CUET UG Physics Booster Test 2-Power Dissipation and Power Factor -

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CUET UG Physics Booster Test 2-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 2-Power Dissipation and Power Factor

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QUESTION 1 OF 20

Expanding the instantaneous power relation p = i_m v_m sin(ωt) sin(ωt + φ), the expression mathematically resolves into:
Statements:
1. A structural form [i_m v_m / 2] [cos φ − cos(2ωt + φ)]
2. A sum of a time-independent term and a time-dependent term
3. A pure sine wave identically oscillating at the original frequency ω
4. A secondary term where the average of cos(2ωt+φ) reduces perfectly to zero over a full cycle

Answer Review

1 Expanding the instantaneous power relation p = i_m v_m sin(ωt) sin(ωt + φ), the expression mathematically resolves into:
Statements:
1. A structural form [i_m v_m / 2] [cos φ − cos(2ωt + φ)]
2. A sum of a time-independent term and a time-dependent term
3. A pure sine wave identically oscillating at the original frequency ω
4. A secondary term where the average of cos(2ωt+φ) reduces perfectly to zero over a full cycle

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

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

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

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

💡 Short Explanation

�� Use trigonometric product identity �� Power contains constant and oscillating parts �� Oscillating term averages to zero over a cycle

✅ Correct Answer Explanation

Using sin A sin B = ½[cos(A−B) − cos(A+B)] we obtain p = (i_m v_m / 2)[cos φ − cos(2ωt + φ)] Hence Statement 1 is correct. The expression consists of a constant term and an oscillating term, so Statement 2 is correct. The oscillating term has frequency 2ω, not ω. Therefore Statement 3 is incorrect. The average value of cos(2ωt + φ) over a complete cycle is zero, so Statement 4 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Contains Statement 3, which is incorrect because the oscillating term has frequency 2ω.
�� Option C → Contains Statement 3, which is incorrect.
�� Option D → Contains Statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate all options containing the incorrect Statement 3.

Final Logic:

�� Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Power oscillates at 2ω

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

2 The instantaneous power p in an AC circuit contains a time-independent term and an oscillating time-dependent term. The time-dependent term oscillates symmetrically about zero, ensuring the average of the ________ half exactly cancels the ________ half over a complete cycle.

A Positive, Negative

B Real, Imaginary

C Sine, Cosine

D Constant, Variable

💡 Short Explanation

�� Oscillating power has positive and negative regions �� Equal positive and negative contributions occur �� Average becomes zero

✅ Correct Answer Explanation

The oscillating component of power alternates above and below zero. Over a complete cycle, the positive area is equal to the negative area. Therefore, the average contribution of the oscillating term becomes zero. Hence the positive half cancels the negative half.

❓ Why Other Options Are Incorrect

�� Option B → Real and imaginary parts are not being averaged.
�� Option C → Sine and cosine are functions, not cancelling halves.
�� Option D → Constant and variable terms do not cancel each other.

🎯 Strategy

Used

Elimination

Application:

�� Identify which pair can physically cancel over a cycle.

Final Logic:

�� Positive and negative portions cancel exactly.

🧠 Memory Trick

+ Area = − Area

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

3 The average power P dissipated in an LCR circuit over a complete cycle, expressed explicitly in terms of rms current I and total impedance Z, is:

A I Z² cos φ

B I² Z cos φ

C I² / (Z cos φ)

D I² Z sin φ

💡 Short Explanation

�� Average power P = VI cosφ �� V = IZ �� Substitute and simplify

✅ Correct Answer Explanation

Average power is P = VI cosφ Since V = IZ, P = I(IZ)cosφ P = I²Z cosφ Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect dimensional form.
�� Option C → Incorrect rearrangement.
�� Option D → Power depends on cosφ, not sinφ.

🎯 Strategy

Used

Substitution

Application:

�� Substitute V = IZ into the average power formula.

Final Logic:

�� P = I²Z cosφ.

🧠 Memory Trick

P = I²Z cosφ

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

4 If an ac circuit possesses a power factor of 0.6, an applied rms voltage of 220 V, and an operating rms current of 5 A, what is the net average power dissipated?

A 1320 W

B 1100 W

C 396 W

D 660 W

💡 Short Explanation

�� Use P = VI cosφ �� Substitute given values �� Calculate average power

✅ Correct Answer Explanation

P = VI cosφ = 220 × 5 × 0.6 = 660 W Thus the average power dissipated is 660 W.

