Q: In a series RLC circuit operating exactly at its resonant frequency, what is the measured net voltage across the two reactive components considered together (inductor in series with capacitor)?

Introduction: Series resonance is a cornerstone concept in AC circuit theory. At resonance, inductive and capacitive reactances are equal in magnitude and opposite in sign. This question examines what happens to the combined voltage across the series inductor and capacitor when the circuit is exactly at resonance. Given Data / Assumptions: Series RLC topology. Operating frequency equals the resonant frequency f0. Ideal components; steady-state sinusoidal excitation. Concept / Approach: At resonance, XL = 2 * pi * f0 * L and XC = 1 / (2 * pi * f0 * C) satisfy XL = XC. Their phasor voltages are equal in magnitude but 180 degrees out of phase. Therefore, the phasor sum across L and C is zero even though each element may have a large individual voltage. Step-by-Step Solution: 1) Recognize resonance condition: XL = XC.2) Phasor voltages: VL leads current by 90 degrees; VC lags current by 90 degrees.3) Since |VL| = |VC| and their phases are opposite, VL + VC = 0 in phasor terms.4) Hence, the net measured voltage across L and C together is zero. Verification / Alternative check: Impedance perspective: ZLC = jXL + (−jXC) = j(XL − XC) = 0 at resonance. A zero reactive impedance means zero net reactive drop across the pair. Why Other Options Are Wrong: applied: The applied source voltage appears across the series combination of R, L, and C; the net across L and C alone is zero at resonance.reactive: Not a magnitude; the correct magnitude of the net is zero.inductive and capacitive: Descriptive words, not the value; they cancel at resonance. Common Pitfalls: Confusing individual element voltages (which can be high with large Q) with their vector sum, which cancels at true resonance. Final Answer: zero

Q: At resonant frequency in a series RLC circuit, what is the net voltage across the two reactive components (inductor and capacitor) when measured together?

Introduction: Understanding voltage distribution at resonance helps in filter design and impedance matching. This question reinforces the vector cancellation of reactive drops in a resonant series RLC circuit. Given Data / Assumptions: Series RLC circuit operating at resonance. Ideal sinusoidal steady-state behavior. Phasor analysis applies. Concept / Approach: At resonance, XL = XC. The inductor voltage phasor leads the current by 90 degrees; the capacitor voltage phasor lags by 90 degrees. Equal magnitudes and opposite phases yield a phasor sum of zero across L and C together. Step-by-Step Solution: 1) Write XL = 2 * pi * f0 * L and XC = 1 / (2 * pi * f0 * C).2) At resonance: XL − XC = 0.3) Phasor addition: VL + VC = 0 (equal magnitude, 180-degree phase difference).4) Therefore, the net voltage across the combined series reactances is zero. Verification / Alternative check: Impedance view: ZLC = jXL + (−jXC) = j(XL − XC) = 0 at f0, so there is no reactive drop across L + C as a pair. Why Other Options Are Wrong: Applied voltage: At resonance most of the source voltage appears across R, not across L + C together.Reactive voltage: Not a numeric value; net reactive drop is zero.VL + VC voltage: The algebraic (phasor) sum equals zero, not a finite nonzero amount. Common Pitfalls: Assuming large circulating voltages imply a large net drop; in reality, large but opposite voltages cancel in phasor sum. Final Answer: Zero voltage

Q: A series circuit resonates at 6 kHz. If at resonance the inductor reactance and capacitor reactance are each 4 kΩ in magnitude and the series resistance is R = 50 Ω, what is the quality factor Q of the circuit?

Introduction: The quality factor Q characterizes the sharpness of resonance and the ratio of stored to dissipated energy per cycle. For a series RLC circuit, Q directly relates reactive impedance to resistance at the resonant frequency. This question checks the simple but crucial Q = X / R relationship at resonance. Given Data / Assumptions: Series RLC at resonance f0 = 6 kHz (frequency value not needed for Q if X is given). Inductive and capacitive reactance magnitudes: XL = XC = 4 kΩ. Series resistance: R = 50 Ω. Idealized small-signal behavior. Concept / Approach: For series resonance, Q = XL / R = XC / R evaluated at f0. Because XL and XC are equal in magnitude at resonance, either may be used. This definition reflects how large the circulating reactive currents are compared to the resistive drop. Step-by-Step Solution: 1) Use Q = X / R at resonance.2) Substitute X = 4000 Ω and R = 50 Ω.3) Compute Q = 4000 / 50 = 80. Verification / Alternative check: Bandwidth relation for series RLC: Q = f0 / BW. A Q of 80 implies BW ≈ 6000 / 80 = 75 Hz, consistent with a sharp resonance for a relatively small R compared to reactance. Why Other Options Are Wrong: 0.001: Orders of magnitude too small; contradicts X ≫ R.50: Would imply X ≈ 2500 Ω with R = 50 Ω, not given here.4.0: Would require X ≈ 200 Ω at resonance, not 4000 Ω. Common Pitfalls: Mixing parallel and series definitions of Q, or mistakenly using total |Z| instead of the reactive magnitude at resonance. Final Answer: 80

Q: In a series RLC circuit driven by an AC source of RMS voltage V, the current magnitude can be determined using which relationship?

