CUET UG Physics Booster Test 2-Foundations of Alternating Current -

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CUET UG Physics Booster Test 2-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 2-Foundations of Alternating Current

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QUESTION 1 OF 20

The AC voltage in an electrical mains supply
1. varies like a sine function with time
2. can be efficiently converted using transformers
3. does not change direction with time
4. is entirely equivalent to a pure battery source

Answer Review

1 The AC voltage in an electrical mains supply
1. varies like a sine function with time
2. can be efficiently converted using transformers
3. does not change direction with time
4. is entirely equivalent to a pure battery source

A (1) and (2)

B (2) and (3)

C (3) and (4)

D (1) and (4)

💡 Short Explanation

�� AC voltage varies sinusoidally with time. �� Transformers work efficiently only with AC. �� AC reverses polarity periodically.

✅ Correct Answer Explanation

Alternating voltage is represented by: v = vm sin(ωt) Hence statement 1 is correct because AC voltage varies sinusoidally with time. Statement 2 is also correct because transformers operate on mutual induction and can efficiently step up or step down AC voltages. Statement 3 is incorrect because AC changes direction periodically. Statement 4 is incorrect because a battery supplies DC, whereas AC continuously changes magnitude and direction. Therefore statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect because AC changes direction periodically.
�� Option C → Statements 3 and 4 are both incorrect.
�� Option D → Statement 4 is incorrect because AC is not equivalent to a battery source.

🎯 Strategy

Used

Elimination

Application:

Identify statements that contradict the basic definition of alternating current.

Final Logic:

Only statements 1 and 2 correctly describe AC voltage.

🧠 Memory Trick

AC = Alternates + Converts

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Nature and Advantages of AC

2 Choose the incorrect statement about AC current driven by an AC voltage:

A It is driven by an alternating voltage in a circuit

B It does not change direction with time

C Most of the electrical devices we use today require it

D It varies harmonically with time

💡 Short Explanation

�� AC reverses direction periodically. �� AC is produced by alternating voltage. �� Sinusoidal AC varies harmonically with time.

✅ Correct Answer Explanation

Alternating current continuously changes both magnitude and direction with time. Therefore statement B is incorrect. Statements A, C, and D correctly describe AC current. AC is produced by alternating voltage sources and usually varies sinusoidally with time.

❓ Why Other Options Are Incorrect

�� Option A → Correct property of AC current.
�� Option C → AC is the standard form of electrical power supply.
�� Option D → Sinusoidal variation is a basic feature of AC.

🎯 Strategy

Used

Odd One Out

Application:

Find the statement that contradicts the definition of AC.

Final Logic:

Only Option B states that AC does not reverse direction.

🧠 Memory Trick

AC = Always Changes direction

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

3 Consider the statements on sinusoidal function properties. Choose the correct statements:
1. The sum of instantaneous values over one complete cycle is zero
2. The voltage and current in a pure resistor reach zero at the same time
3. The average value of sine squared over a cycle is zero
4. The instantaneous value is time-dependent

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

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

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

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

💡 Short Explanation

�� Average value of sine over a cycle is zero. �� Voltage and current are in phase in a resistor. �� Average value of sin²θ is 1/2.

✅ Correct Answer Explanation

Statement 1 is correct because positive and negative halves of a sine wave cancel over a complete cycle. Statement 2 is correct because voltage and current are in phase in a pure resistor and therefore reach maxima, minima, and zero simultaneously. Statement 3 is incorrect because: Average value of sin²θ = 1/2 not zero. Statement 4 is correct because instantaneous voltage and current depend on time. 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 → Includes incorrect statement 3.

🎯 Strategy

Used

Elimination

Application:

Check each statement using standard sinusoidal properties.

Final Logic:

Only statement 3 is false.

🧠 Memory Trick

sin² average = 1/2

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

4 When calculating the instantaneous current in a purely inductive circuit driven by v = vm sin(ωt), the integration constant evaluates to zero because:

A the source has an emf which oscillates symmetrically about zero

B the resistance is extremely high

C time is strictly an independent variable

D the circuit inherently has no power loss

💡 Short Explanation

�� AC source has no DC offset. �� Positive and negative halves are symmetric. �� Integration constant becomes zero.

