CUET UG Physics Booster Test 3-Pure Inductive and Capacitive Circuits -

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CUET UG Physics Booster Test 3-Pure Inductive and Capacitive Circuits

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Pure Inductive and Capacitive Circuits

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QUESTION 1 OF 20

Correct statements on the mathematical derivation of current in an inductor:
1. di/dt = (vm / L) sin(ωt)
2. Integrating di/dt gives a −cos(ωt) term.
3. The time-independent integration constant is evaluated to zero by symmetry.
4. The amplitude of the current is exactly vmωL.

Answer Review

1 Correct statements on the mathematical derivation of current in an inductor:
1. di/dt = (vm / L) sin(ωt)
2. Integrating di/dt gives a −cos(ωt) term.
3. The time-independent integration constant is evaluated to zero by symmetry.
4. The amplitude of the current is exactly vmωL.

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

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

C (1), (3), (4)

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

💡 Short Explanation

�� Voltage across an inductor is proportional to di/dt. �� Integration produces a cosine term. �� No DC component exists in pure AC current.

✅ Correct Answer Explanation

For a pure inductor: v = L(di/dt) Given: v = vm sin(ωt) Therefore: di/dt = (vm/L) sin(ωt) Integrating: i = −(vm/ωL) cos(ωt) + C Since the current oscillates symmetrically about zero, the integration constant C = 0. The amplitude of current is: im = vm/(ωL) not vmωL. 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

Direct Formula Recall

Application:

Use v = L(di/dt) and integrate.

Final Logic:

Current amplitude equals vm/(ωL), not vmωL.

🧠 Memory Trick

Inductor Current = V divided by ωL

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

2 The assumption of an ideal pure inductor with negligible resistance implies:

A it consumes infinite instantaneous power

B all electrical energy is permanently dissipated

C the entire voltage drop matches the self-induced Faraday emf exactly

D the phase angle between voltage and current becomes zero

💡 Short Explanation

�� Resistance is neglected. �� Only self-induced emf balances source voltage. �� Current and voltage remain out of phase.

✅ Correct Answer Explanation

In a pure inductor: v − L(di/dt) = 0 Thus the applied voltage is completely balanced by the self-induced emf. No resistive voltage drop exists. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Instantaneous power remains finite.
�� Option B → No net energy dissipation occurs.
�� Option D → Phase difference remains π/2.

🎯 Strategy

Used

Elimination

Application:

Identify the consequence of neglecting resistance.

Final Logic:

Only self-induced emf balances the source voltage.

🧠 Memory Trick

Pure Inductor = Voltage Opposed by Induced EMF

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

3 If a purely inductive open coil's interior receives an inserted iron rod, the self-induced emf opposing the current changes because:

A The iron rod decreases the self-inductance L.

B The iron rod magnetizes and increases the inductance L.

C The frequency of the ac source increases significantly.

D The capacitive reactance completely dominates the circuit.

💡 Short Explanation

�� Iron has high magnetic permeability. �� Flux linkage increases. �� Self-inductance increases.

✅ Correct Answer Explanation

When an iron core is inserted into a coil, magnetic permeability increases significantly. This increases magnetic flux linkage and therefore increases self-inductance L. Since induced emf equals: ε = −L(di/dt) the self-induced emf increases. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Inductance increases, not decreases.
�� Option C → Frequency is unchanged.
�� Option D → No capacitor is involved.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Relate magnetic permeability to inductance.

Final Logic:

Iron core increases L.

🧠 Memory Trick

Iron Core → More Flux → More L

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

4 The negative sign strictly enforced in the loop equation v − L(di/dt) = 0 is a direct mathematical consequence of

A Ohm's Law

B Faraday's self-induced emf opposing the change in current (Lenz's Law)

C Kirchhoff's Junction Rule

D the alternating polarity of the source voltage

💡 Short Explanation

�� Induced emf opposes current change. �� Lenz's law determines the sign. �� Negative sign represents opposition.

✅ Correct Answer Explanation

The induced emf in an inductor is: ε = −L(di/dt) The negative sign originates from Lenz's law, which states that induced emf opposes the cause producing it. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ohm's law does not determine this sign.
�� Option C → Junction rule concerns current conservation.
�� Option D → AC polarity changes do not create the negative sign.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall Faraday-Lenz law.

Final Logic:

Negative sign comes from opposition to change.

🧠 Memory Trick

Lenz = Negative Sign

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Lenz's Law

5 Iron rod inserted in inductor series circuit statements:
1. The magnetic field inside the coil increases.
2. The inductance L of the coil decreases.
3. The inductive reactance XL of the coil increases.
4. A larger fraction of the applied ac voltage appears across the inductor.

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

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

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

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

💡 Short Explanation

�� Iron increases inductance. �� XL = ωL increases. �� Voltage drop across inductor increases.

