CUET UG Physics Booster Test 3-Power Dissipation and Power Factor -

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CUET UG Physics Booster Test 3-Power Dissipation and Power Factor

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Power Dissipation and Power Factor

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QUESTION 1 OF 20

Identify the incorrect statement concerning the rigorous mathematical expansion of the instantaneous power p = v_m i_m sin(ωt) sin(ωt + φ):

Answer Review

1 Identify the incorrect statement concerning the rigorous mathematical expansion of the instantaneous power p = v_m i_m sin(ωt) sin(ωt + φ):

A The physical expansion directly relies on the standard trigonometric identity manipulating the product of two sine waves.

B The bracketed term cos φ fundamentally represents a static offset relative to the oscillating cos(2ωt + φ) term.

C The instantaneous power equation fundamentally indicates that power never drops mathematically below zero at any micro-point in the AC cycle.

D Both voltage and current frequencies operate intrinsically at ω, generating a resultant power oscillation strictly at frequency 2ω.

💡 Short Explanation

�� Instantaneous power can be positive or negative �� Power oscillates at frequency 2ω �� Average power differs from instantaneous power

✅ Correct Answer Explanation

Using the identity sin A sin B = ½[cos(A−B) − cos(A+B)] the instantaneous power becomes p = (v_m i_m / 2)[cosφ − cos(2ωt + φ)] This expression contains an oscillating term that can make instantaneous power positive or negative depending on time. Therefore, Statement C is incorrect because instantaneous power can fall below zero during parts of the cycle.

❓ Why Other Options Are Incorrect

�� Option A → Correct; the derivation uses the product-to-sum trigonometric identity.
�� Option B → Correct; cosφ is the constant term.
�� Option D → Correct; multiplying two ω-frequency quantities produces a 2ω oscillation.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Distinguish between instantaneous power and average power.

Final Logic:

�� Instantaneous power can become negative even though average power may be positive.

🧠 Memory Trick

Instantaneous ≠ Always Positive

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

2 Regarding the isolated time-dependent term cos(2ωt + φ) structurally inside the generalized power equation:
1. It cleanly integrates identically to zero over any full complete period T.
2. Its driving frequency is geometrically double that of the originally applied AC voltage.
3. It explicitly dictates the actual average real power steadily dissipated in the circuit.
4. The entire positive half-cycle precisely cancels out the entire negative half-cycle.

A 1, 2, 4

B 2, 3, 4

C 1, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Oscillating term averages to zero �� Frequency is 2ω �� Does not contribute to average power

✅ Correct Answer Explanation

The term cos(2ωt + φ) oscillates symmetrically about zero. Statement 1 is correct because its average over a complete cycle is zero. Statement 2 is correct because its frequency is 2ω. Statement 3 is incorrect because average power comes from the constant term involving cosφ. Statement 4 is correct because positive and negative parts cancel over a cycle.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify which statement incorrectly attributes average power to the oscillating term.

Final Logic:

�� The oscillating term averages to zero.

🧠 Memory Trick

2ω Term → Zero Average

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Time-Dependent Power Term

3 By substituting the trigonometric impedance relationship cos φ = R/Z squarely into the formal average power equation P = I² Z cos φ, the final expression for average power structurally simplifies exactly to:

A I² R

B I² Z

C V² Z

D V R

💡 Short Explanation

�� Use cosφ = R/Z �� Substitute into power formula �� Z cancels out

✅ Correct Answer Explanation

Starting from P = I²Z cosφ and cosφ = R/Z Substituting: P = I²Z(R/Z) P = I²R Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ignores substitution.
�� Option C → Incorrect dimensional form.
�� Option D → Not a valid power expression.

🎯 Strategy

Used

Substitution

Application:

�� Replace cosφ with R/Z.

Final Logic:

�� Z cancels, giving P = I²R.

🧠 Memory Trick

Power Ends at Resistance

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Average Power Formula

4 A real series LCR circuit has R = 3 Ω, L = 25.48 mH, C = 796 μF, and is dynamically driven by a physical source of V_rms = (283/√2) V at 50 Hz. If the calculated total impedance resolves to 5 Ω, what is the net exact power dissipated inside the circuit?

