Question 1

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 …

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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 2

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?

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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 3

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?

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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 4

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?

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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 5

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 …

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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°.

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