Question 1

Parallel RL phasor addition: In a parallel RL circuit with 3 A rms in the resistive branch and 3 A rms in the inductive branch, what is the total rms line current?

Accepted Answer

Introduction / Context: In AC analysis, currents in different branches can be out of phase. For a parallel RL, the current through the resistor is in phase with voltage, while the current through the inductor lags the voltage by 90°. Total current is found by vector (phasor) addition, not simple arithmetic sum. Given Data / Assumptions: I_R = 3 A rms (in phase with V). I_L = 3 A rms (lags V by 90°). Parallel RL, both branches share the same applied voltage. Concept / Approach: Because I_R and I_L are 90° apart, the magnitude of the total current is the square root of the sum of squares: I_total = sqrt(I_R^2 + I_L^2). This is basic orthogonal phasor addition. Step-by-Step Solution: I_total = sqrt( I_R^2 + I_L^2 )I_total = sqrt( 3^2 + 3^2 )I_total = sqrt( 9 + 9 ) = sqrt( 18 )I_total ≈ 4.2426 A ≈ 4.24 A Verification / Alternative check: Phasor diagram: one current on the real axis (3 A), one on the negative imaginary axis (3 A). The hypotenuse length is 3 * sqrt(2) ≈ 4.24 A. This confirms the computed value. Why Other Options Are Wrong: 6 A: Simple scalar sum ignores the 90° phase difference. 424 mA: Off by a factor of 10. 42.4 A: Off by a factor of 10 in the other direction. Common Pitfalls: Adding AC currents arithmetically when they are not in phase, or confusing series and parallel behavior. Always use phasor addition for orthogonal components. Final Answer: 4.24 A

Question 2

Parallel RL frequency effect: When the frequency of the applied AC source is decreased, how does the total impedance of a parallel RL circuit change?

Accepted Answer

Introduction / Context: Understanding how frequency affects impedance is key for filter design and AC circuit analysis. In a parallel RL circuit, the resistive branch admittance is constant with frequency, while the inductive branch admittance varies with frequency. Given Data / Assumptions: Parallel RL topology. Inductive reactance XL = 2 * pi * f * L. Admittance of the inductor YL = 1 / (j XL) = -j / XL. Concept / Approach: When frequency decreases, XL decreases. In parallel circuits we add admittances: Y_total = G + jB, where G = 1/R and B = -1/XL. If XL decreases, |B| increases in magnitude, making |Y_total| larger. Since Z_total = 1 / |Y_total|, the total impedance decreases. Step-by-Step Solution: XL = 2 * pi * f * LDecrease f ⇒ decrease XLYL = 1 / (j XL) ⇒ |YL| increases as XL decreasesY_total increases ⇒ Z_total = 1 / |Y_total| decreases Verification / Alternative check: Numerical example: Let R = 100 Ω, L = 0.1 H. At f1 = 100 Hz, XL ≈ 62.8 Ω; at f2 = 50 Hz, XL ≈ 31.4 Ω. The inductive branch admittance doubles in magnitude, so the net admittance increases and net impedance drops, confirming the trend. Why Other Options Are Wrong: Increases: Opposite of parallel behavior driven by inductive branch admittance. Remains constant / not a factor: Frequency directly affects XL, so impedance must change. Common Pitfalls: Applying series-circuit intuition to parallel circuits. In series RL, lowering frequency reduces XL and thus reduces total impedance; in parallel RL, the effect flips because we combine admittances, not impedances. Final Answer: decreases

Question 3

Series RL current control: To increase the current in a series RL circuit, should the excitation frequency be increased, decreased, or kept constant?

Accepted Answer

Introduction / Context: In series RL circuits, current depends on the magnitude of the impedance, which is a function of resistance and inductive reactance. Frequency manipulation is a common way to control current in AC networks and power electronics. Given Data / Assumptions: Series RL circuit with resistance R and inductance L. Inductive reactance XL = 2 * pi * f * L. AC source voltage magnitude is fixed. Concept / Approach: Total impedance magnitude in series RL is Z = sqrt(R^2 + XL^2). Since XL grows with frequency, increasing frequency increases Z, which decreases current I = V / Z. Therefore, to increase current, frequency must be reduced so that XL decreases. Step-by-Step Solution: XL = 2 * pi * f * LZ = sqrt( R^2 + XL^2 )I = V / ZDecrease f ⇒ decrease XL ⇒ decrease Z ⇒ increase I Verification / Alternative check: Example: R = 10 Ω, L = 50 mH. At 100 Hz, XL ≈ 31.4 Ω; at 50 Hz, XL ≈ 15.7 Ω. Z reduces from ≈ 33.9 Ω to ≈ 18.5 Ω, so current nearly doubles for the same applied voltage, validating the conclusion. Why Other Options Are Wrong: Should be increased: Would increase XL and reduce current. Should be constant / cannot be determined: The direction of change is determined from formulas and does not require numeric values. Common Pitfalls: Confusing series and parallel behavior, or forgetting that inductive reactance is proportional to frequency in all linear inductors. Final Answer: should be decreased

