Question 1

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 2

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

Question 3

In a series RL AC circuit, the inductive reactance and resistance are equal in magnitude (XL = 100 Ω and R = 100 Ω). What is the magnitude of the circuit impedance Z?

Accepted Answer

Introduction: Determining the impedance of a series RL circuit is a fundamental AC analysis skill. Because resistance and reactance are orthogonal (one dissipative, the other reactive), they combine vectorially rather than arithmetically. This question checks your ability to compute the magnitude of impedance when the resistor and inductor have equal magnitudes (R = XL). Given Data / Assumptions: Series RL circuit (single loop). R = 100 Ω. XL = 100 Ω. Steady-state sinusoidal excitation; use RMS quantities. Concept / Approach: For a series RL circuit, impedance magnitude is found using the Pythagorean relationship because R and XL are at 90 degrees in the phasor plane: Z = sqrt(R^2 + XL^2). When R = XL, Z becomes R * sqrt(2). Step-by-Step Solution: 1) Write the impedance formula for series RL: Z = sqrt(R^2 + XL^2).2) Substitute R = 100 Ω and XL = 100 Ω.3) Compute: Z = sqrt(100^2 + 100^2) = sqrt(10,000 + 10,000) = sqrt(20,000).4) Evaluate: sqrt(20,000) ≈ 141.4 Ω. Verification / Alternative check: Use the equality case: Z = R * sqrt(2) = 100 * 1.414 ≈ 141.4 Ω, confirming the result quickly without full arithmetic. Why Other Options Are Wrong: 14.14 Ω: misses a factor of 10; likely a decimal-place error.100 Ω: wrongly adds vector components linearly along one axis.200 Ω: simple sum R + XL; not valid for orthogonal phasors. Common Pitfalls: Adding R and XL directly, or confusing series and parallel combinations. Always remember that resistive and inductive components in series combine as a right triangle in the phasor diagram. Final Answer: 141.4 Ω

Question 4

A 24 V(rms) parallel RL circuit has R = 45 Ω and inductive reactance XL = 1100 Ω. What is the magnitude of the phase angle between the source voltage and the total current?

Accepted Answer

Introduction: In a parallel RL circuit, voltage is common to both branches, while currents split through the resistive and inductive paths. The overall current leads or lags the voltage by a small angle determined by the ratio of susceptance to conductance. This problem targets computing the magnitude of that phase angle from component values. Given Data / Assumptions: Parallel RL circuit with ideal components. R = 45 Ω, XL = 1100 Ω. Source voltage magnitude is 24 V(rms); however, the phase angle depends on R and XL, not directly on V. Concept / Approach: Work with admittances. For parallel RL: G = 1 / R and B = −1 / XL (inductive). The magnitude of the phase angle φ between voltage and total current is |φ| = arctan(|B| / G) = arctan( (1 / XL) / (1 / R) ) = arctan(R / XL ). Step-by-Step Solution: 1) Compute G = 1 / 45 ≈ 0.02222 S.2) Compute |B| = 1 / 1100 ≈ 0.0009091 S.3) Ratio |B| / G = (1 / 1100) / (1 / 45) = 45 / 1100 ≈ 0.04091.4) Angle magnitude: |φ| = arctan(0.04091) ≈ 2.35°, which rounds to about 2.3°. Verification / Alternative check: Using |φ| = arctan(R / XL) gives the same result directly: arctan(45 / 1100) ≈ 2.35° → 2.3° (rounded). Why Other Options Are Wrong: 0.001°: unrealistically small; would require XL ≫≫ R.87.6° and 89.9°: imply an almost purely reactive or extremely unbalanced circuit; not supported by R/XL ≈ 0.041. Common Pitfalls: Using series formulas tan φ = XL / R (which applies to series RL) instead of the admittance ratio for parallel RL. Always switch to conductance/susceptance for parallel analysis. Final Answer: 2.3°

Question 5

A 100 mH inductor (XL = 6 kΩ at the operating frequency) is in series with a 1 kΩ resistor and an AC source. What is the magnitude of the phase angle between the source voltage and current?