Introduction: Finding current in a series RLC circuit is a fundamental task in AC analysis. Because resistance and reactance combine vectorially, the current cannot be obtained from the resistance alone unless the reactance is zero. This question ensures familiarity with impedance-based current calculation. Given Data / Assumptions: Series RLC circuit, sinusoidal steady state. Source RMS voltage is V. Impedance magnitude is Z = sqrt(R^2 + (XL − XC)^2). Concept / Approach: Ohm’s law in AC form uses complex impedance: I = V / Z (phasor form). For magnitudes, I = |V| / |Z|. In a series RLC, Z depends on both resistance and the net reactance (XL − XC). Therefore, current is determined by the total impedance, not just R or a simple sum of reactances. Step-by-Step Solution: 1) Compute reactances: XL = 2 * pi * f * L, XC = 1 / (2 * pi * f * C).2) Find net reactance: X = XL − XC.3) Calculate impedance magnitude: Z = sqrt(R^2 + X^2).4) Apply AC Ohm’s law for magnitude: I = V / Z. Verification / Alternative check: At resonance (XL = XC), Z reduces to R, and the formula becomes I = V / R, which is a special case consistent with I = V / Z. Why Other Options Are Wrong: I = V / R: true only when XL = XC (resonance) or when reactance is negligible; not generally correct.I = V / (XL + XC): ignores the vector difference and the resistance component; not an impedance magnitude.I = V * Z: dimensionally incorrect; would increase current with impedance, which contradicts physics. Common Pitfalls: Adding reactances algebraically without considering phase, or forgetting to include resistance in the impedance calculation. Final Answer: I = V / Z

Q: Resonant frequency claim (repaired) — “The resonant frequency f0 is 19 kHz.” With no component values (L and C) or circuit topology given, can this numeric claim be validated?

Introduction / Context: Resonance in RLC networks is a precise function of inductance and capacitance (and topology). Without actual component values or a schematic, any numerical statement such as “f0 is 19 kHz” cannot be verified. This repaired question probes whether you rely on the correct formula rather than accepting an unsupported number. Given Data / Assumptions: No values for L (inductance) or C (capacitance) are specified. No statement of whether the circuit is series or parallel resonance. We assume standard linear, time-invariant components. Concept / Approach: For both series and ideal parallel RLC circuits, the undamped resonant frequency is f0 = 1 / (2 * π * sqrt(L * C)). Resistance affects bandwidth and peak amplitudes but does not set the ideal f0 value by itself. Hence, without L and C, there is no way to compute f0 and compare it with 19 kHz. Step-by-Step Solution: 1) Write the resonance relation: f0 = 1 / (2π√(L*C)).2) Observe the data gap: neither L nor C is provided.3) Conclude that the numeric claim cannot be validated.4) Therefore the only justified choice is that more information is required. Verification / Alternative check: Any attempt to “guess” f0 requires assumed L and C; changing either by a factor of 4 changes f0 by a factor of 2, illustrating the sensitivity and the impossibility of validation here. Why Other Options Are Wrong: “True/False always”: resonance is not a fixed universal value; it depends on L and C.“True only for DC” and “resistance alone sets resonance”: both misunderstand resonance basics. Common Pitfalls: Memorizing example frequencies from textbooks and assuming they generalize without the underlying component values. Final Answer: Cannot be determined from the information provided

Q: Series RLC reactance — when both inductive reactance (XL) and capacitive reactance (XC) are present in the same series circuit, is the magnitude of total reactance equal to the sum of their magnitudes?

Introduction / Context: Understanding how reactances combine in series is essential for predicting a circuit’s impedance and phase. This question distinguishes the correct rule for combining inductive and capacitive reactances in a series path. Given Data / Assumptions: Single series path containing L and C (and possibly R). Steady-state sinusoidal operation at angular frequency ω. Concept / Approach: In series, reactances add algebraically: X_total = XL + (−XC). Therefore, X_total = XL − XC. The magnitude is |X_total| = |XL − XC|, not the sum of magnitudes. At resonance, XL = XC and the net reactance magnitude becomes zero (ideal case), leaving purely resistive impedance. Step-by-Step Solution: 1) Compute XL = ωL and XC = 1/(ωC).2) Sum algebraically: X_total = XL − XC.3) Take magnitude: |X| = |XL − XC|.4) Interpret: if XL > XC, net is inductive; if XC > XL, net is capacitive. Verification / Alternative check: Bode/phasor diagrams show the vector subtraction along the imaginary axis, not a scalar sum of lengths. Why Other Options Are Wrong: Sum of magnitudes: overestimates reactance and ignores the opposite signs of XL and XC.“Only at resonance/only for small R/ideal inductors”: the combination rule is general for sinusoidal steady state. Common Pitfalls: Adding magnitudes instead of algebraic quantities; forgetting that capacitive reactance is negative in phasor notation. Final Answer: False — |X| = |XL − XC|

Question 1

A 60 Hz series RLC circuit has XL = 7.5 ohms, XC = 265.3 ohms, and R = 33 ohms. What is the magnitude of the total impedance Z?

Accepted Answer

Introduction: This problem checks application of series RLC impedance calculation. In a series RLC, the total impedance combines resistance and net reactance vectorially, so you must compute the net reactance first and then the magnitude of Z using the Pythagorean relationship. Given Data / Assumptions: Series RLC at 60 Hz. R = 33 ohms. XL = 7.5 ohms. XC = 265.3 ohms. Concept / Approach: The net reactance is X = XL − XC. The total impedance for a series circuit is Z = R + jX, whose magnitude is |Z| = sqrt(R^2 + X^2). Because XC is much larger than XL here, the circuit is net capacitive and the phase is negative; however, the question asks only for the magnitude of Z. Step-by-Step Solution: Compute X = XL − XC = 7.5 − 265.3 = −257.8 ohms.Form complex impedance Z = 33 + j(−257.8) ohms.Compute magnitude |Z| = sqrt(R^2 + X^2) = sqrt(33^2 + 257.8^2) ohms.|Z| ≈ sqrt(1089 + 66457) ≈ sqrt(67546) ≈ 259.9 ohms. Verification / Alternative check: Since |X| ≫ R, |Z| should be close to |X|. The answer 259.9 ohms is slightly larger than 257.8 ohms due to the added resistive component, which is consistent. Why Other Options Are Wrong: 257.8 ohms: That is |X| alone, not the total |Z|. 290.8 ohms: Overestimates the vector sum; not supported by calculation. 1989.75 ohms: Numerically unrelated to any correct combination here. Common Pitfalls: Using arithmetic addition instead of vector magnitude, forgetting that X can be negative (capacitive), or substituting XL + XC instead of XL − XC for series networks. Final Answer: 259.9 ohms.

