CUET UG Physics Booster Test 2-Series LCR Circuits and Resonance -

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CUET UG Physics Booster Test 2-Series LCR Circuits and Resonance

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2-Series LCR Circuits and Resonance

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QUESTION 1 OF 20

Match List I with List II for a series LCR circuit

List I	List II
1. VR	a. Lags current by π/2
2. VC	b. Ahead of current by π/2
3. VL	c. Same for all elements
4. I	d. Parallel to current phasor

Answer Review

1 Match List I with List II for a series LCR circuit

List I	List II
1. VR	a. Lags current by π/2
2. VC	b. Ahead of current by π/2
3. VL	c. Same for all elements
4. I	d. Parallel to current phasor

A 1-d, 2-a, 3-b, 4-c

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-b, 3-a, 4-c

D 1-c, 2-a, 3-b, 4-d

💡 Short Explanation

�� VR is in phase with current. �� VC lags current by π/2. �� VL leads current by π/2. �� Current is same through all elements.

✅ Correct Answer Explanation

1 → d : Voltage across resistor is parallel to current phasor. 2 → a : Capacitor voltage lags current by π/2. 3 → b : Inductor voltage leads current by π/2. 4 → c : Current is identical through all series elements. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of VR and VC.
�� Option C → Interchanges capacitor and inductor relations.
�� Option D → Incorrect matching of current and resistor voltage.

🎯 Strategy

Used

Option Grouping

Application:

Match phase relations of R, L and C.

Final Logic:

Only Option A correctly matches all four quantities.

🧠 Memory Trick

R-In Phase, L-Leads, C-Lags

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Phasor Diagram of Series LCR Circuit

2 Because there is a uniform current in a series LCR circuit,

A the voltage across each element must also be uniform and equal

B the current phasor I is used as the reference vector in the phasor diagram

C the power dissipation is shared equally among all components

D the net impedance is simply the algebraic sum of individual reactances

💡 Short Explanation

�� Current is common to all elements. �� Phasor diagrams use current as reference. �� Voltages differ across elements.

✅ Correct Answer Explanation

Since the same current flows through R, L and C, the current phasor is chosen as the reference axis. Voltages across individual elements are then drawn relative to this current. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Voltages are generally different.
�� Option C → Only resistor dissipates average power.
�� Option D → Impedance is obtained vectorially.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall construction of phasor diagrams.

Final Logic:

Common current serves as the reference phasor.

🧠 Memory Trick

Same Current → Reference Phasor

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Phasor Representation

3 When applying Kirchhoff's Loop Rule to a series LCR circuit driven by an ac source

A the sum of the peak voltages across all elements equals the peak source voltage algebraically

B the algebraic sum of the instantaneous voltages across the elements is zero over a loop

C the loop rule fails due to the time-varying nature of the magnetic flux in the inductor

D the charge q on the capacitor becomes constant to satisfy the loop rule

💡 Short Explanation

�� Kirchhoff's loop rule remains valid. �� Instantaneous voltages are added. �� Total voltage around loop is zero.

✅ Correct Answer Explanation

For a series LCR circuit: v − vL − vR − vC = 0 or v = vL + vR + vC Thus the algebraic sum of all instantaneous voltages around the closed loop is zero. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Peak voltages are not added algebraically.
�� Option C → Kirchhoff's rule remains applicable.
�� Option D → Charge continuously varies in AC.

🎯 Strategy

Used

Direct Formula Recall

Application:

Apply Kirchhoff's voltage law.

Final Logic:

Instantaneous voltages satisfy loop rule.

🧠 Memory Trick

Loop Sum = Zero

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Kirchhoff's Loop Rule

4 Identify the incorrect statement regarding the sum of voltages in an LCR circuit:

A Instantaneous voltages vL, vR, and vC can be added algebraically to give v.

B The sum of the rms voltages across L, C, and R algebraically gives the rms source voltage.

C The total source voltage must be obtained using the Pythagorean theorem for the voltage phasors.

