CUET UG Physics Booster Test 2-Pure Inductive and Capacitive Circuits -

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CUET UG Physics Booster Test 2-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 2-Pure Inductive and Capacitive Circuits

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QUESTION 1 OF 20

Consider the statements on a purely inductive ac circuit. Choose the correct statements:
1. The equation for current must have a slope di/dt varying sinusoidally.
2. The current is in phase with the voltage.
3. The integration constant for the current equation is zero.
4. No time-independent component of the current exists.

Answer Review

1 Consider the statements on a purely inductive ac circuit. Choose the correct statements:
1. The equation for current must have a slope di/dt varying sinusoidally.
2. The current is in phase with the voltage.
3. The integration constant for the current equation is zero.
4. No time-independent component of the current exists.

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

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

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

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

💡 Short Explanation

�� Current in a pure inductor is AC in nature. �� No DC component exists. �� Current lags voltage by π/2.

✅ Correct Answer Explanation

1. Correct — Since v = L(di/dt) and the applied voltage is sinusoidal, di/dt varies sinusoidally. 2. Incorrect — In a pure inductor, current lags voltage by π/2 and is not in phase. 3. Correct — The integration constant becomes zero because current oscillates symmetrically about zero. 4. Correct — A pure AC current contains no time-independent (DC) component. Therefore statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect statement 2.
�� Option C → Includes incorrect statement 2 and omits statement 1.
�� Option D → Includes incorrect statement 2.

🎯 Strategy

Used

Elimination

Application:

Check each statement using phase relation and current equation of a pure inductor.

Final Logic:

Only statement 2 is incorrect.

🧠 Memory Trick

Inductor = Lag + No DC

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

2 Identify the incorrect statement about inductors in AC circuits:

A Real inductors usually have appreciable resistance in their windings.

B In a purely inductive circuit, winding resistance is assumed to be zero.

C Assuming negligible resistance eliminates the self-induced Faraday emf.

D A pure inductor limits current but does not dissipate average power over a cycle.

💡 Short Explanation

�� Self-induced emf is the defining property of an inductor. �� Negligible resistance does not remove induction. �� Average power is zero.

✅ Correct Answer Explanation

The self-induced emf is given by: −L(di/dt) This emf exists because of changing current, not because of resistance. Assuming zero resistance simplifies the circuit but does not eliminate the self-induced emf. Hence statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Real inductors do have winding resistance.
�� Option B → Standard assumption for a pure inductor.
�� Option D → Average power over a cycle is zero.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting electromagnetic induction.

Final Logic:

Induction remains even when resistance is neglected.

🧠 Memory Trick

No Resistance ≠ No Induction

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

3 Match List I with List II

List I	List II
1. Applied AC voltage	a. 0
2. Self-induced emf	b. vm sin(ωt)
3. Net voltage drop in pure inductor circuit	c. −L(di/dt)
4. Amplitude of slope di/dt	d. vm/L

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

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

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

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

💡 Short Explanation

�� Applied voltage is sinusoidal. �� Self-induced emf opposes change. �� Net voltage around loop is zero.

✅ Correct Answer Explanation

1 → b because applied voltage is vm sin(ωt). 2 → c because self-induced emf is −L(di/dt). 3 → a because Kirchhoff's loop rule gives net voltage as zero. 4 → d because from: di/dt = vm sin(ωt)/L Amplitude of di/dt = vm/L. Thus the correct matching is: 1-b, 2-c, 3-a, 4-d.

❓ Why Other Options Are Incorrect

�� Option A → Multiple mismatches.
�� Option C → Self-induced emf and net voltage are interchanged.
�� Option D → Applied voltage incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each physical quantity with its mathematical expression.

Final Logic:

Only Option B gives all correct pairings.

🧠 Memory Trick

Voltage–EMF–Zero–vm/L

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

4 By combining v = vm sin(ωt) and the Faraday emf based on Lenz's law, we obtain di/dt for a purely inductive circuit. Integrating this yields a constant of integration which is:

A equal to vm/L

B zero, because current oscillates symmetrically about zero

C equal to the peak current im

D equal to L/vm

💡 Short Explanation

�� AC current has zero average value. �� No DC offset exists. �� Integration constant vanishes.

✅ Correct Answer Explanation

After integrating the equation for current in a pure inductor, a constant of integration appears. Since the AC current oscillates equally above and below zero and has no steady component, the integration constant must be zero. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the integration constant.
�� Option C → Peak current is determined separately.
�� Option D → No such relation exists.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall the derivation of current in a pure inductor.

Final Logic:

Symmetric oscillation about zero makes the constant zero.

🧠 Memory Trick

Pure AC → Zero Offset

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

5 A pure inductor of 25.0 mH is connected to a 220 V, 50 Hz source. What is the approximate rms current in the circuit? (π = 3.14)

A 14 A

B 28 A

C 42 A

D 56 A

💡 Short Explanation

�� XL = 2πfL �� Irms = Vrms/XL �� Inductor limits current through reactance.