Question 4

Power factor of a series RL: A 1.5 kΩ resistor in series with an inductor of 2.2 kΩ reactance is driven by an 18 V AC source. What is the circuit power factor?

Accepted Answer

Introduction / Context: Power factor (PF) indicates how effectively current contributes to real power in AC circuits. In a series RL circuit, PF equals cos(theta), where theta is the angle of the impedance. Knowing PF is vital for power delivery and minimizing losses. Given Data / Assumptions: R = 1.5 kΩ = 1500 Ω. XL = 2.2 kΩ = 2200 Ω. Series RL; voltage magnitude is 18 V (value not needed to compute PF). Concept / Approach: For a series RL, PF = cos(theta) = R / |Z|, where |Z| = sqrt(R^2 + XL^2). This follows from the right-triangle relationship of the impedance phasor diagram. Step-by-Step Solution: |Z| = sqrt( R^2 + XL^2 )|Z| = sqrt( 1500^2 + 2200^2 )|Z| = sqrt( 2.25e6 + 4.84e6 ) = sqrt( 7.09e6 ) ≈ 2663.8 ΩPF = R / |Z| = 1500 / 2663.8 ≈ 0.5634 ≈ 0.564 Verification / Alternative check: Angle method: theta = arctan( XL / R ) = arctan( 2200 / 1500 ) ≈ arctan(1.4667) ≈ 55.7°. Then PF = cos(55.7°) ≈ 0.564, matching the above result. Why Other Options Are Wrong: 564, 6.76, 55.7: These are dimensionless numbers linked to intermediate steps (e.g., degrees) but not valid for PF, which must lie between 0 and 1. Common Pitfalls: Using R/XL instead of R/|Z|, or misinterpreting angle in degrees as PF. Remember PF is the cosine of the phase angle and must be ≤ 1. Final Answer: 0.564

Question 5

Purely inductive load: If the reactive power of a load is 12 VAR and the load is purely inductive, what is the apparent power S?

Accepted Answer

Introduction / Context: AC power consists of real power (P), reactive power (Q), and apparent power (S). In purely reactive elements (ideal inductors or capacitors), real power is zero, while reactive power and apparent power are related directly. Given Data / Assumptions: Load is purely inductive (P = 0). Reactive power Q = 12 VAR. Standard power triangle relationships apply. Concept / Approach: For any load, S^2 = P^2 + Q^2. For a purely inductive load, P = 0, so S = |Q|. Therefore, the apparent power equals the magnitude of the reactive power. Step-by-Step Solution: Given P = 0, Q = 12 VARS = sqrt( P^2 + Q^2 )S = sqrt( 0 + 12^2 ) = 12 VA Verification / Alternative check: Power factor PF = P / S = 0 for purely reactive loads. With PF = 0, only reactive power remains, so S = |Q| is consistent and mandatory. Why Other Options Are Wrong: 0 VA: Would imply no current or no voltage; not true for reactive power flow. 6 VA or 24 VA: Do not follow from S = |Q| for P = 0. Common Pitfalls: Confusing the roles of P, Q, and S, or assuming S equals P. In a pure inductor or capacitor, real power is zero but apparent and reactive power are non-zero. Final Answer: 12 VA

Question 6

Power factor and losses: In an RL circuit, which power factor value yields the least energy loss for a given delivered power?

