ALTERNATING CURRENT MCQs for NEET UG PHYSICS — Medium

1 Medium

Analyzing +4 | -1

Assertion (A) : In an

\mathrm{AC}

circuit, potential difference across the capacitor may be greater than the applied voltage.

Reason (R) :

V_C=I X_C

, whereas

V=I Z

and

X_C

can be greater than

Z

also.

View answer and explanation

Correct option: A

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

Option A is correct because the potential difference across a capacitor in an AC circuit can be greater than the applied voltage due to resonance, and the reason correctly explains this with the relationship between capacitive reactance, impedance, and voltage.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and the Reason is correct explanation of Assertion. The assertion is true due to resonance, and the reason explains why.

Step-by-Step Solution:

Step 1: Analyze the Assertion: In an AC circuit, the potential difference across the capacitor can indeed be greater than the applied voltage, especially near resonance.

Step 2: Analyze the Reason: The reason states $V_C = IX_C$ and $V = IZ$ , and $X_C$ can be greater than $Z$ . This is correct. At resonance, the inductive and capacitive reactances can cancel each other out, leading to a small impedance Z, while $X_C$ can still be large, making $V_C > V$ .

Step 3: Determine the relationship: The reason directly explains the assertion. A large $X_C$ compared to $Z$ leads to a larger voltage across the capacitor than the source voltage.

Why other options are wrong:
- Option B: The reason is a correct explanation of the assertion.
- Option C: The assertion is true.
- Option D: The assertion is true, and the reason is also true.

Key Concepts:
- AC circuits
- Capacitive reactance
- Impedance
- Resonance

Memory Aids:
- Remember the formulas: $V_C = IX_C$ , $V = IZ$ , $X_C = \frac{1}{\omega C}$

Quick Tips:
- Consider the effect of resonance in AC circuits.

2 Medium

2 of 84Question permalink

Understanding +4 | -1

When power is drawn from the secondary coil of the transformer, the dynamic resistance

View answer and explanation

Correct option: A

Increases

Option A (Increases) is correct because when power is drawn, the current in the secondary coil increases, which reflects as an increased load on the primary side, effectively increasing the dynamic resistance.

Solution approach

Correct Answer + Core Reason: Increases. When power is drawn, the current increases, leading to an increase in dynamic resistance.

Step-by-Step Solution:

Step 1: Define dynamic resistance: The effective resistance seen by the primary side of the transformer due to the load connected to the secondary side.

Step 2: When power is drawn from the secondary coil, the current in the secondary coil increases.

Step 3: This increased current reflects back to the primary side, effectively increasing the load on the primary side.

Step 4: An increased load corresponds to an increased dynamic resistance.

Why other options are wrong:
- Option B: The dynamic resistance increases, not decreases.
- Option C: The dynamic resistance changes with the load.
- Option D: The change is not erratic but follows the load.

Key Concepts:
- Dynamic resistance
- Transformer loading

Memory Aids:
- More power = More resistance

Quick Tips:
- Remember that the load on the secondary side affects the primary side.

3 Medium

3 of 84Question permalink

Understanding +4 | -1

Average power in the L-C-R circuit depends upon

View answer and explanation

Correct option: D

Current, emf and phase difference

Option D is correct because the average power in an L-C-R circuit depends on the current, emf, and the phase difference between them, as described by the formula P = Vrms * Irms * cos(phi).

Solution approach

Correct Answer + Core Reason: Current, emf and phase difference. Average power depends on all three.

Step-by-Step Solution:

Step 1: Recall the formula for average power in an L-C-R circuit.

Step 2: The average power (P) is given by $P = V_{rms} I_{rms} \cos \phi$ , where $V_{rms}$ is the root mean square voltage, $I_{rms}$ is the root mean square current, and $\phi$ is the phase difference between voltage and current.

Step 3: From the formula, it is clear that the average power depends on the current, emf (voltage), and the phase difference.

Why other options are wrong:
- Option A: Current: While current is a factor, it's not the only factor.
- Option B: phase difference only: While phase difference is a factor, it's not the only factor.
- Option C: Emf: While emf is a factor, it's not the only factor.

Key Concepts:
- Average power in AC circuits
- L-C-R circuit
- RMS voltage and current
- Phase difference

Memory Aids:
- $P = V_{rms} I_{rms} \cos \phi$

Quick Tips:
- Remember the formula for average power in an AC circuit.

4 Medium

4 of 84Question permalink

Understanding +4 | -1

In a choke coil, the resistance

{X_L}

and resistance

R

are such that

View answer and explanation

Correct option: B

{X_L} \gg R

Option B (

{X_L} \gg R

) is correct because a choke coil is designed to have a high inductive reactance and low resistance to minimize power loss.

Solution approach

Correct Answer + Core Reason: ${X_L} \gg R$ . A choke coil has high inductive reactance and low resistance.

Step-by-Step Solution:

Step 1: Understand the purpose of a choke coil.

Step 2: A choke coil is used to reduce AC current without significant power loss.

Step 3: This is achieved by having a high inductive reactance ( $X_L$ ) and a low resistance ( $R$ ).

Why other options are wrong:
- Option A: ${X_L} = R$ - Incorrect, the reactance should be much greater than the resistance.
- Option C: ${X_L} \ll R$ - Incorrect, the reactance should be much greater than the resistance.
- Option D: ${X_L} = \infty$ - Incorrect, while ideally it would be infinite, in reality, it is very large but finite.

Key Concepts:
- Choke coil
- Inductive reactance

Memory Aids:
- Choke coil: High $X_L$ , low R

Quick Tips:
- Remember that a choke coil is used to limit AC current with minimal power loss.