Accepted Answer

Introduction: The phase angle in a series RL circuit indicates how far the current lags the voltage due to inductive reactance. This question verifies your ability to compute the angle from R and XL, a common task in AC circuit design and power calculations. Given Data / Assumptions: Series RL circuit. R = 1 kΩ. XL = 6 kΩ (given explicitly for the operating frequency). Ideal sinusoidal steady state. Concept / Approach: For series RL, the phase angle φ (magnitude) satisfies tan φ = XL / R, since reactive and resistive drops are orthogonal. The current lags the voltage by φ for an inductive circuit. Step-by-Step Solution: 1) Write tan φ = XL / R.2) Substitute values: tan φ = 6000 / 1000 = 6.3) Compute φ = arctan(6) ≈ 80.5°.4) Rounded to the nearest whole-number option, φ ≈ 81.0°. Verification / Alternative check: Use the impedance triangle: Z = sqrt(R^2 + XL^2) and cos φ = R / Z. With R = 1 kΩ and XL = 6 kΩ, Z ≈ sqrt(1^2 + 6^2) kΩ = sqrt(37) kΩ ≈ 6.083 kΩ, giving cos φ ≈ 1 / 6.083 ≈ 0.164 and φ ≈ 80.6°, consistent with 81°. Why Other Options Are Wrong: 0.1° and 9.0°: would require XL ≪ R; not the case here.61.0°: corresponds to a smaller XL/R ratio (tan φ ≈ 1.8), not the given 6. Common Pitfalls: Confusing series and parallel angle relationships or using degrees vs radians incorrectly. Always compute φ with tan φ = XL / R for series RL. Final Answer: 81.0°

Question 6

In a series RL circuit where R = 100 Ω and XL = 100 Ω at the operating frequency, which statement about circuit behavior is correct?

Accepted Answer

Introduction: When resistance and inductive reactance are equal in a series RL circuit, a number of convenient relationships emerge that are useful for hand calculations. This question focuses on identifying the correct qualitative statement from several tempting but incorrect alternatives. Given Data / Assumptions: Series RL configuration. R = 100 Ω and XL = 100 Ω. Ideal sinusoidal operation; RMS quantities implied. Concept / Approach: For series RL, tan φ = XL / R. When R = XL, tan φ = 1, so φ = 45°. The impedance magnitude is Z = sqrt(R^2 + XL^2) = R * sqrt(2), and the power factor PF = cos φ = cos 45° ≈ 0.707, not 1. Voltage drops depend on current and source voltage, not fixed values like 5 V unless specified. Step-by-Step Solution: 1) Compute φ: tan φ = XL / R = 100 / 100 = 1 → φ = 45°.2) Compute Z: Z = sqrt(100^2 + 100^2) ≈ 141.4 Ω (for context).3) Compute PF: PF = cos φ = cos 45° ≈ 0.707, confirming PF ≠ 1.4) Conclude that the only universal statement from the options is that the phase angle equals 45°. Verification / Alternative check: The impedance triangle with equal legs confirms a 45° angle; this is a classic special case in AC circuit theory and appears in standard tables and examples. Why Other Options Are Wrong: Each component drops 5 V: voltage drops depend on current and source value; no 5 V given.Impedance equals 200 Ω: would be true only if R and XL added arithmetically, which they do not.Power factor equals 1: PF would be 0.707 for φ = 45°, not unity. Common Pitfalls: Adding resistive and reactive magnitudes linearly or assuming unity PF when R = XL. Remember that orthogonal phasors lead to 45° and PF ≈ 0.707 in this equality case. Final Answer: The phase angle equals 45°.

Question 7

For a 24 V(rms) parallel RL circuit with R = 45 Ω and XL = 1100 Ω, what is the true (real) power drawn from the source?