Accepted Answer

Introduction / Context: Power factor (PF) quantifies the ratio of real power to apparent power. A higher PF means less reactive current and typically lower I^2R losses in conductors and equipment for the same delivered real power, improving efficiency. Given Data / Assumptions: Comparing several PF values: 1, 0.8, 0.4, 0.2. Same delivered real power is assumed for a fair comparison of losses. Concept / Approach: For fixed real power P, the apparent power S = P / PF. Current I ∝ S (for a given voltage). Lower PF means larger S and larger current, which increases copper losses proportional to I^2. Thus, maximum PF (unity) minimizes losses. Step-by-Step Reasoning: S = P / PFI ∝ S ⇒ I ∝ 1 / PFLoss ∝ I^2 ⇒ Loss ∝ 1 / PF^2Highest PF (1) ⇒ smallest current ⇒ least I^2R loss Verification / Alternative check: Example: For P = 1 kW at 100 V line, PF = 1 ⇒ S = 1 kVA ⇒ I = 10 A. At PF = 0.5 ⇒ S = 2 kVA ⇒ I = 20 A ⇒ losses are quadrupled for the same P. Hence PF = 1 minimizes losses. Why Other Options Are Wrong: 0.8, 0.4, 0.2: Each progressively increases apparent power and current, causing higher losses. Common Pitfalls: Confusing reactive power correction with real power savings; PF correction reduces losses and capacity requirements even if delivered P is unchanged. Final Answer: 1

Question 7

Magnitude of total impedance for a parallel RL: A 140 Ω resistor is in parallel with an inductor having 60 Ω inductive reactance. Both are across a 12 V AC source. What is |Z| of the parallel combination?

Accepted Answer

Introduction / Context: Calculating the total impedance of a parallel RL network requires summing admittances (not impedances). This is a standard AC analysis skill, often needed for filter and load calculations. Given Data / Assumptions: R = 140 Ω (resistive branch). XL = 60 Ω (inductive reactance of L branch). Parallel connection across the same AC source. Ideal components (no resistance in L aside from reactance). Concept / Approach: Total admittance Y_total = Y_R + Y_L. For the resistor, Y_R = 1/R (purely real). For the inductor, Y_L = 1/(j XL) = -j / XL (purely imaginary). Then |Z_total| = 1 / |Y_total| and |Y_total| = sqrt( G^2 + B^2 ). Step-by-Step Solution: G = 1 / R = 1 / 140 ≈ 0.0071429 SB = -1 / XL = -1 / 60 ≈ -0.0166667 S|Y_total| = sqrt( G^2 + B^2 )|Y_total| ≈ sqrt( 0.0071429^2 + 0.0166667^2 ) ≈ sqrt( 0.0000510 + 0.0002778 ) ≈ 0.018129 S|Z_total| = 1 / |Y_total| ≈ 1 / 0.018129 ≈ 55.15 Ω Verification / Alternative check: Back-calculate currents at 12 V: I_R = 12 / 140 ≈ 0.0857 A, I_L = 12 / 60 = 0.2 A at -90°. Total current magnitude ≈ sqrt(0.0857^2 + 0.2^2) ≈ 0.2178 A. Then |Z| = V / I ≈ 12 / 0.2178 ≈ 55.1 Ω, confirming the result. Why Other Options Are Wrong: 5.51 Ω: Off by a factor of 10, likely from inverting at the wrong step. 90 Ω, 200 Ω: Do not match the computed admittance magnitude and are inconsistent with the branch values. Common Pitfalls: Adding impedances instead of admittances for parallel circuits, or forgetting that the inductor admittance is imaginary (negative susceptance). Final Answer: 55.15 Ω

Question 8

Series RL sensitivity: If the frequency is reduced by one-half while the resistance is doubled, how does the total impedance of a series RL circuit change?

Accepted Answer

Introduction / Context: This question examines how the impedance of a series RL circuit responds when both frequency and resistance are changed in opposite directions. It tests understanding of how resistive and inductive parts combine vectorially rather than arithmetically. Given Data / Assumptions: A series RL circuit with resistance R and inductive reactance XL. Frequency is halved ⇒ XL, which depends on frequency, is reduced. Resistance is doubled at the same time. Assume ideal inductor (no winding resistance) and sinusoidal steady state. Concept / Approach: Impedance magnitude for a series RL circuit is Z = sqrt(R^2 + XL^2), where XL = 2 * pi * f * L. If f is halved, XL becomes XL/2. If R is doubled, R becomes 2R. The new impedance is Znew = sqrt((2R)^2 + (XL/2)^2). Whether Znew is larger or smaller than the original depends on the relative sizes of R and XL. Step-by-Step Solution: Original: Zorig = sqrt(R^2 + XL^2).Changes: R → 2R and XL → XL/2.New: Znew = sqrt(4R^2 + (XL^2)/4).Comparison: If R dominates (R >> XL), Znew ≈ 2R (about double). If XL dominates (XL >> R), Znew ≈ XL/2 (about half). Intermediate cases vary. Verification / Alternative check: Try numerical examples: R = 10 Ω, XL = 100 Ω ⇒ Zorig ≈ 100.5 Ω; Znew ≈ 50.0 Ω (halves). Conversely, R = 100 Ω, XL = 10 Ω ⇒ Zorig ≈ 100.5 Ω; Znew ≈ 200.1 Ω (nearly doubles). Why Other Options Are Wrong: 'doubles' / 'halves' / 'remains constant': Each may occur only for specific R–XL ratios, not universally. 'increases by one-fourth': Not supported by the impedance formula. Common Pitfalls: Adding R and XL directly instead of using vector (root-sum-square) combination. Forgetting that XL depends on frequency while R does not. Final Answer: cannot be determined without values