Question 2

In a series RLC circuit, the approximate phase angle when VC = 117 V, VR = 14.5 V, and VL = 3.3 V is closest to which of the following?

Accepted Answer

Introduction: This problem assesses phasor reasoning in a series RLC circuit using the individual component voltage drops. The sign and magnitude of the phase angle come from the balance between inductive and capacitive reactances relative to the resistive drop. Given Data / Assumptions: Series RLC circuit under sinusoidal steady state. Measured voltages: VC = 117 V, VR = 14.5 V, VL = 3.3 V. We seek the overall phase angle phi of the current relative to the source voltage. Concept / Approach: In series RLC, voltages across R, L, and C are mutually perpendicular in the phasor diagram: VR is in phase with current, VL leads current by 90 degrees, and VC lags current by 90 degrees. The reactive phasor is VL − VC, so the tangent of the phase angle is (VL − VC) / VR (sign indicates inductive or capacitive behavior). Step-by-Step Solution: Compute net reactive drop: Vx = VL − VC = 3.3 − 117 = −113.7 V.Compute tangent of phase angle: tan(phi) = Vx / VR = −113.7 / 14.5 ≈ −7.841.Find phi = arctan(−7.841) ≈ −82.7 degrees (capacitive, since negative).Hence the circuit is strongly capacitive and current leads the source voltage by about 82.7 degrees (equivalently, source voltage lags current). Verification / Alternative check: The source voltage magnitude would be Vs ≈ sqrt(VR^2 + (VL − VC)^2) ≈ sqrt(14.5^2 + 113.7^2), consistent with a very large reactive component compared with resistive drop, supporting a phase angle magnitude near 90 degrees but not exactly 90 degrees. Why Other Options Are Wrong: –45.0 degrees: Would imply comparable reactive and resistive drops; contradicted by the much larger |VL − VC| compared to VR. –90.0 degrees: Ideal limit for purely capacitive with negligible resistance; here VR is small but nonzero, so angle is slightly less in magnitude. –172.7 degrees: Physically inconsistent for a simple series RLC driven by a single-frequency source. Common Pitfalls: Adding VL and VC instead of subtracting them, ignoring the sign (which determines lead/lag), or computing tan(phi) using component reactances directly without relating them to the measured voltage drops. Final Answer: –82.7 degrees.

Question 3

For a 100 Hz series RLC circuit with source voltage VS = 20 V, total resistance RT = 66 ohms, and total reactance XT = 47 ohms, what is the circuit current magnitude I?

Accepted Answer

Introduction: This question tests the ability to compute current in a series RLC circuit when total resistance and total reactance are known. The key step is forming the magnitude of the total impedance and then applying Ohm's law with RMS quantities. Given Data / Assumptions: VS = 20 V (RMS). RT = 66 ohms. XT = 47 ohms (net reactance, sign not needed for magnitude). Frequency is 100 Hz; frequency value is not needed once RT and XT are given. Concept / Approach: For a series circuit, total impedance is Z = RT + j * XT with magnitude |Z| = sqrt(RT^2 + XT^2). The RMS current is I = VS / |Z|. If the sign of XT were required for phase, it would affect the angle but not the magnitude of current. Step-by-Step Solution: Compute |Z| = sqrt(RT^2 + XT^2) = sqrt(66^2 + 47^2) ohms.66^2 = 4356; 47^2 = 2209; sum = 6565.|Z| = sqrt(6565) ≈ 81.02 ohms.Compute I = VS / |Z| = 20 / 81.02 ≈ 0.247 A.Express in milliamperes: I ≈ 247 mA. Verification / Alternative check: A quick sanity check: with RT alone, current would be 20 / 66 ≈ 0.303 A. Adding reactance increases |Z| to about 81 ohms, reducing current to about 0.25 A, which is consistent with the detailed calculation. Why Other Options Are Wrong: 1.05 A: Far too high; would require |Z| ≈ 19 ohms, not supported by data. 303 mA: This is 20 / 66 and ignores the reactance contribution. 107 mA: Too low; would imply |Z| ≈ 187 ohms, inconsistent with RT and XT provided. Common Pitfalls: Adding RT and XT arithmetically instead of vectorially, forgetting to take the square root in |Z|, or mixing up RMS and peak values. Also, do not confuse the sign of XT; it affects phase, not current magnitude. Final Answer: 247 mA.

Question 4

In a 30 V series RLC circuit, the total impedance is Z = 20 Ω and the resistance is R = 10 Ω. What is the true (real) power consumed by the circuit under these conditions?

