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

The phase angle will decrease in a series RL circuit if the inductance is increased.

Accepted Answer

False

Question 9

As frequency decreases in a parallel RL circuit the impedance also decreases.

Accepted Answer

True

Question 10

The current lags the source voltage in a series RL circuit.

Accepted Answer

True

Question 11

The circuit in the given circuit is a low-pass filter.

Accepted Answer

True

Question 12

The impedance of the circuit in the given circuit is 110 Ω .

Accepted Answer

False

Question 13

If Y = 20 mS, then Z = 20.

Accepted Answer

False

Question 14

For series ac circuits containing both resistance and inductive reactance, total opposition to current can be found by simply adding ohmic values or oppositions to current.

Accepted Answer

False

Question 15

If a 7.5 kΩ resistance is connected to a 10 kΩ inductive reactance in a series RL circuit, then the impedance equals 12.5 kΩ .

Accepted Answer

True

Question 16

In a series RL circuit, a power factor of 0.5 results in less energy conversion to heat than does a power factor of 0.6.

Accepted Answer

True

Question 17

Phase angle is the difference in phase between the applied voltage (VS) and the circuit current (IT).

Accepted Answer

True

Question 18

As the frequency increases in a series RL circuit the impedance also increases.

Accepted Answer

True

Question 19

The phase angle will increase in a parallel RL circuit if the resistance is increased.

Accepted Answer

True

Question 20

The frequency at which XL = R in an RL circuit is called the _______________.

Accepted Answer

cutoff frequency

RL Circuits Questions