Question 9

Series RL comparison: At a fixed component set, how should frequency be changed so that the resistor's RMS voltage drop is less than the inductor's RMS voltage drop?

Accepted Answer

Introduction / Context: In a series RL circuit, the voltage division between the resistor and inductor depends on reactance, which varies with frequency. This question checks practical intuition about how to make the inductor's drop dominate the resistor's drop. Given Data / Assumptions: Series RL with fixed R and L. AC steady state with variable frequency f. VR = I * R and VL = I * XL with XL = 2 * pi * f * L. Concept / Approach: For a given current I, VR is independent of frequency (proportional to R), whereas VL grows linearly with frequency (proportional to XL). To make VL > VR, increase f so that XL > R in magnitude when scaled by the same current. Step-by-Step Solution: XL = 2 * pi * f * L.As f increases → XL increases.VL = I * XL increases with f; VR = I * R is constant w.r.t. f.Therefore, raising frequency makes VL exceed VR. Verification / Alternative check: At very low f (approaching DC), XL → 0 and VR dominates. At higher f, XL ≫ R, so VL dominates. This matches lab observation of frequency-dependent phasor drops. Why Other Options Are Wrong: 'decreased': Lowers XL and makes VR dominate. 'doubled': Doubling does increase VL but the best generic instruction is simply 'increase frequency'; a specific factor is not required. 'not a factor': Frequency is the key factor here. 'reduced to DC': Makes VL → 0. Common Pitfalls: Assuming VR also depends on frequency in a pure R–L series circuit. Final Answer: increased

Question 10

Find the source phasor: A 12 kΩ resistor in series with a 90 mH inductor is connected to an 8 kHz AC source. If the circuit current is I = 0.3∠0° mA (reference), what is the source voltage in polar form?

Accepted Answer

Introduction / Context: This problem reinforces core AC circuit analysis: compute series RL impedance, then use Ohm’s law in phasor form to find the source voltage given a reference current phasor. Given Data / Assumptions: R = 12 kΩ, L = 90 mH. Frequency f = 8 kHz ⇒ ω = 2 * pi * 8000 rad/s. I = 0.3∠0° mA (phasor reference). Ideal components, steady state sinusoidal. Concept / Approach: Compute XL, then the series impedance Z = R + jXL. The source voltage phasor is V = I * Z. The impedance angle is θ = arctan(XL / R). Step-by-Step Solution: ω = 2 * pi * 8000 ≈ 50265 rad/s.XL = ω * L = 50265 * 0.09 ≈ 4524 Ω.Z = 12000 + j4524 Ω ⇒ |Z| = sqrt(12000^2 + 4524^2) ≈ 12829 Ω, θ ≈ arctan(4524/12000) ≈ 20.6°.|V| = |I| * |Z| = 0.0003 * 12829 ≈ 3.84 V, ∠V = 20.6°. Verification / Alternative check: Component drops: VR = I * R ≈ 0.0003 * 12000 = 3.6 V; VL = I * XL ≈ 0.0003 * 4524 ≈ 1.36 V. Vector sum of 3.6 V (real) and 1.36 V (positive j) gives ≈ 3.84∠20.6°, consistent. Why Other Options Are Wrong: 12.8∠20.6° V and 13.84∠69.4° V: Magnitude errors; do not match I * Z. 0.3∠20.6° V: Units confusion (current magnitude). 3.84∠–20.6° V: Incorrect sign for inductive lead angle. Common Pitfalls: Forgetting to convert milliamps to amps when multiplying by ohms. Mixing rectangular and polar without accounting for the phase angle. Final Answer: 3.84∠20.6° V

Question 11

Frequency effect in series RL: When the frequency of the applied AC voltage is decreased, what happens to the total impedance magnitude of a series RL circuit?