A 1200 W

B 2400 W

C 4800 W

D 9600 W

💡 Short Explanation

�� Use P = I²R �� I = V/Z �� Substitute values

✅ Correct Answer Explanation

Vᵣₘₛ = 283/√2 ≈ 200 V I = V/Z = 200/5 = 40 A Power dissipated: P = I²R = (40)² × 3 = 1600 × 3 = 4800 W Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option B → Half of correct value.
�� Option A → Incorrect current calculation.
�� Option D → Double the correct value.

🎯 Strategy

Used

Substitution

Application:

�� Calculate current first, then use P = I²R.

Final Logic:

�� I = 40 A gives P = 4800 W.

🧠 Memory Trick

Find I First, Then I²R

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Average Power in LCR Circuit

5 Consider the statements regarding power factor. Choose the correct statements.
1. Mathematically represents the exact cosine of the phase difference between voltage and current
2. Is geometrically proven to equal R/Z in any standard impedance diagram
3. Systematically takes its minimum value (zero) exactly at LCR resonance
4. Serves as a definitive measure of how close the active circuit is to expending its absolute maximum power

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Power factor = cosφ �� Also equals R/Z �� Maximum power occurs when power factor approaches unity

✅ Correct Answer Explanation

Statement 1 is correct because power factor is defined as cosφ. Statement 2 is correct because cosφ = R/Z. Statement 3 is incorrect because at resonance, cosφ = 1, not zero. Statement 4 is correct because larger power factor means more effective power transfer.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect resonance statement.

Final Logic:

�� Resonance gives maximum, not minimum, power factor.

🧠 Memory Trick

Resonance → PF = 1

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

6 If X_L > X_C functionally in an LCR circuit, the resulting phase angle is _________ indicating the circuit is predominantly inductive, whereas if X_C > X_L, the resulting phase angle is _________.

A Negative, Positive

B Positive, Negative

C Zero, Infinite

D Infinite, Zero

💡 Short Explanation

�� Inductive circuits have positive phase angle �� Capacitive circuits have negative phase angle �� Sign depends on X_L − X_C

✅ Correct Answer Explanation

For a series LCR circuit, tanφ = (X_L − X_C)/R When X_L > X_C, (X_L − X_C) is positive and therefore φ is positive. The circuit behaves inductively and current lags voltage. When X_C > X_L, (X_L − X_C) is negative and therefore φ is negative. The circuit behaves capacitively and current leads voltage. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reverses the signs of the phase angle.
�� Option C → Phase angle is not zero in general.
�� Option D → Phase angle does not become infinite.

🎯 Strategy

Used

Substitution

Application:

�� Use tanφ = (X_L − X_C)/R and determine the sign.

Final Logic:

�� X_L > X_C ⇒ φ positive, X_C > X_L ⇒ φ negative.

🧠 Memory Trick

L Positive, C Negative

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Phase Angle in LCR Circuit

7 The average evaluated integral of Joule heating (p = i²R) over a cycle is strictly positive, entirely despite the current being mathematically negative during exactly half the cycle, because

A the resistance physically and automatically rectifies the AC current perfectly into DC.

B the physical heating depends strictly on the square of the instantaneous current (i²), which is invariably a positive mathematical quantity.

C the physical inductor completely absorbs the negative power wave and instantly returns it purely as heat.

D the average measured current is fundamentally non-zero in a symmetric AC circuit.

💡 Short Explanation

�� Heating depends on i²R �� Square of current is always positive �� Heat is produced during both half-cycles

✅ Correct Answer Explanation

The instantaneous heating power in a resistor is p = i²R Since i² is always positive regardless of whether current is positive or negative, the resistor continuously dissipates heat. Therefore, the average Joule heating remains positive throughout the cycle.

❓ Why Other Options Are Incorrect

�� Option A → A resistor does not rectify AC into DC.
�� Option C → Inductors do not convert negative power into heat in this manner.
�� Option D → Average current in a symmetrical AC cycle is zero.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Identify the physical quantity responsible for heating.

