CUET UG Physics Booster Test 3-Foundations of Alternating Current -

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CUET UG Physics Booster Test 3-Foundations of Alternating Current

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Foundations of Alternating Current

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QUESTION 1 OF 20

The shape of an alternating mains voltage over time and its average value over a full cycle:

Answer Review

1 The shape of an alternating mains voltage over time and its average value over a full cycle:

A Sine function, Zero

B Sine function, Maximum

C Constant, Zero

D Tangent function, Minimum

💡 Short Explanation

�� AC mains voltage varies sinusoidally with time. �� Positive and negative halves are equal and opposite. �� Average value over one complete cycle is zero.

✅ Correct Answer Explanation

The alternating mains voltage is represented by a sinusoidal function: v = vm sin(ωt) Therefore, the shape of the waveform is sinusoidal. During one complete cycle, the positive half-cycle and negative half-cycle cancel each other exactly. Hence the average value of the voltage over a complete cycle is zero. Thus, "Sine function, Zero" is the correct combination.

❓ Why Other Options Are Incorrect

�� Option B → The waveform is sinusoidal, but the average value is not maximum; it is zero.
�� Option C → AC voltage is not constant; it changes continuously with time.
�� Option D → AC voltage does not follow a tangent function and its average value is not minimum.

🎯 Strategy

Used

Elimination

Application:

Eliminate options that contradict the standard sinusoidal nature of AC voltage.

Final Logic:

AC voltage is sinusoidal and has zero average value over a complete cycle.

🧠 Memory Trick

AC = Sine Wave + Zero Average

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

2 Match List I with List II regarding AC quantities:

List I	List II
1. AC Voltage	a. Zero over one cycle
2. AC Current	b. vm sin(ωt)
3. Average current	c. i²R
4. Instantaneous power	d. im sin(ωt)

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

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

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

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

💡 Short Explanation

�� AC voltage = vm sin(ωt) �� AC current = im sin(ωt) �� Average current over a cycle = 0 �� Instantaneous power = i²R

✅ Correct Answer Explanation

1 → b because AC voltage is represented as vm sin(ωt). 2 → d because AC current is represented as im sin(ωt). 3 → a because the average current over a complete cycle is zero. 4 → c because instantaneous power dissipated in a resistor is i²R. Hence: 1-b, 2-d, 3-a, 4-c

❓ Why Other Options Are Incorrect

�� Option B → AC voltage and AC current expressions are interchanged.
�� Option C → Average current and current expression are incorrectly matched.
�� Option D → Power and voltage expressions are mismatched.

🎯 Strategy

Used

Option Grouping

Application:

Match standard AC expressions with their corresponding quantities.

Final Logic:

Only Option A provides all correct pairings.

🧠 Memory Trick

Voltage–Current–Average–Power = V–I–Zero–i²R

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: AC Quantities

3 A light bulb has a resistance of 484 Ω and operates directly on a 220 V rms ac supply. What is the average power consumed by the bulb over a complete cycle?

A 220 W

B 100 W

C 484 W

D 311 W

💡 Short Explanation

�� RMS values are used in AC power calculations. �� Average power in a resistor is P = V²/R. �� Use RMS voltage directly.

✅ Correct Answer Explanation

Given: Vrms = 220 V R = 484 Ω Average power: P = V²/R P = (220)²/484 P = 48400/484 P = 100 W Therefore, the average power consumed by the bulb is 100 W.

❓ Why Other Options Are Incorrect

�� Option A → Obtained from incorrect substitution.
�� Option C → Confuses resistance value with power.
�� Option D → Approximately equal to peak voltage, not power.

🎯 Strategy

Used

Substitution

Application:

Apply the average power formula using given RMS voltage and resistance.

Final Logic:

P = 220²/484 = 100 W.

🧠 Memory Trick

Power = V²/R

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Average Power in AC Circuits

4 In a purely inductive circuit driven by an instantaneous voltage v = vm sin(ωt),

A the integration constant evaluates to zero because the current oscillates symmetrically about zero

B the integration constant explicitly represents a steady direct current

C the slope di/dt evaluates to a steady constant value

D the instantaneous voltage is practically independent of time

💡 Short Explanation

�� AC current oscillates about zero. �� No DC component exists. �� Integration constant becomes zero.

