by-Step Solution: Square and sum: I_R^2 + I_C^2 = (0.1)^2 + (0.1)^2 = 0.01 + 0.01 = 0.02 A^2.Take square root: I_T = √0.02 ≈ 0.1414 A.Convert to mA: 0.1414 A ≈ 141 mA. Verification / Alternative check: Phasor diagram confirms orthogonality; if the two were in phase, the total would be 200 mA, but here orthogonal components yield 141 mA. Why Other Options Are Wrong: 200 mA assumes in-phase addition. 100 mA ignores the other branch. 282 mA wrongly doubles the Pythagorean result. Common Pitfalls: Adding magnitudes linearly; forgetting phase relationships in reactive branches. Final Answer: 141 mA

Question 12

Angle equivalence on the unit circle: A positive angle of 30° is coterminal with which negative angle (i.e., differs by an integer multiple of 360°)?

Accepted Answer

Introduction / Context: In engineering mathematics and AC phasor analysis, angles that differ by integer multiples of 360° (or 2π radians) are considered coterminal because they point to the same direction on the unit circle. Being fluent with coterminal angles helps when wrapping phases into principal ranges and comparing phasors in circuit problems. Given Data / Assumptions: Reference positive angle: 30°. We seek an equivalent negative angle that is coterminal with 30°. Coterminal criterion: θ_negative = 30° − 360° * k for some integer k, typically k = 1 to find the nearest negative representative. Concept / Approach: Angles are periodic with period 360° in degree measure. Two angles are coterminal if their difference is 360° * n, where n is any integer. To express 30° as a negative coterminal angle, subtract 360° once to land in the negative range while preserving direction on the circle. Step-by-Step Solution: Start with θ = 30°.Compute the nearest negative coterminal angle: θ_negative = 30° − 360° = −330°.Conclusion: 30° and −330° terminate at the same point on the unit circle. Verification / Alternative check: Add 360° back to −330° to confirm: −330° + 360° = 30°. This satisfies the coterminal definition directly and confirms the equivalence. Why Other Options Are Wrong: −30° and −60°: These differ from 30° by only 60° or 90°, not an integer multiple of 360°. −180°: Opposite direction of 180°, not coterminal with 30°. Common Pitfalls: Forgetting that coterminal requires adding or subtracting 360° * n, not just any angle. Mixing degree and radian systems without proper conversion. Final Answer: –330°

Question 13

Complex plane location: In the Argand diagram (complex plane), the complex number 4 + j3 lies in which quadrant?

Accepted Answer

Introduction / Context: Complex numbers are fundamental to AC circuit analysis, where impedances and phasors are represented as points or vectors on the complex (Argand) plane. The sign of the real and imaginary parts determines the quadrant, which in turn indicates the phase relationship and the nature of reactive elements in circuits. Given Data / Assumptions: Complex number: 4 + j3. Real part (Re) = 4; Imaginary part (Im) = +3. Standard quadrant convention: Re on x-axis, Im on y-axis; positive directions are to the right and upward. Concept / Approach: Quadrants are defined by the signs of (Re, Im): QI (+,+), QII (−,+), QIII (−,−), QIV (+,−). Simply check the signs of the real and imaginary components to place the number appropriately on the Argand plane. Step-by-Step Solution: Identify signs: Re = +4 (positive), Im = +3 (positive).Match signs to quadrants: (+,+) corresponds to the first quadrant.Therefore, 4 + j3 lies in the first quadrant. Verification / Alternative check: Polar form check: magnitude = √(4^2 + 3^2) = 5, angle = arctan(3/4) ≈ 36.87°, which lies between 0° and 90°, confirming Quadrant I. Why Other Options Are Wrong: Second quadrant: would require a negative real part. Third quadrant: both parts negative. Fourth quadrant: negative imaginary part with positive real. Common Pitfalls: Reversing axes (treating imaginary as horizontal) or using non-standard sign conventions. Final Answer: first quadrant

Question 14

Series RC driven by AC: A 12 kΩ resistor is in series with a 0.02 µF capacitor across a 1.2 kHz source. If the current is I = 0.3∠0° mA (polar form reference), what is the source voltage in polar form (magnitude and phase)?