Final Logic:

�� Heating depends on i², not the sign of i.

🧠 Memory Trick

Square Removes Sign

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Joule Heating in AC Circuits

8 Match List I (Electrical Circuit Condition) with List II (Analytical Power Dissipation Profile):

List I	List II
1. Purely resistive operational mode (φ = 0)	a. Theoretical maximum possible power dissipation
2. Purely inductive operational mode (φ = π/2)	b. Analytically zero power dissipation
3. Operational LCR directly at resonance (X_C = X_L)	c. P = I²R (Maximum internal dissipation via Z collapsing to R)
4. Operational LCR entirely off-resonance (X_C ≠ X_L)	d. Partial fragmented power dissipation (P = I²Z cosφ)

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

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

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

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

💡 Short Explanation

�� Resistive circuits dissipate maximum power �� Pure inductors dissipate zero average power �� Resonance gives P = I²R

✅ Correct Answer Explanation

1 → a because a purely resistive circuit has power factor equal to 1 and maximum power dissipation. 2 → b because a pure inductor has cos90° = 0 and therefore zero average power. 3 → c because at resonance Z = R and P = I²R. 4 → d because off resonance, P = I²Z cosφ, leading to partial power dissipation. Thus the correct matching is: 1-a, 2-b, 3-c, 4-d.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching for resistive and inductive cases.
�� Option C → Incorrect resonance and off-resonance assignments.
�� Option D → Incorrect matching of all major conditions.

🎯 Strategy

Used

Option Grouping

Application:

�� Fix the obvious matches first (inductor → zero power, resonance → P = I²R).

Final Logic:

�� Only Option A satisfies all physical conditions.

🧠 Memory Trick

Resistor-Max, Inductor-Zero, Resonance-I²R

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Power Dissipation Conditions

9 Choose the correct statements regarding the formal mathematical proof of zero average power in pure inductors
Statements:
1. The instantaneous power expands correctly to p = -i_m v_m sin(ωt) cos(ωt)
2. The expression simplifies cleanly to -(i_m v_m / 2) sin(2ωt) via trigonometric identity
3. The temporal average of sin(2ωt) calculated over a complete cycle mathematically equals zero
4. The evaluated power factor cos(π/2) evaluates physically to 1

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Pure inductors have phase difference of 90° �� Instantaneous power contains sin(2ωt) term �� Average value becomes zero

✅ Correct Answer Explanation

For a pure inductor, i = i_m sin(ωt − π/2) and v = v_m sinωt This gives p = -i_m v_m sinωt cosωt Hence Statement 1 is correct. Using 2sinθcosθ = sin2θ gives p = -(i_m v_m/2)sin(2ωt) Hence Statement 2 is correct. The average of sin(2ωt) over one cycle is zero, making Statement 3 correct. Since cos(π/2) = 0, Statement 4 is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 4.
�� Option C → Contains incorrect Statement 4.
�� Option A → Contains incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect power-factor statement.

Final Logic:

�� cos(π/2) = 0, not 1.

🧠 Memory Trick

Inductor → 90° → Zero Power

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Wattless Current in Inductor

10 The moving current passing through a purely capacitive AC circuit is systematically ahead of the voltage by exactly π/2. If the instantaneous structural voltage is analytically v_m sin(ωt), what is the exact instantaneous power specifically at the moment ωt = π/2?

A (1/2) i_m v_m

B i_m v_m

C 0

D -i_m v_m

💡 Short Explanation

�� Current leads voltage by 90° �� Evaluate voltage and current at ωt = π/2 �� Instantaneous power becomes zero

✅ Correct Answer Explanation

For a pure capacitor, i = i_m sin(ωt + π/2) At ωt = π/2 Voltage: v = v_m sin(π/2) = v_m Current: i = i_m sinπ = 0 Instantaneous power: p = vi = v_m × 0 = 0 Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option B → Requires current to be maximum.
�� Option A → Not obtained from the power expression.
�� Option D → Current is zero, so power cannot be negative.

🎯 Strategy

Used

Substitution

Application:

�� Substitute ωt = π/2 directly into voltage and current equations.