5 Medium

5 of 84Question permalink

Analyzing +4 | -1

Assertion: With increase in frequency of an A.C supply, the impedance of an L-C-R series circuit, decreases at first, becomes minimum and then increases.

Reason: In L-C-R series circuit, impedance (z)

z=\sqrt{R^2+\left(X_L-X_C\right)^2}

.

View answer and explanation

Correct option: A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

Option A is correct because both the assertion and the reason are true, and the reason correctly explains the assertion. The impedance of an L-C-R series circuit decreases, reaches a minimum at resonance, and then increases as frequency increases. The impedance formula explains this behavior.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and the reason is the correct explanation of the assertion. The impedance formula explains the behavior of the LCR circuit with changing frequency.

Step-by-Step Solution:

Step 1: Analyze the Assertion: With increase in frequency of an A.C supply, the impedance of an L-C-R series circuit, decreases at first, becomes minimum and then increases.

Step 2: Consider the behavior of impedance in an L-C-R series circuit as frequency changes. At low frequencies, the capacitive reactance ( $X_C = \frac{1}{\omega C}$ ) is high, and the inductive reactance ( $X_L = \omega L$ ) is low. As frequency increases, $X_C$ decreases and $X_L$ increases. At resonance, $X_L = X_C$ , and the impedance is minimum (equal to R). Beyond resonance, $X_L$ becomes greater than $X_C$ , and the impedance increases. So, the assertion is true.

Step 3: Analyze the Reason: In L-C-R series circuit, impedance (z) $z=\sqrt{R^2+\left(X_L-X_C\right)^2}$ .

Step 4: Examine the impedance formula. The formula $z=\sqrt{R^2+\left(X_L-X_C\right)^2}$ correctly represents the impedance of an L-C-R series circuit. This formula shows that when $X_L = X_C$ , the impedance is minimum (z = R). Also, it shows that as the difference between $X_L$ and $X_C$ increases (either $X_L > X_C$ or $X_C > X_L$ ), the impedance increases. So, the reason is true and explains the assertion.

Step 5: Conclusion: Both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Why other options are wrong:
- Option B: If both assertion and reason are true but reason is not the correct explanation of the assertion: The reason is the correct explanation.
- Option C: If assertion is true but reason is false: The reason is true.
- Option D: If assertion is false but reason is true: The assertion is true.

Key Concepts:
- Impedance in AC circuits
- L-C-R series circuit
- Inductive reactance ( $X_L = \omega L$ )
- Capacitive reactance ( $X_C = \frac{1}{\omega C}$ )
- Resonance

Memory Aids:
- Impedance formula: $z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
- Resonance condition: $X_L = X_C$

Quick Tips:
- Remember the impedance formula for an L-C-R series circuit.
- Understand how frequency affects inductive and capacitive reactance.
- Recall the resonance condition.

6 Medium

6 of 84Question permalink

Remembering +4 | -1

Which type of losses donot occur in the transformer?

View answer and explanation

Correct option: C

Mechanical losses

Option C (Mechanical losses) is correct because transformers are static devices and do not have moving parts, thus they do not experience mechanical losses.

Solution approach

Correct Answer + Core Reason: Mechanical losses. Transformers are static devices and do not have moving parts.

Step-by-Step Solution:

Step 1: Identify the types of losses in a transformer.

Step 2: Iron losses (core losses) are due to hysteresis and eddy currents in the core.

Step 3: Copper losses are due to the resistance of the windings.

Step 4: Flux leakage is a loss of magnetic flux that doesn't link both windings.

Step 5: Mechanical losses are associated with rotating machines due to friction and windage.

Step 6: Since transformers are static devices, they do not have mechanical losses.

Why other options are wrong:
- Option A: Iron losses are present in the transformer core due to hysteresis and eddy currents.
- Option B: Copper losses occur in the windings due to their resistance.
- Option D: Flux leakage does occur in transformers, reducing efficiency.

Key Concepts:
- Transformer Losses
- Iron Losses (Hysteresis and Eddy Current)
- Copper Losses
- Flux Leakage

Memory Aids:
- Transformers don't \"move\", so they don't have mechanical losses.

Quick Tips:
- Remember the different types of losses and their causes to quickly identify them in problems.

7 Medium

7 of 84Question permalink

Understanding +4 | -1

There is a current of

20 \mathrm{~mA}

passing through an ideal inductor when connected to a

100 \mathrm{~V}, 20 \mathrm{~Hz}

supply. The power consumed is

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because the power consumed by an ideal inductor in an AC circuit is zero due to the 90-degree phase difference between voltage and current.

Solution approach

Correct Answer + Core Reason: Zero. The power consumed by an ideal inductor is zero.

Step-by-Step Solution:

Step 1: Recall the formula for power in an AC circuit: $P = VI\cos \phi$ , where $V$ is the voltage, $I$ is the current, and $\phi$ is the phase angle between the voltage and the current.

Step 2: For an ideal inductor, the phase angle between the voltage and the current is 90 degrees ( $\phi = 90^\circ$ ).

Step 3: Calculate the power: $P = VI\cos (90^\circ) = VI \times 0 = 0$ .

Step 4: Final Answer: Zero

Why other options are wrong:
- Option A: This value is incorrect because the power consumed by an ideal inductor is zero.
- Option B: This value is incorrect because the power consumed by an ideal inductor is zero.
- Option C: This value is incorrect because the power consumed by an ideal inductor is zero.

Key Concepts:
- Power in AC circuits
- Phase angle
- Ideal inductor

Memory Aids:
- Remember that the power consumed by an ideal inductor is zero because the voltage and current are 90 degrees out of phase.

Quick Tips:
- In a purely inductive or capacitive circuit, the average power consumed over a complete cycle is zero.

