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: Quality factor (Q) of a resonant circuit — is it equal to the ratio of true power to apparent power (i.e., the power factor)?

Introduction / Context: Quality factor and power factor are distinct concepts that apply to different aspects of AC and resonant circuit behavior. Confusing them leads to errors in filter design and bandwidth estimation. Given Data / Assumptions: Sinusoidal steady-state conditions. Q refers to resonant circuit quality factor, not reactive power symbol Q in power systems. PF denotes power factor P/S. Concept / Approach: Power factor is PF = P / S = cos(phi). Quality factor of a simple inductor or capacitor at a given frequency equals reactance divided by resistance (Q = X / R). For a resonant network, Q relates stored reactive energy to energy dissipated per cycle and also to bandwidth via Q = f0 / BW. Step-by-Step Solution: Define PF: PF = P / S, a value between 0 and 1.Define Q for inductor: Q_L = omega * L / R. For capacitor: Q_C = 1 / (omega * C * R_series).At resonance: Q = 2 * pi * (energy stored per cycle) / (energy dissipated per cycle).Therefore Q is not P/S and does not equal the power factor. Verification / Alternative check: Bandwidth relation: Q = f0 / BW shows Q determines selectivity, a property unrelated to PF. Devices can have high Q but low PF depending on the load and phase angle. Why Other Options Are Wrong: Limiting statements (parallel, series, high-Q) do not change the definition of Q. Negligible losses increase Q but do not convert Q into PF. Common Pitfalls: Confusing the symbol Q for reactive power in power engineering with quality factor Q in resonance theory. Always check context. Final Answer: False

Q: Terminology of resonant circuits — is a series resonant circuit commonly called a “tank circuit”?

Introduction / Context: The term “tank circuit” is widely used in RF and analog electronics. Proper terminology matters for communication among engineers and for understanding oscillator and filter topologies. Given Data / Assumptions: Two common resonant topologies: series LC and parallel LC (both possibly with resistance). Industry usage and standard textbooks typically associate “tank” with the parallel form. Concept / Approach: A tank circuit is most commonly a parallel LC network that can store and exchange energy between the inductor and capacitor, maintaining oscillation when lightly loaded. The high impedance at resonance in a parallel LC and its energy storage behavior inspire the tank analogy. Although both series and parallel LC resonate, the term “tank circuit” is not commonly used for the series resonant form. Step-by-Step Solution: Identify standard usage: “tank” refers to a parallel resonant LC network.Series LC at resonance has minimum impedance and is used for impedance matching or selective current peaking, not typically called a tank.Therefore, stating that a series resonant circuit is commonly called a tank circuit is not correct. Verification / Alternative check: Oscillator schematics and RF textbooks show LC tanks as parallel networks feeding active devices, further confirming common usage. Why Other Options Are Wrong: Conditions like very high Q or negligible resistance do not change the conventional naming. The RF versus power qualifier is unnecessary; the terminology is consistent across domains. Common Pitfalls: Assuming any LC pair is a tank. Be precise: series LC is typically a notch or series resonator; parallel LC is the canonical tank. Final Answer: False

Q: Selectivity and bandwidth — does a lower quality factor (Q) produce a narrower bandwidth in a resonant circuit?

Introduction / Context: Quality factor Q and bandwidth are inversely related in linear resonant circuits. Designers rely on this relation to meet selectivity specifications in filters and tuned amplifiers. Given Data / Assumptions: Single, lightly loaded resonant network with center frequency f0. Half-power bandwidth definition applies: BW = f2 − f1 where response is 3 dB down from the peak. Concept / Approach: The standard relation is Q = f0 / BW. Therefore BW = f0 / Q. As Q decreases, BW increases; as Q increases, BW decreases. Hence a lower Q gives a wider bandwidth, not a narrower one. Step-by-Step Solution: Start with Q = f0 / BW.Rearrange: BW = f0 / Q.Decrease Q → denominator smaller → BW increases.Therefore, the statement claiming narrower bandwidth for lower Q is false. Verification / Alternative check: Simulation or lab measurements of an RLC circuit show that increasing series resistance reduces Q and broadens the resonance curve, confirming the inverse relationship between Q and BW. Why Other Options Are Wrong: The inverse relation holds for both series and parallel resonant forms under linear conditions. Zero resistance is unrealistic; however, practical resistance simply sets Q and hence BW by the same formula. Frequency range does not change the mathematical relationship. Common Pitfalls: Confusing sharpness of the peak (selectivity) with gain. Lower Q flattens and widens the resonance, even if peak amplitude also changes due to loading. Final Answer: False

Q: Reactances in RLC circuits — do inductive reactance (XL) and capacitive reactance (XC) act in opposition to each other?