Final Logic:

�� Current becomes zero, so power becomes zero.

🧠 Memory Trick

If i = 0, then P = 0

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Power in Pure Capacitor

11 Incorrect analytical statement regarding physical power distribution in a functioning LCR circuit:

A The total effective power supplied actively by the source practically perfectly equals the power dissipated physically in the resistor.

B Physical capacitors and inductors cleanly cyclically store and release internal energy without any net cumulative consumption.

C The isolated term cos φ mathematically and structurally isolates the resistive portion (R/Z) to determine net actual power loss.

D Purely ideal inductors experience massive permanent internal energy dissipation directly due to their purely reactive running currents.

💡 Short Explanation

�� Ideal inductors do not dissipate average power �� L and C store and return energy �� Only resistance causes real power loss

✅ Correct Answer Explanation

In an LCR circuit, average power is dissipated only in the resistor. An ideal inductor stores energy in its magnetic field and returns it to the source. Its average power consumption is zero because the phase difference is 90°. Therefore, Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct; source power equals resistive power dissipation.
�� Option B → Correct; ideal L and C only store and return energy.
�� Option C → Correct because cosφ = R/Z determines the resistive contribution.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement contradicting the concept of wattless current.

Final Logic:

�� Ideal inductors do not consumes average power.

🧠 Memory Trick

Ideal L = No Real Loss

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Power Dissipation in LCR Circuit

12 At the calculated natural resonant frequency ω₀ = 1/√LC of an active LCR series circuit,

A the total physical impedance drops fully to zero, instantly yielding theoretically infinite power dissipation.

B the varying reactances X_L and X_C cancel fundamentally completely, driving the power factor equation cos φ = R/Z to become exactly 1.

C the physical current severely lags the moving voltage by π/2, generating a totally wattless circuit condition.

D the circuit dynamically and inherently rejects the external source frequency, driving all running current instantly to zero.

💡 Short Explanation

�� Resonance occurs when XL = XC �� Net reactance becomes zero �� Power factor becomes unity

✅ Correct Answer Explanation

At resonance, XL = XC Therefore net reactance becomes zero and Z = R Hence cosφ = R/Z = 1 The circuit behaves as a purely resistive circuit and current becomes maximum. Thus Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Impedance becomes R, not zero.
�� Option C → Phase angle becomes zero, not π/2.
�� Option D → Current becomes maximum, not zero.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Recall resonance conditions in a series LCR circuit.

Final Logic:

�� XL = XC ⇒ Z = R ⇒ cosφ = 1.

🧠 Memory Trick

Resonance → Z = R

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Resonance in LCR Circuit

13 To pragmatically mitigate dangerously large power loss (I²R) in extensive transmission lines while still securely delivering specific power P:
Statements:
1. The physical transmission voltage must predictably be stepped up.
2. The engineered power factor cos φ must strictly be forced to approach 1.
3. Reactive line components actively causing low power factors must be thoroughly compensated.
4. The physical current must actively be reduced by heavily stepping down the voltage.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� High voltage lowers current �� High power factor improves efficiency �� Reactive current should be minimized

✅ Correct Answer Explanation

Statement 1 is correct because stepping up voltage reduces current for the same power. Statement 2 is correct because a power factor close to unity minimizes current demand. Statement 3 is correct because compensation reduces reactive current. Statement 4 is incorrect because stepping down voltage would increase current and losses.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 4.
�� Option C → Contains incorrect Statement 4.
�� Option D → Contains incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing the incorrect transmission statement.

Final Logic:

�� Efficient transmission requires high voltage and high power factor.

🧠 Memory Trick

High V, Low I, Low Loss

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Power Transmission

14 If I_q is exactly the lagging wattless current introduced heavily by a massive inductive load, a parallel compensating capacitor is mathematically chosen such that it reliably produces an exact leading wattless current I′_q. The final target optimization condition guaranteeing neutralization is structurally represented by:

A I_q + I′_q = 1

B I_q = I′_q

C I_q × I′_q = 0

D I_q / I′_q = R

💡 Short Explanation

�� Capacitor supplies leading reactive current �� Equal magnitudes cancel �� Power factor improves

✅ Correct Answer Explanation

For complete compensation, Lagging reactive current = Leading reactive current Therefore I_q = I′_q Under this condition the net reactive current becomes zero and the power factor approaches unity.