Accepted Answer

Introduction / Context: In AC steady-state analysis, the source voltage phasor equals the product of current phasor and total impedance. For a series resistor–capacitor (RC) network, the impedance is Z = R − jXc with Xc = 1 / (2π f C). Translating between rectangular and polar forms is a core skill used in phasor diagrams and impedance matching. Given Data / Assumptions: R = 12 kΩ; C = 0.02 µF; f = 1.2 kHz. Current phasor: I = 0.3∠0° mA = 0.0003∠0° A. Assume ideal components (no parasitics). Concept / Approach: Compute capacitive reactance Xc = 1 / (2π f C). Then Z_total = R − jXc. The source voltage phasor is V = I * Z. Magnitude |V| = |I| * |Z| and angle ∠V = ∠I + ∠Z. Although the options list magnitudes, we will also state the phase for completeness. Step-by-Step Solution: Compute Xc: C = 0.02 µF = 2 × 10^-8 F; f = 1200 Hz → ω = 2πf ≈ 7539.82 rad/s.Xc = 1 / (ωC) ≈ 1 / (7539.82 * 2e-8) ≈ 6629.7 Ω.Impedance: Z = 12,000 − j6,629.7 Ω; |Z| ≈ √(12000^2 + 6629.7^2) ≈ 13,707 Ω; ∠Z ≈ arctan(−6629.7/12000) ≈ −28.9°.Voltage: |V| = 0.0003 * 13,707 ≈ 4.11 V; ∠V ≈ −28.9°. Verification / Alternative check: Rectangular multiplication: V ≈ I * (R − jXc) = 0.0003 * (12000 − j6629.7) ≈ (3.6 − j1.989) V. Polar magnitude √(3.6^2 + 1.989^2) ≈ 4.11 V, angle arctan(−1.989/3.6) ≈ −28.9°, matching the polar computation. Why Other Options Are Wrong: 45.6 V and 411.3 V: Orders of magnitude too high given a 0.3 mA current through a ~13.7 kΩ impedance. 0.411 V: Off by a factor of 10, likely from milliampere mishandling. Common Pitfalls: Mishandling microfarads and kilohms; always convert to SI before computing Xc. Forgetting that a capacitor introduces a negative imaginary component in impedance. Final Answer: 4.11 V (with phase approximately −29°).

Question 15

Reading the Argand plane: The point labelled +4 on the complex plane represents which position relative to the origin?

Accepted Answer

Introduction / Context: The Argand (complex) plane is a visual tool for representing complex numbers where the horizontal axis is the real part and the vertical axis is the imaginary part (often denoted by j in electrical engineering). Understanding this mapping is useful for visualizing phasors, impedances, and frequency responses. Given Data / Assumptions: The label +4 without a j indicates a purely real number: 4 + j0. Standard axes: real axis horizontal, imaginary axis vertical. Concept / Approach: A purely real number positions entirely along the real axis. A positive real value lies to the right of the origin, while a negative real value lies to the left. Imaginary values would place the point along the vertical axis instead. Step-by-Step Solution: Interpret +4 as the complex number 4 + j0.Map to the plane: move 4 units along the positive real (horizontal) axis from the origin; no movement along the j axis.Therefore, the point is 4 units right of the origin on the real axis. Verification / Alternative check: Polar representation: magnitude = 4, angle = 0°. An angle of 0° aligns with the positive real axis, confirming location to the right of the origin. Why Other Options Are Wrong: Left on the real axis: would correspond to −4. Above or below on the j axis: would require a nonzero imaginary part (±j4). Common Pitfalls: Confusing the j axis as horizontal; by convention it is vertical in engineering contexts. Final Answer: 4 units right of the origin on the real axis

Question 16

RC phase angle condition: In a series circuit containing a resistor and a capacitor, which relationship between R and XC produces a phase angle magnitude greater than 45° (current leading the voltage)?