Accepted Answer

Introduction: True power in AC circuits refers to the real power dissipated as heat in resistive elements. In a series RLC circuit, reactive components (inductor and capacitor) exchange energy with the source but do not consume net real power over a cycle. This problem tests using impedance, current, and power relationships to compute real power. Given Data / Assumptions: Source RMS voltage: V = 30 V. Total series impedance magnitude: Z = 20 Ω. Resistive part: R = 10 Ω. Ideal sinusoidal steady state; reactive elements consume no average real power. Concept / Approach: Key formulas: current magnitude I = V / Z. Real (true) power P is given by P = I^2 * R = V * I * cos(phi). In a series RLC, cos(phi) = R / Z, so an equivalent single-step form is P = V^2 * (R / Z^2). Any of these consistent formulas will yield the same result. Step-by-Step Solution: 1) Compute current: I = V / Z = 30 / 20 = 1.5 A.2) Use P = I^2 * R = (1.5)^2 * 10 = 2.25 * 10 = 22.5 W.3) Cross-check with P = V^2 * (R / Z^2) = 30^2 * (10 / 20^2) = 900 * (10 / 400) = 900 * 0.025 = 22.5 W. Verification / Alternative check: Power factor cos(phi) = R / Z = 10 / 20 = 0.5. Then P = V * I * cos(phi) = 30 * 1.5 * 0.5 = 22.5 W, confirming consistency among methods. Why Other Options Are Wrong: 15.0 watts: would require a smaller current or resistance; not supported by the given numbers.30.0 watts: equals V * I without power factor; ignores that cos(phi) < 1.45.0 watts: equals V * I with cos(phi) = 1; impossible because reactive components make cos(phi) < 1. Common Pitfalls: Forgetting to include power factor or mixing instantaneous values with RMS. Another frequent error is using Z instead of R in P = I^2 * R; only the resistive part dissipates real power. Final Answer: 22.5 watts

Question 5

In a series RLC circuit operating exactly at its resonant frequency, what is the measured net voltage across the two reactive components considered together (inductor in series with capacitor)?

Accepted Answer

Introduction: Series resonance is a cornerstone concept in AC circuit theory. At resonance, inductive and capacitive reactances are equal in magnitude and opposite in sign. This question examines what happens to the combined voltage across the series inductor and capacitor when the circuit is exactly at resonance. Given Data / Assumptions: Series RLC topology. Operating frequency equals the resonant frequency f0. Ideal components; steady-state sinusoidal excitation. Concept / Approach: At resonance, XL = 2 * pi * f0 * L and XC = 1 / (2 * pi * f0 * C) satisfy XL = XC. Their phasor voltages are equal in magnitude but 180 degrees out of phase. Therefore, the phasor sum across L and C is zero even though each element may have a large individual voltage. Step-by-Step Solution: 1) Recognize resonance condition: XL = XC.2) Phasor voltages: VL leads current by 90 degrees; VC lags current by 90 degrees.3) Since |VL| = |VC| and their phases are opposite, VL + VC = 0 in phasor terms.4) Hence, the net measured voltage across L and C together is zero. Verification / Alternative check: Impedance perspective: ZLC = jXL + (−jXC) = j(XL − XC) = 0 at resonance. A zero reactive impedance means zero net reactive drop across the pair. Why Other Options Are Wrong: applied: The applied source voltage appears across the series combination of R, L, and C; the net across L and C alone is zero at resonance.reactive: Not a magnitude; the correct magnitude of the net is zero.inductive and capacitive: Descriptive words, not the value; they cancel at resonance. Common Pitfalls: Confusing individual element voltages (which can be high with large Q) with their vector sum, which cancels at true resonance. Final Answer: zero

Question 6

At resonant frequency in a series RLC circuit, what is the net voltage across the two reactive components (inductor and capacitor) when measured together?

Accepted Answer

Introduction: Understanding voltage distribution at resonance helps in filter design and impedance matching. This question reinforces the vector cancellation of reactive drops in a resonant series RLC circuit. Given Data / Assumptions: Series RLC circuit operating at resonance. Ideal sinusoidal steady-state behavior. Phasor analysis applies. Concept / Approach: At resonance, XL = XC. The inductor voltage phasor leads the current by 90 degrees; the capacitor voltage phasor lags by 90 degrees. Equal magnitudes and opposite phases yield a phasor sum of zero across L and C together. Step-by-Step Solution: 1) Write XL = 2 * pi * f0 * L and XC = 1 / (2 * pi * f0 * C).2) At resonance: XL − XC = 0.3) Phasor addition: VL + VC = 0 (equal magnitude, 180-degree phase difference).4) Therefore, the net voltage across the combined series reactances is zero. Verification / Alternative check: Impedance view: ZLC = jXL + (−jXC) = j(XL − XC) = 0 at f0, so there is no reactive drop across L + C as a pair. Why Other Options Are Wrong: Applied voltage: At resonance most of the source voltage appears across R, not across L + C together.Reactive voltage: Not a numeric value; net reactive drop is zero.VL + VC voltage: The algebraic (phasor) sum equals zero, not a finite nonzero amount. Common Pitfalls: Assuming large circulating voltages imply a large net drop; in reality, large but opposite voltages cancel in phasor sum. Final Answer: Zero voltage

Question 7

A series circuit resonates at 6 kHz. If at resonance the inductor reactance and capacitor reactance are each 4 kΩ in magnitude and the series resistance is R = 50 Ω, what is the quality factor Q of the circuit?