❓ Why Other Options Are Incorrect

�� Option A → No physical basis.
�� Option C → Reactive currents need not individually be zero.
�� Option D → No relation exists with resistance.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Identify the condition for perfect reactive-current cancellation.

Final Logic:

�� Equal and opposite reactive currents cancel.

🧠 Memory Trick

Lag = Lead

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

15 The specific component of alternating current that solely dictates actual net energy transfer is exactly the one operating purely in phase with the applied voltage. If the total calculated current magnitude is strictly I and the specific phase angle is φ, this power-bearing component is functionally represented by ________, resulting in continuous power ________.

A I cos φ, I_pV

B I sin φ, Zero

C I tan φ, Maximum

D I / cos φ, Minimum

💡 Short Explanation

�� In-phase current produces real power �� I_p = I cosφ �� Power equals I_pV

✅ Correct Answer Explanation

The current component in phase with voltage is I_p = I cosφ Average power becomes P = VI cosφ = VI_p Therefore, the power-bearing component is I cosφ and the corresponding power is I_pV.

❓ Why Other Options Are Incorrect

�� Option B → I sinφ is reactive current.
�� Option C → Not a valid current component.
�� Option D → Incorrect expression.

🎯 Strategy

Used

Substitution

Application:

�� Use the standard current-component decomposition.

Final Logic:

�� I_p = I cosφ is responsible for real power.

🧠 Memory Trick

Power Uses cosφ

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Current Components

16 Consider the statements regarding complex analysis of the reactive component of current (I_q) statements. Choose the correct statements:
1. It is structurally defined via trigonometry as I sin φ
2. It geometrically causes absolutely zero average physical power loss over any complete cycle
3. It continuously oscillates identically out of phase with the primary power component
4. It is technically entirely responsible for driving real actual mechanical work in a physical motor

A 2, 3, 4

B 1, 3, 4

C 1, 2, 3

D 1, 2, 4

💡 Short Explanation

�� I_q = I sinφ �� Reactive current does not transfer net energy �� Real work is done by I_p

✅ Correct Answer Explanation

Statement 1 is correct because I_q = I sinφ. Statement 2 is correct because reactive current contributes zero average power. Statement 3 is correct because it is the quadrature component of current. Statement 4 is incorrect because real mechanical work depends on the power component I_p.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 4.
�� Option A → Contains incorrect Statement 4.
�� Option D → Contains incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify which statement incorrectly attributes real work to reactive current.

Final Logic:

�� Reactive current contributes no average power.

🧠 Memory Trick

I_q = No Real Work

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Reactive Current

17 Identify the correct advanced analytical statements detailing the severe systemic hazards of operating a low power factor
1. Mathematical equation P = IV cos φ means that a low cos φ practically demands a disproportionately high operating current I
2. Real I²R transmission power losses scale aggressively and quadratically with the increased baseline current
3. The physical infrastructure (including wires and core transformers) experiences far greater thermal/heating stress
4. The true measured physical power delivered paradoxically increases unconditionally

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Low power factor increases current requirement �� Higher current increases I²R losses �� Equipment heating becomes more severe

✅ Correct Answer Explanation

From P = VI cosφ for a fixed power P, a lower cosφ requires a higher current I. Therefore Statement 1 is correct. Since transmission losses are proportional to I²R, the increased current significantly raises power losses. Hence Statement 2 is correct. The increased losses generate additional heating in conductors, transformers, and associated equipment. Therefore Statement 3 is correct. Low power factor does not automatically increase useful power delivered. Hence Statement 4 is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 4.
�� Option C → Contains incorrect Statement 4.
�� Option D → Contains incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement that contradicts the effects of low power factor.

Final Logic:

�� Low power factor increases current and losses, not useful power.

