NEETAceToday's NEET Daily

LCR series circuit, resonance and AC power: NEET MCQs with answers

50 questions on this topic of Electromagnetic Induction and Alternating Currents (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Electromagnetic Induction and Alternating Currents freeAll Electromagnetic Induction and Alternating Currents questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. A series LCR circuit has an inductor of inductance L, a capacitor of capacitance C, and a resistor of resistance R. If the value of L is doubled and the value of C is halved while R is kept constant, what happens to the quality factor Q and the resonant frequency ω₀?

  1. Both ω₀ and Q remain unchanged
  2. ω₀ is halved, Q remains unchanged
  3. ω₀ remains unchanged, Q is doubled
  4. ω₀ remains unchanged, Q is halved
Show answer and explanation

Answer: (C) ω₀ remains unchanged, Q is doubled

Resonant angular frequency is ω₀ = 1/√(LC). When L' = 2L and C' = C/2, the product L'C' = (2L)(C/2) = LC, so ω₀ remains unchanged. The quality factor is Q = (1/R)√(L/C). With the new values, √(L'/C') = √(2L / (C/2)) = √(4L/C) = 2√(L/C). Therefore, the new quality factor is Q' = 2Q, which is doubled.

medium

2. An alternating voltage v = 200√2 sin(100t) V is applied across a series combination of a resistor R = 30 Ω and an inductor L = 0.4 H. The average power dissipated in the circuit over a complete cycle is:

  1. 1200 W
  2. 960 W
  3. 240 W
  4. 480 W
Show answer and explanation

Answer: (D) 480 W

Inductive reactance XL = ωL = 100 × 0.4 = 40 Ω. Impedance Z = √(R² + XL²) = √(30² + 40²) = 50 Ω. The rms voltage is Vrms = 200 V, so Irms = Vrms / Z = 200 / 50 = 4 A. Power dissipated is P = (Irms)²R = 4² × 30 = 16 × 30 = 480 W.

hard

3. In a series LCR circuit, at an angular frequency ω₁ = 400 rad/s, the current lags behind the applied voltage by 45°. At an angular frequency ω₂ = 100 rad/s, the current leads the applied voltage by 45°. The resonant angular frequency ω₀ of the circuit is:

  1. 200 rad/s
  2. 250 rad/s
  3. 150 rad/s
  4. 300 rad/s
Show answer and explanation

Answer: (A) 200 rad/s

At ω₁, tan(45°) = 1 = (ω₁L - 1/(ω₁C))/R, giving ω₁L - 1/(ω₁C) = R. At ω₂, tan(-45°) = -1 = (ω₂L - 1/(ω₂C))/R, giving 1/(ω₂C) - ω₂L = R. Equating the two yields (ω₁ + ω₂)L = (1/C)(1/ω₁ + 1/ω₂) = (ω₁ + ω₂)/(ω₁ω₂C). Thus, LC = 1/(ω₁ω₂). Since ω₀ = 1/√(LC), we get ω₀ = √(ω₁ω₂) = √(400 × 100) = 200 rad/s.

medium

4. A series LCR circuit has R = 12 Ω, XL = 18 Ω, and XC = 13 Ω when connected to an AC source of 130 V (rms). The potential difference across the inductor is:

  1. 180 V
  2. 130 V
  3. 120 V
  4. 234 V
Show answer and explanation

Answer: (A) 180 V

The impedance is Z = √[R² + (XL - XC)²] = √[12² + (18 - 13)²] = √(144 + 25) = 13 Ω. The rms current is Irms = V / Z = 130 / 13 = 10 A. The potential difference across the inductor is VL = Irms × XL = 10 × 18 = 180 V.

easy

5. In a series LCR circuit, the voltage across the source is given by V = V₀ sin(ωt). At resonance, the phase difference between the voltage across the inductor VL and the voltage across the capacitor VC is:

  1. 0 rad
  2. π/2 rad
  3. 2π rad
  4. π rad
Show answer and explanation

Answer: (D) π rad

In a series LCR circuit, the current I is common to all components. The voltage across the inductor VL leads the current by π/2, while the voltage across the capacitor VC lags behind the current by π/2. Therefore, the phase difference between VL and VC is π/2 - (-π/2) = π radians, regardless of whether the circuit is at resonance or not.

hard

6. In a series LCR circuit, when the capacitor is removed, the current lags behind the applied voltage by an angle of 60°. When the inductor is removed instead, the current leads the voltage by 60°. When all three components (L, C, and R) are connected in series to the same source, the power factor of the circuit is:

  1. 0
  2. 0.5
  3. 0.866
  4. 1.0
Show answer and explanation

Answer: (D) 1.0

Without C, tan 60° = XL / R. Without L, tan 60° = XC / R. Since the phase shift magnitudes are identical, XL = XC. When all three are connected in series, the circuit is in resonance because XL - XC = 0, which gives an impedance Z = R and a power factor cos φ = 1.0.

easy

7. The average power consumed in a purely inductive or purely capacitive AC circuit is:

  1. Zero
  2. Maximum
  3. Vrms Irms
  4. Infinite
Show answer and explanation

Answer: (A) Zero

For pure L or C, current and voltage differ in phase by 90°, so cos φ = 0 and the average power is zero (wattless current).

NCERT Class 12

hard

8. A series LCR circuit has an inductor L = 0.2 H, a capacitor C = 20 μF, and a resistor R. When an AC voltage of frequency f = 50 Hz is applied, the phase angle between the current and the voltage is 45°. Taking π ≈ 3.14, the value of the resistance R is approximately:

  1. 32 Ω
  2. 63 Ω
  3. 159 Ω
  4. 96 Ω
Show answer and explanation

Answer: (D) 96 Ω

At f = 50 Hz, ω = 2πf = 100π ≈ 314 rad/s. XL = ωL = 314 × 0.2 = 62.8 Ω. XC = 1/(ωC) = 1/(314 × 20 × 10⁻⁶) ≈ 159.2 Ω. Net reactance |XL - XC| = |62.8 - 159.2| = 96.4 Ω. Since tan φ = tan 45° = 1 = |XL - XC| / R, we find R = 96.4 Ω ≈ 96 Ω.

medium

9. Consider the following two statements regarding a series LCR circuit: Statement I: The potential difference across the inductor or capacitor can be significantly larger than the applied source voltage at resonance. Statement II: The quality factor Q represents the voltage magnification factor of the series LCR circuit at resonance.

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

At resonance, current I = V/R, so the voltage across the inductor is VL = I × XL = (V/R) × ω₀L = Q × V. If Q > 1, VL (and VC) exceeds the source voltage V. Thus, Q directly quantifies the voltage magnification of the circuit.

medium

10. An AC voltage source V = 220 sin(100πt) V is applied across a series LCR circuit having R = 22 Ω, XL = 36 Ω, and XC = 14 Ω. The average power consumed in the circuit over a full cycle is:

  1. 1100 W
  2. 550 W
  3. 275 W
  4. 2200 W
Show answer and explanation

Answer: (B) 550 W

The net reactance is X = XL - XC = 36 - 14 = 22 Ω. The impedance of the circuit is Z = √(R² + X²) = √(22² + 22²) = 22√2 Ω. The rms voltage is Vrms = V₀/√2 = 220/√2 V. The rms current is Irms = Vrms/Z = (220/√2)/(22√2) = 220/44 = 5 A. The average power dissipated is P = Irms² × R = (5)² × 22 = 25 × 22 = 550 W.

easy

11. A series LCR circuit has R = 10 Ω, L = 2 mH, and C = 5 μF. Which of the following modifications will double the Quality factor (Q) of the circuit?

  1. Doubling R while keeping L and C unchanged
  2. Doubling L while keeping R and C unchanged
  3. Doubling C while keeping R and L unchanged
  4. Halving R while keeping L and C unchanged
Show answer and explanation

Answer: (D) Halving R while keeping L and C unchanged

The Quality factor of a series LCR circuit is Q = (1/R)√(L/C). Q is inversely proportional to R. Therefore, halving the resistance R (R → R/2) will double the quality factor Q.

easy

12. In a series LCR circuit, the voltage across each element is measured by an AC voltmeter: VR = 40 V, VL = 90 V, and VC = 60 V. The power factor of the circuit is:

  1. 0.80
  2. 0.60
  3. 0.75
  4. 0.50
Show answer and explanation

Answer: (A) 0.80

The net reactive voltage is VL - VC = 90 - 60 = 30 V. The net applied voltage is V = √[VR² + (VL - VC)²] = √(40² + 30²) = 50 V. Thus, the power factor is cos φ = VR / V = 40 / 50 = 0.80.

Keep going with Electromagnetic Induction and Alternating Currents. All 324 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Electromagnetic Induction and Alternating Currents