8 Medium

8 of 84Question permalink

Analyzing +4 | -1

Assertion: Average of sinusoidal A. C. can never be zero for half cycle.

Reason: Impedance given by inductance does not depends on frequency.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False

Option C is correct because the average of a sinusoidal AC over a half cycle is non-zero, while the impedance due to inductance depends on frequency.

Solution approach

Correct Answer + Core Reason: If Assertion is True but the Reason is False. The assertion is true, but the reason is false because inductive impedance depends on frequency.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The average of a sinusoidal AC over a complete cycle is zero, but over a half cycle, it is non-zero. Therefore, the assertion is true.

Step 2: Analyze the Reason: The impedance due to inductance is given by $X_L = \omega L = 2\pi fL$ , which clearly depends on the frequency $f$ . Therefore, the reason is false.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- AC circuits
- Average value of sinusoidal AC
- Inductive impedance

Memory Aids:
- Remember that $X_L = \omega L$ .

Quick Tips:
- Understand the frequency dependence of inductive and capacitive reactances.

9 Medium

9 of 84Question permalink

Application +4 | -1

The potential difference V and the current i flowing through an instrument in an ac circuit of frequency

f

are given by V=5cos⁡ωt volts and I=2sin⁡ωt amperes (where ω=2πf). The power dissipated in the instrument is

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because the voltage and current are 90 degrees out of phase, resulting in zero average power dissipation.

Solution approach

Correct Answer + Core Reason: Zero. The voltage and current are out of phase by 90 degrees, leading to zero power dissipation.

Step-by-Step Solution:

Step 1: Identify the given information:\nVoltage, $V = 5 \cos(\omega t)$ \nCurrent, $I = 2 \sin(\omega t)$

Step 2: Recall the formula for instantaneous power:
$P(t) = V(t) \cdot I(t)$

Step 3: Calculate the instantaneous power:
$P(t) = 5 \cos(\omega t) \cdot 2 \sin(\omega t) = 10 \cos(\omega t) \sin(\omega t) = 5 \sin(2\omega t)$

Step 4: Calculate the average power:\nThe average power over a complete cycle is the time average of the instantaneous power.\nSince the average value of $\sin(2\omega t)$ over a complete cycle is zero, the average power is zero.
$P_{avg} = \frac{1}{T} \int_0^T P(t) dt = \frac{1}{T} \int_0^T 5 \sin(2\omega t) dt = 0$

Why other options are wrong:
- Option B: This option gives a power of 10W, which is incorrect. It might result from multiplying the peak voltage and peak current without considering the phase difference.
- Option C: This option gives a power of 5W, which is incorrect. It might result from a miscalculation or misunderstanding of the phase difference.
- Option D: This option gives a power of 2.5W, which is incorrect. It might result from a miscalculation or misunderstanding of the phase difference.

Key Concepts:
- Instantaneous power
- Average power in AC circuits
- Phase difference between voltage and current

Memory Aids:
- Remember that if voltage and current are 90 degrees out of phase, the average power is zero.

Quick Tips:
- Always consider the phase difference between voltage and current when calculating power in AC circuits.
- The average power is zero when the phase difference is 90 degrees.

10 Medium

10 of 84Question permalink

Understanding +4 | -1

A 10 ohm resistance, 5 mH coil and 10 μF capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency

View answer and explanation

Correct option: C

Remains unchanged

Option C is correct because the resonance frequency depends only on the inductance (L) and capacitance (C) values, not on the resistance (R). Therefore, halving the resistance does not change the resonance frequency.

Solution approach

Correct Answer + Core Reason: Remains unchanged. Resonance frequency depends only on L and C.

Step-by-Step Solution:

Step 1: Recall the formula for the resonance frequency in an LCR circuit.

Step 2: The resonance frequency (f) is given by $f = \frac{1}{2 \pi \sqrt{LC}}$ , where L is the inductance and C is the capacitance.

Step 3: Observe that the resonance frequency does not depend on the resistance (R).

Step 4: Therefore, if the resistance is halved, the resonance frequency remains unchanged.

Why other options are wrong:
- Option A: Is halved: The resonance frequency does not depend on the resistance.
- Option B: Is doubled: The resonance frequency does not depend on the resistance.
- Option D: In quadrupled: The resonance frequency does not depend on the resistance.

Key Concepts:
- Resonance frequency
- Inductance
- Capacitance

Memory Aids:
- $f = \frac{1}{2 \pi \sqrt{LC}}$

Quick Tips:
- Remember that the resonance frequency depends only on the inductance and capacitance values.

11 Medium

11 of 84Question permalink

Understanding +4 | -1

For a series L-C-R circuit at resonance, the statement which is not true is

View answer and explanation

Correct option: D

Power factor is zero

Option D (Power factor is zero) is correct because at resonance in a series LCR circuit, the power factor is 1, not 0.

Solution approach

Correct Answer + Core Reason: Power factor is zero. At resonance, the power factor is 1, not 0.

Step-by-Step Solution:

Step 1: Recall the conditions for resonance in a series LCR circuit.

Step 2: At resonance, the inductive reactance ( $X_L$ ) equals the capacitive reactance ( $X_C$ ).

Step 3: This means the impedance is purely resistive, and the phase angle between voltage and current is zero.

Step 4: The power factor is $\cos\phi$ , where $\phi$ is the phase angle. Since $\phi = 0$ , $\cos\phi = 1$ .

Why other options are wrong:
- Option A: Peak energy stored by a capacitor = peak energy stored by an inductor - Correct statement at resonance.
- Option B: Average power = apparent power - Correct statement at resonance because the power factor is 1.
- Option C: Wattles current is zero - Correct statement at resonance because the circuit is purely resistive.