✅ Correct Answer Explanation

Given: L = 25 mH = 0.025 H f = 50 Hz Vrms = 220 V XL = 2πfL = 2 × 3.14 × 50 × 0.025 = 7.85 Ω Irms = 220/7.85 ≈ 28 A Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Approximately half the correct value.
�� Option C → Overestimation.
�� Option D → Excessively large.

🎯 Strategy

Used

Substitution

Application:

Apply XL = 2πfL and I = V/XL.

Final Logic:

220/7.85 ≈ 28 A.

🧠 Memory Trick

Inductor Current = V/XL

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

6 The dimension of inductive reactance is mathematically the same as that of

A Inductance

B Capacitance

C Resistance

D Frequency

💡 Short Explanation

�� Reactance opposes AC current. �� Resistance opposes current generally. �� Both have unit ohm.

✅ Correct Answer Explanation

Inductive reactance: XL = ωL Its SI unit is ohm (Ω), the same as resistance. Therefore inductive reactance has the same dimensions as resistance.

❓ Why Other Options Are Incorrect

�� Option A → Inductance has unit henry.
�� Option B → Capacitance has unit farad.
�� Option D → Frequency has unit hertz.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare SI units.

Final Logic:

Reactance and resistance share identical dimensions.

🧠 Memory Trick

Reactance Acts Like Resistance

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

7 Choose the correct statements about the phasor diagram of a purely inductive circuit:
1. The current phasor I is π/2 behind the voltage phasor V.
2. Both phasors rotate with frequency ω counter-clockwise.
3. The phasors are in the exact same direction at all times.
4. The vertical components of the phasors generate the sinusoidal variation.

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

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

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

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

💡 Short Explanation

�� Current lags voltage by π/2. �� Phasors rotate together. �� Vertical projection gives instantaneous value.

✅ Correct Answer Explanation

1. Correct — Current lags voltage by π/2. 2. Correct — Both rotate counter-clockwise with angular frequency ω. 3. Incorrect — They maintain a phase difference of π/2. 4. Correct — Vertical projections produce sinusoidal quantities. 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:

Use phase relation in phasor representation.

Final Logic:

Only statement 3 is false.

🧠 Memory Trick

Inductor → I Behind V

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

8 The current reaches its maximum value later than the voltage by one-fourth of a period. This time interval T/4 in terms of angular frequency ω is

A π/(2ω)

B π/ω

C 2π/ω

D ω/(2π)

💡 Short Explanation

�� T = 2π/ω �� Quarter cycle = T/4 �� Substitute directly.

✅ Correct Answer Explanation

T = 2π/ω Therefore: T/4 = (2π/ω)/4 = π/(2ω) Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Equals half the period.
�� Option C → Equals the full period.
�� Option D → Incorrect dimensional form.

🎯 Strategy

Used

Substitution

Application:

Use T = 2π/ω.

Final Logic:

T/4 = π/(2ω).

🧠 Memory Trick

Quarter Cycle = π/2ω

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Phase Difference and Time Lag

9 In deriving the instantaneous power p for an inductor, the trigonometric identity sin(2ωt) arises strictly from the product of:

A sin(ωt) and sin(ωt)

B cos(ωt) and cos(ωt)

C sin(ωt) and −cos(ωt)

D sin(ωt) and sin(ωt − π/2)

💡 Short Explanation

�� p = vi �� Current contains −cos(ωt). �� Double-angle identity appears.

✅ Correct Answer Explanation

For an inductor: i = im sin(ωt − π/2) = −im cos(ωt) Therefore: p ∝ sin(ωt) × [−cos(ωt)] Using: 2sinθcosθ = sin2θ the term sin(2ωt) appears. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Produces sin² term.
�� Option B → Produces cos² term.
�� Option D → Equivalent expression exists, but the identity specifically arises after converting to −cos(ωt).

🎯 Strategy

Used

Substitution

Application:

Substitute the phase-shifted current expression.

Final Logic:

sin(ωt) × [−cos(ωt)] generates sin(2ωt).

🧠 Memory Trick

sin × cos → Double Angle

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

10 The average power supplied to an inductor over one complete cycle

A depends heavily on the values of vm and im

B is permanently negative due to Lenz's law

C is exactly zero because sin(2ωt) averages to zero

D equals the power supplied to a pure resistor

💡 Short Explanation

�� Instantaneous power oscillates. �� Positive and negative contributions cancel. �� Average power becomes zero.

✅ Correct Answer Explanation

For a pure inductor: p = −(imvm/2) sin(2ωt) The average value of sin(2ωt) over one complete cycle is zero. Hence the average power supplied to the inductor is zero. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Average power remains zero regardless of amplitudes.
�� Option B → Power alternates between positive and negative values.
�� Option D → A resistor dissipates positive average power.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use the expression for average power in a pure inductor.