✅ Correct Answer Explanation

For a pure inductor: v = L(di/dt) Integrating introduces a constant of integration. Since the current in an AC circuit oscillates equally above and below zero and contains no steady DC component, the integration constant must be zero. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → No steady direct current exists in the circuit.
�� Option C → di/dt varies continuously with time.
�� Option D → Voltage explicitly depends on time through sin(ωt).

🎯 Strategy

Used

Elimination

Application:

Reject statements inconsistent with sinusoidal AC behavior.

Final Logic:

The absence of a DC component makes the integration constant zero.

🧠 Memory Trick

Pure AC → No DC Offset

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: AC Through Inductor

5 The voltage amplitude in an AC circuit is,
1. is symbolically denoted by vm
2. is exactly equal to the rms voltage multiplied by 1.414
3. is strictly independent of the rms current
4. represents the peak value of the oscillating potential difference
Choose the correct option:

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

B (1), (2), and (4)

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

D (1), (3), and (4)

💡 Short Explanation

�� vm denotes peak voltage. �� Vm = √2 Vrms. �� Amplitude means maximum value.

✅ Correct Answer Explanation

1. Correct — vm is the symbol for voltage amplitude. 2. Correct — Vm = √2Vrms ≈ 1.414Vrms. 3. Incorrect — In a resistive circuit, RMS current depends on voltage amplitude through Ohm's law. 4. Correct — Amplitude represents the maximum value of the oscillating voltage. Hence statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 3.
�� Option C → Includes incorrect statement 3.
�� Option D → Includes incorrect statement 3.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using RMS relations and Ohm's law.

Final Logic:

Only statement 3 is incorrect.

🧠 Memory Trick

Amplitude = Peak = 1.414 × RMS

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

6 Consider the statements regarding current amplitude in a pure resistor. Choose the correct statements.
1. It is analytically calculated by dividing voltage amplitude by resistance
2. It is computationally equal to the rms current divided by 0.707
3. It predictably reaches its maximum value exactly when the voltage reaches its minimum
4. It actively determines the peak instantaneous power along with voltage amplitude

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

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

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

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

💡 Short Explanation

�� im = vm/R �� im = Irms/0.707 �� Peak power depends on peak voltage and current.

✅ Correct Answer Explanation

1. Correct — From Ohm's law, im = vm/R. 2. Correct — Since Irms = im/√2, im = Irms/0.707. 3. Incorrect — In a pure resistor, voltage and current are in phase and reach maxima simultaneously. 4. Correct — Instantaneous power depends on both voltage and current amplitudes. Thus 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 → Includes incorrect statement 3.

🎯 Strategy

Used

Elimination

Application:

Use phase relationships in a pure resistor.

Final Logic:

Statement 3 contradicts the in-phase nature of voltage and current.

🧠 Memory Trick

Resistor → V and I Peak Together

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Current Amplitude in Resistors

7 Identify the incorrect statement about the effective or rms voltage:

A It provides a robust way to express ac power in the same form as dc power

B It is mathematically equal to the peak voltage divided by √2

C The household line voltage of 220 V is definitively a peak voltage, not an rms value

D It is closely related to rms current by V = IR in a resistive circuit

💡 Short Explanation

�� Household voltage is RMS voltage. �� Vrms = Vm/√2. �� RMS values simplify power calculations.

✅ Correct Answer Explanation

Domestic supply voltage ratings such as 220 V refer to RMS values, not peak values. Peak voltage: Vm = √2 × 220 ≈ 311 V Therefore statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement regarding AC power.
�� Option B → Correct RMS relation.
�� Option D → Correct for a resistive circuit.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with RMS voltage definitions.

Final Logic:

220 V is an RMS value, not a peak value.

🧠 Memory Trick

Home Voltage = RMS Voltage

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

8 Identify the correct statements about the calculation of peak values from RMS:
1. The peak voltage is invariably larger than the rms voltage by a factor of √2
2. A 220 V rms supply yields a peak voltage of approximately 311 V
3. The established relation holds reliably true for purely sinusoidal voltages
4. Peak voltage is mathematically smaller than rms voltage

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

B (1), (2), (4)

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

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

💡 Short Explanation

�� Vm = √2Vrms. �� 220 V RMS gives 311 V peak. �� Formula applies to sinusoidal waves.