✅ Correct Answer Explanation

While integrating the voltage equation for a pure inductor, an integration constant appears. Since the source voltage oscillates symmetrically about zero and has no constant DC component, the integration constant must be zero. The remaining options do not affect the value of the integration constant.

❓ Why Other Options Are Incorrect

�� Option B → Resistance does not determine the integration constant.
�� Option C → Time being an independent variable is unrelated.
�� Option D → Power loss does not affect the integration constant.

🎯 Strategy

Used

Elimination

Application:

Remove options unrelated to the mathematical integration process.

Final Logic:

Symmetry about zero ensures zero integration constant.

🧠 Memory Trick

Pure AC → Zero Offset

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

5 In a purely resistive AC circuit, the voltage amplitude

A determines the peak current amplitude through Ohm's law as im = vm/R

B is equal to the RMS voltage at all times

C constantly changes the phase angle between current and voltage

D is always zero over a complete cycle

💡 Short Explanation

�� Ohm's law applies to AC resistor circuits. �� Voltage and current remain in phase. �� Peak values satisfy im = vm/R.

✅ Correct Answer Explanation

For a pure resistor: v = iR Using peak values: vm = imR Therefore: im = vm/R Hence voltage amplitude determines current amplitude. RMS voltage is not equal to peak voltage. The phase angle remains zero and voltage amplitude is not zero.

❓ Why Other Options Are Incorrect

�� Option B → RMS voltage equals vm/√2.
�� Option C → Phase difference remains zero.
�� Option D → Voltage amplitude is the maximum value, not zero.

🎯 Strategy

Used

Substitution

Application:

Apply Ohm's law directly to peak values.

Final Logic:

im = vm/R directly proves Option A.

🧠 Memory Trick

Peak V ÷ R = Peak I

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

6 Match List I with List II:

List I	List II
1. Peak current im	a. vm / R
2. Instantaneous current i	b. Zero
3. RMS current I	c. im / √2
4. Average current over a cycle	d. im sin(ωt)

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

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

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

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

💡 Short Explanation

�� Peak current = vm/R. �� Instantaneous current = im sin(ωt). �� RMS current = im/√2. �� Average current over a cycle = 0.

✅ Correct Answer Explanation

1 → a because im = vm/R 2 → d because i = im sin(ωt) 3 → c because I = im/√2 4 → b because average AC current over one complete cycle is zero. Therefore: 1-a, 2-d, 3-c, 4-b

❓ Why Other Options Are Incorrect

�� Option B → Incorrect mapping of all major quantities.
�� Option C → Peak and instantaneous current are interchanged.
�� Option D → RMS and instantaneous current are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match standard AC expressions with their definitions.

Final Logic:

Only Option A gives all four correct pairings.

🧠 Memory Trick

Peak–Instant–RMS–Average = V/R–Sin–√2–Zero

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Peak, RMS and Average Values

7 Choose the correct statements about effective voltage:
1. It is also called root mean square voltage
2. It is equal to 0.707 vm
3. It is customarily used to measure and specify AC quantities like household line voltage
4. It is fundamentally twice the peak voltage

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

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

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

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

💡 Short Explanation

�� Effective voltage means RMS voltage. �� RMS voltage equals 0.707 Vm. �� Household voltage ratings use RMS values.

✅ Correct Answer Explanation

Statement 1 is correct because effective voltage and RMS voltage are the same. Statement 2 is correct because: Vrms = Vm/√2 = 0.707Vm Statement 3 is correct because domestic supply voltages are specified using RMS values. Statement 4 is incorrect because RMS voltage is smaller than peak voltage.

❓ 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 RMS relations to verify each statement.

Final Logic:

Only statement 4 is false.

🧠 Memory Trick

RMS = 70.7% of Peak

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

8 If the rms voltage across a resistor is 100 V, what will be the peak voltage of the source?

A 70.7 V

B 141.4 V

C 200 V

D 100 V

💡 Short Explanation

�� RMS and peak values differ. �� Vm = √2 Vrms. �� √2 ≈ 1.414.