D The voltages across the inductor and capacitor are out of phase by π.

💡 Short Explanation

�� RMS voltages are phasor quantities. �� Algebraic addition is incorrect. �� Vector addition is required.

✅ Correct Answer Explanation

The source RMS voltage is obtained from phasor addition: V = √[VR² + (VL − VC)²] Therefore, RMS voltages cannot be added algebraically. Hence Option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → True for instantaneous values.
�� Option C → Correct phasor method.
�� Option D → VL and VC differ by π radians.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Differentiate algebraic addition and phasor addition.

Final Logic:

RMS voltages must be added vectorially.

🧠 Memory Trick

RMS Voltages → Phasor Addition

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Voltage Phasors

5 What is the total circuit impedance if the resistance is 3.0 Ω, inductive reactance is 8.0 Ω, and capacitive reactance is 4.0 Ω?

A 5.0 Ω

B 15.0 Ω

C 7.0 Ω

D 1.0 Ω

💡 Short Explanation

�� Z = √[R² + (XL − XC)²] �� XL − XC = 4 Ω �� Use Pythagoras theorem.

✅ Correct Answer Explanation

Given: R = 3 Ω XL = 8 Ω XC = 4 Ω Z = √[3² + (8 − 4)²] = √(9 + 16) = √25 = 5 Ω Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not obtained from impedance formula.
�� Option C → Incorrect calculation.
�� Option D → Net reactance not zero.

🎯 Strategy

Used

Substitution

Application:

Substitute values into impedance formula.

Final Logic:

Z = √25 = 5 Ω.

🧠 Memory Trick

Z = √(R² + X²)

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Impedance Calculation

6 In the impedance triangle, the relation between the base and the perpendicular determines the phase angle. The quantities corresponding to Base and Perpendicular are:

A R, (XC − XL)

B Z, R

C (XC − XL), R

D XL, XC

💡 Short Explanation

�� Base represents resistance. �� Perpendicular represents net reactance. �� Phase angle depends on their ratio.

✅ Correct Answer Explanation

In the impedance triangle: Base = R Perpendicular = XC − XL Hypotenuse = Z Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Z is hypotenuse.
�� Option C → Reversed assignment.
�� Option D → Not impedance triangle sides.

🎯 Strategy

Used

Diagram Recall

Application:

Recall impedance triangle structure.

Final Logic:

Base = R and perpendicular = XC − XL.

🧠 Memory Trick

Base-R, Height-X

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Impedance Triangle

7 If the circuit is predominantly inductive (XL > XC), the net phase angle formula will be

A tan φ = (XC − XL)/R, resulting in negative φ

B tan φ = (XL − XC)/R, resulting in negative φ

C tan φ = R/(XL − XC), resulting in positive φ

D tan φ = (XC + XL)/R, resulting in zero φ

💡 Short Explanation

�� NCERT uses tanφ = (XC − XL)/R. �� XL > XC makes numerator negative. �� Therefore φ becomes negative.

✅ Correct Answer Explanation

For a series LCR circuit: tanφ = (XC − XL)/R If XL > XC: (XC − XL) < 0 Therefore φ is negative, indicating an inductive circuit where current lags voltage. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives positive value.
�� Option C → Incorrect formula.
�� Option D → Incorrect relation.

🎯 Strategy

Used

Formula Recall

Application:

Substitute inductive condition XL > XC.

Final Logic:

Negative numerator gives negative φ.

🧠 Memory Trick

Inductive → Negative φ (NCERT Convention)

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Phase Angle

8 Choose the correct statements about a predominantly capacitive series LCR circuit
1. The voltage across the capacitor is greater than the voltage across the inductor.
2. The current leads the applied voltage.
3. The phase angle φ is negative.
4. Decreasing the source frequency makes it even more capacitive.

A (1), (2), (4)

B (2), (3)

C (1), (2), (3)

D (1), (4)

💡 Short Explanation

�� Capacitive dominance means XC > XL. �� Current leads voltage. �� Lower frequency increases XC.