✅ Correct Answer Explanation

1. Correct — Vm = √2Vrms. 2. Correct — Vm = 1.414 × 220 ≈ 311 V. 3. Correct — The relation is valid for sinusoidal AC. 4. Incorrect — Peak voltage is always greater than RMS voltage. Therefore statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Apply RMS–peak relationships.

Final Logic:

Only statement 4 is incorrect.

🧠 Memory Trick

Peak = 1.414 × RMS

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Peak and RMS Voltage

9 The value of the integration of sin²(ωt) over one complete cycle divided by the period T, which is structurally used to derive the RMS current, is calculated precisely to:

A 0

B 1

C 1/2

D 1/√2

💡 Short Explanation

�� RMS derivation uses sin². �� Average value of sin² over a cycle is 1/2. �� Basis of RMS formulas.

✅ Correct Answer Explanation

The RMS current derivation uses: (1/T) ∫ sin²(ωt) dt over one complete cycle. The average value of sin² over a full cycle is: 1/2 Therefore the required value is 1/2.

❓ Why Other Options Are Incorrect

�� Option A → Average of sin² is not zero.
�� Option B → Average never reaches unity.
�� Option D → This value appears after taking the square root in RMS derivation.

🎯 Strategy

Used

Direct Formula Recall

Application:

Recall the standard trigonometric average value.

Final Logic:

Average value of sin² over a complete cycle is 1/2.

🧠 Memory Trick

sin² Average = 1/2

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: RMS Current Derivation

10 If an alternating current has a peak value of 1.414 A, what is the equivalent DC current that would produce the identical average Joule heating in a given resistor?

A 1.0 A

B 2.0 A

C 0.5 A

D 1.414 A

💡 Short Explanation

�� Equivalent DC current equals RMS current. �� Irms = Im/√2. �� RMS current gives same heating effect.

✅ Correct Answer Explanation

Given: Im = 1.414 A RMS current: Irms = Im/√2 Irms = 1.414/1.414 Irms = 1.0 A The equivalent DC current producing the same average heating effect is therefore 1.0 A.

❓ Why Other Options Are Incorrect

�� Option B → Exceeds peak current.
�� Option C → Incorrect RMS calculation.
�� Option D → Peak current is not equivalent DC current.

🎯 Strategy

Used

Substitution

Application:

Use the RMS current formula directly.

Final Logic:

Irms = 1.414/√2 = 1.0 A.

🧠 Memory Trick

Equivalent DC = RMS

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: RMS Current and Heating Effect

11 Consider the statements regarding rotating vector (phasor) analysis. Choose the correct statements.
1. Phasors rotate counter-clockwise with an angular frequency ω
2. The length of the phasor is explicitly proportional to the rms or peak value of the alternating quantity
3. Phasors adeptly represent scalar physical quantities mathematically
4. Phasors strictly represent physical vectors navigating in 3D space

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

B (1), (2), (4)

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

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

💡 Short Explanation

�� Phasors rotate with angular frequency ω. �� Phasor length represents amplitude. �� Phasors are mathematical representations.

✅ Correct Answer Explanation

1. Correct — A phasor is represented as a rotating vector moving counter-clockwise with angular velocity ω. 2. Correct — The phasor length is proportional to the peak or RMS value of the AC quantity. 3. Correct — Voltage and current are scalar quantities, but phasors provide a convenient mathematical representation. 4. Incorrect — Phasors are not physical vectors moving in three-dimensional space; they are mathematical tools. Hence statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Remove the statement that incorrectly interprets phasors as physical vectors.

Final Logic:

Only statement 4 is incorrect.

🧠 Memory Trick

Phasor = Mathematical Rotator

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

12 Choose the correct statements about the projection of phasors:
1. The vertical projection of the voltage phasor distinctly represents vm sin(ωt)
2. The projection gives the tangible value of the voltage at that specific instant
3. As the phasor physically rotates, the projection effectively generates a sinusoidal curve
4. The vertical component remains strictly constant over progressing time

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

B (1), (2), (4)

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

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

💡 Short Explanation

�� Vertical projection gives instantaneous value. �� Projection varies sinusoidally. �� Projection is not constant.

✅ Correct Answer Explanation

1. Correct — The vertical projection of a rotating voltage phasor equals vm sin(ωt). 2. Correct — This projection represents the instantaneous value of voltage. 3. Correct — Continuous rotation produces a sinusoidal variation with time. 4. Incorrect — The vertical projection changes continuously and is not constant. Therefore statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify the statement that contradicts sinusoidal variation.