Accepted Answer

Introduction: True power (real power) in AC circuits is the portion of power actually dissipated as heat in resistive elements. In parallel RL circuits, voltage is the same across each branch, so the real power is determined entirely by the resistor branch, while the inductor branch stores and returns energy without net real power consumption. Given Data / Assumptions: Parallel RL circuit with ideal elements. R = 45 Ω (resistive branch). XL = 1100 Ω (inductive branch). Source voltage: V = 24 V(rms). Concept / Approach: In a parallel circuit, true power P is P = V^2 / R for the resistive branch. The inductor branch has reactive power only (Q), contributing no average real power. Therefore, total real power equals the resistor's real power. Step-by-Step Solution: 1) Use P = V^2 / R for the resistive branch.2) Substitute V = 24 V, R = 45 Ω: P = 24^2 / 45.3) Compute 24^2 = 576.4) Compute P = 576 / 45 = 12.8 W. Verification / Alternative check: Admittance method: G = 1 / R = 0.02222 S, so I_R = V * G = 24 * 0.02222 ≈ 0.5333 A. Then P = I_R^2 * R ≈ (0.5333)^2 * 45 ≈ 12.8 W, matching the V^2 / R result. Why Other Options Are Wrong: 313.45 W: far exceeds V^2 / R; impossible with a 24 V source and 45 Ω resistor.44.96 W and 22.3 W: inconsistent with P = V^2 / R and with the computed branch current. Common Pitfalls: Including reactive power in real power calculations or mistakenly using series formulas. In parallel circuits, compute real power per branch and sum; inductors contribute zero average real power. Final Answer: 12.8 W

Question 8

Series RL circuit — phase behavior with changing inductance: Consider a series RL circuit driven by a sinusoidal source. If the inductance L is increased (R fixed and frequency fixed), the phase angle between source voltage and current will ______.

Accepted Answer

Introduction / Context: In AC circuits, inductors cause current to lag voltage. The phase angle between source voltage and current depends on the ratio of reactance to resistance. Understanding how this angle changes with L helps in power factor correction and filter design. Given Data / Assumptions: Series RL circuit at fixed frequency f. Resistance R is constant. Inductance L increases. Concept / Approach: The impedance is Z = R + jX_L with X_L = 2πfL. The phase angle φ (current lag) satisfies tan φ = X_L / R. As L increases, X_L increases linearly, so tan φ increases and thus φ increases (i.e., the current lags by a larger angle). Therefore, the angle does not decrease; it grows toward 90 degrees as the inductive reactance dominates. Step-by-Step Solution: Write X_L = 2πfL.Compute tan φ = X_L / R.Increase L → increase X_L → increase tan φ → φ increases.Conclusion: Phase lag grows with L. Verification / Alternative check: Phasor diagrams show the reactive voltage drop vector lengthening with larger X_L, rotating the current vector further behind the voltage vector. Why Other Options Are Wrong: Decrease/remain zero/lead: contradict the inductive nature of the circuit at constant frequency. Common Pitfalls: Confusing series RL (current lags) with RC (current leads); forgetting that φ depends on X_L/R, not on absolute voltage levels. Final Answer: increase

Question 9

Parallel RL circuit — impedance vs frequency trend: As frequency decreases in a parallel RL circuit (R in parallel with an inductor), the magnitude of the circuit impedance will ______.

Accepted Answer

Introduction / Context: Parallel RL networks are common as loads and bias elements. Their input impedance varies with frequency because the inductor's reactance changes with ω. Recognizing the trend helps you predict loading and filter behavior. Given Data / Assumptions: Topology: R in parallel with L. Linear components at small-signal levels. We examine how |Z_in| changes as frequency decreases. Concept / Approach: Inductive reactance X_L = 2πfL. At low frequency (f → 0), X_L → 0 Ω, so the inductor behaves like a short. The parallel of any finite R with nearly 0 Ω approaches 0 Ω, so the total impedance decreases toward zero. At high frequency, X_L becomes large (approaching open circuit), so the parallel pair approaches R. Thus, as frequency decreases from high to low, |Z| decreases from approximately R toward 0 Ω. Step-by-Step Solution: Compute X_L = 2πfL.At low f: X_L ≈ 0 → Z_total ≈ 0 (short-dominated).At high f: X_L ≫ R → Z_total ≈ R (resistor-dominated).Therefore, decreasing frequency decreases |Z|. Verification / Alternative check: Plot |Z| vs f using Z = (jωL * R) / (R + jωL). The curve drops toward zero as f approaches DC. Why Other Options Are Wrong: Increase/constant: contradicts the limiting behavior at f → 0.Unpredictable: the dependence is monotonic and well-defined. Common Pitfalls: Confusing series and parallel behaviors; forgetting that inductors look like shorts at DC. Final Answer: decrease

Question 10

Series RL circuit — lead/lag identification: In a series RL circuit driven by a sinusoidal source, the current waveform lags the source voltage waveform. Is this statement correct?