❓ Why Other Options Are Incorrect

�� Option B → Obtained by ignoring power factor.
�� Option C → Incorrect multiplication.
�� Option A → Exceeds apparent power.

🎯 Strategy

Used

Substitution

Application:

�� Direct numerical substitution.

Final Logic:

�� 220 × 5 × 0.6 = 660 W.

🧠 Memory Trick

Real Power = Apparent Power × PF

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

5 Identify the incorrect statement about the mathematical power factor (cos φ):

A It can be accurately calculated as the ratio of resistance to impedance (R/Z).

B It natively evaluates to zero for pure inductors and pure capacitors.

C It is constantly equal to 1 in any varying LCR circuit, regardless of the supplied frequency.

D It inherently determines the final fraction of maximum power dissipated in an AC circuit.

💡 Short Explanation

�� Power factor depends on phase angle �� Changes with frequency �� Unity only at resonance

✅ Correct Answer Explanation

Power factor is cosφ = R/Z In an LCR circuit, impedance varies with frequency. Therefore power factor also changes with frequency. Power factor becomes 1 only at resonance where XL = XC. Hence Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct relation.
�� Option B → Pure inductors and capacitors have cosφ = 0.
�� Option D → Average power depends directly on cosφ.

🎯 Strategy

Used

Extreme Word Filter

Application:

�� Words like "constantly" and "regardless" usually indicate an incorrect statement.

Final Logic:

�� Power factor is not always 1.

🧠 Memory Trick

PF = 1 only at Resonance

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

6 In a series LCR circuit, if the capacitive reactance X_C is significantly greater than the inductive reactance X_L,

A the phase angle φ is explicitly positive, and the current inherently lags the source voltage.

B the phase angle φ is explicitly positive, and the current inherently leads the source voltage.

C the phase angle φ is negative, and the circuit's behavior is predominantly inductive.

D the phase angle φ becomes exactly zero, resulting in zero overall power dissipation.

💡 Short Explanation

�� XC>XLX_C > X_LXC>XL implies capacitive behavior. �� Current leads voltage. �� Option B is the closest available answer.

✅ Correct Answer Explanation

For a series LCR circuit, tan φ = (XL − XC)/R When XC > XL then XL − XC < 0 Therefore, tan φ < 0 which implies φ < 0 Hence, the circuit is predominantly capacitive and the current leads the applied voltage. Conclusion: XC > XL ⇒ Capacitive Circuit ⇒ Current Leads Voltage ⇒ φ < 0 (NCERT convention) Hence the circuit behaves predominantly as a capacitive circuit and the current leads the applied voltage. Although the standard NCERT convention gives a negative phase angle for a capacitive circuit, among the given options only Option B correctly identifies that the current leads the source voltage. Therefore, Option B is the nearest correct answer.

❓ Why Other Options Are Incorrect

�� Option A → Current lags voltage, which occurs in an inductive circuit.
�� Option C → Phase angle sign is consistent with a capacitive circuit, but the circuit is incorrectly described as inductive.
�� Option D → Phase angle becomes zero only at resonance (XL=XC)(X_L = X_C)(XL=XC).

🎯 Strategy

Used: Conceptual Elimination

Application:

�� Determine whether the circuit is inductive or capacitive.
�� XC>XLX_C > X_LXC>XL indicates capacitive dominance.
�� Current must lead voltage.

Final Logic:

�� Only Option B contains the correct physical behavior.

🧠 Memory Trick

Inductor → Current Lags Voltage

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Series LCR Circuit and Phase Relationship.

7 Consider the statements regarding Joule heating in resistors in applied AC circuits. Choose the correct statements:
1. It relies on the parameter i², which is always a positive mathematical quantity
2. It physically occurs regardless of whether the instantaneous current is positive or negative
3. It inherently results in zero average power dissipation over time
4. It mandates continuous dissipation of electrical energy into heat

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

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

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

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

8 Identify the correct statements detailing the maximum power dissipation case
1. It predictably occurs exactly at the resonant frequency in an LCR circuit
2. It necessitates the reactive condition where X_C − X_L = 0
3. It functionally results in cos φ = 0
4. The average power mathematically reduces to P = I²R

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Maximum power occurs at resonance �� Net reactance becomes zero �� Circuit behaves purely resistive

✅ Correct Answer Explanation

In a series LCR circuit, maximum power dissipation occurs at resonance where XL = XC or XC − XL = 0 Therefore Statements 1 and 2 are correct. At resonance, Z = R and P = I²R Hence Statement 4 is correct. At resonance, the power factor is unity: cosφ = 1 not zero. Therefore Statement 3 is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 3. At resonance, cosφ = 1, not 0.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect statement regarding power factor at resonance.

