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

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

The resonant frequency fo is 19 kHz.

Accepted Answer

False

Question 10

When inductive and capacitive reactance are present in the same series circuit, the magnitude of the total reactance is equal to the sum of the magnitudes of the individual reactances.

Accepted Answer

False

Question 11

The quality factor of a parallel circuit at resonance is VC/VS or VL/VS.

Accepted Answer

True

Question 12

The less selective a filter is, the steeper the slope of its curves at the cutoff frequencies.

Accepted Answer

False

Question 13

In a series RLC circuit, at above resonance the circuit is more capacitive than it is inductive.

Accepted Answer

False

Question 14

The bandwidth of a filter is the range of frequencies between the half-power frequencies.

Accepted Answer

True

Question 15

With the frequency given, the circuit in the given circuit is operating above resonance.

Accepted Answer

True

Question 16

If the output voltage were taken from across the resistor in the given circuit, the circuit would be a band-pass filter.

Accepted Answer

True

Question 17

The impedance is 1177 Ω.

Accepted Answer

True

Question 18

The current through the inductor is 5 amps.

Accepted Answer

False

Question 19

If the given circuit were at resonance, the total current would be at its maximum.

Accepted Answer

True

Question 20

If the given circuit in were operating at resonance, the total current would be maximum.

Accepted Answer

False

RLC Circuits and Resonance Questions