✅ Correct Answer Explanation

Given: Vrms = 100 V Peak voltage: Vm = √2 × 100 Vm = 141.4 V Therefore the peak voltage is 141.4 V.

❓ Why Other Options Are Incorrect

�� Option A → Less than RMS value.
�� Option C → Calculation error.
�� Option D → RMS and peak values are not equal.

🎯 Strategy

Used

Substitution

Application:

Substitute the given RMS value into the standard formula.

Final Logic:

Vm = 1.414 × 100 = 141.4 V.

🧠 Memory Trick

Peak = 1.414 × RMS

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

9 The root mean square (rms) current in an AC circuit is,
1. is mathematically defined as im / √2
2. produces the same average power loss as an equivalent DC current
3. is strictly equal to the peak current
4. is always zero over a complete cycle
Choose correct:

A (1) and (2)

B (2) and (3)

C (3) and (4)

D (1) and (4)

💡 Short Explanation

�� RMS current equals im/√2. �� RMS value gives DC-equivalent heating effect. �� RMS current is not zero.

✅ Correct Answer Explanation

Statement 1 is correct because: Irms = im/√2 Statement 2 is correct because RMS current is defined as the value of DC current producing the same average power loss in a resistor. Statement 3 is incorrect because RMS current is less than peak current. Statement 4 is incorrect because RMS current has a finite positive value. Therefore statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Both statements are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

Use RMS definition and heating effect concept.

Final Logic:

Only statements 1 and 2 are correct.

🧠 Memory Trick

RMS = Heating Equivalent DC

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

10 To express AC power in the same formulaic form as DC power (P = I²R), what specific value of the alternating current is utilized?

A Instantaneous current

B Peak current

C Average current

D Effective or RMS current

💡 Short Explanation

�� RMS current gives equivalent heating effect. �� AC power calculations use RMS values. �� Average current over a cycle is zero.

✅ Correct Answer Explanation

The average power dissipated in a resistor carrying alternating current is: P = I²R where I represents the RMS current. The RMS value is chosen because it produces the same heating effect as an equivalent DC current. Therefore AC power expressions can be written in the same form as DC power expressions.

❓ Why Other Options Are Incorrect

�� Option A → Changes continuously with time.
�� Option B → Does not represent average heating effect.
�� Option C → Average AC current over a cycle is zero.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the quantity that makes AC and DC power formulas equivalent.

Final Logic:

RMS current uniquely represents DC-equivalent power dissipation.

🧠 Memory Trick

Power Uses RMS

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Average Power and RMS Values

11 Voltage and current quantities in an AC circuit, and the mathematical way they combine using phasors are respectively:

A Scalar, Vectorial

B Vector, Scalar

C Scalar, Scalar

D Vector, Vector

💡 Short Explanation

�� Voltage and current are scalar physical quantities. �� Phasors provide a vector-like representation. �� Phasor addition follows vector rules.

✅ Correct Answer Explanation

Voltage and current possess magnitude but no spatial direction; hence they are scalar quantities. However, AC voltages and currents are conveniently represented by rotating vectors called phasors. The addition and subtraction of phasors follow vector algebra. Therefore: Voltage and current → Scalar Phasor combination → Vectorial Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Voltage and current are not vector quantities.
�� Option C → Phasor addition is vectorial, not scalar.
�� Option D → Voltage and current are scalars.

🎯 Strategy

Used

Conceptual Classification (Odd One Out)

Application:

Separate physical quantities from their mathematical representation.

Final Logic:

Quantities are scalar, but phasor operations are vectorial.

🧠 Memory Trick

Physical Scalar, Phasor Vector

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

12 In the application of phasor diagrams to an AC circuit,

A the horizontal components exclusively represent the instantaneously varying quantities

B the vertical components of phasors V and I represent the sinusoidally varying quantities v and i

C the length of the phasor exactly represents the instantaneous value

D the phasors remain stationary while the reference axes rotate

💡 Short Explanation

�� Phasors rotate with angular velocity ω. �� Vertical projection gives instantaneous value. �� Phasor length gives amplitude.