Key Concepts:
- Resonance in LCR circuits
- Power factor

Memory Aids:
- At resonance, $X_L = X_C$ and $\phi = 0$ .

Quick Tips:
- Remember that at resonance, the circuit behaves like a purely resistive circuit.

12 Medium

12 of 84Question permalink

Analyzing +4 | -1

Given below are two statements:
Statement-I: The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.
Statement-II: In ac circuit, the average power delivered by the source never becomes zero.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: C

Statement I is true but Statement II is false

Option C (Statement I is true but Statement II is false) is correct because reactance can be zero with both inductor and capacitor, and average power can be zero in an AC circuit.

Solution approach

Correct Answer + Core Reason: Statement I is true but Statement II is false. Reactance can be zero if inductive and capacitive reactances cancel each other, and average power can be zero in a purely capacitive or inductive circuit.

Step-by-Step Solution:

Step 1: Analyze Statement I. The reactance of an AC circuit is given by $X = X_L - X_C$ , where $X_L = \omega L$ and $X_C = \frac{1}{{\omega C}}$ . If $X_L = X_C$ , then $X = 0$ . This is possible when both a capacitor and an inductor are present in the circuit. Therefore, Statement I is true.

Step 2: Analyze Statement II. The average power delivered by the source in an AC circuit is given by $P_{avg} = V_{rms} I_{rms} \cos{\phi}$ , where $\phi$ is the phase angle between the voltage and current. If the circuit is purely capacitive or purely inductive, $\phi = \frac{\pi}{2}$ , and $\cos{\phi} = 0$ . Therefore, $P_{avg} = 0$ . Thus, Statement II is false.

Why other options are wrong:
- Option A (Both Statement I and Statement II are true): Statement II is false, so this option is incorrect.
- Option B (Both Statement I and Statement II are false): Statement I is true, so this option is incorrect.
- Option D (Statement I is false but Statement II is true): Statement I is true, and Statement II is false, so this option is incorrect.

Key Concepts:
- Reactance: $X = X_L - X_C$
- Inductive Reactance: $X_L = \omega L$
- Capacitive Reactance: $X_C = \frac{1}{{\omega C}}$
- Average Power in AC circuit: $P_{avg} = V_{rms} I_{rms} \cos{\phi}$

Memory Aids:
- Remember that in a purely capacitive or inductive circuit, the phase angle is 90 degrees, and the average power is zero.

Quick Tips:
- Understand the conditions for zero reactance and zero average power in AC circuits.

13 Medium

13 of 84Question permalink

Analyzing +4 | -1

Assertion: In higher power transmission,

240 \mathrm{~V}

is preferred over

120 \mathrm{~V}

.
Reason: Power loss is more in case of 120 Vas compared to 240 V.

View answer and explanation

Correct option: D

If both Assertion & Reason are false

Option D (If both Assertion & Reason are false) is correct because higher voltage is preferred for power transmission to reduce power loss, and the reason provided is the opposite of the truth.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are false. Higher voltage is preferred for power transmission to reduce power loss.

Step-by-Step Solution:

Step 1: Analyze the assertion. Higher voltage is preferred for power transmission to minimize current and reduce power loss.

Step 2: Analyze the reason. Power loss is given by $P_{loss} = I^2 R$ , where I is the current and R is the resistance of the transmission line. To transmit the same power P at a higher voltage V, the current I must be lower ( $P = VI$ ). Therefore, the power loss is less at higher voltages.

Step 3: Since higher voltage is preferred and power loss is less at higher voltages, both the assertion and the reason are false.

Why other options are wrong:
- Option A: Both the assertion and reason are false, so this option is incorrect.
- Option B: Both the assertion and reason are false, so this option is incorrect.
- Option C: The assertion is false, so this option is incorrect.

Key Concepts:
- Power transmission
- Power loss
- Voltage
- Current
- Resistance

Memory Aids:
- Remember that power loss is proportional to the square of the current: $P_{loss} = I^2 R$ .

Quick Tips:
- Higher voltage is used for long-distance power transmission to reduce power loss.

14 Medium

14 of 84Question permalink

Application +4 | -1

In a LCR circuit having

L = 8.0\;henry,\;C = 0.5\;\mu F

and

R = 100\;ohm

in series. The resonance frequency in per second is

View answer and explanation

Correct option: C

500 radian

Option C (500 radian) is correct because the resonance frequency is calculated using the formula

\omega = \frac{1}{\sqrt{LC}}

, resulting in 500 rad/s.

Solution approach

Correct Answer + Core Reason: 500 radian. The resonance frequency is calculated using the formula $\omega = \frac{1}{\sqrt{LC}}$ .

Step-by-Step Solution:

Step 1: Identify the given information:\nInductance, $L = 8.0 \text{ H}$ \nCapacitance, $C = 0.5 \mu \text{F} = 0.5 \times 10^{-6} \text{ F}$

Step 2: Recall the formula for resonance frequency:
$\omega = \frac{1}{\sqrt{LC}}$

Step 3: Substitute the given values:
$\omega = \frac{1}{\sqrt{8.0 \times 0.5 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-6}}} = \frac{1}{2 \times 10^{-3}} = 500 \text{ rad/s}$

Why other options are wrong:
- Option A: This option gives a frequency of 700 rad/s, which is incorrect. It might result from a miscalculation.
- Option B: This option gives a frequency of 600 Hz, which is incorrect and also in the wrong units (Hz instead of rad/s).
- Option D: This option gives a frequency of 500 Hz, which is incorrect because the question asks for the answer in radians per second.

Key Concepts:
- Resonance frequency: $\omega = \frac{1}{\sqrt{LC}}$

Memory Aids:
- Remember the formula for resonance frequency: "One over root LC"

Quick Tips:
- Pay attention to the units required in the answer (rad/s vs Hz).
- Ensure correct substitution of values.

