Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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

Question 9

Series RL impedance magnitude — calculation practice: In a series RL circuit used in basic electronics, the impedance magnitude |Z| is found from the resistor and inductive reactance as |Z| = sqrt(R^2 + X_L^2). If R = 15 kΩ and X_L = 10 kΩ, what is …

Accepted Answer

Introduction / Context: Series RL circuits appear in filters, chokes, and transient-limiting networks. The combined opposition to current is not a simple sum because the resistor (R) and the inductor's reactance (X_L) act at right angles in the impedance plane. This question checks your ability to compute the impedance magnitude |Z| from given R and X_L values and to pick the correct approximate answer from options. Given Data / Assumptions: Resistor R = 15 kΩ. Inductive reactance X_L = 10 kΩ (at the operating frequency). Series combination; standard phasor relationships apply. We need the magnitude |Z|, not the angle. Concept / Approach: For a series RL circuit, impedance is Z = R + jX_L. The magnitude is |Z| = sqrt(R^2 + X_L^2). Because R and X_L are perpendicular components (real and imaginary), the Pythagorean relationship applies. Units must be consistent (kΩ with kΩ). Step-by-Step Solution: 1) Write |Z| = sqrt(R^2 + X_L^2). 2) Substitute R = 15 kΩ and X_L = 10 kΩ. 3) Compute inside the radical: 15^2 + 10^2 = 225 + 100 = 325 (kΩ)^2. 4) Take the square root: sqrt(325) ≈ 18.027 → approximately 18 kΩ. Verification / Alternative check: A quick bound check helps: if R = 15 and X_L = 10, |Z| must be larger than the larger component (15 kΩ) but smaller than their sum (25 kΩ). The value 18 kΩ lies within this reasonable range, confirming the calculation. Why Other Options Are Wrong: 5 kΩ / 6 kΩ: too small; magnitude cannot be less than the largest individual component in series. 25 kΩ: that is the arithmetic sum, not the vector magnitude. 20 kΩ: plausible-sounding estimate but higher than the correct Pythagorean result. Common Pitfalls: Adding R and X_L arithmetically; forgetting that reactance is orthogonal to resistance; mixing Ω with kΩ; rounding prematurely before comparing with options. Final Answer: 18 kΩ

Question 10

Effect of source amplitude in a linear RL series circuit (frequency fixed): If the source voltage magnitude V_S increases while component values and frequency remain unchanged, which statement about the circuit's behavior is correct?

Accepted Answer

Introduction / Context: In linear AC circuits (such as a series RL network), the relationship between source voltage, current, and impedance at a fixed frequency follows a direct proportionality: I = V_S / |Z|. This question tests recognition of which quantities change when the source voltage is varied while the circuit's impedance stays constant because R, L, and f are unchanged. Given Data / Assumptions: Series RL circuit with fixed R and L. Operating frequency is fixed, so X_L = 2 * pi * f * L is constant. Linear components; no saturation or nonlinearity. We consider steady-state sinusoidal operation. Concept / Approach: The circuit impedance magnitude |Z| = sqrt(R^2 + X_L^2) is constant when R, L, and f do not change. Therefore, increasing the source voltage magnitude V_S increases current magnitude proportionally, I = V_S / |Z|. The phase angle theta = arctan(X_L / R) is also constant because it depends only on R and X_L at the given frequency. Real power P = I^2 * R increases with I, not decreases. Step-by-Step Solution: 1) Compute |Z| from R and X_L; note it is independent of V_S. 2) Use I = V_S / |Z|. When V_S increases, I increases linearly. 3) Recognize theta = arctan(X_L / R) is unchanged (R, L, f fixed). 4) Real power P = I^2 * R; since I increases, P increases (it does not decrease). Verification / Alternative check: Doubling V_S while holding |Z| constant doubles I. With P proportional to I^2 for a fixed R, power quadruples, consistent with basic AC power relations. Why Other Options Are Wrong: Impedance increases: |Z| depends on R and X_L, not V_S. Real power decreases: incorrect; as I rises, P = I^2 * R rises. Phase angle increases: theta depends on R and X_L only. Reactive power magnitude decreases: Q = I^2 * X_L rises with I. Common Pitfalls: Assuming source voltage influences impedance; confusing voltage-scaling effects with frequency changes; forgetting that both real and reactive power scale with I^2 for fixed R, X_L. Final Answer: Current increases

Question 11

Frequency change in a series RL circuit (R and L fixed): If the operating frequency decreases, which of the following statements is correct for the circuit's parameters and responses?

Accepted Answer

Introduction / Context: In a series RL circuit, several quantities depend on the operating frequency. The most direct dependency is the inductive reactance X_L = 2 * pi * f * L. Understanding how X_L and related responses change with frequency is vital for filter design and power systems where frequency variations can shift current, voltage drops, and phase angle. Given Data / Assumptions: Series RL network with fixed R and L. Steady-state sinusoidal operation. We analyze qualitative change when frequency f decreases. Concept / Approach: The most fundamental effect is on inductive reactance: X_L = 2 * pi * f * L. If f decreases, X_L decreases proportionally. Consequences include a smaller impedance magnitude |Z| = sqrt(R^2 + X_L^2), larger current I = V_S / |Z| (for fixed V_S), and a smaller phase angle theta = arctan(X_L / R). Among the choices, the statement that is unambiguously and directly true for any values is that X_L decreases. Step-by-Step Solution: 1) Start with X_L = 2 * pi * f * L. 2) Decrease f → X_L decreases in direct proportion. 3) Recognize the ripple effects: |Z| decreases; current increases; phase angle decreases. 4) Identify the option that matches this primary dependency: “Inductive reactance decreases.” Verification / Alternative check: Pick L = 100 mH. At 50 Hz, X_L ≈ 31.4 Ω; at 25 Hz, X_L ≈ 15.7 Ω. The halved frequency halves X_L, confirming the rule. Why Other Options Are Wrong: I_T decreases: opposite; current increases as |Z| falls (for fixed V_S). I_R increases / I_L increases: in a series circuit I_R = I_L = I_T, but stating only one branch increases is incomplete and ambiguous; the fundamental certain statement concerns X_L directly. The phase angle increases: it actually decreases because X_L/R becomes smaller. Common Pitfalls: Confusing series and parallel cases; assuming only the inductive branch current changes; overlooking that all series branch currents are identical and follow I = V_S / |Z|. Final Answer: Inductive reactance decreases

