CUET UG Physics Booster Test 3-Series LCR Circuits and Resonance -

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CUET UG Physics Booster Test 3-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 3-Series LCR Circuits and Resonance

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QUESTION 1 OF 20

Consider the statements regarding the phasor addition in a series LCR arrangement. Choose the correct statements:
1. VC and VL are always along the same line and in opposite directions.
2. The combined phasor is (VC + VL) with magnitude |VCm - VLm|.
3. The source voltage V is the hypotenuse of the right triangle formed by VR and (VC + VL).
4. VR and I can sometimes be out of phase.

Answer Review

1 Consider the statements regarding the phasor addition in a series LCR arrangement. Choose the correct statements:
1. VC and VL are always along the same line and in opposite directions.
2. The combined phasor is (VC + VL) with magnitude |VCm - VLm|.
3. The source voltage V is the hypotenuse of the right triangle formed by VR and (VC + VL).
4. VR and I can sometimes be out of phase.

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

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

C (1), (3)

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

💡 Short Explanation

�� VL and VC are 180° apart. �� Net reactive voltage is their difference. �� Source voltage is obtained by phasor addition.

✅ Correct Answer Explanation

1. Correct — VL leads current by π/2 and VC lags current by π/2, so they lie on the same line in opposite directions. 2. Correct — Their resultant magnitude is |VCm − VLm|. 3. Correct — V is the phasor sum of VR and the net reactive voltage, forming the hypotenuse of the voltage triangle. 4. Incorrect — Voltage across a resistor is always in phase with current. Hence statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application:

Use standard LCR phasor relations.

Final Logic:

VR always remains in phase with current.

🧠 Memory Trick

R In Phase, L Up, C Down

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

2 Choose the incorrect statement concerning the uniform current in a series LCR circuit

A Because the current is uniform, it serves as the most convenient reference phasor.

B The instantaneous current is identical for all three elements.

C Uniform current implies that the instantaneous voltages across the three components must also reach their maximum values simultaneously.

D The amplitude of the current is independent of the positional order of L, C, and R in the series.

💡 Short Explanation

�� Current is common throughout. �� Voltages have phase differences. �� Maxima occur at different times.

✅ Correct Answer Explanation

Although the current is identical through all components, voltages across R, L and C are phase shifted relative to the current and each other. Therefore their maxima generally do not occur simultaneously. Hence Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct use of current as reference phasor.
�� Option B → Correct for series circuits.
�� Option D → Series order does not affect current amplitude.

🎯 Strategy

Used

Elimination

Application:

Separate current behavior from voltage behavior.

Final Logic:

Same current does not imply same voltage phase.

🧠 Memory Trick

Same Current ≠ Same Voltage Timing

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Uniform Current in Series Circuit

3 While formulating Kirchhoff's loop rule for the series LCR circuit, the self-induced Faraday emf in the inductor

A acts as a driving source equal to the external ac voltage

B is represented as a voltage drop L(di/dt) which algebraically adds with iR and q/C to equal the source voltage

C creates a dc offset current that prevents true alternating behavior

D must strictly be in phase with the charge q on the capacitor

💡 Short Explanation

�� Inductor contributes L(di/dt). �� Kirchhoff's rule includes all drops. �� Sum equals source voltage.

✅ Correct Answer Explanation

For a series LCR circuit: L(di/dt) + iR + q/C = v The inductor contributes a voltage drop L(di/dt), which combines with resistor and capacitor voltage drops. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Induced emf opposes changes, not acts as the source.
�� Option C → No DC offset is produced.
�� Option D → Charge and inductor voltage are not necessarily in phase.

🎯 Strategy

Used

Direct Formula Recall

Application:

Apply Kirchhoff's voltage law.

Final Logic:

All voltage drops add to the source voltage.