✅ Correct Answer Explanation

1. Correct — Iron increases magnetic flux and field strength. 2. Incorrect — Inductance increases, not decreases. 3. Correct — XL = ωL, so XL increases. 4. Correct — Larger reactance means a larger AC voltage drop across the inductor. 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

Elimination

Application:

Evaluate the effect of increasing inductance.

Final Logic:

Iron core increases both L and XL.

🧠 Memory Trick

Iron Core → Bigger L → Bigger XL

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

6 Inductive reactance limits the current in a ______ manner to resistance, and dimensionally Lω is equivalent to ______ .

A identical, Capacitance

B similar, Resistance

C proportional, Current

D opposite, Inductance

💡 Short Explanation

�� Reactance opposes AC current. �� Resistance also limits current. �� Both have unit ohm.

✅ Correct Answer Explanation

Inductive reactance: XL = ωL It limits AC current in a manner similar to resistance. Also, XL has the same dimensions and SI unit (ohm) as resistance. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not dimensionally capacitance.
�� Option C → Current has different dimensions.
�� Option D → Lω is not inductance.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare units of reactance and resistance.

Final Logic:

Lω behaves dimensionally as resistance.

🧠 Memory Trick

XL Acts Like R

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

7 Incorrect statement regarding the phase relation i = im sin(ωt − π/2) in an inductor:

A The current phasor is π/2 behind the voltage phasor.

B The current leads the voltage by one-quarter cycle.

C Deriving it requires the integration of a sine function to yield a negative cosine function.

D It proves the current reaches its maximum value later than the voltage.

💡 Short Explanation

�� Current lags voltage. �� Lag equals π/2. �� Maximum occurs later.

✅ Correct Answer Explanation

The expression: i = im sin(ωt − π/2) shows that current is delayed relative to voltage by π/2. Therefore current lags by one-quarter cycle, not leads. Hence Option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct phasor interpretation.
�� Option C → Correct mathematical derivation.
�� Option D → Correct timing interpretation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Interpret the physical meaning of the phase term.

Final Logic:

A negative phase angle indicates lagging current.

🧠 Memory Trick

L = Lag

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

8 Match List I (Mathematical/Timing Representations) with List II (Interpretations for an Inductor)

List I	List II
1. sin(ωt)	a. Time lag of current maximum behind voltage maximum
2. −cos(ωt)	b. Phase representation showing a π/2 lag
3. sin(ωt − π/2)	c. Basic waveform of the applied voltage
4. T/4	d. Result of directly integrating the slope di/dt

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

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

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

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

💡 Short Explanation

�� Voltage is sinusoidal. �� Integration produces −cos term. �� T/4 represents lag.

✅ Correct Answer Explanation

1 → c because applied voltage is sin(ωt). 2 → d because integrating sin(ωt) gives −cos(ωt). 3 → b because sin(ωt − π/2) shows a π/2 lag. 4 → a because T/4 is the time lag between voltage and current maxima. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect mappings.
�� Option C → Incorrect interpretation of all terms.
�� Option D → Incorrectly matches −cos(ωt).

🎯 Strategy

Used

Option Grouping

Application:

Match mathematical expressions with physical meaning.

Final Logic:

Only Option A gives correct correspondence.

🧠 Memory Trick

sin → Voltage, −cos → Current

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

9 In an ac circuit with a pure inductor, if the peak voltage is 311 V and the peak current is 1.47 A, what is the maximum possible value of the instantaneous power supplied to the inductor?

A 0 W

B 228.6 W

C 457.2 W

D 114.3 W

💡 Short Explanation

�� p = −(imvm/2) sin(2ωt) �� Maximum occurs when |sin(2ωt)| = 1. �� Use peak values.

✅ Correct Answer Explanation

Maximum instantaneous power: pmax = (imvm)/2 = (1.47 × 311)/2 = 228.585 W ≈ 228.6 W Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Average power is zero, not maximum power.
�� Option C → Equal to imvm, not imvm/2.
�� Option D → Half the required value.

🎯 Strategy

Used

Substitution

Application:

Substitute values into pmax = imvm/2.

Final Logic:

Maximum power equals 228.6 W.

🧠 Memory Trick

Max Inductor Power = imvm/2

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

10 Correct logic statements for zero average power in an inductor:
1. Energy supplied during one quarter cycle is mathematically balanced in the next.
2. The integral of sin(2ωt) over [0, T] is exactly zero.
3. The instantaneous power is permanently zero at every instant.
4. The current and voltage are π/2 out of phase.

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

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

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

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

💡 Short Explanation

�� Energy is stored and returned. �� Average power is zero. �� Instantaneous power varies with time.