Accepted Answer

Introduction / Context: This item tests understanding that the inductive reactance in a series RL circuit depends on frequency, directly influencing total impedance. Given Data / Assumptions: Series RL with fixed R and L. AC steady state, frequency f changes. Concept / Approach: Impedance magnitude is Z = sqrt(R^2 + XL^2). Since XL = 2 * pi * f * L, reducing f reduces XL, thereby reducing the vector sum unless R completely dominates already (in which case Z tends toward R). In all cases with finite L, lowering f cannot increase XL, so Z does not increase. Step-by-Step Solution: XL = 2 * pi * f * L.Decrease f ⇒ decrease XL.Z = sqrt(R^2 + XL^2) shrinks toward R as XL decreases.Therefore, the impedance magnitude decreases. Verification / Alternative check: Limits: as f → 0, XL → 0 and Z → R. As f → ∞, Z → ∞ (inductor dominating). The trend with decreasing f is consistent: Z goes down. Why Other Options Are Wrong: 'increases': Opposite of XL behavior. 'does not change': Would require L = 0 (no inductor), contrary to setup. 'cannot be determined': Determinable from formula. Common Pitfalls: Confusing series with parallel RL behavior. Final Answer: decreases

Question 12

Convert to polar impedance: A 47 Ω resistor in series with an inductive reactance of 120 Ω is connected to an AC source. What is the total impedance in polar form?

Accepted Answer

Introduction / Context: Converting a series R and jXL into polar form is a staple AC analysis skill, linking rectangular and polar representations of impedance for phasor calculations. Given Data / Assumptions: R = 47 Ω, XL = 120 Ω (inductive; positive imaginary part). Series connection; Z = R + jXL. Concept / Approach: Compute magnitude |Z| by root-sum-square and phase angle θ = arctan(XL / R). For an inductor in series with a resistor, the impedance angle is positive (current lags voltage). Step-by-Step Solution: |Z| = sqrt(47^2 + 120^2) = sqrt(2209 + 14400) = sqrt(16609) ≈ 129 Ω.θ = arctan(120 / 47) ≈ arctan(2.553) ≈ 68.6°.Therefore, Z ≈ 129∠68.6° Ω. Verification / Alternative check: Rectangular check: 129∠68.6° ≈ 47 + j120 when converted back, confirming consistency. Why Other Options Are Wrong: 47∠68.6° Ω and 120∠68.6° Ω: Use component magnitudes rather than the combined magnitude. 129∠31.4° Ω: Complementary angle; wrong quadrant for inductive series. 127∠60° Ω: Rounded magnitude/angle do not match computed values. Common Pitfalls: Using θ = arctan(R/XL) instead of XL/R. Forgetting the positive angle sign for inductive reactance in series. Final Answer: 129∠68.6° Ω

Question 13

Parallel RL total current: A 3.3 kΩ resistor and a 120 mH inductor are connected in parallel across a 2 kHz, 12 V AC source. What is the total RMS current drawn from the source?

Accepted Answer

Introduction / Context: This problem blends branch calculations in a parallel RL circuit with phasor addition of currents that are not in phase. It is a practical exercise in using both magnitude and phase information to get the total current. Given Data / Assumptions: R = 3.3 kΩ, L = 120 mH. Source: 12 V RMS at f = 2 kHz. Ideal inductor; line frequency steady state. Concept / Approach: Calculate branch currents: IR = V / R (in phase with V), IL = V / XL (lagging by 90°) where XL = 2 * pi * f * L. The total current is the vector sum: I = sqrt(IR^2 + IL^2) because the two branches are orthogonal in phase for ideal RL. Step-by-Step Solution: XL = 2 * pi * 2000 * 0.12 ≈ 1508 Ω.IR = 12 / 3300 ≈ 0.00364 A = 3.64 mA.IL = 12 / 1508 ≈ 0.00796 A = 7.96 mA (−90° relative to V).IT = sqrt(IR^2 + IL^2) ≈ sqrt(3.64^2 + 7.96^2) mA ≈ 8.74 mA. Verification / Alternative check: Admittance method: Y = 1/R − j(1/XL) = 0.000303 − j0.000663 S, |Y| ≈ 0.000728 S ⇒ IT ≈ 12 * 0.000728 ≈ 8.74 mA (same result). Why Other Options Are Wrong: 874 mA and 8.74 µA: Scale errors by factors of 100 or 1000. 874 µA: Off by exactly 10×. 3.64 mA: Resistor branch only; ignores inductive branch. Common Pitfalls: Treating branch currents as algebraic sums instead of vector sums. Neglecting the 90° phase difference in an ideal inductor branch. Final Answer: 8.74 mA