🧠 Memory Trick

L + R + C = Source

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

4 Correct statements about the sum of voltages in the LCR circuit paradox
1. The algebraic sum of rms voltages can exceed the rms source voltage.
2. Instantaneous voltages cannot exceed the peak source voltage.
3. The total rms voltage must be evaluated using vector addition (Pythagorean theorem).
4. Phase difference prevents simple arithmetic addition of rms values.

A (1), (3), (4)

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

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

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

💡 Short Explanation

�� RMS voltages are phasors. �� Vector addition is required. �� Individual voltages can exceed source voltage near resonance.

✅ Correct Answer Explanation

1. Correct — VL and VC can individually be large, making algebraic sums exceed source voltage. 2. Incorrect — Instantaneous voltages across L or C can exceed source voltage under resonance conditions. 3. Correct — RMS voltages are added vectorially. 4. Correct — Phase differences prevent arithmetic addition. Hence statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Conceptual Reasoning

Application:

Use phasor relationships in resonance.

Final Logic:

Voltage addition in AC is vectorial.

🧠 Memory Trick

AC Voltages Add as Vectors

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Voltage Addition in LCR Circuit

5 What will be the total circuit impedance if an LCR circuit operates at 50 Hz with R = 200 Ω, an absent inductor (L = 0), and a capacitor C = 15.0 × 10⁻⁶ F? (Use π = 3.14)

A 212.3 Ω

B 291.5 Ω

C 412.3 Ω

D 200.0 Ω

💡 Short Explanation

�� XL = 0. �� XC = 1/(2πfC). �� Use impedance formula.

✅ Correct Answer Explanation

XC = 1/(2πfC) = 1/[2 × 3.14 × 50 × 15 × 10⁻⁶] ≈ 212.3 Ω Z = √(R² + XC²) = √(200² + 212.3²) ≈ 291.5 Ω Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Only XC value.
�� Option C → Incorrect computation.
�� Option D → Ignores capacitive reactance.

🎯 Strategy

Used

Substitution

Application:

Calculate XC then Z.

Final Logic:

Z ≈ 291.5 Ω.

🧠 Memory Trick

No L ⇒ Z from R and XC

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

6 In the impedance triangle diagram, if the frequency of the source is increased indefinitely

A the hypotenuse Z eventually becomes identical to R

B the perpendicular side becomes dominated by XL, extending infinitely upwards

C the diagram collapses to a point because XL and XC both approach zero

D the angle φ approaches zero from a negative value

💡 Short Explanation

�� XL ∝ f. �� XC ∝ 1/f. �� High frequency makes XL dominant.

✅ Correct Answer Explanation

As frequency increases: XL = ωL → ∞ XC = 1/(ωC) → 0 Therefore the net reactance becomes strongly inductive and the perpendicular side of the impedance triangle grows upward. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → True only at resonance.
�� Option C → XL does not approach zero.
�� Option D → Phase approaches ±90°, not zero.

🎯 Strategy

Used

Formula Analysis

Application:

Examine frequency dependence of reactances.

Final Logic:

XL dominates at very high frequencies.

🧠 Memory Trick

High f → High XL

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Frequency Dependence of Impedance

7 For an LCR circuit, the expression for the power factor derived from the net phase angle formula will be

A R / √[R² + (XC − XL)²]

B (XC − XL)/R

C √[R² + (XC − XL)²]/R

D R/(XC − XL)

💡 Short Explanation

�� Power factor = cosφ. �� cosφ = R/Z. �� Substitute impedance.

✅ Correct Answer Explanation

Since: Z = √[R² + (XC − XL)²] Power factor: cosφ = R/Z = R / √[R² + (XC − XL)²] Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents tanφ numerator relation.
�� Option C → Reciprocal expression.
�� Option D → Incorrect formula.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use cosφ = R/Z.

Final Logic:

Power factor equals resistance divided by impedance.

🧠 Memory Trick

Power Factor = R/Z

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Power Factor

8 Under predominantly capacitive conditions, the sign of phase angle φ and the relationship of current to voltage are:

A Negative, Lags

B Positive, Leads

C Positive, Lags

D Negative, Leads

💡 Short Explanation

�� XC > XL. �� Current leads voltage. �� φ is positive in NCERT convention.