Final Logic:

�� Resonance gives maximum power because cosφ = 1 and Z = R.

🧠 Memory Trick

Resonance = Maximum Power

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Resonance and Maximum Power Dissipation

9 Match List I with List II for a pure inductor connected into an AC circuit:

List I	List II
1. Phase difference value (φ)	a. Zero
2. Resulting power factor	b. −π/2 (current lags)
3. Instantaneous power integral over a complete cycle	c. ωL
4. Fundamental inductive reactance formula	d. 0

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

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

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

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

💡 Short Explanation

�� Current lags voltage by 90° in an inductor �� Power factor is zero �� Average power consumed is zero

✅ Correct Answer Explanation

For a pure inductor: 1 → b because current lags voltage by π/2. 2 → d because power factor cos 90° = 0. 3 → a because average power over one cycle is zero. 4 → c because inductive reactance is XL = ωL. Thus the correct matching is: 1-b, 2-d, 3-a, 4-c.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of phase angle, power factor and reactance.
�� Option A → Incorrect assignment of all major inductive properties.
�� Option D → Incorrect matching of phase difference and reactance.

🎯 Strategy

Used

Option Grouping

Application:

�� First identify the unique formula XL = ωL and the 90° phase lag.

Final Logic:

�� Once 1-b and 4-c are fixed, only Option A remains.

🧠 Memory Trick

Inductor: Lag 90°, PF 0, XL = ωL

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

10 If an ideal 15.0 μF pure capacitor is independently connected to a 220 V, 50 Hz source, what is the exact average power intrinsically consumed by the capacitor over a complete cycle?

A 100 W

B 0 W

C 220 W

D 311 W

💡 Short Explanation

�� Capacitor is a purely reactive element �� Voltage and current differ by 90° �� Average power consumed is zero

✅ Correct Answer Explanation

For a pure capacitor, φ = 90° and average power is P = VI cosφ Since cos 90° = 0, P = 0 Although energy is alternately stored and returned to the source, no net energy is dissipated over a complete cycle. Therefore, the average power consumed is 0 W.

❓ Why Other Options Are Incorrect

�� Option A → A pure capacitor does not dissipate average power.
�� Option C → Incorrect because reactive elements consume zero average power.
�� Option D → Confuses voltage values with power dissipation.

🎯 Strategy

Used

Conceptual Elimination

Application:

�� Recall that pure capacitors are wattless devices.

Final Logic:

�� cos90° = 0 ⇒ Average power = 0.

🧠 Memory Trick

Pure Capacitor = Wattless

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

11 Even though an active LCR series circuit contains an inductor and a capacitor, the steady power is exclusively dissipated in the resistor because

A the inductor and capacitor perfectly reflect the moving current back to the source at all frequencies.

B the phase difference between applied voltage and current for L and C is π/2, driving their specific power factors to zero.

C the physical resistance automatically becomes zero mathematically when L and C are simultaneously present.

D the resistor actively absorbs the reactive components of current to create false power.

💡 Short Explanation

�� Inductor and capacitor do not consume average power �� Their power factor is zero �� Only the resistor dissipates energy

✅ Correct Answer Explanation

In a pure inductor or pure capacitor, the phase difference between voltage and current is 90°. Therefore, P = VI cosφ and cos90° = 0 Hence the average power consumed by L and C is zero. The resistor alone converts electrical energy into heat and dissipates average power.

❓ Why Other Options Are Incorrect

�� Option A → L and C store and return energy, but do not simply reflect current.
�� Option C → Presence of L and C does not make resistance zero.
�� Option D → Resistors do not absorb reactive current to create false power.

🎯 Strategy

Used

Elimination

Application:

�� Identify the component responsible for real power consumption.

Final Logic:

�� Only the resistor has non-zero average power dissipation.

🧠 Memory Trick

R Burns, L & C Return

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

12 Consider the statements on LCR resonance power factor. Choose the correct statements.
1. The mathematical impedance Z structurally simplifies to equal R
2. The calculated power factor immediately becomes unity (1)
3. The instantaneous out-of-phase voltages across L and C entirely cancel each other
4. The average power geometrically collapses to zero

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 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 is zero and Z = R Hence Statement 1 is correct. Since Z = R, cosφ = R/Z = 1 Therefore Statement 2 is correct. Voltages across L and C are equal and opposite, so they cancel. Statement 3 is correct. Average power is maximum at resonance, not zero. 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 incorrect resonance statement.