Accepted Answer

Introduction: The quality factor Q characterizes the sharpness of resonance and the ratio of stored to dissipated energy per cycle. For a series RLC circuit, Q directly relates reactive impedance to resistance at the resonant frequency. This question checks the simple but crucial Q = X / R relationship at resonance. Given Data / Assumptions: Series RLC at resonance f0 = 6 kHz (frequency value not needed for Q if X is given). Inductive and capacitive reactance magnitudes: XL = XC = 4 kΩ. Series resistance: R = 50 Ω. Idealized small-signal behavior. Concept / Approach: For series resonance, Q = XL / R = XC / R evaluated at f0. Because XL and XC are equal in magnitude at resonance, either may be used. This definition reflects how large the circulating reactive currents are compared to the resistive drop. Step-by-Step Solution: 1) Use Q = X / R at resonance.2) Substitute X = 4000 Ω and R = 50 Ω.3) Compute Q = 4000 / 50 = 80. Verification / Alternative check: Bandwidth relation for series RLC: Q = f0 / BW. A Q of 80 implies BW ≈ 6000 / 80 = 75 Hz, consistent with a sharp resonance for a relatively small R compared to reactance. Why Other Options Are Wrong: 0.001: Orders of magnitude too small; contradicts X ≫ R.50: Would imply X ≈ 2500 Ω with R = 50 Ω, not given here.4.0: Would require X ≈ 200 Ω at resonance, not 4000 Ω. Common Pitfalls: Mixing parallel and series definitions of Q, or mistakenly using total |Z| instead of the reactive magnitude at resonance. Final Answer: 80

Question 8

In a series RLC circuit driven by an AC source of RMS voltage V, the current magnitude can be determined using which relationship?

Accepted Answer

Introduction: Finding current in a series RLC circuit is a fundamental task in AC analysis. Because resistance and reactance combine vectorially, the current cannot be obtained from the resistance alone unless the reactance is zero. This question ensures familiarity with impedance-based current calculation. Given Data / Assumptions: Series RLC circuit, sinusoidal steady state. Source RMS voltage is V. Impedance magnitude is Z = sqrt(R^2 + (XL − XC)^2). Concept / Approach: Ohm’s law in AC form uses complex impedance: I = V / Z (phasor form). For magnitudes, I = |V| / |Z|. In a series RLC, Z depends on both resistance and the net reactance (XL − XC). Therefore, current is determined by the total impedance, not just R or a simple sum of reactances. Step-by-Step Solution: 1) Compute reactances: XL = 2 * pi * f * L, XC = 1 / (2 * pi * f * C).2) Find net reactance: X = XL − XC.3) Calculate impedance magnitude: Z = sqrt(R^2 + X^2).4) Apply AC Ohm’s law for magnitude: I = V / Z. Verification / Alternative check: At resonance (XL = XC), Z reduces to R, and the formula becomes I = V / R, which is a special case consistent with I = V / Z. Why Other Options Are Wrong: I = V / R: true only when XL = XC (resonance) or when reactance is negligible; not generally correct.I = V / (XL + XC): ignores the vector difference and the resistance component; not an impedance magnitude.I = V * Z: dimensionally incorrect; would increase current with impedance, which contradicts physics. Common Pitfalls: Adding reactances algebraically without considering phase, or forgetting to include resistance in the impedance calculation. Final Answer: I = V / Z

Question 9

Resonant frequency claim (repaired) — “The resonant frequency f0 is 19 kHz.” With no component values (L and C) or circuit topology given, can this numeric claim be validated?

Accepted Answer

Introduction / Context: Resonance in RLC networks is a precise function of inductance and capacitance (and topology). Without actual component values or a schematic, any numerical statement such as “f0 is 19 kHz” cannot be verified. This repaired question probes whether you rely on the correct formula rather than accepting an unsupported number. Given Data / Assumptions: No values for L (inductance) or C (capacitance) are specified. No statement of whether the circuit is series or parallel resonance. We assume standard linear, time-invariant components. Concept / Approach: For both series and ideal parallel RLC circuits, the undamped resonant frequency is f0 = 1 / (2 * π * sqrt(L * C)). Resistance affects bandwidth and peak amplitudes but does not set the ideal f0 value by itself. Hence, without L and C, there is no way to compute f0 and compare it with 19 kHz. Step-by-Step Solution: 1) Write the resonance relation: f0 = 1 / (2π√(L*C)).2) Observe the data gap: neither L nor C is provided.3) Conclude that the numeric claim cannot be validated.4) Therefore the only justified choice is that more information is required. Verification / Alternative check: Any attempt to “guess” f0 requires assumed L and C; changing either by a factor of 4 changes f0 by a factor of 2, illustrating the sensitivity and the impossibility of validation here. Why Other Options Are Wrong: “True/False always”: resonance is not a fixed universal value; it depends on L and C.“True only for DC” and “resistance alone sets resonance”: both misunderstand resonance basics. Common Pitfalls: Memorizing example frequencies from textbooks and assuming they generalize without the underlying component values. Final Answer: Cannot be determined from the information provided

Question 10

Series RLC reactance — when both inductive reactance (XL) and capacitive reactance (XC) are present in the same series circuit, is the magnitude of total reactance equal to the sum of their magnitudes?

Accepted Answer

Introduction / Context: Understanding how reactances combine in series is essential for predicting a circuit’s impedance and phase. This question distinguishes the correct rule for combining inductive and capacitive reactances in a series path. Given Data / Assumptions: Single series path containing L and C (and possibly R). Steady-state sinusoidal operation at angular frequency ω. Concept / Approach: In series, reactances add algebraically: X_total = XL + (−XC). Therefore, X_total = XL − XC. The magnitude is |X_total| = |XL − XC|, not the sum of magnitudes. At resonance, XL = XC and the net reactance magnitude becomes zero (ideal case), leaving purely resistive impedance. Step-by-Step Solution: 1) Compute XL = ωL and XC = 1/(ωC).2) Sum algebraically: X_total = XL − XC.3) Take magnitude: |X| = |XL − XC|.4) Interpret: if XL > XC, net is inductive; if XC > XL, net is capacitive. Verification / Alternative check: Bode/phasor diagrams show the vector subtraction along the imaginary axis, not a scalar sum of lengths. Why Other Options Are Wrong: Sum of magnitudes: overestimates reactance and ignores the opposite signs of XL and XC.“Only at resonance/only for small R/ideal inductors”: the combination rule is general for sinusoidal steady state. Common Pitfalls: Adding magnitudes instead of algebraic quantities; forgetting that capacitive reactance is negative in phasor notation. Final Answer: False — |X| = |XL − XC|

Question 11

Parallel resonance quality factor — is it correct that the Q of a parallel resonant circuit at resonance equals VC/VS or VL/VS (ratio of reactive branch voltage to source voltage)?