✅ Correct Answer Explanation

For a predominantly capacitive circuit: XC > XL tanφ = (XC − XL)/R > 0 Therefore φ is positive and current leads the voltage. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Both relations incorrect.
�� Option C → Current does not lag.
�� Option D → Wrong sign of φ.

🎯 Strategy

Used

Formula Recall

Application:

Apply capacitive condition XC > XL.

Final Logic:

Positive φ corresponds to leading current.

🧠 Memory Trick

Capacitor Leads

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

9 Match List I with List II regarding LCR analytical solutions

List I	List II
1. Phasor diagram solution	a. Fails to yield transient behavior
2. Transient solution	b. Dies out after a sufficiently long time
3. General solution	c. Accurate only after a long time
4. Steady-state solution	d. Needed to satisfy arbitrary initial conditions

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

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

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

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

💡 Short Explanation

�� Phasors describe steady state only. �� Transients decay with time. �� General solution includes initial conditions.

✅ Correct Answer Explanation

1 → a : Phasor method does not include transient effects. 2 → b : Transient solution dies out. 3 → d : General solution satisfies arbitrary initial conditions. 4 → c : Steady-state solution is valid after transients disappear. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Options B, C, D → Incorrect matching of transient and steady-state concepts.

🎯 Strategy

Used

Option Grouping

Application:

Match definitions with solution types.

Final Logic:

Only Option A gives all correct correspondences.

🧠 Memory Trick

Phasor → Steady State Only

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

10 When an LCR circuit is first energized, its immediate behavior differs from its long-term behavior because

A the applied voltage frequency slowly ramps up to its stated value

B the steady-state solution requires time to build up to its maximum amplitude

C an exponential transient component must decay before the pure sinusoidal steady-state solution dominates

D the resistor takes time to heat up and reach its nominal resistance

💡 Short Explanation

�� Initial behavior includes transients. �� Transients decay exponentially. �� Long-term response is sinusoidal.

✅ Correct Answer Explanation

The complete response consists of: Current = Steady-state solution + Transient solution The transient component decays exponentially with time, leaving only the sinusoidal steady-state response. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Source frequency is fixed.
�� Option B → Steady-state exists from the start mathematically.
�� Option D → Not the fundamental reason.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Separate transient and steady-state behavior.

Final Logic:

Decay of transient response causes the difference.

🧠 Memory Trick

Transient Dies → Steady Remains

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

11 Correct statements about a system's natural frequency tendency
1. It is the frequency at which the system tends to oscillate when undisturbed.
2. For an LCR circuit, it depends on L and C.
3. It requires periodic pulling or driving to exist.
4. Energy transfer between components defines this frequency.

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

B (2), (3)

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

D (3), (4)

💡 Short Explanation

�� Natural frequency is an inherent property. �� It depends on L and C. �� Energy oscillates between magnetic and electric fields.

✅ Correct Answer Explanation

1. Correct — Natural frequency is the frequency at which a system tends to oscillate on its own. 2. Correct — For an LCR circuit, ( \omega_0 = 1/\sqrt{LC} ), so it depends on L and C. 3. Incorrect — A natural frequency exists even without an external driving force. 4. Correct — Energy exchange between the inductor and capacitor determines this frequency. 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 correct statements 1 and 2.

🎯 Strategy

Used

Elimination

Application:

Identify the incorrect statement regarding external driving.

Final Logic:

Natural frequency exists inherently and does not require continuous driving.

🧠 Memory Trick

Natural = Own Frequency

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

12 To achieve maximum current amplitude conditions in a given LCR circuit

A the source frequency must be adjusted to infinity

B the resistance must be completely removed

C the applied frequency must perfectly match the circuit's natural frequency

D the phase difference between voltage and current must be π/2

💡 Short Explanation

�� Maximum current occurs at resonance. �� Resonance occurs when driving frequency equals natural frequency. �� Impedance becomes minimum.