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Application +4 | -1

A coil of wire of certain radius has 100 turns and a self inductance of 15 mH. The self inductance of a second similar coil of 500 turns will be

View answer and explanation

Correct option: B

375 mH

Option B (t = 1.57 ms) is correct because the energy is completely magnetic when the capacitor is fully discharged, which occurs at T/4.

Solution approach

Correct Answer + Core Reason: t = 1.57 ms. The energy is completely magnetic when the capacitor is fully discharged, which occurs at T/4.

Step-by-Step Solution:

Step 1: Identify the given information:\nL = 20 mH = 0.02 H\nC = 50 $\mu$ F = 50 x $10^{-6}$ F\nQ = 10 mC = 0.01 C

Step 2: Calculate the angular frequency:
$\omega = \frac{{1}}{{\sqrt{{LC}}}} = \frac{{1}}{{\sqrt{{0.02 * 50 * 10^{-6}}}}} = \frac{{1}}{{\sqrt{{10^{-6}}}}} = 1000 rad/s$

Step 3: Calculate the time period:\nT = $\frac{{2\pi}}{{\omega}} = \frac{{2\pi}}{{1000}} = 2\pi * 10^{-3} s \approx 6.28 ms$

Step 4: The energy is completely magnetic at T/4:\nt = $\frac{{T}}{{4}} = \frac{{6.28}}{{4}} ms = 1.57 ms$

Step 5: Final Answer:\nt = 1.57 ms

Why other options are wrong:
- Option A (t = 0): This is incorrect because at t=0, the energy is completely electrical, stored in the capacitor.
- Option C (t = 3.14 ms): This is incorrect because this is T/2, when the energy is again completely electrical but with opposite polarity.
- Option D (t = 6.28 ms): This is incorrect because this is the full time period T, when the energy is back to its initial state (completely electrical).

Key Concepts:
- LC circuit oscillations
- Angular frequency: $\omega = \frac{{1}}{{\sqrt{{LC}}}}$
- Time period: T = $\frac{{2\pi}}{{\omega}}$
- Energy transfer between capacitor and inductor

Memory Aids:
- Remember the formulas for angular frequency and time period in an LC circuit.

Quick Tips:
- The energy is completely electrical at t=0 and t=T, and completely magnetic at t=T/4 and t=3T/4.

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Application +4 | -1

The intensity of EMW at a distance of 1 km from a source of power 12.56 KW is

View answer and explanation

Correct option: A

{10^{ - 3}}\,w{m^2}

Option A (

{10^{ - 3}}\,w{m^2}

) is correct because intensity is power per unit area, and the area is the surface area of a sphere with radius 1 km.

Solution approach

Correct Answer + Core Reason: ${10^{ - 3}}\,w{m^2}$ . Intensity is power divided by area.

Step-by-Step Solution:

Step 1: Define the intensity formula: $I = \frac{P}{A}$

Step 2: Identify the given values: Power $P = 12.56 \text{ kW} = 12.56 \times 10^3 \text{ W}$ , distance $r = 1 \text{ km} = 10^3 \text{ m}$

Step 3: Calculate the surface area of a sphere with radius r: $A = 4 \pi r^2$

Step 4: Substitute the value of r: $A = 4 \pi (10^3 \text{ m})^2 = 4 \pi \times 10^6 \text{ m}^2$

Step 5: Approximate $4\pi \approx 4 \times 3.14 = 12.56$ , so $A = 12.56 \times 10^6 \text{ m}^2$

Step 6: Calculate the intensity: $I = \frac{12.56 \times 10^3 \text{ W}}{12.56 \times 10^6 \text{ m}^2} = 10^{-3} \text{ W/m}^2$

Why other options are wrong:
- Option B: ${10^{ - 3}}\,aw{m^2}$ . This option has incorrect units (aw/m²).
- Option C: ${10^{ - 3}}\,wc{m^2}$ . This option has incorrect units (w/cm²).
- Option D: ${10^{ - 3}}\,wm{m^2}$ . This option has incorrect units (w/mm²).

Key Concepts:
- Intensity of electromagnetic waves
- Surface area of a sphere

Memory Aids:
- Intensity = Power / Area

Quick Tips:
- Always check the units and make sure they are consistent.
- Remember the formula for the surface area of a sphere.

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Understanding +4 | -1

The ratio of secondary to primary turns is

\mathrm{9 : 4}

. If power input is

P

, what will be the ratio of power output (neglect all losses) to power input

View answer and explanation

Correct option: D

\mathrm{1 : 1}

Option D (1 : 1) is correct because in an ideal transformer, the power input equals the power output, regardless of the turns ratio.

Solution approach

Correct Answer + Core Reason: $\mathrm{1 : 1}$ . In an ideal transformer, power input equals power output.

Step-by-Step Solution:

Step 1: State the assumption for an ideal transformer: No power losses.

Step 2: Therefore, power input = power output.

Step 3: The ratio of power output to power input is 1:1.

Why other options are wrong:
- Option A: The turns ratio affects voltage and current, but not power in an ideal transformer.
- Option B: The turns ratio affects voltage and current, but not power in an ideal transformer.
- Option C: The turns ratio affects voltage and current, but not power in an ideal transformer.

Key Concepts:
- Ideal transformer
- Power conservation

Memory Aids:
- Ideal = Input = Output

Quick Tips:
- For ideal transformers, power in = power out.