Final Logic:

The vertical projection changes with time and cannot remain constant.

🧠 Memory Trick

Projection = Instantaneous Value

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

13 Choose the incorrect statement about angular frequency and frequency:

A ω is the angular frequency of the oscillating potential difference

B The phasor rotates about the origin with angular speed ω

C The inductive reactance visibly limits current and is inversely proportional to frequency

D The capacitive reactance fundamentally remains inversely proportional to the frequency

💡 Short Explanation

�� XL = ωL = 2πfL. �� Inductive reactance increases with frequency. �� Capacitive reactance decreases with frequency.

✅ Correct Answer Explanation

Inductive reactance is given by: XL = ωL = 2πfL Hence XL is directly proportional to frequency, not inversely proportional. Therefore statement C is incorrect. Statements A and B correctly describe angular frequency and phasor motion. Statement D is correct because: XC = 1/(ωC) which is inversely proportional to frequency.

❓ Why Other Options Are Incorrect

�� Option A → Correct definition of angular frequency.
�� Option B → Correct description of phasor rotation.
�� Option D → Correct relation for capacitive reactance.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use reactance formulas directly.

Final Logic:

XL ∝ f, not 1/f.

🧠 Memory Trick

Inductor Likes Frequency

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

14 Regarding the specific phase angle between voltage and current in basic AC circuits:
1. In a pure resistor, the phase angle is exactly zero
2. In a pure inductor, the current consistently lags the voltage by π/2
3. In a pure capacitor, the current predictably leads the voltage by π/2
4. The phase angle primarily dictates the amount of wattless current
Choose correct:

A (1) and (2) only

B (2) and (3) only

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

D (3) and (4) only

💡 Short Explanation

�� Resistor → phase difference zero. �� Inductor → current lags by π/2. �� Capacitor → current leads by π/2.

✅ Correct Answer Explanation

1. Correct — Voltage and current are in phase in a resistor. 2. Correct — In a pure inductor, current lags voltage by π/2. 3. Correct — In a pure capacitor, current leads voltage by π/2. 4. Though phase angle influences reactive behavior, the option set provided identifies only statements 1, 2 and 3 as the standard NCERT relationships. Hence Option C is the correct answer.

❓ Why Other Options Are Incorrect

�� Option A → Omits statement 3.
�� Option B → Omits statement 1.
�� Option D → Excludes two correct statements.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall standard phase relationships in R, L and C circuits.

Final Logic:

Statements 1, 2 and 3 are the fundamental phase-angle relations.

🧠 Memory Trick

R = 0, L = Lag, C = Lead

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

15 Which of the following was explicitly NOT mentioned as a defining invention or conception of Nikola Tesla?

A The rotating magnetic field

B The induction motor

C The polyphase system of ac power

D The modern electric battery

💡 Short Explanation

�� Tesla pioneered AC technology. �� Developed induction motor. �� Conceived rotating magnetic field.

✅ Correct Answer Explanation

Tesla is associated with the rotating magnetic field, induction motor and polyphase AC power systems. The modern electric battery was not one of Tesla's defining inventions. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → A major Tesla contribution.
�� Option B → Tesla invented the induction motor.
�� Option C → Tesla pioneered polyphase AC systems.

🎯 Strategy

Used

Odd One Out

Application:

Identify the invention unrelated to Tesla's AC innovations.

Final Logic:

Battery technology is not a defining Tesla invention.

🧠 Memory Trick

Tesla = AC, Not Battery

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Historical Development of AC

16 The primary electrical systems advocated enthusiastically by George Westinghouse and Thomas Edison respectively were:

A Alternating current, Direct current

B Direct current, Alternating current

C Polyphase systems, Transformers

D Direct current, Batteries

💡 Short Explanation

�� Westinghouse supported AC. �� Edison supported DC. �� This conflict formed the "War of Currents."

✅ Correct Answer Explanation

George Westinghouse strongly promoted alternating current systems because of their advantages in transmission and voltage conversion. Thomas Edison advocated direct current systems. Hence the correct pairing is: Alternating current, Direct current.

❓ Why Other Options Are Incorrect

�� Option B → Reverses their positions.
�� Option C → Not electrical systems.
�� Option D → Incorrect association.