Question 14

Series RL impedance magnitude: A 1.2 kΩ resistor is in series with a 15 mH inductor across a 10 kHz AC source. What is the total impedance magnitude?

Accepted Answer

Introduction / Context: Here we compute the magnitude of series RL impedance by combining resistance and inductive reactance via the root-sum-square relationship, a routine but essential AC task. Given Data / Assumptions: R = 1.2 kΩ = 1200 Ω. L = 15 mH; f = 10 kHz. Ideal components, sinusoidal steady state. Concept / Approach: XL = 2 * pi * f * L. Total impedance magnitude is |Z| = sqrt(R^2 + XL^2). Step-by-Step Solution: XL = 2 * pi * 10000 * 0.015 = 2 * pi * 150 ≈ 942.48 Ω.|Z| = sqrt(1200^2 + 942.48^2) ≈ sqrt(1,440,000 + 888,276) ≈ sqrt(2,328,276) ≈ 1,526 Ω.Therefore, the impedance magnitude is about 1.53 kΩ. Verification / Alternative check: Angle check: θ = arctan(XL / R) ≈ arctan(942.5 / 1200) ≈ 38.9°, reasonable for RL with R slightly dominant. This supports the computed magnitude. Why Other Options Are Wrong: 152.6 Ω and 942 Ω: Confuse steps (either divide by 10 or use XL alone). 1,200 Ω: Ignores inductive reactance contribution. 1,920 Ω: Overestimates by summing instead of using root-sum-square. Common Pitfalls: Adding R and XL directly instead of vectorially. Final Answer: 1,526 Ω

Question 15

Phase angle trend: As the frequency of the applied AC voltage increases in a series RL circuit, how does the current–voltage phase angle (|θ|) change?

Accepted Answer

Introduction / Context: The phase angle between source voltage and current in a series RL depends on the ratio XL/R. Understanding how this ratio shifts with frequency is crucial for filter design and transient response intuition. Given Data / Assumptions: Series RL with fixed R and L. Frequency f increases. θ = arctan(XL / R) (current lags voltage by θ in inductive circuits). Concept / Approach: Since XL = 2 * pi * f * L, increasing f increases XL while R is constant. Therefore, XL / R increases, and θ = arctan(XL / R) increases monotonically toward 90° as frequency becomes very large. Step-by-Step Solution: At low f: XL small ⇒ θ small (close to 0°, nearly resistive).As f increases: XL grows ⇒ θ increases.At very high f: XL ≫ R ⇒ θ approaches 90° (strongly inductive). Verification / Alternative check: Bode intuition: The RL corner frequency fc = R / (2 * pi * L) marks where θ ≈ 45°. Above fc, the circuit becomes increasingly inductive, confirming the trend of increasing |θ|. Why Other Options Are Wrong: 'decreases' or 'does not change': Contradict the XL dependence on frequency. 'cannot be determined': The direction is determinable independent of specific values. 'becomes zero immediately': Zero angle happens only at DC (f → 0) for ideal RL. Common Pitfalls: Confusing series and parallel behavior; in parallel RL, current components behave differently but θ of branch currents is defined separately. Final Answer: increases

Question 16

Parallel RL across 15 V AC: A 470 Ω resistor and an inductor with 125 Ω inductive reactance are connected in parallel across a 15 V (rms) source. What is the rms current through the inductor branch?