Accepted Answer

Introduction / Context: Quality factor (Q) measures selectivity and energy storage relative to loss. Its definition and practical expressions differ between series and parallel resonant circuits. This item checks a common misconception about using a simple voltage ratio for Q in parallel resonance. Given Data / Assumptions: Parallel RLC at resonance (admittance is purely real). Reactive branch voltages equal the applied source voltage in parallel topology. Realistic losses exist in R, L, and C. Concept / Approach: In a series resonant circuit, reactive element voltages can be much larger than the source voltage, and ratios like VL/VR can relate to Q (e.g., Q = XL/R). In a parallel resonant circuit, each branch is directly across the source, so VC ≈ VS and VL ≈ VS at resonance; the ratios VC/VS or VL/VS are about 1, not equal to Q. For parallel resonance, Q is more naturally expressed via currents (reactive branch current to source current) or via reactive/real power in the tank. Step-by-Step Solution: 1) Recognize parallel topology: branch voltages equal source voltage.2) Therefore VC/VS ≈ 1 and VL/VS ≈ 1 at resonance.3) Q in parallel is often Q = I_reactive / I_source (or related to susceptance vs conductance).4) Conclude the proposed voltage ratios do not represent Q for parallel circuits. Verification / Alternative check: Tank circuits exhibit large circulating reactive currents at resonance while source current is minimal; this current ratio aligns with Q, not a voltage ratio of ~1. Why Other Options Are Wrong: “True/True only…”: conflict with parallel topology where branch voltages equal the source.“Depends solely on ESR”: Q depends on aggregate losses, not exclusively capacitor ESR. Common Pitfalls: Transferring series-circuit voltage relationships directly to parallel tanks. Final Answer: False

Question 12

Filter selectivity — “The less selective a filter is, the steeper the slope (sharper the roll-off) at its cutoff frequencies.” Is this statement accurate?

Accepted Answer

Introduction / Context: Selectivity describes how well a filter separates passband from stopband. It is tightly linked to roll-off steepness, transition bandwidth, and quality factor (Q). This question tests whether you can correctly relate selectivity and slope at the cutoff region. Given Data / Assumptions: Standard low-pass/high-pass/band-pass/band-stop filters. Slope discussed in dB/decade or dB/octave per order. Concept / Approach: Greater selectivity corresponds to sharper transition regions and steeper skirts. Higher-order filters (or higher Q for resonant responses) achieve steeper roll-off, which improves selectivity. Therefore, a filter that is “less selective” has gentler slopes and broader transition regions, not steeper ones. Step-by-Step Solution: 1) Define selectivity: ability to attenuate unwanted frequencies near the cutoff.2) Recognize slope per order: first order ~20 dB/decade, second order ~40 dB/decade, etc.3) Higher order → steeper slope → more selective, not less.4) Conclude the statement as written is false. Verification / Alternative check: Bode plots of Butterworth/Chebyshev/Elliptic filters show that tighter selectivity comes with steeper roll-off (or ripples/zeros) near cutoff. Why Other Options Are Wrong: “True/only for active/only below 100 Hz/only for high-Q”: selectivity principles are frequency-scale invariant and do not depend on “active vs passive” per se (though implementation choices affect practical limits). Common Pitfalls: Confusing passband flatness with selectivity; assuming “less selective” equals “more slope,” which reverses the real relationship. Final Answer: False

Question 13

Series RLC above resonance — “In a series RLC circuit, at frequencies above resonance the circuit behaves more capacitive than inductive.” Is this correct?

Accepted Answer

Introduction / Context: Knowing how a series RLC behaves relative to its resonant frequency helps in filter and matching design. The sign of net reactance determines whether current leads or lags voltage and which component dominates energy storage. Given Data / Assumptions: Series RLC (single current path). Sinusoidal steady state. Concept / Approach: Net series reactance is X = XL − XC, where XL = ωL and XC = 1/(ωC). At resonance, XL = XC and X = 0 (purely resistive). For frequencies above resonance, ω is larger, so XL grows while XC shrinks; therefore XL > XC and X is positive (inductive). Thus the circuit is more inductive above resonance and more capacitive below resonance. Step-by-Step Solution: 1) Identify resonance condition: XL = XC.2) Increase frequency above f0: XL increases, XC decreases.3) Compute X = XL − XC > 0 → inductive behavior.4) Conclude the statement is false. Verification / Alternative check: Phasor diagrams show current lagging voltage above resonance (inductive), and leading below resonance (capacitive). Why Other Options Are Wrong: Any “True” variants contradict the algebraic combination of reactances for series RLC. Common Pitfalls: Remembering the rule in reverse; mixing series with parallel behavior. Final Answer: False

Question 14

Filter bandwidth — by definition, the bandwidth (BW) of a band-pass (or band-stop) filter is the span between its two half-power (−3 dB) frequencies. Is this correct?