✅ Correct Answer Explanation

In phasor representation, voltage and current are shown as rotating vectors. The instantaneous values are obtained from the vertical projection of the rotating phasors. Thus: v = vertical component of V i = vertical component of I Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Instantaneous values are represented by vertical components.
�� Option C → Phasor length represents amplitude, not instantaneous value.
�� Option D → Reference axes are fixed while phasors rotate.

🎯 Strategy

Used

Elimination

Application:

Compare each statement with standard phasor-diagram conventions.

Final Logic:

Only vertical projections represent instantaneous sinusoidal values.

🧠 Memory Trick

Vertical Projection = Instantaneous Value

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

13 Match List I with List II regarding terms in v = vm sin(ωt)

List I	List II
1. ω	a. Time instant
2. t	b. Voltage amplitude
3. ωt	c. Phase angle parameter
4. vm	d. Angular frequency

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

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

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

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

💡 Short Explanation

�� ω denotes angular frequency. �� t denotes time. �� ωt represents phase angle. �� vm denotes maximum voltage.

✅ Correct Answer Explanation

In v = vm sin(ωt) 1 → d because ω is angular frequency. 2 → a because t is time. 3 → c because ωt represents phase angle. 4 → b because vm is voltage amplitude. Therefore: 1-d, 2-a, 3-c, 4-b

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly assigns ω and vm.
�� Option C → Interchanges time and phase angle.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Option Grouping

Application:

Match each symbol with its physical meaning.

Final Logic:

Only Option A gives all correct correspondences.

🧠 Memory Trick

ω-Time-Phase-Peak

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

14 Choose the incorrect statement about phase angle in a purely resistive AC circuit:

A The phase angle between voltage and current is zero

B Voltage and current reach their maxima at the same time

C The current phasor is π/2 behind the voltage phasor

D Phasors V and I are in the same direction for all times

💡 Short Explanation

�� In a resistor, voltage and current are in phase. �� Phase difference equals zero. �� Maxima occur simultaneously.

✅ Correct Answer Explanation

For a pure resistor: v = Vm sin(ωt) i = Im sin(ωt) Both quantities have identical phase. Therefore phase angle is zero. Statements A, B and D are correct. Statement C is incorrect because a phase difference of π/2 occurs in pure inductive or capacitive circuits, not in a pure resistor.

❓ Why Other Options Are Incorrect

�� Option A → Correct property of a pure resistor.
�� Option B → Current and voltage peak simultaneously.
�� Option D → Phasors remain aligned.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with in-phase behavior.

Final Logic:

Only Option C introduces a non-zero phase difference.

🧠 Memory Trick

Resistor = Zero Phase

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

15 Consider the statements on Nikola Tesla contributions. Choose the correct statements.
1. He conceived the idea of the rotating magnetic field
2. He invented the induction motor
3. He opposed the use of alternating current
4. He invented the high frequency induction coil

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

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

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

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

💡 Short Explanation

�� Tesla pioneered AC technology. �� Invented induction motor. �� Developed Tesla coil.

✅ Correct Answer Explanation

Statement 1 is correct because Tesla developed the rotating magnetic field concept. Statement 2 is correct because he invented the induction motor. Statement 4 is correct because he developed the high-frequency induction coil known as the Tesla coil. Statement 3 is incorrect because Tesla strongly supported alternating current. 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:

Remove the historically incorrect statement.

Final Logic:

Tesla supported AC, making statement 3 false.

🧠 Memory Trick

Tesla = AC + Motor + Coil

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

16 Choose the correct statements about George Westinghouse:
1. He was a leading proponent of alternating current over direct current
2. He enlisted the services of Nikola Tesla and other inventors
3. He advocated direct current alongside Thomas Edison
4. He pioneered in large scale lighting utilizing AC

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

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

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

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

💡 Short Explanation

�� Westinghouse promoted AC systems. �� Worked with Tesla. �� Helped establish large-scale AC distribution.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are historically correct. Westinghouse became a major supporter of AC power systems, collaborated with Tesla, and promoted large-scale AC transmission and lighting projects. Statement 3 is incorrect because Thomas Edison supported DC, whereas Westinghouse promoted AC.