Accepted Answer

Introduction / Context: Lead/lag relationships are crucial for power factor and control systems. Inductors store energy in magnetic fields and cause current to lag voltage, whereas capacitors cause current to lead. Recognizing the RL case prevents sign errors in phasor analysis. Given Data / Assumptions: Series RL circuit, linear and time-invariant. Sinusoidal steady-state operation. Finite R and L values. Concept / Approach: Impedance is Z = R + jX_L with X_L = 2πfL > 0. The current I = V / Z lags the voltage by angle φ = arctan(X_L / R) > 0 (lag). This positive phase angle means the current's zero crossings occur later in time than the voltage's, a hallmark of inductive circuits. Step-by-Step Solution: Express Z and compute φ = arctan(X_L / R).Since X_L > 0, φ > 0 → current lags voltage.Thus, the statement is correct.If L were zero (pure R), φ = 0 and current would be in phase. Verification / Alternative check: Oscilloscope with current probe shows current waveform delayed relative to voltage in an RL network; the delay grows with larger X_L / R. Why Other Options Are Wrong: “Incorrect” would contradict the physics of inductive reactance. Common Pitfalls: Confusing RL with RC, where current leads voltage; mixing up the sign convention for phase angles. Final Answer: Correct

Question 11

RL filter identification (series network): In a simple series R–L circuit driven by a sinusoidal source, if the output is measured across the resistor (voltage across R), does the network behave as a low-pass filter that passes low frequencies and a…

Accepted Answer

Introduction / Context: First-order RL circuits are the magnetic dual of RC filters. Whether they act as low-pass or high-pass depends entirely on where the output is taken. In practice, the series RL measured across the resistor is a classic one-pole low-pass used for current sensing and simple anti-alias filtering. Given Data / Assumptions: Single loop: sinusoidal source feeding a series resistor R and inductor L. Output node is the voltage across R (not across L). Linear, time-invariant operation with ideal components. Concept / Approach: The inductor’s reactance is XL = 2 * pi * f * L. As frequency increases, XL increases and the series current decreases. Since v_R = i * R, the output magnitude falls with frequency. Conversely, at low frequency XL is small, current is larger, and v_R approaches the input, which is the hallmark of a low-pass response. The cutoff frequency (−3 dB) occurs at f_c = R / (2 * pi * L). Step-by-Step Solution: Write transfer: H(jf) = V_R / V_in = R / sqrt(R^2 + (2 * pi * f * L)^2).At f → 0, H → 1 (pass low frequencies).At f → ∞, H → 0 (attenuate high frequencies).Cutoff when R = XL ⇒ f_c = R / (2 * pi * L). Verification / Alternative check: Bode magnitude shows −20 dB/decade roll-off above f_c, and the phase of V_R lags less than 90 degrees, approaching 0 degrees at low frequency—typical of a one-pole low-pass across a resistive element in a series RL path. Why Other Options Are Wrong: High-pass behavior occurs when the output is taken across L, not R. Limiting L to zero removes filtering. The response is broadband (a full passband and stopband), not a single-frequency effect. Common Pitfalls: Confusing where the output is measured; swapping to the inductor output flips the function to high-pass. Forgetting that the cutoff depends on R/L, not on R or L alone. Final Answer: Applies (low-pass across the series resistor)

Question 12

RL impedance calculation (numerical): A series RL circuit has resistance R = 60 Ω and inductive reactance XL = 92 Ω at a certain frequency. What is the magnitude of the total impedance Z?