Final Logic:

�� Average power is maximum at resonance, not zero.

🧠 Memory Trick

Resonance → Z = R → PF = 1

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

13 To pragmatically minimize large I²R losses during the large-scale industrial transmission of electrical energy:
1. The grid voltage is heavily stepped up using an appropriate transformer.
2. The line current is proportionally reduced by the transformer logic.
3. The power factor is deliberately engineered to approach 0.
4. A generally high power factor is consistently maintained.

A 2, 3, 4

B 1, 2, 3

C 1, 2, 4

D 1, 3, 4

💡 Short Explanation

�� High voltage lowers current �� Lower current reduces I²R loss �� High power factor is desirable

✅ Correct Answer Explanation

For a given power, P = VI Increasing voltage decreases current. Since transmission loss is I²R, lower current greatly reduces power loss. Power systems are also operated with high power factor for efficient transmission. Therefore Statements 1, 2 and 4 are correct. Statement 3 is incorrect because power factor approaching zero increases transmission problems.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Remove options containing the incorrect low-power-factor statement.

Final Logic:

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

🧠 Memory Trick

Step Up V, Cut I, Cut Loss

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

14 Identify the incorrect statement detailing the technique of improving a system's power factor using a shunt capacitor:

A It is strategically connected in parallel to the main circuit grid.

B It specifically supplies an equal leading wattless current I′q.

C It explicitly aims to make the total circuit impedance Z functionally tend toward R.

D It works by massively multiplying the reactive inductive current Iq already present in the circuit.

💡 Short Explanation

�� Capacitor provides leading current �� Reactive current is neutralized �� Power factor improves

✅ Correct Answer Explanation

A shunt capacitor supplies leading reactive current. This leading current cancels the lagging reactive current of inductive loads and improves power factor. It does not increase the inductive reactive current. Therefore Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct method of connection.
�� Option B → Correct function of capacitor.
�� Option C → Improved power factor makes Z approach R.

🎯 Strategy

Used

Conceptual Elimination

Application:

�� Identify the statement opposite to power-factor correction.

Final Logic:

�� Capacitor neutralizes reactive current, not increases it.

🧠 Memory Trick

Capacitor Cancels Lag

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

15 The main current vector in an AC circuit phasor diagram can be systematically resolved into two unique components. The ________ component fundamentally operates in phase with the voltage, while the ________ component strictly runs perpendicular to it.

A Power, Wattless

B Wattless, Power

C Leading, Lagging

D Reactive, Resistive

💡 Short Explanation

�� Power component is in phase �� Wattless component is perpendicular �� Only in-phase component contributes to power

✅ Correct Answer Explanation

Current can be resolved into: Ip = Power component (in phase with voltage) Iq = Wattless or reactive component (90° to voltage) Only the power component contributes to average power. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the components.
�� Option C → Leading and lagging do not define the decomposition.
�� Option D → Reversed terminology.

🎯 Strategy

Used

Conceptual Matching

Application:

�� Match phasor components with their physical roles.

Final Logic:

�� Ip is in phase, Iq is perpendicular.

🧠 Memory Trick

P for Power = Parallel

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

16 If I_p is mathematically isolated as the power component and I_q is isolated as the reactive component, the total effective power P (when properly neutralized by an equal leading current) is securely expressed as:

A I_p V

B I_q V

C (I_p + I_q) V

D (I_p - I_q) V

💡 Short Explanation

�� I_p is the in-phase component �� Only I_p contributes to real power �� I_q contributes no average power

✅ Correct Answer Explanation

The current in an AC circuit can be resolved into: Power component (I_p), which is in phase with voltage Reactive component (I_q), which is perpendicular to voltage Only the in-phase component contributes to average power. Therefore, P = VI_p Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → I_q is the reactive component and does not contribute to average power.
�� Option C → Reactive current does not contribute to real power.
�� Option D → Real power is not obtained by subtracting current components.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Match the physical meaning of the power component with the power expression.

Final Logic:

�� Only the in-phase current component contributes to average power.