Accepted Answer

Introduction / Context: This problem checks your ability to compute branch current in an AC parallel RL network using reactance. Because the branches are in parallel, each branch sees the full source voltage, and currents are determined independently by their branch impedances. Given Data / Assumptions: AC source: 15 V (rms). Resistive branch: R = 470 Ω (not needed for the inductor current). Inductive branch: XL = 125 Ω. Ideal components; frequency is such that the given reactance applies. Concept / Approach: For any single reactive branch, the magnitude of branch current is I = V / |Z_branch|. For a pure inductor branch, |Z| = XL. Since the branches are in parallel, the inductor sees the full 15 V (rms). Step-by-Step Solution: Given: V_rms = 15 V, XL = 125 ΩI_L = V_rms / XLI_L = 15 / 125 = 0.12 AI_L = 120 mA (rms) Verification / Alternative check: If instead you computed total current, you would need phasor addition of the resistor and inductor currents. However, the question asks only for the inductor branch current, which is simply V / XL = 120 mA. Why Other Options Are Wrong: 152 mA, 32 mA, 12 mA: Do not match V / XL for 15 V and 125 Ω. Common Pitfalls: Confusing parallel with series behavior, or accidentally using the resistor value or total impedance instead of the inductor’s reactance for the branch current. Final Answer: 120 mA

Question 17

Series RL vector addition: In a series RL circuit, 12 V (rms) is across the resistor and 14 V (rms) is across the inductor. What is the peak value of the source voltage?

Accepted Answer

Introduction / Context: This question evaluates AC phasor skills for series RL circuits. The resistor voltage is in phase with current, while the inductor voltage leads the current by 90°. The source voltage is the phasor (vector) sum of the two branch drops. Given Data / Assumptions: VR = 12 V (rms) across R. VL = 14 V (rms) across L. Series RL, ideal components. Asked: Source voltage in peak (not rms). Concept / Approach: For series RL: V_s(rms) = sqrt(VR^2 + VL^2). Convert that rms result to peak by multiplying by sqrt(2). Step-by-Step Solution: V_s(rms) = sqrt(12^2 + 14^2)V_s(rms) = sqrt(144 + 196) = sqrt(340) ≈ 18.439 VV_s(peak) = V_s(rms) * sqrt(2)V_s(peak) ≈ 18.439 * 1.414 ≈ 26.08 V ≈ 26.0 V Verification / Alternative check: Power triangle or impedance triangle gives the same phasor relationship; the orthogonal nature of VR and VL requires a Pythagorean sum for rms values. Why Other Options Are Wrong: 18.4 V: That is the rms magnitude, not peak. 20 V, 2 V: Not consistent with the phasor sum and peak conversion. Common Pitfalls: Adding VR and VL arithmetically, or forgetting to convert rms to peak. Always perform vector addition first, then scale to the requested unit (peak or rms). Final Answer: 26.0 V

Question 18

Series RL phase behavior: When the resistor voltage becomes less than the inductor voltage in a series RL circuit, how does the overall phase angle change?

Accepted Answer

Introduction / Context: Phase angle in a series RL circuit indicates how far the current lags the source voltage. It depends on the ratio of inductive reactance to resistance, and it can be inferred by comparing the voltage drops across the inductor and the resistor. Given Data / Assumptions: Series RL circuit. Observed: VR < VL. Ideal linear components. Concept / Approach: In series RL, VR:VL equals R:XL because the same current flows through both. If VR < VL, then R < XL, which means the impedance angle theta = arctan(XL / R) is larger. A larger ratio XL/R yields a larger theta (more lag). Step-by-Step Reasoning: Given VR < VL ⇒ R < XLtheta = arctan(XL / R)If XL/R increases, then theta increasesTherefore, the phase angle increases Verification / Alternative check: Phasor diagram: the vertical component (inductive) becomes longer than the horizontal (resistive). The angle between the impedance vector and the real axis grows, confirming an increased phase angle. Why Other Options Are Wrong: Decreases / not affected: Contradicts theta = arctan(XL/R). Cannot be determined: It can be determined directly from the voltage ratio in series. Common Pitfalls: Confusing series and parallel intuition or thinking VL and VR can be compared arithmetically without considering their phase relationship with current. In series, magnitudes directly reflect R and XL because current is common. Final Answer: increases

Question 19

RC lag network behavior: Which statement correctly describes how phase shift and output magnitude of a lag (low-pass RC) network change with frequency?