Accepted Answer

Introduction / Context: In AC analysis, impedance combines resistance and reactance as perpendicular (orthogonal) components on the complex plane. For a series RL circuit, the magnitude of impedance is not the arithmetic sum R + XL but the vector sum. This question reinforces the correct computation for Z given R and XL. Given Data / Assumptions: Series RL network with R = 60 Ω. Inductive reactance XL = 92 Ω at the operating frequency. Ideal components and sinusoidal steady state. Concept / Approach: Because resistance (real axis) and inductive reactance (positive imaginary axis) are orthogonal, the magnitude of the series impedance is |Z| = sqrt(R^2 + XL^2). This follows directly from Pythagoras on the impedance triangle. The phase angle is theta = arctan(XL / R), but only the magnitude is requested here. Step-by-Step Solution: Compute R^2 = 60^2 = 3600.Compute XL^2 = 92^2 = 8464.Sum: 3600 + 8464 = 12064.Magnitude: |Z| = sqrt(12064) ≈ 109.87 Ω ≈ 110 Ω. Verification / Alternative check: If you incorrectly add R + XL = 152 Ω, you overestimate the impedance because you ignored the quadrature relation. Phasor diagrams or complex arithmetic (Z = 60 + j92) confirm |Z| ≈ 110 Ω. Why Other Options Are Wrong: 102 Ω and 92 Ω underestimate the vector length; 152 Ω is the erroneous arithmetic sum, not the magnitude. Common Pitfalls: Adding magnitudes linearly; forgetting units; mixing up series vs parallel combinations (parallel requires admittance addition, not impedance). Final Answer: 110 Ω

Question 13

Admittance–impedance conversion: If the admittance of a branch is Y = 20 mS (millisiemens), what is the magnitude of the equivalent impedance |Z|?

Accepted Answer

Introduction / Context: Admittance (Y) is the reciprocal of impedance (Z). In many AC problems it is convenient to switch between these domains—especially for parallel networks where admittances add directly. This question checks the unit handling and reciprocal operation using millisiemens. Given Data / Assumptions: Admittance Y = 20 mS = 20 * 10^-3 S = 0.02 S. We are asked for the magnitude of impedance |Z| = 1 / |Y|. Assume purely resistive magnitude for the numeric conversion (reactive angle not specified). Concept / Approach: The reciprocal relationship is Z = 1 / Y. Be careful with milli- prefix: 1 mS = 10^-3 S. Thus 20 mS = 0.02 S, and its reciprocal gives an impedance magnitude in ohms. This is a straightforward unit inversion often tripped up by prefix mistakes. Step-by-Step Solution: Convert units: 20 mS = 0.02 S.Compute reciprocal: |Z| = 1 / 0.02 = 50.Attach unit: 50 Ω.Select the option: 50 Ω. Verification / Alternative check: Check dimensional consistency: siemens is 1/ohm; the reciprocal returns ohms. 1 / (20 * 10^-3) = 1 / 0.02 = 50 confirms the arithmetic. Why Other Options Are Wrong: 20 Ω corresponds to 50 mS, not 20 mS. 5 Ω would require 200 mS. 0.02 Ω is the inverse of 50 S, not 20 mS. Common Pitfalls: Forgetting to convert milli- to base SI units; mixing ohms and siemens; applying the reciprocal to each component separately in non-parallel cases (reciprocals add only for admittances in parallel). Final Answer: 50 Ω

Question 14

Total opposition in a series RL circuit: For a series AC circuit containing both resistance and inductive reactance, how should the total opposition to current (impedance magnitude) be found?