Introduction / Context: RLC behavior depends on how reactive components combine in the impedance domain. Inductive and capacitive effects oppose each other in phase, which determines net reactance and the condition for resonance. Given Data / Assumptions: Sinusoidal steady-state operation. Inductor reactance XL = omega * L (positive imaginary). Capacitor reactance XC = 1 / (omega * C) and impedance is negative imaginary. Concept / Approach: In phasor form, inductive reactance contributes +jXL, while capacitive reactance contributes −jXC. The net reactive term is j(XL − XC) in series, or the difference of susceptances in parallel. This sign opposition is the mathematical expression of their opposing effects. Step-by-Step Solution: Write series impedance: Z = R + jXL − jXC = R + j(XL − XC).If XL > XC, circuit is net inductive; if XC > XL, net capacitive.Resonance occurs when XL = XC, so the net reactive term cancels.Therefore, XL and XC oppose each other in their effect on total reactance. Verification / Alternative check: Bode plots of a series RLC show phase crossing zero at the frequency where XL equals XC, confirming the cancellation between the opposing reactances. Why Other Options Are Wrong: They do not always add; their signs are opposite in the imaginary axis. The opposition exists at all frequencies, not only at resonance. Presence of resistance does not change the sign opposition, only the damping. Common Pitfalls: Using magnitudes only (XL and XC) without considering sign in phasor arithmetic. Always keep the j sign to avoid mistakes. Final Answer: True

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

Quality factor (Q) of a resonant circuit — is it equal to the ratio of true power to apparent power (i.e., the power factor)?

Accepted Answer

Introduction / Context: Quality factor and power factor are distinct concepts that apply to different aspects of AC and resonant circuit behavior. Confusing them leads to errors in filter design and bandwidth estimation. Given Data / Assumptions: Sinusoidal steady-state conditions. Q refers to resonant circuit quality factor, not reactive power symbol Q in power systems. PF denotes power factor P/S. Concept / Approach: Power factor is PF = P / S = cos(phi). Quality factor of a simple inductor or capacitor at a given frequency equals reactance divided by resistance (Q = X / R). For a resonant network, Q relates stored reactive energy to energy dissipated per cycle and also to bandwidth via Q = f0 / BW. Step-by-Step Solution: Define PF: PF = P / S, a value between 0 and 1.Define Q for inductor: Q_L = omega * L / R. For capacitor: Q_C = 1 / (omega * C * R_series).At resonance: Q = 2 * pi * (energy stored per cycle) / (energy dissipated per cycle).Therefore Q is not P/S and does not equal the power factor. Verification / Alternative check: Bandwidth relation: Q = f0 / BW shows Q determines selectivity, a property unrelated to PF. Devices can have high Q but low PF depending on the load and phase angle. Why Other Options Are Wrong: Limiting statements (parallel, series, high-Q) do not change the definition of Q. Negligible losses increase Q but do not convert Q into PF. Common Pitfalls: Confusing the symbol Q for reactive power in power engineering with quality factor Q in resonance theory. Always check context. Final Answer: False

Question 9

Terminology of resonant circuits — is a series resonant circuit commonly called a “tank circuit”?

Accepted Answer

Introduction / Context: The term “tank circuit” is widely used in RF and analog electronics. Proper terminology matters for communication among engineers and for understanding oscillator and filter topologies. Given Data / Assumptions: Two common resonant topologies: series LC and parallel LC (both possibly with resistance). Industry usage and standard textbooks typically associate “tank” with the parallel form. Concept / Approach: A tank circuit is most commonly a parallel LC network that can store and exchange energy between the inductor and capacitor, maintaining oscillation when lightly loaded. The high impedance at resonance in a parallel LC and its energy storage behavior inspire the tank analogy. Although both series and parallel LC resonate, the term “tank circuit” is not commonly used for the series resonant form. Step-by-Step Solution: Identify standard usage: “tank” refers to a parallel resonant LC network.Series LC at resonance has minimum impedance and is used for impedance matching or selective current peaking, not typically called a tank.Therefore, stating that a series resonant circuit is commonly called a tank circuit is not correct. Verification / Alternative check: Oscillator schematics and RF textbooks show LC tanks as parallel networks feeding active devices, further confirming common usage. Why Other Options Are Wrong: Conditions like very high Q or negligible resistance do not change the conventional naming. The RF versus power qualifier is unnecessary; the terminology is consistent across domains. Common Pitfalls: Assuming any LC pair is a tank. Be precise: series LC is typically a notch or series resonator; parallel LC is the canonical tank. Final Answer: False

Question 10

Selectivity and bandwidth — does a lower quality factor (Q) produce a narrower bandwidth in a resonant circuit?