✅ Correct Answer Explanation

1. Correct — XC > XL implies VC > VL. 2. Correct — Current leads voltage in capacitive circuits. 3. Incorrect — Using NCERT convention, capacitive circuits give positive φ. 4. Correct — Lower frequency increases XC further. Hence statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.
�� Option D → Omits statement 2.

🎯 Strategy

Used

Elimination

Application:

Apply properties of capacitive circuits.

Final Logic:

XC > XL gives leading current and positive φ.

🧠 Memory Trick

Capacitor Leads

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Capacitive LCR Circuit

9 When analyzing the steady state vs transient solutions of an LCR circuit

A the transient solution is completely derived from the phasor diagram

B the steady-state solution accounts for the initial conditions of the circuit at t = 0

C the general solution is essential immediately after the circuit is switched on

D the steady-state solution is only valid when the voltage v = 0

💡 Short Explanation

�� Initial response includes transient effects. �� General solution combines both parts. �� Steady state dominates later.

✅ Correct Answer Explanation

Immediately after switching on the circuit, both transient and steady-state components exist. Therefore the complete behavior is described by the general solution. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Phasors represent steady state only.
�� Option B → Initial conditions affect transient solution.
�� Option D → Steady-state applies for AC operation.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Distinguish transient and steady-state behavior.

Final Logic:

Initial response requires the full solution.

🧠 Memory Trick

Start-Up → General Solution

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Transient and Steady-State Response

10 Consider the statements on long-term circuit behaviour. Choose the correct statements:
1. Transient effects die out.
2. Steady-state solution governs the behavior.
3. The current becomes constant and non-alternating.
4. Phasor analysis becomes perfectly applicable.

A (1), (2), (4)

B (1), (2), (3)

C (2), (3), (4)

D (1), (3)

💡 Short Explanation

�� Transients disappear. �� AC current remains alternating. �� Phasor analysis applies to steady state.

✅ Correct Answer Explanation

1. Correct — Transient effects decay with time. 2. Correct — Steady-state behavior remains. 3. Incorrect — Current remains alternating in AC circuits. 4. Correct — Phasor analysis describes steady-state AC behavior. Hence statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.
�� Option D → Omits statement 2 and includes statement 3.

🎯 Strategy

Used

Elimination

Application:

Identify the incorrect statement.

Final Logic:

AC current never becomes steady DC.

🧠 Memory Trick

Transient Dies, AC Survives

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Long-Term Behavior of LCR Circuits

11 The natural frequency tendency of an LCR circuit:

A is dependent on the resistance R driving the oscillations

B dictates the frequency at which the system naturally stores and transfers energy between L and C

C implies the circuit will always produce radio waves

D requires an external dc source to manifest

💡 Short Explanation

�� Energy oscillates between L and C. �� Natural frequency is an inherent property. �� It governs resonance behavior.

✅ Correct Answer Explanation

The natural frequency of an LCR circuit is determined primarily by L and C. At this frequency, energy oscillates between the magnetic field of the inductor and the electric field of the capacitor. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Natural frequency depends mainly on L and C, not on R.
�� Option C → Natural frequency does not imply radio-wave generation.
�� Option D → No external DC source is required.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Identify the physical meaning of natural frequency.

Final Logic:

Natural frequency describes energy exchange between L and C.

🧠 Memory Trick

L ↔ C Energy Exchange

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Natural Frequency

12 Choose the correct statements about maximum amplitude conditions at resonance:
1. Current amplitude depends only on source voltage and resistance.
2. Power dissipation is maximum.
3. Impedance equals zero.
4. Power factor becomes 1.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3)

D (1), (2), (3), (4)

💡 Short Explanation

�� At resonance Z = R. �� Current becomes maximum. �� Power factor becomes unity.