Accepted Answer

Introduction / Context: Impedance combines resistance (real part) and reactance (imaginary part). In a series RL circuit, these components are orthogonal on the complex plane, so the total opposition to AC current is obtained by vector addition, not simple arithmetic addition. This is foundational when computing currents, voltages, and power factors. Given Data / Assumptions: Series connection of R and L with sinusoidal excitation. Reactance is inductive only: XL = 2 * pi * f * L. Goal: impedance magnitude for current calculation via Ohm’s law I = V / |Z|. Concept / Approach: Represent the impedance as Z = R + jXL. The magnitude is |Z| = sqrt(R^2 + XL^2) from the impedance triangle. Simple sums (R + XL) ignore the 90-degree phase difference and produce systematic errors, especially at higher reactance ratios. The correct approach preserves orthogonality via Pythagorean addition. Step-by-Step Solution: Express Z in rectangular form: Z = R + jXL.Compute magnitude: |Z| = sqrt(R^2 + XL^2).Use I = V / |Z| for current magnitude once |Z| is known.Find phase if needed: theta = arctan(XL / R). Verification / Alternative check: Phasor diagrams show the resistive and reactive drops at right angles; the hypotenuse length equals |Z|. Experimental measurements with an LCR meter corroborate the vector relationship across frequencies. Why Other Options Are Wrong: Arithmetic sum and average ignore phase. Subtraction could apply to reactive cancellation (XL − XC) but not to R with XL in magnitude computation. Common Pitfalls: Adding R and XL directly; forgetting that only reactances combine algebraically when both are reactive; confusing magnitude with real part. Final Answer: Use the vector magnitude: Z = sqrt(R^2 + XL^2)

Question 15

Compute impedance of a given series RL network (numerical): A 7.5 kΩ resistor is in series with an inductive reactance of 10 kΩ. What is the magnitude of the impedance Z of the series RL circuit?

Accepted Answer

Introduction / Context: Many quick design and exam questions focus on computing the magnitude of impedance for first-order RL series networks. The result determines current draw and power factor at a given frequency. The calculation uses the impedance triangle, not simple arithmetic addition. Given Data / Assumptions: Series elements: R = 7.5 kΩ and XL = 10 kΩ. Ideal components and sinusoidal steady state. Asked for magnitude |Z| only. Concept / Approach: For series RL, Z = R + jXL. Therefore |Z| = sqrt(R^2 + XL^2). With values in kilo-ohms, you can keep the kΩ factor outside the square-root arithmetic for convenience, then apply it back to the result. Step-by-Step Solution: Compute squares in kΩ^2: 7.5^2 = 56.25 and 10^2 = 100.Add: 56.25 + 100 = 156.25.Take square root: sqrt(156.25) = 12.5.Attach unit: |Z| = 12.5 kΩ. Verification / Alternative check: Using rectangular to polar conversion: Z = 7.5 + j10 (kΩ). Magnitude √(7.5^2 + 10^2) = 12.5 kΩ; angle theta = arctan(10/7.5) ≈ 53.1 degrees (not asked but confirms consistency). Why Other Options Are Wrong: 17.5 kΩ is the incorrect arithmetic sum R + XL. 10.6 kΩ results from an arithmetic/rounding mistake. 7.5 kΩ ignores reactance entirely. Common Pitfalls: Switching units mid-calculation; forgetting to use vector sum; mistaking reactance for resistance when estimating power. Final Answer: 12.5 kΩ

Question 16

Power factor and real power in a series RL circuit (concept check): At a fixed RMS source voltage, does a power factor of 0.5 result in less average real power (heat in R) than a power factor of 0.6?

Accepted Answer

Introduction / Context: Power factor (pf) describes the fraction of apparent power that becomes real power (heat and useful work). In a series RL circuit supplied by a fixed RMS voltage, the higher the power factor, the larger the real power delivered to the resistor for a given apparent power. This item compares pf = 0.5 and pf = 0.6 under constant supply voltage. Given Data / Assumptions: Sinusoidal steady state with fixed RMS source voltage V. Series RL load; real power dissipates only in R. Power factor pf = cos(theta) with theta = arctan(XL / R). Concept / Approach: Real power P is P = V * I * pf. At fixed V, apparent power S = V * I, while pf determines how much of S becomes P. For the same network at different operating points (or different L/R combinations) resulting in pf = 0.5 vs 0.6, the case with pf = 0.6 converts a larger portion of the apparent power into real power, hence more heat in R and a higher meter reading on a wattmeter. Step-by-Step Solution: Define: P = V * I * pf and S = V * I.At identical V and comparable operating conditions, higher pf increases P for the same S.Thus pf = 0.5 ⇒ lower P than pf = 0.6.Conclusion: the statement that 0.5 yields less real power than 0.6 is correct. Verification / Alternative check: With normalized S = 1 VA, pf = 0.5 gives P = 0.5 W; pf = 0.6 gives P = 0.6 W. The trend is general: increasing pf (approaching unity) increases real power for a given apparent power or given supply voltage with similar loading. Why Other Options Are Wrong: “More power at 0.5” contradicts P = S * pf. “Same power” is only true if other variables change to offset pf, which is not implied here. “Need L” is unnecessary for the qualitative comparison. Common Pitfalls: Confusing fixed current with fixed voltage. At fixed current, P = I^2 * R and pf discussion differs; the question specifies fixed voltage supply context. Final Answer: Applies (pf = 0.5 yields less real power than pf = 0.6)