Accepted Answer

Introduction / Context: Quality factor Q and bandwidth are inversely related in linear resonant circuits. Designers rely on this relation to meet selectivity specifications in filters and tuned amplifiers. Given Data / Assumptions: Single, lightly loaded resonant network with center frequency f0. Half-power bandwidth definition applies: BW = f2 − f1 where response is 3 dB down from the peak. Concept / Approach: The standard relation is Q = f0 / BW. Therefore BW = f0 / Q. As Q decreases, BW increases; as Q increases, BW decreases. Hence a lower Q gives a wider bandwidth, not a narrower one. Step-by-Step Solution: Start with Q = f0 / BW.Rearrange: BW = f0 / Q.Decrease Q → denominator smaller → BW increases.Therefore, the statement claiming narrower bandwidth for lower Q is false. Verification / Alternative check: Simulation or lab measurements of an RLC circuit show that increasing series resistance reduces Q and broadens the resonance curve, confirming the inverse relationship between Q and BW. Why Other Options Are Wrong: The inverse relation holds for both series and parallel resonant forms under linear conditions. Zero resistance is unrealistic; however, practical resistance simply sets Q and hence BW by the same formula. Frequency range does not change the mathematical relationship. Common Pitfalls: Confusing sharpness of the peak (selectivity) with gain. Lower Q flattens and widens the resonance, even if peak amplitude also changes due to loading. Final Answer: False

Question 11

Reactances in RLC circuits — do inductive reactance (XL) and capacitive reactance (XC) act in opposition to each other?

Accepted Answer

Introduction / Context: RLC behavior depends on how reactive components combine in the impedance domain. Inductive and capacitive effects oppose each other in phase, which determines net reactance and the condition for resonance. Given Data / Assumptions: Sinusoidal steady-state operation. Inductor reactance XL = omega * L (positive imaginary). Capacitor reactance XC = 1 / (omega * C) and impedance is negative imaginary. Concept / Approach: In phasor form, inductive reactance contributes +jXL, while capacitive reactance contributes −jXC. The net reactive term is j(XL − XC) in series, or the difference of susceptances in parallel. This sign opposition is the mathematical expression of their opposing effects. Step-by-Step Solution: Write series impedance: Z = R + jXL − jXC = R + j(XL − XC).If XL > XC, circuit is net inductive; if XC > XL, net capacitive.Resonance occurs when XL = XC, so the net reactive term cancels.Therefore, XL and XC oppose each other in their effect on total reactance. Verification / Alternative check: Bode plots of a series RLC show phase crossing zero at the frequency where XL equals XC, confirming the cancellation between the opposing reactances. Why Other Options Are Wrong: They do not always add; their signs are opposite in the imaginary axis. The opposition exists at all frequencies, not only at resonance. Presence of resistance does not change the sign opposition, only the damping. Common Pitfalls: Using magnitudes only (XL and XC) without considering sign in phasor arithmetic. Always keep the j sign to avoid mistakes. Final Answer: True

Question 12

Series resonance condition — at series resonance, are the inductive and capacitive reactances never equal?

Accepted Answer

Introduction / Context: Resonance in a series RLC circuit is defined by the balance between inductive and capacitive reactances. This balance governs impedance magnitude and phase, and it sets the peak current condition for a given source voltage. Given Data / Assumptions: Series RLC under sinusoidal steady state. XL = omega * L, XC = 1 / (omega * C). Resonant frequency denoted as omega0. Concept / Approach: By definition of series resonance, XL equals XC in magnitude at omega0. The imaginary parts cancel, leaving net impedance Z = R (minimum magnitude). This is the basis for current maximization and purely resistive phase at resonance. Step-by-Step Solution: Set condition for resonance: XL = XC.This gives omega0 * L = 1 / (omega0 * C).Solve for frequency: omega0 = 1 / sqrt(L * C).At omega0, Z = R + j(XL − XC) = R, confirming equality and cancellation of reactances. Verification / Alternative check: Measured current peaks at resonance and the phase angle between source voltage and current is zero, both confirming that reactive terms are equal and opposite. Why Other Options Are Wrong: Claims that equality never occurs contradict the definition of series resonance. Parallel resonance is a different topology; the equality condition discussed here is for series resonance. Non-ideal losses do not alter the equality condition; they only broaden the resonance and reduce current magnitude. Common Pitfalls: Confusing equality of reactances with equality of impedances. The magnitudes of reactances are equal at resonance; the net impedance is resistive, not zero. Final Answer: False

Question 13

Parallel RLC behavior — does the smaller reactance branch determine the net reactance of the overall circuit?