✅ Correct Answer Explanation

1. Correct — Energy absorbed is returned during another part of the cycle. 2. Correct — Average of sin(2ωt) over a cycle is zero. 3. Incorrect — Instantaneous power continuously changes and is not always zero. 4. Correct — Voltage and current differ in phase by π/2. 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:

Separate instantaneous power from average power.

Final Logic:

Instantaneous power is not always zero.

🧠 Memory Trick

Store → Return → Zero Average Power

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

11 By strictly applying Kirchhoff's loop rule to a purely capacitive circuit, the relationship between source voltage vm sin(ωt) and capacitor charge q is:

A q = Cvm cos(ωt)

B vm sin(ωt) = q/C

C q/C = −vm sin(ωt)

D vm sin(ωt) = C/q

💡 Short Explanation

�� Voltage across capacitor is q/C. �� Source voltage equals capacitor voltage. �� Kirchhoff's loop rule applies.

✅ Correct Answer Explanation

For a pure capacitor: v = q/C Since the applied voltage is: v = vm sin(ωt) Therefore: vm sin(ωt) = q/C Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents neither charge nor voltage correctly.
�� Option C → Wrong sign convention.
�� Option D → Incorrect capacitor relation.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use v = q/C.

Final Logic:

Applied voltage equals capacitor voltage.

🧠 Memory Trick

Capacitor: V = q/C

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: AC Applied to Capacitor

12 Identify the incorrect statement concerning charge on a capacitor in an ac circuit:

A Charge q at any time t exactly equals Cvm sin(ωt).

B Current is obtained as the time derivative of charge dq/dt.

C The capacitor prevents all flow of charge uniformly from the moment the ac source is connected.

D The capacitor is alternately charged and discharged during each cycle.

💡 Short Explanation

�� Capacitors allow alternating charge flow. �� Charging and discharging occur continuously. �� Current equals dq/dt.

✅ Correct Answer Explanation

A capacitor connected to an AC source repeatedly charges and discharges. Therefore charge flow occurs continuously in alternating directions. Hence statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct since q = Cv = Cvm sin(ωt).
�� Option B → Correct definition of current.
�� Option D → Correct description of capacitor behavior.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting AC capacitor behavior.

Final Logic:

A capacitor does not completely block AC.

🧠 Memory Trick

AC Keeps Charging and Discharging

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

13 If the capacitive reactance of a circuit is 100 Ω at a frequency of 50 Hz, what will be the exact capacitive reactance if the frequency is halved to 25 Hz?

A 50 Ω

B 100 Ω

C 200 Ω

D 400 Ω

💡 Short Explanation

�� XC = 1/(ωC). �� XC is inversely proportional to frequency. �� Halving frequency doubles XC.

✅ Correct Answer Explanation

Since: XC ∝ 1/f If frequency is reduced from 50 Hz to 25 Hz: XC(new) = 2 × 100 = 200 Ω Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Occurs if frequency doubles.
�� Option B → Assumes no frequency dependence.
�� Option D → Would require quarter frequency.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use inverse proportionality between XC and f.

Final Logic:

Half frequency → Double reactance.

🧠 Memory Trick

Half f → Double XC

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

14 The algebraic expression im = vm/(1/ωC) demonstrates that for a given voltage amplitude, the current amplitude im is:

A inversely proportional to both ω and C

B directly proportional to both ω and C

C inversely proportional to ω but directly proportional to C

D directly proportional to ω but inversely to C

💡 Short Explanation

�� im = ωCvm. �� Increasing ω increases current. �� Increasing C increases current.

✅ Correct Answer Explanation

Since: im = vm/(1/ωC) = ωCvm Therefore current amplitude is directly proportional to both angular frequency and capacitance. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite relation.
�� Option C → Incorrect dependence on frequency.
�� Option D → Incorrect dependence on capacitance.

🎯 Strategy

Used

Algebraic Simplification

Application:

Rewrite the given expression.

Final Logic:

im = ωCvm.

🧠 Memory Trick

More ω, More C → More Current

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

15 In a purely capacitive circuit, the derivative derivation
i = ωCvm cos(ωt)
is rewritten using the mathematical relation
cos(ωt) = sin(ωt + π/2)
to explicitly show that

A current and voltage are perfectly in phase

B current leads the voltage by π/2

C current lags the voltage by π/2

D current frequency is twice the voltage frequency

💡 Short Explanation

�� Current contains a positive phase shift. �� Positive phase shift means leading. �� Phase difference is π/2.

✅ Correct Answer Explanation

Substituting: cos(ωt) = sin(ωt + π/2) gives: i = im sin(ωt + π/2) Thus current is ahead of voltage by π/2. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not in phase.
�� Option C → Opposite phase relation.
�� Option D → Frequency remains unchanged.