Accepted Answer

Introduction / Context: Bandwidth is a fundamental descriptor of how wide a filter’s pass (or stop) region extends around a center frequency. The industry-standard definition uses the half-power (−3 dB) points for linear time-invariant systems unless otherwise specified. Given Data / Assumptions: Band-pass or band-stop context. Power proportional to voltage squared in constant-impedance systems. Half-power corresponds to 1/√2 in voltage magnitude (−3 dB). Concept / Approach: The half-power frequencies f1 and f2 are where the output power is half the peak (or where |H(jω)| drops by 3 dB from the reference). The bandwidth BW is defined as f2 − f1. This convention is ubiquitous in communications and filter design, enabling consistent comparison of selectivity and Q (Q ≈ f0 / BW for narrowband band-pass filters). Step-by-Step Solution: 1) Identify f1 and f2 as the −3 dB points around the center.2) Compute BW = f2 − f1.3) For narrowband, Q ≈ f0 / BW; for broader responses, use exact definitions.4) Conclude that the statement is correct as a general definition. Verification / Alternative check: Standard texts and datasheets define bandwidth using −3 dB points unless a different metric (e.g., ripple limits) is explicitly stated. Why Other Options Are Wrong: “False/only Butterworth/only high Q/depends solely on ripple” are inconsistent with the general, technology-neutral definition. Common Pitfalls: Confusing −3 dB (voltage) with absolute voltage drop; mixing alternative definitions used in specific standards without noting the context. Final Answer: True

Question 15

Operating relative to resonance (repaired) — “With the stated frequency, the given circuit is operating above resonance.” Without a schematic and component values, can we determine whether operation is above resonance?

Accepted Answer

Introduction / Context: Determining whether an RLC circuit operates above or below resonance requires knowing the actual resonant frequency f0 and the applied frequency f. Without the circuit values or topology, any claim about “above resonance” is unsupported. This repaired stem evaluates your reasoning about required data. Given Data / Assumptions: No values for L and C are given; R may be present but does not set f0. Topology (series vs parallel) is unspecified. A single operating frequency is referenced but not related to f0. Concept / Approach: Resonant frequency is f0 = 1/(2π√(L*C)) for ideal RLC (series or parallel). To decide if operation is “above resonance,” you must compare f to f0. Without L and C, f0 is unknown; thus the statement cannot be validated or refuted objectively. Step-by-Step Solution: 1) Recognize that deciding “above/below” requires both f and f0.2) Note that f0 depends on L and C only (ideal), not R.3) Since L and C are missing, f0 is unknown.4) Therefore, the claim cannot be determined from the given data. Verification / Alternative check: Example: if L and C produce f0 = 10 kHz, then f = 20 kHz is above resonance; if L and C produce f0 = 100 kHz, the same f = 20 kHz is below resonance. The conclusion flips with different L and C. Why Other Options Are Wrong: “Always above/below”: incorrect generalizations.Topology-based answers: both series and parallel use the same f0 relation.“Resistance sets resonance”: f0 is not set by R. Common Pitfalls: Assuming “high frequency” implies “above resonance” without checking component values. Final Answer: Cannot be determined from the information provided

Question 16

Output tapping and filter type (repaired) — for a series RLC network driven by a source, if the output is taken across the resistor (R), does the resulting two-port behave as a band-pass filter?

Accepted Answer

Introduction / Context: Filter behavior in simple RLC networks depends on where you probe or take the output. A single series RLC path can realize different transfer functions depending on whether the output is measured across L, C, or R. This question clarifies the case when output is taken across R. Given Data / Assumptions: Single series RLC driven by a voltage source. Output is the voltage across the resistor, VR. Linear time-invariant conditions; moderate Q. Concept / Approach: The magnitude of VR is low at both very low and very high frequencies: far below resonance, current is limited by large XC; far above resonance, current is limited by large XL. At resonance (where XL = XC), impedance is minimum and current peaks, so VR = I * R peaks. This “low at extremes, high at center” behavior is characteristic of a band-pass filter centered at f0. Step-by-Step Solution: 1) Write |I(jω)| = |V_source| / |Z_series|, with Z_series minimum at resonance.2) Compute VR = I * R; since I peaks at f0, so does VR.3) Observe that VR diminishes away from f0 in both directions.4) Conclude that the transfer |VR/Vs| exhibits a band-pass profile. Verification / Alternative check: Bode plots for |VR/Vs| in series RLC show a peak at f0 with attenuation on both sides, confirming band-pass behavior. Why Other Options Are Wrong: “False” or “always low-pass”: contradicts the current-peaking nature at resonance when measuring across R.“Only if Q>5/ideal components”: while Q affects sharpness and peak gain, the qualitative band-pass shape persists for finite Q with real components. Common Pitfalls: Assuming the network type (LPF/HPF/BPF) is intrinsic to the parts rather than the measurement node. Final Answer: True

Question 17

Impedance value claim (repaired) — “The impedance is 1177 Ω.” Without a schematic, frequency, and component values, can this numeric impedance be verified?

Accepted Answer

Introduction / Context: Impedance (Z) in AC circuits depends on frequency and component values. A bare numerical claim like “Z = 1177 Ω” is meaningless without context. This repaired item checks whether you recognize the missing data necessary to compute or confirm impedance. Given Data / Assumptions: No circuit diagram or element values (R, L, C) are supplied. No frequency is given; reactances depend on ω. We assume linear components; parasitics ignored. Concept / Approach: For a general network, Z(jω) = R + j(XL − XC) for simple series RLC, or more complex functions for other topologies. The magnitude |Z| and phase ∠Z vary with ω. Without ω and component values, any specific value such as 1177 Ω cannot be computed or validated. Therefore, the only defensible choice is that the claim cannot be determined from the given information. Step-by-Step Solution: 1) Recognize that impedance is frequency-dependent whenever L or C is present.2) Note that even for purely resistive networks, you still need resistor values and connections to compute Z.3) Since neither the schematic nor the values nor the frequency are given, Z is indeterminate.4) Conclude: the numeric claim cannot be verified. Verification / Alternative check: Provide any two different hypothetical sets of R, L, C and frequencies that yield different |Z| values; the same number 1177 Ω will not be universally correct. Why Other Options Are Wrong: “Always 1177 Ω / always resistive”: physically incorrect generalizations.“True only for DC”: still need resistor values and topology.“False — cannot be 1177 Ω”: it could be 1177 Ω for some specific case; we simply lack the data to assert it. Common Pitfalls: Accepting numeric statements without verifying the underlying parameters; forgetting the role of frequency in impedance. Final Answer: Cannot be determined from the information provided

Question 18

Series RLC — same current in all series elements: Assume a single-loop series RLC circuit in steady-state AC where a clamp-on ammeter at the source lead reads 5 A (rms). What is the inductor current (rms)?