Accepted Answer

Introduction / Context: In parallel reactive networks, branch currents depend on branch reactances. Understanding which branch dominates the net behavior is important for filter design and tuning around resonance. Given Data / Assumptions: Two reactive branches in parallel, one inductive and one capacitive, with finite resistive loss neglected for the conceptual argument. Sinusoidal steady state and linear components. Concept / Approach: In parallel, admittances add. The branch with smaller reactance magnitude has larger susceptance magnitude and therefore draws more current for the same applied voltage. Consequently, that branch dominates the net reactive behavior, effectively determining whether the total appears predominantly inductive or capacitive away from exact resonance. Step-by-Step Solution: Compute branch susceptances: B_L = −1 / XL and B_C = +1 / XC (signs per convention).Total susceptance: B_total = B_C + B_L.If |XC| < |XL| (capacitor reactance smaller), then |B_C| > |B_L| and the circuit is net capacitive.If |XL| < |XC|, the inductor dominates and the circuit is net inductive. Verification / Alternative check: The equivalent reactance magnitude of parallel reactances is less than the smaller of the branch magnitudes. Measurements will show current primarily in the smaller reactance branch, confirming dominance. Why Other Options Are Wrong: Dominance is not restricted to resonance; it holds for any frequency where branch magnitudes differ. Nonzero resistance modifies Q but does not reverse which branch dominates reactively. Both inductive and capacitive branches can dominate depending on frequency. Common Pitfalls: Interpreting “determine” as “equal to.” The smaller reactance does not set the exact value; it dominates the net behavior. The exact equivalent must be computed from admittances. Final Answer: True

Question 14

Series RLC at resonance — is the current minimum at resonance in a series circuit?

Accepted Answer

Introduction / Context: Resonance in series and parallel circuits has opposite current characteristics. Recognizing which topology exhibits maximum current at resonance prevents design mistakes in filters and matching networks. Given Data / Assumptions: Series RLC with source voltage V_rms. XL = omega * L, XC = 1 / (omega * C). Resonance at omega0 where XL = XC. Concept / Approach: At series resonance, the reactive terms cancel, yielding Z = R (minimum magnitude). With Z minimized, current is maximized for a given source voltage by Ohm’s law. Therefore the current is at its maximum, not minimum, at series resonance. Step-by-Step Solution: Write Z_series = R + j(XL − XC).At resonance XL = XC ⇒ Z_series = R (minimum impedance).Current I = V_rms / |Z_series| is maximum when |Z_series| is minimum.Hence the statement that current is minimum is false. Verification / Alternative check: Contrast with parallel resonance, where input current from the source can be minimum at resonance due to high input impedance, while large circulating currents can exist within the branches. Why Other Options Are Wrong: High Q increases the current peak but does not change the fact that the peak occurs at resonance. The condition “only if XL equals XC” defines resonance; even then the current is maximum, not minimum. Common Pitfalls: Confusing series and parallel resonance properties. Remember: series resonance minimizes impedance and maximizes current; parallel resonance maximizes impedance and minimizes source current. Final Answer: False

Question 15

Reactive voltage cancellation — at resonance, do the inductor and capacitor voltage drops (VL and VC) cancel each other?

Accepted Answer

Introduction / Context: At resonance, reactive effects in RLC circuits balance. The concept of reactive voltage cancellation explains why the net reactive component of impedance disappears at the resonant frequency. Given Data / Assumptions: Series RLC circuit under sinusoidal steady state (the explanation also extends to reactive power in parallel resonance). Inductive and capacitive reactances are equal in magnitude at resonance. Finite resistance is present, so current is finite. Concept / Approach: In a series RLC at resonance, XL = XC. The inductor drop is VL = I * XL, and the capacitor drop is VC = I * (−XC) in phasor form. These two phasors are equal in magnitude and 180 degrees apart, so their sum is zero. The source therefore only supplies the resistive drop across R. Step-by-Step Solution: Set XL = XC at resonance.Compute VL = I * jXL and VC = I * (−jXC).Since XL = XC, VL + VC = I * jXL + I * (−jXL) = 0.Thus, reactive voltages cancel and only the resistive component remains in the loop sum. Verification / Alternative check: Applying Kirchhoff’s Voltage Law around the loop at resonance yields V_source = V_R, with the vector sum of VL and VC equal to zero. Oscilloscope measurements with current sensing show opposing inductor and capacitor waveforms that cancel vectorially. Why Other Options Are Wrong: Cancellation does not require zero resistance; it requires equality of reactances. It is not restricted to parallel RLC; in parallel circuits, reactive powers cancel at resonance as well. Infinite Q is unnecessary; finite Q simply limits peak magnitudes. Common Pitfalls: Interpreting “cancel” as “each voltage is zero.” VL and VC can be large individually at high Q, yet their phasor sum is zero at resonance. Final Answer: True