18 Medium

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Understanding +4 | -1

The instantaneous value of current and emf in an A.C. circuit are

I = \frac{1}{{\sqrt 2 }}\sin 314t

amp and

E = \sqrt 2 \sin \left( {314t - \frac{\pi }{6}} \right)V,

with respect to I) will be

View answer and explanation

Correct option: A

- \frac{\pi }{6}\mathrm{ rad}

Option A is correct. The phase difference is calculated by subtracting the phase of the current from the phase of the voltage, resulting in

-\frac{\pi}{6}

rad.

Solution approach

Correct Answer + Core Reason: $- \frac{\pi }{6}\mathrm{ rad}$ . The phase difference is the voltage phase minus the current phase.

Step-by-Step Solution:

Step 1: Identify the phase of the current: $I = \frac{1}{{\sqrt 2 }} \sin 314t$ has a phase of 0.

Step 2: Identify the phase of the voltage: $E = \sqrt 2 \sin \left( {314t - \frac{\pi }{6}} \right)V$ has a phase of $-\frac{\pi }{6}$ .

Step 3: Calculate the phase difference: Phase difference = Voltage phase - Current phase = $-\frac{\pi }{6} - 0 = -\frac{\pi }{6}$ rad.

Why other options are wrong:
- Option B: This is incorrect. The phase difference is $-\frac{\pi }{6}$ .
- Option C: This is incorrect. The phase difference is negative.
- Option D: This is incorrect. The phase difference is $-\frac{\pi }{6}$ .

Key Concepts:
- Phase difference in AC circuits

Memory Aids:
- Phase difference = Voltage phase - Current phase.

Quick Tips:
- Pay attention to the signs when calculating phase differences.

19 Medium

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Analyzing +4 | -1

The voltage of an AC source varies with time according to the relation

E = 200\sin \left( {100\pi t} \right)\cos \left( {100\pi t} \right)

Select correct statement.

View answer and explanation

Correct option: D

The frequency of source voltage is 100 Hz

Option D (The frequency of source voltage is 100 Hz) is correct because using the trigonometric identity, the given equation simplifies to

E = 100\sin(200\pi t)

, indicating a frequency of 100 Hz.

Solution approach

Correct Answer + Core Reason: The frequency of source voltage is 100 Hz. The frequency is determined by simplifying the equation using trigonometric identities.

Step-by-Step Solution:

Step 1: Simplify the given equation using the trigonometric identity: $2 \sin \theta \cos \theta = \sin 2\theta$
$E = 200 \sin(100 \pi t) \cos(100 \pi t) = 100 \times 2 \sin(100 \pi t) \cos(100 \pi t) = 100 \sin(200 \pi t)$

Step 2: Compare the simplified equation with the general form of a sinusoidal voltage: $E = E_0 \sin(\omega t) = E_0 \sin(2 \pi f t)$

Step 3: Identify the angular frequency: $\omega = 200 \pi$

Step 4: Calculate the frequency (f): $f = \frac{\omega}{2 \pi} = \frac{200 \pi}{2 \pi} = 100 Hz$

Step 5: Determine the peak voltage: $E_0 = 100 V$

Final Answer: The frequency of source voltage is 100 Hz

Why other options are wrong:
- Option A: The peak voltage of source is 200 V - Incorrect, the peak voltage is 100 V.
- Option B: The peak voltage of source is 400 V - Incorrect, the peak voltage is 100 V.
- Option C: The frequency of source voltage is 50 Hz - Incorrect, the frequency is 100 Hz.

Key Concepts:
- Trigonometric identities: $2 \sin \theta \cos \theta = \sin 2\theta$
- Sinusoidal voltage equation: $E = E_0 \sin(\omega t)$
- Relationship between angular frequency and frequency: $\omega = 2 \pi f$

Memory Aids:
- Remember the trigonometric identity for simplifying the equation.

Quick Tips:
- Always simplify trigonometric expressions before analyzing the equation.
- Remember the relationship between angular frequency and frequency.

20 Medium

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Analyzing +4 | -1

Statement (A): An electric lamp is connected in series with a long solenoid of copper with air core and then connected to AC source. If an iron rod is inserted in solenoid the lamp will become dim.

Statement (B): If iron rod is inserted in solenoid, the induction of solenoid increases.

View answer and explanation

Correct option: A

Both (A) and (B) are true

Option A (Both (A) and (B) are true) is correct because inserting an iron rod increases the solenoid's inductance, which increases impedance and dims the lamp.

Solution approach

Correct Answer + Core Reason: Both (A) and (B) are true. Inserting an iron rod increases inductance, increasing impedance and dimming the lamp.

Step-by-Step Solution:

Step 1: Analyze Statement (A). The electric lamp is connected in series with a solenoid. The impedance of the circuit is $Z = \sqrt{{R^2 + X_L^2}}$ , where $R$ is the resistance of the lamp and $X_L = \omega L$ is the inductive reactance of the solenoid. The current in the circuit is $I = \frac{V}{Z}$ . If an iron rod is inserted into the solenoid, the inductance $L$ increases. This increases $X_L$ , which increases $Z$ . As a result, the current $I$ decreases, and the lamp becomes dimmer. Therefore, Statement (A) is true.

Step 2: Analyze Statement (B). The inductance of a solenoid is given by $L = \mu_0 n^2 A l$ , where $n$ is the number of turns per unit length, $A$ is the cross-sectional area, and $l$ is the length of the solenoid. When an iron rod is inserted, the permeability changes from $\mu_0$ to $\mu = \mu_r \mu_0$ , where $\mu_r$ is the relative permeability of the iron rod. Since $\mu_r > 1$ , the inductance $L$ increases. Therefore, Statement (B) is true.

Why other options are wrong:
- Option B ((A) is true but (B) is false): Statement (B) is also true, so this option is incorrect.
- Option C ((A) is false but (B) is true): Statement (A) is also true, so this option is incorrect.
- Option D (Both (A) and (B) are false): Both statements are true, so this option is incorrect.