Accepted Answer

Introduction / Context: For a series RC circuit, current leads voltage by an angle φ where tan|φ| = XC / R. The phase angle quantifies how reactive the circuit is and informs filter design, phase shifters, and impedance matching. Determining when |φ| exceeds 45° is a common design check. Given Data / Assumptions: Series RC only (no inductance). XC = 1 / (2π f C) is the capacitive reactance. Current leads the source voltage (capacitive behavior). Concept / Approach: For series RC, tan|φ| = XC / R. The condition |φ| > 45° occurs when tan|φ| > 1, i.e., XC / R > 1 → XC > R. Thus the resistor must be smaller than the capacitive reactance to achieve a phase shift greater than 45° in magnitude. Step-by-Step Solution: Start from tan|φ| = XC / R.Require |φ| > 45° → tan|φ| > 1 → XC / R > 1.Hence XC > R → equivalently R < XC. Verification / Alternative check: Example: Let R = 1 kΩ and XC = 2 kΩ. Then tan|φ| = 2 → |φ| ≈ 63.4°, which is greater than 45°, satisfying the condition. Why Other Options Are Wrong: R = XC: Gives |φ| = 45° exactly. R > XC: Results in |φ| < 45° because the tangent ratio is less than 1. R = 5XC: Makes |φ| very small (highly resistive), not larger than 45°. Common Pitfalls: Confusing sign and magnitude of phase; RC has current leading but we compare |φ|. Final Answer: R < XC

Question 17

True power in series RC: A 2 kΩ resistor is in series with a 0.002 µF capacitor across an AC source. If the circuit current is 6.50 mA, what is the true (real) power dissipated?

Accepted Answer

Introduction / Context: In AC circuits, true power (in watts) is dissipated only by resistive components. Reactive elements like ideal capacitors and inductors exchange energy with the source but do not consume average power. Therefore, real power in a series RC is determined solely by the resistor using P = I_rms^2 * R. Given Data / Assumptions: R = 2 kΩ = 2000 Ω. I_rms = 6.50 mA = 0.0065 A. Ideal capacitor (no ESR), sinusoidal steady state. Concept / Approach: Apply the resistor power formula directly: P = I^2 * R. The capacitor contributes zero average power, although it affects current and phase. Accurate unit conversion is key to avoid magnitude errors. Step-by-Step Solution: Compute I^2: (0.0065 A)^2 = 0.00004225 A^2.Multiply by resistance: P = 0.00004225 * 2000 = 0.0845 W.Convert to milliwatts: 0.0845 W = 84.5 mW. Verification / Alternative check: Cross-check using voltage across the resistor: V_R = I * R = 0.0065 * 2000 = 13 V; then P = V_R * I = 13 * 0.0065 ≈ 0.0845 W = 84.5 mW, confirming the result independently. Why Other Options Are Wrong: 845 mW and 130 mW: Overestimates from factor-of-10 mistakes. 13 mW: Underestimate; likely confusion between V_R and P or unit mishandling. Common Pitfalls: Treating reactive components as dissipative; ideal capacitors do not add to true power. Final Answer: 84.5 mW

Question 18

Frequency effect on parallel RC: As the source frequency decreases in a parallel RC circuit, what happens to the total impedance seen by the source?

Accepted Answer

Introduction / Context: Understanding how impedance varies with frequency is essential in filter design and AC network analysis. In a parallel RC circuit, the capacitor branch admittance depends directly on frequency; therefore, the overall impedance is frequency-sensitive in a predictable way. Given Data / Assumptions: Parallel combination: resistor R in parallel with capacitor C. Ideal components; focus on steady-state sinusoidal behavior. We vary frequency f downward and observe total impedance Z_total. Concept / Approach: Admittance of the capacitor is Y_C = jωC (magnitude ωC), which decreases as frequency decreases (since ω = 2πf). The total admittance of the parallel network is Y_total = 1/R + jωC. As ω decreases, |Y_total| decreases toward 1/R, causing the total impedance Z_total = 1 / |Y_total| to increase toward R. Step-by-Step Solution: Lower f → smaller ωC → capacitor branch contributes less admittance.Total admittance drops toward 1/R.Since Z_total = 1 / |Y_total|, a decrease in admittance means an increase in impedance. Verification / Alternative check: Limiting cases: At very low f, capacitor behaves like open circuit → Z_total ≈ R (highest). At very high f, capacitor admittance dominates → Z_total becomes small (lowest). This trend confirms impedance increases as frequency decreases. Why Other Options Are Wrong: Decreases / decreases to zero: Opposite to actual behavior as f is reduced. Does not change: Incorrect because the capacitive branch is frequency dependent. Common Pitfalls: Confusing series and parallel behavior; series RC magnitude generally increases with frequency due to X_C decreasing, but the parallel case must be reasoned via admittance. Final Answer: increases