Question 12

Phase relationship in a series RL circuit: Considering standard phasor conventions, what is the correct relationship between the resistor voltage V_R and the inductor voltage V_L in a series RL branch driven by a sinusoidal source?

Accepted Answer

Introduction / Context: Understanding voltage phasors in series RL circuits is essential for predicting waveform timing, interpreting oscillograph traces, and constructing accurate phasor diagrams. While the current is common in a series branch, the individual element voltages are shifted relative to that current and, therefore, relative to one another. Given Data / Assumptions: Series RL circuit with sinusoidal steady-state excitation. Standard passive sign convention. Ideal components (no parasitic capacitance or resistance in the inductor beyond modeled R). Concept / Approach: In a resistor, voltage and current are in phase. In an inductor, voltage leads current by 90 degrees (v_L = L * di/dt corresponds to a +90° phase shift of voltage relative to current in phasor form). Because the series current is the same through both elements, V_R aligns with the current, while V_L leads the current by +90°. Therefore, V_L leads V_R by +90°, or equivalently, V_R lags V_L by 90°. Step-by-Step Solution: 1) Note I is common to both elements in series. 2) For the resistor: V_R is in phase with I. 3) For the inductor: V_L leads I by +90°. 4) Therefore V_L leads V_R by +90°, meaning V_R lags V_L. Verification / Alternative check: Construct a phasor diagram: draw I along the horizontal axis; V_R on the same axis; V_L upward (positive imaginary axis). The source voltage phasor V_S is the vector sum of V_R and V_L, sitting at an angle between them, verifying the lag relationship. Why Other Options Are Wrong: V_R leads V_L: reverses the correct order. V_R is in phase with V_L: not in an RL branch; they are 90° apart. All in phase: impossible when an inductor is present with nonzero reactance. V_R leads by 180°: no such inversion occurs in linear passive RL. Common Pitfalls: Confusing voltage-current relationships of R and L; thinking all voltages in one series branch must be in phase; mixing up the sign of the 90° shift. Final Answer: V_R lags V_L

Question 13

RL series circuit — phase angle when resistance dominates: In a series R–L AC circuit, if the resistance R is larger than the inductive reactance X_L (i.e., R > X_L), what is the approximate range of the circuit phase angle φ between source voltage …

Accepted Answer

Introduction / Context: In AC circuit analysis, the phase angle φ between voltage and current in a series R–L network indicates how inductive the circuit is. Recognizing how φ changes as the resistance R and the inductive reactance X_L vary is essential for power factor correction, load characterization, and transformer/motor drive applications. This question asks for the qualitative range of φ when R dominates, i.e., when R > X_L. Given Data / Assumptions: Series R–L circuit driven by a sinusoidal source. Resistance R and inductive reactance X_L = 2 * pi * f * L. Phase angle φ defined as the angle by which voltage leads current in an R–L series circuit (current lags by φ). Linear, time-invariant, steady-state conditions. Concept / Approach: The impedance of a series R–L circuit is Z = R + j X_L. The phase angle is φ = arctan(X_L / R). When R > X_L, the ratio X_L / R is less than 1, which directly implies 0° < φ < 45°. As R increases relative to X_L, φ approaches 0°, meaning the circuit becomes more resistive with improved power factor. Conversely, if X_L dominates (X_L > R), φ approaches 90° (strongly inductive). Step-by-Step Solution: Write φ = arctan(X_L / R) for a series R–L circuit.Use the given condition R > X_L ⇒ X_L / R < 1.Since arctan of a positive number smaller than 1 lies between 0° and 45°, conclude 0° < φ < 45°.Interpretation: the circuit is inductive (positive φ), but only mildly so because resistance dominates. Verification / Alternative check: Try numeric examples. If R = 10 Ω and X_L = 5 Ω, φ = arctan(0.5) ≈ 26.6°, which indeed lies between 0° and 45°. If R approaches X_L (e.g., both 10 Ω), φ ≈ 45°. If R ≫ X_L, φ tends to 0°, reflecting a nearly resistive load. Why Other Options Are Wrong: less than 0° (capacitive lead): A negative φ indicates capacitive behavior, which is not possible with only R and L in series. exactly 45°: Occurs only when R = X_L, not when R > X_L. between 45° and 90°: Requires X_L > R (inductance dominating). Common Pitfalls: Confusing magnitude dominance (R vs X_L) with sign of φ; forgetting that inductors cause current to lag (positive φ) while capacitors cause current to lead (negative φ); mixing up series versus parallel cases. Final Answer: between 0° and 45°.

RL Circuits Questions