🎯 Strategy

Used

Substitution

Application:

Convert cosine into sine form.

Final Logic:

+π/2 indicates current leads.

🧠 Memory Trick

Capacitor: Current Comes First

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

16 Consider the statements on interpreting the phasor diagram for a purely capacitive circuit:
1. Phasor I rotate counterclockwise geometrically ahead of V.
2. Current reaches its zero value exactly when voltage is at a maximum.
3. Current reaches its maximum exactly when voltage is zero.
4. The geometric angle between I and V is π.

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

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

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

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

💡 Short Explanation

�� Current leads voltage by π/2. �� Current maximum occurs at voltage zero. �� Phase difference is not π.

✅ Correct Answer Explanation

1. Correct — Current phasor leads voltage phasor. 2. Correct — When voltage is maximum, current becomes zero. 3. Correct — Current maximum occurs when voltage passes through zero. 4. Incorrect — Phase difference is π/2, not π. 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 π/2 phase relationship.

Final Logic:

The phase angle is π/2, not π.

🧠 Memory Trick

Voltage Zero → Current Peak

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

17 The true instantaneous power supplied to a capacitor
Choose correct:

A is strictly non-negative at all measured times

B oscillates dynamically at twice the frequency of the voltage source

C is derived and given by pc = imvm sin²(ωt)

D reaches its maximum exactly when voltage is at its peak

💡 Short Explanation

�� pc = (imvm/2) sin(2ωt). �� Frequency doubles. �� Power alternates positive and negative.

✅ Correct Answer Explanation

For a capacitor: pc = (imvm/2) sin(2ωt) The factor 2ω shows that instantaneous power oscillates at twice the source frequency. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Power becomes both positive and negative.
�� Option C → Incorrect expression.
�� Option D → Power is not maximum when voltage is maximum.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use the expression for capacitive instantaneous power.

Final Logic:

Power varies with frequency 2ω.

🧠 Memory Trick

Power Frequency = Double Source Frequency

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

18 The analytical fact that average power supplied to a capacitor over one complete cycle is zero implies that the electrical energy transferred to the capacitor during charging is completely

A dissipated continuously as heat

B converted directly to mechanical work

C returned safely to the source during discharging

D transformed strictly into inductive reactance

💡 Short Explanation

�� Capacitor stores energy. �� Energy is not dissipated. �� Stored energy returns to the source.

✅ Correct Answer Explanation

A pure capacitor stores electrical energy in its electric field during charging. During discharging, the same energy is returned to the source. Therefore, average power over a cycle becomes zero. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → No continuous heat dissipation occurs.
�� Option B → Mechanical work is not produced.
�� Option D → Reactance is not a form of energy.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Understand energy storage and return.

Final Logic:

Stored energy returns during discharge.

🧠 Memory Trick

Store → Return → Zero Average Power

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

19 Choose the incorrect statement about current limitation by basic components:

A Both pure resistors and capacitors effectively limit current in ac circuits.

B Ideal resistance regulates current independent of the applied frequency.

C Capacitive reactance regulates current but fundamentally depends on frequency.

D A capacitor limits current in an ac circuit by permanently stopping the flow of charge.

💡 Short Explanation

�� Capacitors limit but do not stop AC. �� Reactance depends on frequency. �� Resistance is frequency independent.

✅ Correct Answer Explanation

A capacitor connected to AC continuously charges and discharges. Thus it does not permanently stop charge flow. Instead, it limits current through capacitive reactance. Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct for an ideal resistor.
�� Option C → Correct because XC = 1/(ωC).

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting AC capacitor behavior.

Final Logic:

Capacitors limit current but do not completely block AC.

🧠 Memory Trick

Capacitor Limits, Not Stops

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

20 Identify the correct statements if the frequency of an AC source connected to a pure capacitor is exactly doubled:
1. The angular frequency ω doubles.
2. The capacitive reactance XC is perfectly halved.
3. The peak current im in the circuit is exactly doubled.
4. The average power dissipated becomes non-zero.

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

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

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

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

💡 Short Explanation

�� ω = 2πf. �� XC = 1/(ωC). �� im = ωCvm.

✅ Correct Answer Explanation

1. Correct — Doubling frequency doubles ω. 2. Correct — XC = 1/(ωC), therefore XC becomes half. 3. Correct — im = ωCvm, therefore current doubles. 4. Incorrect — Average power in a pure capacitor remains zero. 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

Direct Formula Recall

Application:

Apply XC = 1/(ωC) and im = ωCvm.

Final Logic:

Doubling frequency halves XC and doubles current.

🧠 Memory Trick

Double f → Half XC → Double i

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Frequency Effects in Capacitive Circuits

CUET UG Physics Booster Test 3-Pure Inductive and Capacitive Circuits

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