✅ Correct Answer Explanation

Current amplitude is maximum when the circuit reaches resonance. At resonance: ω = ω₀ XL = XC Z = R (minimum) Therefore current becomes maximum. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Infinite frequency does not produce resonance.
�� Option B → Resistance affects current but is not the resonance condition.
�� Option D → At resonance phase difference is zero.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall resonance condition.

Final Logic:

Maximum current occurs when driving frequency equals natural frequency.

🧠 Memory Trick

Resonance ⇒ Maximum Current

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

13 In the phenomenon of resonant reactance cancellation:

A the actual physical inductors and capacitors are removed from the circuit

B the equal and opposite voltages across L and C sum to zero, effectively removing their combined opposition to the current

C the inductive reactance completely destroys the capacitor's ability to store charge

D the resistance is amplified to counteract the reactances

💡 Short Explanation

�� XL = XC at resonance. �� Reactive effects cancel. �� Circuit behaves like a pure resistor.

✅ Correct Answer Explanation

At resonance: XL = XC Therefore: VL = VC The voltages are equal in magnitude and opposite in phase, producing cancellation of reactive effects. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Components remain physically present.
�� Option C → Capacitor continues storing charge.
�� Option D → Resistance does not increase.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Apply resonance condition XL = XC.

Final Logic:

Opposite reactances cancel each other.

🧠 Memory Trick

XL = XC ⇒ Cancellation

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Resonant Reactance Cancellation

14 The formula for the resonant linear frequency ν₀ (in Hz) will be

A 1 / (2π(LC)^(1/2))

B 1 / (LC)^(1/2)

C 2π / (LC)^(1/2)

D (LC)^(1/2) / 2π

💡 Short Explanation

�� Resonant angular frequency is ω₀. �� ν₀ = ω₀ / 2π. �� Substitute ω₀ expression.

✅ Correct Answer Explanation

Since: ω₀ = 1/√LC and ν₀ = ω₀/(2π) Therefore: ν₀ = 1/(2π√LC) Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Missing factor 2π.
�� Option C → Incorrect placement of 2π.
�� Option D → Inverted formula.

🎯 Strategy

Used

Formula Recall

Application:

Convert angular frequency into linear frequency.

Final Logic:

ν₀ = 1/(2π√LC).

🧠 Memory Trick

f = ω / 2π

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

15 Identify the incorrect statement regarding the state of minimum impedance at resonance

A At this state, Z becomes exactly equal to R.

B The power factor cos(φ) becomes 1.

C The circuit behaves as a purely resistive circuit.

D The total impedance drops to zero if the driving frequency matches ω₀.

💡 Short Explanation

�� At resonance Z = R. �� Impedance is minimum but not zero. �� Circuit becomes purely resistive.

✅ Correct Answer Explanation

At resonance: XL = XC Net reactance = 0 Therefore: Z = R The impedance becomes minimum, not zero. Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct.
�� Option B → Correct because φ = 0.
�� Option C → Correct because reactive effects cancel.

🎯 Strategy

Used

Elimination

Application:

Apply resonance conditions.

Final Logic:

Only reactance becomes zero, not total impedance.

🧠 Memory Trick

Resonance: Z = R, Not 0

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

16 In an LCR circuit at resonance where R = 3.0 Ω and the peak source voltage is 283 V, what will be the rms current at resonance?

A 94.3 A

B 66.7 A

C 47.1 A

D 13.3 A

💡 Short Explanation

�� At resonance Z = R. �� Convert peak voltage to rms voltage. �� Apply Ohm's law.

✅ Correct Answer Explanation

Vrms = Vm/√2 = 283/1.414 ≈ 200 V At resonance: Irms = Vrms/R = 200/3 ≈ 66.7 A Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uses peak voltage directly.
�� Option C → Calculation error.
�� Option D → Incorrect substitution.