🧠 Memory Trick

Low PF → High I → High Loss

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Effects of Low Power Factor

18 Match List I (Deep Neutralization Strategy) with List II (Predictable Circuit Outcome):

List I	List II
1. Intentionally introduce I′q leading current	a. Practically uses a massive parallel shunt capacitor
2. Mathematically ensure I′q = Iq	b. Physically cancels the lagging wattless current completely
3. Leave primary Ip fundamentally unaffected	c. Securely maintains pure real power delivery (IpV)
4. Final adjusted Power factor dynamically approaches	d. 1 (Strict Unity)

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

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

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

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

💡 Short Explanation

�� Capacitor supplies leading reactive current �� Reactive currents cancel each other �� Power factor approaches unity

✅ Correct Answer Explanation

1 → a because leading reactive current is supplied using a shunt capacitor. 2 → b because when I′q = Iq the lagging and leading reactive currents cancel. 3 → c because the power component Ip remains unchanged and continues delivering real power. 4 → d because after compensation the power factor approaches unity. Thus, the correct matching is: 1-a, 2-b, 3-c, 4-d.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of compensation process and outcomes.
�� Option C → Incorrect assignment of unity power factor and cancellation.
�� Option D → Incorrect matching of all major concepts.

🎯 Strategy

Used

Option Grouping

Application:

�� First identify capacitor → leading current and unity power factor relationships.

Final Logic:

�� Only Option A correctly matches all compensation outcomes.

🧠 Memory Trick

Capacitor → Cancel → Unity PF

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

19 A fully physical, non-ideal actual step-up transformer has a documented running efficiency of exactly 95%. If the active primary voltage is fixed at 220 V, the primary line current is stable at 10 A, and the secondary line voltage steps identically to 440 V, what is the exact operational current flowing in the secondary circuit?

A 10.0 A

B 5.0 A

C 4.5 A

D 4.75 A

💡 Short Explanation

�� Efficiency = Output Power/Input Power �� Calculate output power first �� Use P = VI for secondary side

✅ Correct Answer Explanation

Input power: Pᵢ = VᵢIᵢ = 220 × 10 = 2200 W Efficiency: η = 95% Output power: Pₒ = 0.95 × 2200 = 2090 W Secondary current: Iₛ = Pₒ/Vₛ = 2090/440 = 4.75 A Hence Option D is correct.

❓ Why Other Options Are Incorrect

�� Option B → Assumes 100% efficiency.
�� Option C → Incorrect calculation.
�� Option A → Violates transformer power relationship.

🎯 Strategy

Used

Substitution

Application:

�� Compute output power using efficiency, then find current.

Final Logic:

�� Iₛ = (0.95 × 2200)/440 = 4.75 A.

🧠 Memory Trick

Efficiency First, Current Next

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Transformer Efficiency

20 In an idealized hypothetical transformer, the core assumption that primary input power strictly equals secondary output power inherently and mathematically relies directly on the strict physical conditions that

A the primary physical resistance is completely zero, absolutely all core flux flawlessly links directly to the secondary, and there is physically no hysteresis or eddy current losses inside the core.

B the primary and secondary coils have completely identical physical numbers of wire turns, fully neutralizing all voltage variations.

C the base frequency of the AC source is permanently stepped down precisely by the strict physical turn ratio.

D the inductive core reactance perfectly and dynamically matches the capacitive line reactance deep within the core structure.

💡 Short Explanation

�� Ideal transformer has no losses �� Entire magnetic flux links both coils �� Input power equals output power

✅ Correct Answer Explanation

For an ideal transformer: Winding resistance is zero. Flux leakage is absent. Eddy current losses are absent. Hysteresis losses are absent. All magnetic flux links both coils. Under these conditions, Input Power = Output Power Therefore, Option A correctly states the assumptions required for an ideal transformer.

❓ Why Other Options Are Incorrect

�� Option B → Equal turns are not required for ideal operation.
�� Option C → Transformer does not change frequency.
�� Option D → No such condition exists in transformer theory.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Identify the standard assumptions of an ideal transformer.

Final Logic:

�� No losses and complete flux linkage are required for 100% efficiency.

🧠 Memory Trick

Ideal = No Loss + Full Flux

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

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