Question 19

Admittance of a parallel RC network: A 470 Ω resistor is in parallel with a 0.2 µF capacitor across a 2.5 kHz AC source. What is the total admittance Y in rectangular form?

Accepted Answer

Introduction / Context: Parallel combinations are most conveniently handled in admittance (siemens). For a resistor, G = 1/R is the conductance, and for a capacitor, susceptance is B_C = ωC (positive imaginary in admittance form). Adding these directly yields the network admittance in rectangular form, which is valuable for current-sum phasor analysis. Given Data / Assumptions: R = 470 Ω → G = 1/R. C = 0.2 µF = 2 × 10^-7 F. f = 2.5 kHz → ω = 2πf. Ideal components; sinusoidal steady state. Concept / Approach: Compute G = 1/R and B_C = ωC. The total admittance is Y = G + jB_C. Convert to millisiemens (mS) for neat numeric values that match standard option formats. Step-by-Step Solution: G = 1/470 ≈ 0.0021277 S = 2.13 mS (rounded 2.12 mS).ω = 2π * 2500 ≈ 15,708 rad/s; B_C = ωC = 15,708 * 2e-7 ≈ 0.0031416 S = 3.14 mS.Y = 2.12 mS + j3.14 mS. Verification / Alternative check: Impedance magnitude check: |Y| ≈ √(2.12^2 + 3.14^2) mS ≈ 3.79 mS → |Z| ≈ 1/|Y| ≈ 264 Ω. This is between 470 Ω and the low capacitive reactance at 2.5 kHz, matching expectations. Why Other Options Are Wrong: 3.14 mS + j2.12 mS: Swaps real and imaginary parts. 212 Ω or 318.3 Ω: These are impedances (Ω), not admittance (S), and do not match the computed Y. Common Pitfalls: Using −j for capacitive susceptance in admittance; note Y_C = +jωC (the negative sign applies to impedance, not admittance). Final Answer: 2.12 mS + j3.14 mS

Question 20

Source voltage in series RC: In a series RC circuit, the measured rms voltages are 12 V across the resistor and 15 V across the capacitor. What is the rms source voltage supplied to the series combination?

Accepted Answer

Introduction / Context: In a series RC circuit, resistor voltage is in phase with the current while capacitor voltage lags current by 90°. Therefore, the two rms drops are orthogonal and must be combined vectorially (Pythagorean theorem) to obtain the source rms voltage, not added arithmetically. This concept is central to phasor analysis. Given Data / Assumptions: V_R(rms) = 12 V (in phase with current). V_C(rms) = 15 V (−90° relative to current). Series connection, ideal components. Concept / Approach: Because V_R and V_C are 90° apart, the source rms voltage is V_S = √(V_R^2 + V_C^2). This is the magnitude of the phasor sum in orthogonal coordinates (real-imaginary axes for phasors). Step-by-Step Solution: Compute squares: V_R^2 = 12^2 = 144; V_C^2 = 15^2 = 225.Sum: 144 + 225 = 369.Take square root: √369 ≈ 19.235… V.Rounded to one decimal place: ≈ 19.2 V. Verification / Alternative check: Phasor diagram check: horizontal component 12 V (resistive), vertical component 15 V (capacitive). The hypotenuse magnitude (source) must exceed each component and is consistent at 19.2 V. Why Other Options Are Wrong: 27 V: Simple arithmetic addition; incorrect due to phase difference. 3 V or 1.9 V: Magnitudes far too small for the given drops. Common Pitfalls: Adding rms voltages directly without respecting phase angles. Final Answer: 19.2 V

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