Question 17

Phase angle definition in AC circuit analysis: What is meant by the phase angle in a series RL circuit driven by a sinusoidal source?

Accepted Answer

Introduction / Context: Phase angle is a cornerstone of AC analysis, capturing the timing relationship between voltage and current in reactive networks. In a series RL circuit, the current lags the source voltage by an angle determined by the ratio of inductive reactance to resistance. Properly defining the phase angle avoids confusion in power factor and impedance calculations. Given Data / Assumptions: Sinusoidal steady state. Series RL elements only. Standard phasor conventions (voltage reference unless stated otherwise). Concept / Approach: The phase angle theta in a series RL is defined by tan(theta) = XL / R with theta positive (current lags voltage). In words, it is the angular difference between the applied source voltage phasor and the circuit current phasor. This angle directly determines power factor: pf = cos(theta). The concept generalizes to any linear AC network but is especially transparent in first-order cases. Step-by-Step Solution: Express impedance: Z = R + jXL.Compute angle: theta = arctan(XL / R).Interpret: current phasor lags voltage by theta.Relate: pf = cos(theta) ties angle to real power flow. Verification / Alternative check: Oscilloscope Lissajous plots or dual-channel measurements show current zero crossings delayed relative to voltage when an inductor is present, confirming the lag and the phase relation. Why Other Options Are Wrong: “Between any two node voltages” is ambiguous and not the standard definition. The magnetic-field/heat-flow comparison is not an electrical phase definition. DC has no meaningful phase angle between power vectors because frequency is zero and reactive effects vanish. Common Pitfalls: Swapping lead/lag sign; confusing the angle of impedance with the angle between two unrelated node voltages. Final Answer: The angular difference between the applied source voltage and the circuit current

Question 18

Frequency dependence of impedance in a series RL: As frequency increases in a series RL circuit, how does the magnitude of the impedance |Z| change?

Accepted Answer

Introduction / Context: Understanding frequency dependence is critical when predicting how circuits respond to different signals. In a series RL circuit, the reactive part grows with frequency, changing both the magnitude and the phase of the impedance. This directly affects current draw and filter behavior. Given Data / Assumptions: Series RL, ideal components. Inductive reactance XL = 2 * pi * f * L. Interest is in impedance magnitude |Z|. Concept / Approach: Impedance magnitude for a series RL is |Z| = sqrt(R^2 + XL^2). Because XL is proportional to frequency f, increasing f increases XL and therefore increases |Z|. At low frequency XL is small and |Z| ≈ R; at high frequency XL dominates and |Z| grows approximately linearly with f (when XL ≫ R). Step-by-Step Solution: Start with |Z| = sqrt(R^2 + (2 * pi * f * L)^2).Differentiate qualitatively with respect to f: |Z| increases as f increases.Consider limits: f → 0 ⇒ |Z| → R; f → ∞ ⇒ |Z| ≈ 2 * pi * f * L.Therefore, the trend is monotonic increase with frequency. Verification / Alternative check: Bode plots show magnitude rising at +20 dB/decade beyond the frequency where XL surpasses R. Bench measurements with an LCR meter exhibit the same trend in the measured impedance. Why Other Options Are Wrong: Inductors do not become shorts at high frequency; they impede more. Constant magnitude equal to R applies only at f ≈ 0. There is no resonance in a simple RL (no capacitor), so no oscillatory behavior in |Z|. Common Pitfalls: Confusing the RL series with RC trends; thinking resonance exists without a capacitive element. Final Answer: It increases because XL = 2 * pi * f * L increases

Question 19

Parallel RL phase behavior: In a parallel RL circuit excited by a sinusoidal voltage, what happens to the phase angle between the total current and the applied voltage if the resistance R is increased (with L and frequency fixed)?