🎯 Strategy

Used

Historical Recall

Application:

Recall the War of Currents.

Final Logic:

Westinghouse → AC, Edison → DC.

🧠 Memory Trick

West = AC, Edison = DC

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: History of AC Power

17 Match List I with List II regarding transformers and power transmission:

List I	List II
1. Step-up transformer	a. Output coil
2. Step-down transformer	b. Reduces voltage
3. Primary coil	c. Increases voltage
4. Secondary coil	d. Input coil

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

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

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

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

💡 Short Explanation

�� Step-up increases voltage. �� Step-down reduces voltage. �� Primary is input coil. �� Secondary is output coil.

✅ Correct Answer Explanation

1 → c because a step-up transformer increases voltage. 2 → b because a step-down transformer decreases voltage. 3 → d because the primary coil receives input power. 4 → a because the secondary coil delivers output power. Therefore: 1-c, 2-b, 3-d, 4-a

❓ Why Other Options Are Incorrect

�� Option B → Step-up and step-down functions reversed.
�� Option C → Primary and secondary coils interchanged.
�� Option D → Multiple incorrect matches.

🎯 Strategy

Used

Option Grouping

Application:

Match transformer components with functions.

Final Logic:

Only Option A provides all correct pairings.

🧠 Memory Trick

Primary = Input, Secondary = Output

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

18 In an AC distribution network, when electrical energy is transmitted economically over long distances,

A the voltage is typically stepped up to substantially reduce current and minimize I²R power losses

B the voltage is stepped down to completely ensure the wires do not overheat

C direct current is exclusively used to categorically prevent phase lagging

D transformers operate optimally at low voltages to purposely maximize current flow

💡 Short Explanation

�� Power loss = I²R. �� High voltage means low current. �� Lower current reduces losses.

✅ Correct Answer Explanation

For a given power: P = VI Increasing voltage reduces current. Since transmission losses are: Ploss = I²R reducing current greatly decreases power loss. Therefore high-voltage transmission is economically preferred.

❓ Why Other Options Are Incorrect

�� Option B → Voltage is stepped up, not down, during transmission.
�� Option C → AC transmission is widely used.
�� Option D → Low voltage increases current and losses.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use P = VI and I²R loss equations.

Final Logic:

High voltage minimizes current and transmission loss.

🧠 Memory Trick

High V → Low I → Low Loss

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Long Distance Transmission

19 Choose the correct statements about AC voltage applied to a pure resistor:
1. The voltage and current are exactly in phase with each other
2. The minima, zero, and maxima of voltage and current naturally occur at the same respective times
3. Ohm's law fundamentally works equally well for AC and DC voltages in resistors
4. The average power dissipated predictably averages to zero

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

B (1), (2), (4)

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

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

💡 Short Explanation

�� Voltage and current are in phase. �� Ohm's law applies. �� Average power is not zero.

✅ Correct Answer Explanation

1. Correct — Voltage and current remain in phase. 2. Correct — Both reach maxima, minima and zero simultaneously. 3. Correct — Ohm's law applies to a pure resistor for both AC and DC. 4. Incorrect — Average power is positive and equals I²R. Therefore statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Use the power relation in a resistor.

Final Logic:

Average power is not zero; statement 4 is false.

🧠 Memory Trick

Resistor = In Phase + Positive Power

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: AC Through Resistor

20 Identify the incorrect statement about the measurable current passing through a pure resistor in an AC circuit:

A It observably varies sinusoidally with time

B Its amplitude is reliably given by im = vm/R

C It inherently exhibits a phase difference of π/2 relative to the applied voltage

D The instantaneous power dissipated directly depends on the square of the instantaneous current

💡 Short Explanation

�� Current and voltage are in phase. �� im = vm/R. �� Power depends on i²R.

✅ Correct Answer Explanation

In a pure resistor: i = im sin(ωt) and v = vm sin(ωt) Both quantities are in phase, meaning the phase difference is zero. Therefore statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct property of AC current.
�� Option B → Correct expression from Ohm's law.
�� Option D → Instantaneous power equals i²R.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with resistive AC circuits.

Final Logic:

A pure resistor has zero phase difference, not π/2.

🧠 Memory Trick

Pure Resistor = Zero Phase Angle

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Current in a Pure Resistor

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