Key Concepts:
- Inductance of a solenoid: $L = \mu_0 n^2 A l$
- Inductive Reactance: $X_L = \omega L$
- Impedance in an L-R circuit: $Z = \sqrt{{R^2 + X_L^2}}$

Memory Aids:
- Inserting an iron core increases inductance.
- Increased inductance leads to increased impedance and decreased current.

Quick Tips:
- Remember that the inductance of a solenoid increases when a ferromagnetic material is inserted into its core.
- Understand how changes in inductance affect the impedance and current in an AC circuit.

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Understanding +4 | -1

Alternating current cannot be measured by DC ammeter because

View answer and explanation

Correct option: C

Average value of current for complete cycle is zero

Option C (Average value of current for complete cycle is zero) is correct because the average value of AC over a complete cycle is zero, so a DC ammeter, which measures average current, will read zero.

Solution approach

Correct Answer + Core Reason: Average value of current for complete cycle is zero. A DC ammeter measures average current, which is zero for AC.

Step-by-Step Solution:

Step 1: Understand how a DC ammeter works.

Step 2: A DC ammeter measures the average value of current.

Step 3: The average value of alternating current over a complete cycle is zero because the positive and negative halves cancel each other out.

Why other options are wrong:
- Option A: AC cannot pass through DC ammeter - Incorrect, AC can pass through a DC ammeter, but it will not give a meaningful reading.
- Option B: AC changes direction - While true, this is not the primary reason. The average value being zero is the main reason.
- Option D: DC ammeter will get damaged - Incorrect, it will simply read zero.

Key Concepts:
- Average value of AC
- DC ammeter operation

Memory Aids:
- Average AC over a cycle is zero.

Quick Tips:
- Remember that DC ammeters measure average current.

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Understanding +4 | -1

If the capacity of a condenser is 1 F, then its resistance in a DC circuit will be

View answer and explanation

Correct option: B

infinity

Option B (infinity) is correct because a capacitor blocks DC current after it is fully charged, effectively acting as an open circuit with infinite resistance.

Solution approach

Correct Answer + Core Reason: Infinity. A capacitor blocks DC current after being fully charged.

Step-by-Step Solution:

Step 1: Understand the behavior of a capacitor in a DC circuit.

Step 2: A capacitor initially allows current to flow as it charges. However, once fully charged, it blocks the flow of DC current.

Step 3: In a DC circuit, after the capacitor is fully charged, no more current flows through it. This is equivalent to an open circuit.

Step 4: An open circuit has infinite resistance.

Step 5: Therefore, the resistance of a 1 F capacitor in a DC circuit after it is fully charged is infinite.

Why other options are wrong:
- Option A: Zero resistance would imply a short circuit, which is not the case for a fully charged capacitor in a DC circuit.
- Option C: 1 Ω is an arbitrary value and does not represent the behavior of a capacitor in a DC circuit.
- Option D: $\frac{{1}}{{2}}$ Ω is also an arbitrary value and does not represent the behavior of a capacitor in a DC circuit.

Key Concepts:
- Capacitor behavior in DC circuits
- Resistance
- Open circuit
- DC current

Memory Aids:
- Capacitors block DC (after charging)

Quick Tips:
- Remember that capacitors act as open circuits in DC after they are fully charged.

23 Medium

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Remembering +4 | -1

The power factor of good choke coil is

View answer and explanation

Correct option: A

Nearly zero

Option A is correct because a good choke coil is designed to have a high inductance and low resistance, resulting in a power factor that is nearly zero.

Solution approach

Correct Answer + Core Reason: Nearly zero. A choke coil ideally has high inductance and negligible resistance.

Step-by-Step Solution:

Step 1: Understand the purpose of a choke coil. A choke coil is an inductor used to block high-frequency AC while allowing low-frequency AC or DC to pass.

Step 2: Recall the characteristics of an ideal choke coil. An ideal choke coil has high inductance (L) and zero resistance (R).

Step 3: Recall the formula for the power factor: $\cos \phi = \frac{R}{Z}$ , where R is the resistance and Z is the impedance.

Step 4: In a choke coil, the impedance is dominated by the inductive reactance ( $X_L = \omega L$ ), so $Z \approx X_L$ .

Step 5: Therefore, the power factor is approximately $\cos \phi \approx \frac{R}{X_L} = \frac{R}{\omega L}$ .

Step 6: Since a good choke coil has a very low resistance (R) and a high inductance (L), the power factor is nearly zero.

Step 7: Conclusion: The power factor of a good choke coil is nearly zero.

Why other options are wrong:
- Option B: Exactly zero: While ideal, real choke coils have some resistance, so the power factor is nearly zero, not exactly zero.
- Option C: Nearly one: A power factor of nearly one would mean the circuit is mostly resistive, which is not the case for a choke coil.
- Option D: Exactly one: A power factor of exactly one means the circuit is purely resistive, which is not the case for a choke coil.

Key Concepts:
- Choke coil
- Inductance (L)
- Resistance (R)
- Impedance (Z)
- Power factor ( $\cos \phi$ )
- Inductive reactance ( $X_L = \omega L$ )

Memory Aids:
- Power factor: $\cos \phi = \frac{R}{Z}$

Quick Tips:
- Remember that a choke coil is designed to have high inductance and low resistance.

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Application +4 | -1

An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency ω. Power dissipated in the circuit is

View answer and explanation

Correct option: B

\frac{{{V^2}R}}{{\left( {{R^2} + {\omega ^2}{L^2}} \right)}}

Option B (

\frac{{{V^2}R}}{{\left( {{R^2} + {\omega ^2}{L^2}} \right)}}

) is correct because it represents the power dissipated in an RL circuit connected to an AC source.