✅ Correct Answer Explanation

1. Correct — At resonance, im = vm/R. 2. Correct — Maximum current causes maximum power dissipation. 3. Incorrect — Impedance equals R, not zero. 4. Correct — Phase angle becomes zero, so cosφ = 1. Hence statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3 and omits correct statements.
�� Option D → Includes incorrect statement 3.

🎯 Strategy

Used

Elimination

Application:

Apply resonance conditions.

Final Logic:

At resonance Z = R, not zero.

🧠 Memory Trick

Resonance → Maximum I, Unity Power Factor

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Resonance Characteristics

13 Match List I with List II for conditions at resonant reactance cancellation

List I	List II
1. XC − XL	a. R
2. Impedance Z	b. 1
3. Phase angle φ	c. 0
4. Power factor cos(φ)	d. Same as applied voltage phase (0 radians)

A 1-c, 2-a, 3-d, 4-b

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

�� Reactances cancel at resonance. �� Z becomes R. �� Phase difference vanishes.

✅ Correct Answer Explanation

1 → c : XC − XL = 0 2 → a : Z = R 3 → d : Phase angle becomes 0 radians. 4 → b : cosφ = 1 Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of impedance and phase angle.
�� Option C → Multiple incorrect correspondences.
�� Option D → Incorrect matching of reactance cancellation.

🎯 Strategy

Used

Option Grouping

Application:

Apply resonance conditions to each quantity.

Final Logic:

Only Option A satisfies all resonance properties.

🧠 Memory Trick

Resonance: X=0, Z=R, φ=0, PF=1

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Resonance Conditions

14 If L = 25.48 × 10⁻³ H and C = 796 × 10⁻⁶ F, what will be the resonant angular frequency ω₀?

A 222.1 rad s⁻¹

B 35.4 rad s⁻¹

C 50.0 rad s⁻¹

D 314.1 rad s⁻¹

💡 Short Explanation

�� ω₀ = 1/√LC �� Substitute L and C. �� Evaluate numerically.

✅ Correct Answer Explanation

ω₀ = 1/√(LC) = 1/√[(25.48 × 10⁻³)(796 × 10⁻⁶)] ≈ 222.1 rad s⁻¹ Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Underestimated value.
�� Option C → Incorrect substitution.
�� Option D → Overestimated value.

🎯 Strategy

Used

Substitution

Application:

Apply resonance formula directly.

Final Logic:

ω₀ = 222.1 rad s⁻¹.

🧠 Memory Trick

ω₀ = 1/√LC

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Resonant Angular Frequency

15 At resonance, the values of imaginary impedance (reactance) and total real impedance are:

A Zero, Maximum

B Zero, Minimum

C Maximum, Minimum

D Minimum, Maximum

💡 Short Explanation

�� Net reactance becomes zero. �� Impedance reduces to R. �� R is the minimum impedance.

✅ Correct Answer Explanation

At resonance: XL = XC Therefore net reactance = 0. Thus: Z = R Since impedance cannot become smaller than R, the total impedance is minimum. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Real impedance is not maximum.
�� Option C → Reactance is not maximum.
�� Option D → Real impedance is minimum, not maximum.

🎯 Strategy

Used

Direct Formula Recall

Application:

Apply resonance condition XL = XC.

Final Logic:

Zero reactance leaves minimum impedance R.

🧠 Memory Trick

Resonance → Reactance Zero, Z Minimum

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Minimum Impedance

16 The formula for maximum current amplitude im at resonance will be

A vm / √[R² + (XC − XL)²]

B vm / R

C vm / (XC − XL)

D (vm / Z) × cosφ

💡 Short Explanation

�� At resonance Z = R. �� Current reaches maximum value. �� Ohm's law applies directly.

✅ Correct Answer Explanation

At resonance: Z = R Therefore: im = vm/Z = vm/R Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → General formula, not resonance-specific.
�� Option C → Undefined at resonance.
�� Option D → Not the expression for current amplitude.

🎯 Strategy

Used

Direct Formula Recall

Application:

Replace Z by R at resonance.