Accepted Answer

Introduction / Context: In any single-loop series circuit (R, L, and/or C in series), the same instantaneous current flows through every element because there is only one current path. This question tests the core idea of series topology and how it applies to an inductor’s branch current reading. Given Data / Assumptions: Single-loop series RLC driven by a sinusoidal source. The measured circuit (source-lead) current is 5 A rms. Ideal lumped elements; parasitics and measurement burden are negligible. Concept / Approach: Kirchhoff’s Current Law applied to a series loop implies a single branch; therefore the current is identical through R, L, and C at every instant. Phasor relationships change how voltage distributes (phase shifts), but not the series current equality. Thus the inductor current must equal the source current. Step-by-Step Solution: Recognize a single series path → same current in all components.Given I_source(rms) = 5 A → I_L(rms) = 5 A by series rule.Voltage magnitudes will differ (V_R = IR, V_L = IX_L, V_C = I*X_C), but the current is common. Verification / Alternative check: Measure with an ammeter inserted in series before the inductor and again between L and C; both readings match the source current (within instrument tolerance). Why Other Options Are Wrong: “Depends only on L” or “greater than 5 A”: violates series current continuity.“Zero at any frequency”: only true if the loop is open; here the loop is closed.“Cannot be determined”: the series rule determines it directly. Common Pitfalls: Confusing series (same current) with parallel (same voltage). Also, mixing up current magnitude with reactive voltage magnitudes, which can be large at resonance without changing current equality. Final Answer: 5 A

Question 19

Resonance in a series RLC — total current behavior: If the given circuit operates exactly at series resonance, is the total current drawn from the source at its maximum value (for a fixed source voltage)?

Accepted Answer

Introduction / Context: Series resonance is a cornerstone concept in AC circuit theory and filter/tank design. It occurs when the inductive reactance equals the capacitive reactance in magnitude (X_L = X_C), causing reactances to cancel so the impedance is purely resistive and minimal. Given Data / Assumptions: Series RLC circuit driven by a fixed RMS voltage. Ideal L and C, with a finite series resistance R. Frequency is adjustable and can be set to the resonant frequency. Concept / Approach: Impedance of a series RLC is Z = R + j(X_L − X_C). At resonance, X_L = X_C → Z = R (minimum magnitude). Ohm’s law shows I = V/Z is largest when |Z| is smallest (for fixed V), hence current peaks at resonance. Large reactive voltages may appear across L and C individually, but the net reactive effect in the loop is zero at resonance. Step-by-Step Solution: Write Z(f) = √(R^2 + (X_L − X_C)^2).At f = f_0, X_L = X_C → |Z| = R (minimum possible for that circuit).Thus I(f_0) = V/R is the maximum attainable loop current for that applied V. Verification / Alternative check: Plot I versus frequency (a resonance curve). The peak occurs at f_0. Bandwidth relates to quality factor Q = X_L/R at resonance. Why Other Options Are Wrong: “Incorrect”: contradicts Z minimization at resonance.“Only if R = 0”: even with finite R, current is maximal at resonance for that V.“Parallel resonance”: refers to a different topology where current dips in the feed branch.“XL = 2XC”: not the resonance condition. Common Pitfalls: Confusing series and parallel resonance characteristics; focusing on large VL or VC and missing that total impedance is minimized, not maximized. Final Answer: Correct

Question 20

Resonance duplicate (clarified wording): For a series RLC circuit at its resonant frequency (where XL = XC), the total current from a fixed-voltage source reaches a maximum. Is this statement accurate?

Accepted Answer

Introduction / Context: This item reaffirms the behavior of a series RLC at resonance, emphasizing the equality of reactances and the resulting purely resistive impedance that dictates the current peak for a given source voltage. Given Data / Assumptions: Series RLC network, sinusoidal steady state. Source RMS voltage held constant. Losses modeled by a series resistance R. Concept / Approach: At resonance, |Z| = R (minimum), so I = V/|Z| is largest. Although VL and VC can be large and cancel phasorially, they do not increase |Z| at resonance; instead, they produce internal voltage magnification characterized by Q, while the source sees a small resistive |Z|. Step-by-Step Solution: Impose XL = XC → Z = R.Compare current off-resonance: |Z| grows as |XL − XC| increases → current falls.Therefore, the maximum current occurs at the resonant frequency. Verification / Alternative check: Measure current while sweeping frequency; the plot yields a bell-shaped resonance curve with peak at f0 and −3 dB bandwidth linked to Q. Why Other Options Are Wrong: Requiring L = C numerically is irrelevant; resonance depends on ωL = 1/(ωC).Parallel resonance property is different (feed current minimum).Q affects sharpness of the peak, not whether a peak exists. Common Pitfalls: Confusing internal reactive voltages with source current; overlooking that the source only “sees” R at resonance. Final Answer: Accurate

RLC Circuits and Resonance Questions