❓ 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:

Identify the statement conflicting with AC history.

Final Logic:

Westinghouse supported AC, not DC.

🧠 Memory Trick

Westinghouse = West for AC

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

17 The primary device for AC voltage conversion and the fundamental principle it prominently utilises:

A Transformer, Mutual induction

B Resistor, Joule heating

C Capacitor, Energy storage

D Inductor, Self-induction

💡 Short Explanation

�� Transformers change AC voltage. �� Mutual induction links two coils. �� Widely used in power transmission.

✅ Correct Answer Explanation

A transformer operates using mutual induction between primary and secondary windings. Changing current in the primary coil produces changing magnetic flux which induces emf in the secondary coil. Therefore, AC voltage conversion is achieved using transformers based on mutual induction.

❓ Why Other Options Are Incorrect

�� Option B → Resistors cannot transform voltage.
�� Option C → Capacitors store energy but do not perform voltage transformation.
�� Option D → Self-induction alone is insufficient for transformer action.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the device with its operating principle.

Final Logic:

Transformer operation fundamentally depends on mutual induction.

🧠 Memory Trick

Transformer → Mutual Transfer

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

18 If an ideal step-up transformer at a power station changes a 220 V input at 10 A to an output of 440 V, what will be the output current transmitted over the long distance?

A 20 A

B 5 A

C 10 A

D 40 A

💡 Short Explanation

�� Ideal transformer conserves power. �� Input power equals output power. �� Current decreases when voltage increases.

✅ Correct Answer Explanation

For an ideal transformer: VpIp = VsIs 220 × 10 = 440 × Is 2200 = 440Is Is = 5 A Therefore, output current equals 5 A.

❓ Why Other Options Are Incorrect

�� Option A → Would violate power conservation.
�� Option C → Current must decrease when voltage doubles.
�� Option D → Impossible under ideal transformer conditions.

🎯 Strategy

Used

Substitution

Application:

Apply transformer power relation directly.

Final Logic:

220 × 10 = 440 × Is gives Is = 5 A.

🧠 Memory Trick

Voltage ↑ ⇒ Current ↓

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

19 When AC voltage is applied to a pure resistor, the sum of the instantaneous current values over one complete cycle is zero. This functionally implies that:

A the average power consumed is ultimately zero

B there is absolutely no dissipation of electrical energy

C Joule heating still occurs and depends on i²

D the resistor acts entirely as an open circuit

💡 Short Explanation

�� Average current over a cycle is zero. �� Heating depends on i²R. �� Positive and negative currents both produce heat.

✅ Correct Answer Explanation

Although the algebraic sum of current values over one cycle is zero, power dissipation depends on: P = i²R Since i² is always positive, electrical energy continues to be converted into heat during both halves of the cycle. Therefore Joule heating occurs continuously.

❓ Why Other Options Are Incorrect

�� Option A → Average power is not zero.
�� Option B → Heat is continuously produced.
�� Option D → Current clearly flows through the resistor.

🎯 Strategy

Used

Odd One Out

Application:

Distinguish current averaging from power averaging.

Final Logic:

Heating depends on i², not on the average value of i.

🧠 Memory Trick

Current Cancels, Heat Doesn't

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Heating Effect of AC

20 The current in a pure resistor driven by an AC voltage
1. varies sinusoidally
2. has corresponding positive and negative values during each cycle
3. strictly lags the applied voltage by π/2
4. has an average value of zero over one cycle
Choose the correct options:

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

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

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

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

💡 Short Explanation

�� Current follows sinusoidal voltage. �� Positive and negative halves occur. �� Average current over a cycle is zero.

✅ Correct Answer Explanation

For a pure resistor: i = Im sin(ωt) Hence statement 1 is correct. Statement 2 is correct because sinusoidal current assumes both positive and negative values. Statement 4 is correct because the positive and negative halves cancel over a complete cycle. Statement 3 is incorrect because current and voltage are in phase; there is no phase lag. Therefore 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:

Apply the in-phase property of a pure resistor.

Final Logic:

Only statement 3 is false.

🧠 Memory Trick

Pure Resistor = Zero Lag

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

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