Accepted Answer

Introduction / Context: Parallel RL networks are often analyzed more cleanly in the admittance domain. The total current leads the applied voltage by an angle set by the ratio of susceptance to conductance. Changing the resistance changes conductance, which directly affects the phase angle between total current and voltage. Given Data / Assumptions: Parallel branches: a resistor R and an inductor L. Frequency and inductance are fixed; excitation is sinusoidal. Interest is the angle between total current and applied voltage. Concept / Approach: Work with admittance: Y = G − jB, where G = 1 / R and B = 1 / XL = 1 / (2 * pi * f * L). The phase of the total current relative to voltage is given by tan(phi) = B / G. Increasing R decreases G. With B fixed, B / G therefore increases, so |phi| increases. In a parallel RL, current leads voltage (inductive susceptance), so the lead increases in magnitude as R increases. Step-by-Step Solution: Write Y = 1/R − j(1/XL).Define tan(phi) = B / G = (1/XL) / (1/R) = R / XL.Increase R ⇒ tan(phi) increases ⇒ |phi| increases.Therefore, the total current’s lead over the voltage grows in magnitude. Verification / Alternative check: Vector diagrams of branch currents show the resistive branch current shrinking with higher R, leaving the inductive branch current more dominant, so the phasor sum tilts further toward the inductive axis. Why Other Options Are Wrong: “Angle decreases” contradicts tan(phi) = R / XL. “Unchanged” ignores that G depends on R. “Resonant” behavior requires a capacitor; an RL parallel circuit cannot exhibit resonance. Common Pitfalls: Applying series RL intuition to a parallel network; mixing impedance and admittance phase formulas. Final Answer: The magnitude of the phase angle increases (more inductive current lead over conductance)

Question 20

First-order RL breakpoint terminology: In a first-order RL network used as a filter, the frequency at which XL equals R is called the ________.

Accepted Answer

Introduction / Context: First-order RL and RC filters have a single breakpoint that defines where the response transitions from passband to stopband. For an RL filter, this occurs where the reactive impedance equals the resistive component. The terminology for this frequency is widely standardized in filter theory and instrumentation. Given Data / Assumptions: First-order RL network (either low-pass or high-pass depending on output pick-off). Sinusoidal steady state, ideal components. Definition of inductive reactance XL = 2 * pi * f * L. Concept / Approach: The cutoff (also called corner or break) frequency f_c occurs when the magnitude of the reactive impedance equals the resistance: XL = R. Solving gives f_c = R / (2 * pi * L). At this frequency, the magnitude of the transfer function is 1/sqrt(2) of the passband value (−3 dB point) for the standard one-pole filter. “Resonant frequency” does not apply because resonance requires energy exchange between inductance and capacitance, and an RL alone does not resonate. Step-by-Step Solution: Set XL = R with XL = 2 * pi * f * L.Solve: f_c = R / (2 * pi * L).Interpret: at f_c the output is 0.707 of the low-frequency (or high-frequency) asymptote depending on topology.Use this f_c to plot Bode magnitude and phase. Verification / Alternative check: Measurement of an RL low-pass across the resistor shows −3 dB amplitude at f = R / (2 * pi * L), confirming the break-frequency definition. Why Other Options Are Wrong: “Optimum frequency” is not a standard filter term. “Bandwidth characteristic” is a graph, not a frequency. “Resonant frequency” requires an LC network. Common Pitfalls: Calling any notable frequency “resonant” even when no capacitor is present; mixing up RL with RC formulas (RC has f_c = 1 / (2 * pi * R * C)). Final Answer: cutoff frequency

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