🎯 Strategy

Used

Substitution

Application:

Convert peak values to rms values before calculation.

Final Logic:

Irms = 200/3 = 66.7 A.

🧠 Memory Trick

RMS First, Then Ohm's Law

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

17 Consider the statements of radio tuning mechanism. Choose the correct statements:
1. The antenna receives signals of many frequencies simultaneously.
2. The tuning circuit is driven by these multiple signals.
3. Varying C changes the resonant frequency to match a specific station.
4. At resonance, the current amplitude for the desired signal becomes minimum to reduce noise.

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

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

C (1), (3)

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

💡 Short Explanation

�� Antenna receives many frequencies. �� Tuning selects one frequency. �� Resonance maximizes desired signal.

✅ Correct Answer Explanation

1. Correct — Many radio frequencies are intercepted. 2. Correct — These frequencies drive the tuning circuit. 3. Correct — Changing capacitance changes resonant frequency. 4. Incorrect — Current amplitude becomes maximum, not minimum, at resonance. 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 principle in radio tuning.

Final Logic:

Desired signal is amplified because current is maximum at resonance.

🧠 Memory Trick

Tune C to Select Station

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

18 In an airport metal detector, the presence of metal alters the coil's property, which in turn shifts the circuit out of resonance. The altered property and the resulting change in current are:

A Inductance, Significant change

B Capacitance, No change

C Resistance, Minimum change

D Inductance, Zero current

💡 Short Explanation

�� Metal affects inductance. �� Resonance condition changes. �� Circuit current changes noticeably.

✅ Correct Answer Explanation

Metal entering the magnetic field changes the effective inductance of the detector coil. This shifts the resonant frequency and causes a measurable change in circuit current. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Inductance is primarily affected.
�� Option C → Resistance change is not the operating principle.
�� Option D → Current changes but does not become zero.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Apply resonance detector principle.

Final Logic:

Metal changes inductance and therefore impedance.

🧠 Memory Trick

Metal → L Changes

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Airport Metal Detector

19 Match List I with List II regarding components in resonance

List I	List II
1. Inductor (L)	a. Determines maximum current amplitude vm/R
2. Capacitor (C)	b. Voltage lags current by π/2
3. Resistor (R)	c. Essential for voltage cancellation
4. Both L and C	d. Voltage leads current by π/2

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

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

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

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

💡 Short Explanation

�� VL leads current. �� VC lags current. �� Resistance determines resonance current.

✅ Correct Answer Explanation

1 → d : Inductor voltage leads current by π/2. 2 → b : Capacitor voltage lags current by π/2. 3 → a : Maximum current at resonance is vm/R. 4 → c : Both L and C are needed for voltage cancellation. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Options B, C and D contain incorrect phase relations and resonance matching.

🎯 Strategy

Used

Option Grouping

Application:

Match phase relationships and resonance roles.

Final Logic:

Only Option A gives all correct correspondences.

🧠 Memory Trick

L Leads, C Lags

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

20 The fundamental constraint preventing resonance in a purely RL circuit is that

A the inductive reactance always remains strictly greater than the resistance

B there is no opposing capacitive reactance to cancel the inductive reactance and bring the voltage and current into phase

C the magnetic field generated by the inductor completely suppresses the steady-state solution

D the resistance continuously increases with frequency, overpowering the inductance

💡 Short Explanation

�� Resonance requires XL = XC. �� RL circuit lacks capacitance. �� Reactance cancellation cannot occur.

✅ Correct Answer Explanation

Resonance occurs when inductive and capacitive reactances are equal and opposite. An RL circuit contains no capacitor, so XC is absent and cancellation is impossible. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → XL need not always exceed R.
�� Option C → Steady-state solution still exists.
�� Option D → Resistance is approximately frequency independent.

🎯 Strategy

Used

Direct Concept Recall

Application:

Apply 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: Limitation of RL and RC Circuits for Resonance

CUET UG Physics Booster Test 3-Series LCR Circuits and Resonance

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