Question 16

Series RLC behavior: In a series R–L–C circuit, the larger reactance (either inductive XL or capacitive XC) determines the sign and overall net reactance Xnet = XL − XC of the circuit. State whether this statement is true or false.

Accepted Answer

Introduction / Context: This statement examines how inductive reactance (XL) and capacitive reactance (XC) combine in a series RLC circuit. Understanding which term dominates determines whether the circuit behaves overall as inductive or capacitive, which directly affects current phase, impedance magnitude, and resonance behavior. Given Data / Assumptions: Series connection of R, L, and C. Inductive reactance: XL = 2 * pi * f * L. Capacitive reactance: XC = 1 / (2 * pi * f * C). Net series reactance: Xnet = XL − XC (sign matters). Concept / Approach: Because L and C are in series, their reactances algebraically subtract. The magnitude of the larger of XL or XC determines not only the sign of Xnet but also the overall reactive nature of the circuit. If XL > XC, the circuit is net inductive; if XC > XL, it is net capacitive. At resonance, XL = XC and Xnet = 0, leaving only the resistive part R. Step-by-Step Solution: Compute XL = 2 * pi * f * L.Compute XC = 1 / (2 * pi * f * C).Form Xnet = XL − XC.If XL > XC ⇒ Xnet > 0 ⇒ net inductive; if XC > XL ⇒ Xnet < 0 ⇒ net capacitive. Verification / Alternative check: Plot XL and XC versus frequency. The curves cross at f0 where XL = XC. On either side of f0, the curve with the larger magnitude dictates the sign and magnitude of Xnet, confirming that the “larger reactance” determines the net reactive behavior. Why Other Options Are Wrong: “False” would imply the smaller term or some other operation sets Xnet, which contradicts Xnet = XL − XC in series. Common Pitfalls: Confusing series and parallel combination rules; in parallel, susceptances add instead. Also, ignoring sign leads to wrong conclusions about leading/lagging current. Final Answer: True

Question 17

Series RLC resonance definition: Resonance occurs when the inductive and capacitive reactances are equal in magnitude (XL = XC), causing the net reactance to be zero. True or false?

Accepted Answer

Introduction / Context: Resonance in a series RLC circuit is a cornerstone concept in AC analysis, filters, and tuned amplifiers. At resonance, reactive effects cancel, leaving purely resistive behavior that maximizes current for a given source voltage. Given Data / Assumptions: Series R, L, C circuit driven by a sinusoidal source. XL = 2 * pi * f * L and XC = 1 / (2 * pi * f * C). Resonant frequency often denoted by f0. Concept / Approach: In series RLC, the net reactance is Xnet = XL − XC. Resonance is defined by Xnet = 0, which requires XL = XC. Solving XL = XC yields f0 = 1 / (2 * pi * sqrt(L * C)). At f0, impedance is purely resistive (equal to R), current is in phase with voltage, and reactive power exchange between L and C is internal and equal. Step-by-Step Solution: Set XL = 2 * pi * f * L.Set XC = 1 / (2 * pi * f * C).At resonance: XL = XC ⇒ 2 * pi * f0 * L = 1 / (2 * pi * f0 * C).Solve for f0 to confirm the equality criterion. Verification / Alternative check: Measure phase angle between source voltage and current. At resonance, phase angle is zero and the power factor is unity, confirming XL = XC and Xnet = 0. Why Other Options Are Wrong: “False” would deny the established definition of series resonance (XL equals XC), which is fundamental in circuit theory. Common Pitfalls: Mixing series and parallel resonance definitions. In parallel resonance, the condition is different when expressed in terms of susceptances and involves R, but the equality XL = XC in magnitude still indicates reactive cancellation. Final Answer: True

Question 18

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 19

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 20

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

RLC Circuits and Resonance Questions