Final Logic:

Average value of sin(2ωt) over a cycle is zero.

🧠 Memory Trick

Pure Inductor = Zero Average Power

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

11 A capacitor in a dc circuit will ______ current flow when fully charged, but in an ac circuit it ______ the current as it reverses each half cycle.

A prevent, limits

B increase, prevents

C limits, prevents

D maintain, limits

💡 Short Explanation

�� A fully charged capacitor blocks steady DC. �� In AC, charge and discharge occur repeatedly. �� Capacitive reactance limits current.

✅ Correct Answer Explanation

In a DC circuit, after a capacitor becomes fully charged, current falls to zero and the capacitor prevents further current flow. In an AC circuit, the voltage reverses every half cycle, causing repeated charging and discharging. The capacitor does not completely block AC but limits the current through its capacitive reactance. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → A capacitor does not increase DC current and does not completely prevent AC current.
�� Option C → A fully charged capacitor prevents, rather than merely limits, DC current.
�� Option D → A capacitor does not maintain DC current.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare capacitor behavior in DC and AC circuits.

Final Logic:

DC → Prevents current; AC → Limits current.

🧠 Memory Trick

DC Stops, AC Limits

📚 Class XII – Physics Part I – Chapter: Alternating Current – Topic: Capacitor in AC and DC Circuits

12 When an ac source is applied to a capacitor, the instantaneous voltage v across the capacitor at any time t, given charge q and capacitance C, is

A v = q/C

B v = C/q

C v = qC

D v = dq/dt

💡 Short Explanation

�� Charge and voltage are proportional. �� Capacitance is charge stored per volt. �� Fundamental capacitor relation.

✅ Correct Answer Explanation

The basic relation for a capacitor is: q = Cv Therefore: v = q/C This equation is valid at every instant, whether the source is AC or DC. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Inverse of the correct relation.
�� Option C → Dimensionally incorrect.
�� Option D → dq/dt represents current.

🎯 Strategy

Used

Direct Formula Recall

Application:

Recall the fundamental capacitor equation.

Final Logic:

v = q/C.

🧠 Memory Trick

Q = CV

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

13 The amplitude of the oscillating current in a purely capacitive circuit is im = ωCvm. The capacitive reactance XC, which plays the role of resistance, is expressed as:

A ωC

B 1/(ωC)

C ω/C

D C/ω

💡 Short Explanation

�� Capacitive reactance opposes AC. �� Depends on frequency and capacitance. �� Inverse relationship.

✅ Correct Answer Explanation

From: im = ωCvm Comparing with Ohm's law form: im = vm/XC Therefore: XC = 1/(ωC) Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reciprocal of the correct expression.
�� Option C → Incorrect dimensional form.
�� Option D → Incorrect expression.

🎯 Strategy

Used

Substitution

Application:

Compare capacitive current equation with Ohm's law.

Final Logic:

XC = 1/(ωC).

🧠 Memory Trick

Capacitor = One Over ωC

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

14 If a lamp is connected in series with a capacitor to an ac source
Choose correct:

A reducing capacitance increases reactance and the lamp shines less brightly

B reducing capacitance decreases reactance and the lamp shines more brightly

C increasing capacitance increases reactance

D changing capacitance has no effect on lamp brightness

💡 Short Explanation

�� XC = 1/(ωC) �� Smaller capacitance means larger reactance. �� Larger reactance reduces current.

✅ Correct Answer Explanation

Capacitive reactance is: XC = 1/(ωC) If capacitance decreases, XC increases. A larger reactance reduces current through the lamp, decreasing its brightness. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Opposite of the actual relationship.
�� Option C → Increasing capacitance decreases reactance.
�� Option D → Capacitance directly affects current and brightness.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use XC = 1/(ωC).

Final Logic:

Lower C → Higher XC → Lower Current.

🧠 Memory Trick

Small C = Big XC

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

15 Consider the statements about a purely capacitive ac circuit. Choose the correct statements:
1. The current is π/2 ahead of the voltage.
2. The current phasor I is π/2 ahead of voltage phasor V.
3. The current reaches its maximum value later than the voltage.
4. The voltage across the source and the capacitor are equal.

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

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

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

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

💡 Short Explanation

�� Current leads voltage by π/2. �� Phasor I is ahead of V. �� Source voltage equals capacitor voltage in a pure capacitive circuit.

✅ Correct Answer Explanation

1. Correct — Current leads voltage by π/2. 2. Correct — The current phasor is ahead of the voltage phasor by π/2. 3. Incorrect — Current reaches maximum earlier, not later. 4. Correct — In a pure capacitive circuit, the source voltage appears across the capacitor. 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:

Use phase relation of a pure capacitor.