🧠 Memory Trick

P comes from Iₚ

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

17 Choose the correct statements defining the physical hazards of maintaining a persistently low power factor
Statements:
1. It drastically requires a significantly higher current I to deliver the exact same target power P
2. It inherently provokes massive quadratic I²R power loss inside the cables
3. It causes the electrical infrastructure to experience large unnecessary transmission drops
4. It can be instantly rectified by removing all resistive loads from the grid

A 1, 2, 4

B 2, 3, 4

C 1, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Low power factor increases current �� Higher current increases I²R losses �� Voltage drops become larger

✅ Correct Answer Explanation

For a fixed power, P = VI cosφ If cosφ decreases, current must increase to deliver the same power. Therefore Statement 1 is correct. Higher current increases transmission loss proportional to I²R, making Statement 2 correct. The increased current also causes larger voltage drops in transmission lines, making Statement 3 correct. Removing resistive loads does not solve low power factor problems. Hence 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 statement regarding power-factor correction.

Final Logic:

�� Statements 1, 2 and 3 describe genuine consequences of low power factor.

🧠 Memory Trick

Low PF → High I → High Loss

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

18 In order to completely neutralize the lagging wattless current I_q and permanently improve the industrial power factor,

A an equal leading wattless current I′_q is successfully introduced via a parallel capacitor.

B the source voltage is forcefully halved to reduce the raw power component.

C an external inductor is permanently placed in series to wildly increase total impedance.

D the main frequency of the driving AC source is actively set to zero.

💡 Short Explanation

�� Capacitor supplies leading reactive current �� Lagging reactive current gets neutralized �� Power factor improves

✅ Correct Answer Explanation

Industrial loads are generally inductive and draw lagging reactive current. A shunt capacitor supplies an equal leading reactive current I′_q. These currents cancel each other, reducing the reactive component and increasing the power factor. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Lowering voltage does not eliminate reactive current.
�� Option C → Additional inductance worsens the power factor.
�� Option D → Setting frequency to zero is impractical and unrelated to power-factor correction.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Match the standard industrial power-factor correction technique.

Final Logic:

�� Capacitors provide leading reactive current that cancels lagging reactive current.

🧠 Memory Trick

Capacitor Cancels Lag

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

19 Match List I (Core Transformer Losses) with List II (Engineering Reduction Methods):

List I	List II
1. Internal Eddy currents	a. Designing and using a laminated core
2. Resistance of physical windings	b. Using physically thick wire for high-current segments
3. Magnetic Hysteresis	c. Utilizing specific magnetic material featuring low hysteresis loss
4. Environmental Flux leakage	d. Accurately winding primary and secondary directly over each other

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

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

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

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

💡 Short Explanation

�� Laminations reduce eddy currents �� Thick wire reduces copper loss �� Special core materials reduce hysteresis loss

✅ Correct Answer Explanation

1 → a because laminating the core reduces eddy current loops. 2 → b because thicker wires have lower resistance and reduce copper losses. 3 → c because low-hysteresis materials reduce energy loss during repeated magnetization. 4 → d because closely wound coils reduce magnetic flux leakage. Hence the correct matching is: 1-a, 2-b, 3-c, 4-d.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect matching of eddy current and winding losses.
�� Option C → Incorrect assignment of all major transformer-loss reduction methods.
�� Option D → Incorrect matching of leakage flux and hysteresis reduction.

🎯 Strategy

Used

Option Grouping

Application:

�� First identify the standard transformer-loss reduction techniques and match them systematically.

Final Logic:

�� Each loss has a unique and well-known reduction method.

🧠 Memory Trick

Eddy-Lamination, Copper-Thick Wire, Hysteresis-Special Core

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

20 In a completely theoretical 100% efficient step-up transformer obeying conservation, if the primary input power is exactly 2200 W at 220 V and the secondary output voltage steps to 440 V, what is the resulting secondary current?

A 5.0 A

B 10.0 A

C 2.5 A

D 20.0 A

💡 Short Explanation

�� Ideal transformer conserves power �� Input power = Output power �� Use P = VI

✅ Correct Answer Explanation

For an ideal transformer, Pₚ = Pₛ Given: Pₛ = 2200 W Vₛ = 440 V Using P = VI Iₛ = Pₛ / Vₛ = 2200 / 440 = 5 A Therefore the secondary current is 5.0 A.

❓ Why Other Options Are Incorrect

�� Option B → Would imply output power of 4400 W.
�� Option C → Would imply output power of 1100 W.
�� Option D → Would imply output power of 8800 W.

🎯 Strategy

Used

Substitution

Application:

�� Apply conservation of power and substitute values directly.

Final Logic:

�� Iₛ = 2200/440 = 5 A.

🧠 Memory Trick

Ideal Transformer: Power Same

📚 Class XII – Physics Part II – Chapter: Alternating Current – Topic: Ideal Transformer and Power Conservation

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