Final Logic:

Maximum current occurs when Z = R.

🧠 Memory Trick

At Resonance: I = V/R

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Maximum Current at Resonance

17 Identify the incorrect statement about radio tuning mechanisms

A The antenna picks up signals from many broadcasting stations.

B Tuning changes the resistance of the circuit to select the station frequency.

C The circuit is driven to resonance at the frequency of the desired radio signal.

D Varying the capacitance changes the resonant frequency of the tuning circuit.

💡 Short Explanation

�� Tuning uses resonance. �� Capacitance is varied. �� Resistance is not the tuning parameter.

✅ Correct Answer Explanation

Radio tuning is achieved by adjusting capacitance so that the resonant frequency matches the desired station frequency. Resistance does not determine the selected frequency. Hence Option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of antenna operation.
�� Option C → Correct resonance principle.
�� Option D → Correct tuning method.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with resonance tuning.

Final Logic:

Frequency selection is achieved by varying C, not R.

🧠 Memory Trick

Tune by C, Not by R

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Radio Tuning

18 When a person carrying metal walks through an airport metal detector, it fundamentally alters the

A capacitance of the completely rigid doorway

B resistance of the person's body

C effective inductance of the detector coil, thus changing the circuit's impedance

D frequency of the external power grid

💡 Short Explanation

�� Metal interacts with magnetic fields. �� Effective inductance changes. �� Resonance condition shifts.

✅ Correct Answer Explanation

The detector contains a resonant AC circuit. Metal objects alter magnetic coupling and effective inductance, causing a change in impedance that is detected electronically. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Capacitance change is not the primary mechanism.
�� Option B → Body resistance is irrelevant.
�� Option D → Power-grid frequency remains unchanged.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Connect resonance circuits with metal detection.

Final Logic:

Metal changes inductance and impedance.

🧠 Memory Trick

Metal → Inductance Change

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Applications of Resonance

19 Choose the statements on the requirement for L and C in resonance
1. Voltages across L and C cancel each other.
2. Both elements provide phase shifts in opposite directions.
3. Total source voltage appears across R.
4. Resistance is effectively infinite.

A (1), (2), (3)

B (2), (3), (4)

C (1), (3)

D (1), (4)

💡 Short Explanation

�� L and C provide opposite reactances. �� Their voltages cancel at resonance. �� Circuit behaves like a resistor.

✅ Correct Answer Explanation

1. Correct — VL and VC cancel each other. 2. Correct — L and C produce opposite phase shifts. 3. Correct — Net reactive voltage becomes zero, so source voltage appears across R. 4. Incorrect — Resistance remains finite. Hence statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect statement 4.
�� Option C → Omits statement 2.
�� Option D → Includes incorrect statement 4.

🎯 Strategy

Used

Elimination

Application:

Apply resonance properties.

Final Logic:

Only statement 4 is false.

🧠 Memory Trick

L Cancels C → R Remains

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Requirement of L and C for Resonance

20 The inability of an RL circuit to exhibit resonance is primarily because

A the resistor dissipates too much energy

B there is no capacitor to provide a leading voltage to cancel the lagging voltage of the inductor

C inductors do not operate on alternating currents

D the impedance of an inductor is inherently zero

💡 Short Explanation

�� Resonance requires XL = XC. �� RL circuit has no capacitor. �� Reactance cancellation is impossible.

✅ Correct Answer Explanation

Resonance occurs only when inductive and capacitive reactances cancel each other. An RL circuit contains no capacitor, so no capacitive reactance exists to balance XL. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Energy loss is not the fundamental reason.
�� Option C → Inductors operate effectively with AC.
�� Option D → Inductive impedance is not zero.

🎯 Strategy

Used

Direct Concept Recall

Application:

Apply the resonance condition XL = XC.

Final Logic:

No capacitor means no resonance.

🧠 Memory Trick

No C → No Resonance

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Resonance Requirement in AC Circuits

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