Final Logic:

Current leads voltage; therefore statement 3 is false.

🧠 Memory Trick

Capacitor = Current Comes First

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

16 Choose the incorrect statement regarding the timing of maxima in a purely capacitive circuit:

A The current reaches its maximum value earlier than the voltage.

B The time difference between the current and voltage maxima is one-fourth of a period.

C The current phasor rotates clockwise relative to voltage to reach the maximum first.

D Phasors rotate counter-clockwise and I is geometrically ahead of V.

💡 Short Explanation

�� Phasors rotate counter-clockwise. �� Current leads by π/2. �� Lead corresponds to T/4.

✅ Correct Answer Explanation

In standard phasor representation, all phasors rotate counter-clockwise. The current phasor is ahead of the voltage phasor by π/2, but it does not rotate clockwise. Therefore, statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct for a capacitor.
�� Option B → π/2 corresponds to T/4.
�� Option D → Correct phasor description.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement violating phasor conventions.

Final Logic:

Phasors rotate counter-clockwise, not clockwise.

🧠 Memory Trick

All Phasors Rotate CCW

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

17 The instantaneous power pc supplied to a capacitor is
pc = imvm cos(ωt) sin(ωt)
Using a fundamental trigonometric identity, this simplifies directly to:

A (imvm/2) cos(2ωt)

B (imvm/2) sin(2ωt)

C imvm sin²(2ωt)

D (imvm) sin(2ωt)

💡 Short Explanation

�� Power = voltage × current. �� Use double-angle identity. �� Instantaneous power oscillates.

✅ Correct Answer Explanation

Using: 2sinθcosθ = sin2θ Therefore: sin(ωt)cos(ωt) = ½sin(2ωt) Hence: pc = (imvm/2) sin(2ωt) Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uses incorrect trigonometric identity.
�� Option C → Wrong functional form.
�� Option D → Missing factor 1/2.

🎯 Strategy

Used

Substitution

Application:

Apply the double-angle identity directly.

Final Logic:

sinθ cosθ = ½ sin2θ.

🧠 Memory Trick

sin × cos = Half sin Double Angle

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

18 The mathematical fact that average power supplied to a capacitor over one complete cycle is zero occurs because the average value of

A sin(ωt) is 1

B sin(2ωt) is zero over a complete cycle

C cos(2ωt) is 1/2

D voltage amplitude vm is zero

💡 Short Explanation

�� Average capacitor power is zero. �� Instantaneous power contains sin(2ωt). �� Positive and negative halves cancel.

✅ Correct Answer Explanation

For a pure capacitor: pc = (imvm/2) sin(2ωt) The average value of sin(2ωt) over a complete cycle is zero. Therefore average power is zero. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Average value of sin(ωt) is zero.
�� Option C → Not the reason for zero average power.
�� Option D → Voltage amplitude is not zero.

🎯 Strategy

Used

Direct Formula Recall

Application:

Use the power expression for a capacitor.

Final Logic:

Average of sin(2ωt) over a cycle equals zero.

🧠 Memory Trick

sin(2ωt) Averages to Zero

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

19 Choose the correct statements about capacitive reactance limiting current:
1. It limits current amplitude in a purely capacitive circuit just as resistance does in a resistive circuit.
2. It is directly proportional to frequency.
3. It is inversely proportional to capacitance.
4. Its SI unit is the Ohm.

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

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

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

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

💡 Short Explanation

�� XC limits current. �� XC = 1/(ωC). �� Unit is ohm.

✅ Correct Answer Explanation

1. Correct — Reactance limits current like resistance. 2. Incorrect — XC is inversely proportional to frequency. 3. Correct — XC ∝ 1/C. 4. Correct — SI unit is ohm. 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:

Use XC = 1/(ωC).

Final Logic:

Statement 2 contradicts the reactance formula.

🧠 Memory Trick

XC Falls When f Rises

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

20 A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. The peak current is 1.47 A. If the frequency is doubled to 100 Hz, what will be the new peak current?

A 0.735 A

B 1.47 A

C 2.94 A

D 5.88 A

💡 Short Explanation

�� im = ωCvm �� Current is directly proportional to frequency. �� Doubling frequency doubles current.

✅ Correct Answer Explanation

For a capacitor: im = ωCvm Since: ω = 2πf Current is directly proportional to frequency. When frequency doubles: im(new) = 2 × 1.47 = 2.94 A Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents halving current.
�� Option B → Assumes no frequency dependence.
�� Option D → Assumes quadrupling current.

🎯 Strategy

Used

Substitution

Application:

Use im ∝ f.

Final Logic:

Doubling frequency doubles peak current.

🧠 Memory Trick

Double f → Double im

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

CUET UG Physics Booster Test 2-Pure Inductive and Capacitive Circuits

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