Accepted Answer

Introduction: Lag networks—typically simple RC low-pass filters—are ubiquitous in analog signal conditioning, control systems, and power electronics. Understanding how their phase shift and amplitude response vary with frequency is essential for stability analysis, timing, and noise attenuation. Given Data / Assumptions: Single-pole RC low-pass network with output taken across the capacitor or resistor depending on definition; here, “lag network” denotes the classic configuration producing negative phase (output lags input). Linear, time-invariant behavior; small-signal sinusoidal steady state. Cutoff frequency fc = 1 / (2 * pi * R * C) defines the transition band. Concept / Approach: The transfer function magnitude of a first-order low-pass is |H(jω)| = 1 / sqrt(1 + (ω/ωc)^2). The phase shift is φ(ω) = −arctan(ω/ωc). As frequency increases, the magnitude decreases and the phase becomes more negative, approaching −90 degrees as ω → ∞. Conversely, as frequency decreases toward zero, |H| approaches 1 and φ approaches 0 degrees (negligible lag). Therefore, the correct qualitative statement is that increasing frequency increases phase lag in magnitude (more negative angle). Step-by-Step Solution: Define ωc = 2 * pi * fc = 1 / (R * C).Compute phase: φ(ω) = −arctan(ω/ωc); as ω increases, arctan term increases → φ becomes more negative.Compute magnitude: |H(jω)| = 1 / sqrt(1 + (ω/ωc)^2); as ω increases, denominator grows → magnitude decreases.Relate limits: ω ≪ ωc → φ ≈ 0°, |H| ≈ 1; ω ≫ ωc → φ ≈ −90°, |H| ≈ 0. Verification / Alternative check: Bode plots show a −20 dB/decade magnitude slope above fc and a monotonic phase transition from 0° toward −90°. Experimental step-response also indicates longer apparent delay at higher frequency components (more lag in the frequency domain corresponds to time-domain smoothing). Why Other Options Are Wrong: An increase in frequency causes an increase in the magnitude of the output voltage: Opposite of low-pass behavior; magnitude falls with frequency beyond fc. A decrease in frequency causes an increase in phase lag: At low frequency, phase tends toward 0°, not more lag. A decrease in frequency causes a decrease in the magnitude of the output voltage: At low frequency, magnitude approaches unity, so this is false. Common Pitfalls: Confusing lead and lag networks; assuming phase is constant; forgetting that the −3 dB point occurs at fc with φ = −45°. Also, some texts define “lag network” by where the output is taken; regardless, the low-pass case clearly shows increasing phase lag with increasing frequency. Final Answer: An increase in frequency causes an increase in phase lag.

Question 20

RL impedance with frequency: As the excitation frequency increases, how do the magnitudes of the equivalent impedances of series RL and parallel RL circuits change?

Accepted Answer

Introduction: Frequency dependence of impedance is a core idea in AC circuit design and filter selection. Resistor–inductor (RL) networks exhibit different behaviors in series and parallel configurations, but both are governed by the inductor's reactance, which grows with frequency. This question asks you to identify the overall trend for each topology as frequency rises. Given Data / Assumptions: Ideal components: R is frequency independent; L has reactance XL = ω * L. Small-signal sinusoidal steady state. We are concerned with impedance magnitude trends, not phase. Concept / Approach: Series RL: Zs = R + jXL → |Zs| = sqrt(R^2 + (ωL)^2); as ω increases, |Zs| increases monotonically. Parallel RL: The inductor branch impedance XL increases with frequency, so its branch current diminishes; the equivalent impedance approaches R from below as frequency rises, meaning |Zp| increases toward R (and can exceed the low-frequency value set by the parallel of R and a small XL). Step-by-Step Solution: Series RL: XL = ω * L; compute |Zs| = sqrt(R^2 + (ωL)^2) → increases with ω.Parallel RL: Equivalent admittance Yp = 1/R + 1/jXL; as ω increases, 1/jXL → 0.Therefore |Zp| = 1/|Yp| increases from a low value (strong shunt through L at low ω) toward R at high ω.Conclude both series and parallel RL impedances increase with frequency. Verification / Alternative check: Plotting |Zs| vs frequency shows a rising curve; plotting |Zp| vs frequency shows a rising asymptote approaching R. SPICE simulations confirm reduced inductor current at high frequency in the parallel branch, raising total impedance. Why Other Options Are Wrong: Both decrease: Contradicts XL = ωL growth. Series decreases / parallel increases: Series behavior is opposite; |Zs| increases. Series increases / parallel decreases: Parallel impedance does not decrease; it grows toward R as the inductor stops shunting current. Common Pitfalls: Confusing high-frequency inductor as a short (that is true at low frequency); forgetting to analyze admittance for the parallel case; overlooking that we ask for magnitudes, not phase. Final Answer: both series and parallel RL impedance increase

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