Solution approach

Correct Answer + Core Reason: $\frac{{{V^2}R}}{{\left( {{R^2} + {\omega ^2}{L^2}} \right)}}$ . This is the formula for power dissipated in an RL circuit.

Step-by-Step Solution:

Step 1: Recall the formula for power dissipated in an AC circuit:
$P = V_{rms} I_{rms} \cos \phi$ , where $\cos \phi$ is the power factor.

Step 2: In an RL circuit, the impedance is given by:
$Z = \sqrt{R^2 + (\omega L)^2}$

Step 3: The RMS current is given by:
$I_{rms} = \frac{V_{rms}}{Z} = \frac{V}{\sqrt{R^2 + (\omega L)^2}}$

Step 4: The power factor is given by:
$\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$

Step 5: Substitute $I_{rms}$ and $\cos \phi$ into the power formula:
$P = V \times \frac{V}{\sqrt{R^2 + (\omega L)^2}} \times \frac{R}{\sqrt{R^2 + (\omega L)^2}} = \frac{V^2 R}{R^2 + (\omega L)^2}$

Why other options are wrong:
- Option A: This option is incorrect as it has the inverse relationship and is missing the square on the voltage.
- Option C: This option is incorrect as it is missing the voltage squared and the resistance in the numerator.
- Option D: This option is incorrect as it involves the square root of the impedance divided by the voltage squared, which is not the correct formula for power dissipation.

Key Concepts:
- Power dissipation in AC circuits
- Impedance in RL circuits
- Power factor

Memory Aids:
- Remember the power formula: "P equals V squared R over Z squared"

Quick Tips:
- Always remember to use RMS values for voltage and current when calculating power in AC circuits.
- Understand the relationship between impedance, resistance, and reactance.

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Remembering +4 | -1

For an R-L-C circuit driven with voltage of amplitude

{v_m}

and frequency

{\omega _0} = \frac{1}{{\sqrt {LC} }},

the current exhibits resonance. The quality factor, Q is given by

View answer and explanation

Correct option: A

\frac{{{\omega _0}L}}{R}

Option A is correct because the quality factor Q for an R-L-C circuit at resonance is given by

\frac{{\omega _0L}}{R}

.

Solution approach

Correct Answer + Core Reason: $\frac{{\omega _0L}}{R}$ . The quality factor is a measure of the sharpness of resonance.

Step-by-Step Solution:

Step 1: Recall the definition of the quality factor (Q) for an R-L-C circuit at resonance.

Step 2: The quality factor is defined as the ratio of the inductive reactance (or capacitive reactance) to the resistance at resonance.

Step 3: Express the quality factor mathematically: $Q = \frac{X_L}{R} = \frac{\omega_0 L}{R}$ or $Q = \frac{X_C}{R} = \frac{1}{\omega_0 CR}$ .

Step 4: Since ${\omega _0} = \frac{1}{{\sqrt {LC} }}$ , we can also write $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ .

Step 5: From the given options, the correct expression for the quality factor is $Q = \frac{{\omega _0L}}{R}$ .

Why other options are wrong:
- Option B: $\frac{{\omega _0}R}{L}$ : This is the inverse of the correct expression.
- Option C: $\frac{R}{{\omega _0}C}$ : This is the inverse of the correct expression involving capacitance.
- Option D: $\frac{{CR}}{{\omega _0}}$ : This is not a standard expression for the quality factor.

Key Concepts:
- Quality factor (Q) in AC circuits
- Resonance in AC circuits
- Inductive reactance ( $X_L = \omega L$ )
- Capacitive reactance ( $X_C = \frac{1}{\omega C}$ )
- Resonance frequency ( ${\omega _0} = \frac{1}{{\sqrt {LC} }}$ )

Memory Aids:
- Quality factor: $Q = \frac{\omega_0 L}{R}$

Quick Tips:
- Remember the formula for the quality factor in an R-L-C circuit at resonance.

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Analyzing +4 | -1

Assertion (A): An inductance and a resistance are connected in series with an AC circuit. In this circuit, the current and the potential difference across the resistance logs behind potential difference across the inductance by an angle

\frac{\pi}{2}.

Reason (R): In L - R circuit, voltage leads the current by phase angle which depends on the value of inductance and resistance both.

View answer and explanation

Correct option: B

Both assertion and reason are true but reason is not the correct explanation of assertion

Option B is correct because both the assertion and the reason are true, but the reason does not correctly explain the assertion. The phase difference in an LR circuit depends on both L and R, but the assertion's specific angle is incorrect.

Solution approach

Correct Answer + Core Reason: Both assertion and reason are true but reason is not the correct explanation of assertion. The assertion is partially incorrect about the phase difference being exactly $\frac{\pi}{2}$ , but the reason correctly states the dependency on L and R.

Step-by-Step Solution:

Step 1: Analyze the Assertion: In an L-R circuit, the current lags behind the voltage across the inductor by $\frac{\pi}{2}$ . However, the assertion states the current lags behind the *potential difference across the resistance* by $\frac{\pi}{2}$ , which is incorrect. The current is in phase with the voltage across the resistor.

Step 2: Analyze the Reason: The reason states that the voltage leads the current by a phase angle that depends on both L and R. This is a correct statement.

Step 3: Determine the relationship: While both statements are true, the reason does not explain why the assertion is (partially) true. The phase difference is due to the inductor's reactance, but the reason doesn't directly explain the incorrect angle mentioned in the assertion.

Why other options are wrong:
- Option A: The reason is not a correct explanation of the assertion.
- Option C: The assertion is not entirely true, as the phase difference is not specifically between the current and the potential difference across the resistance.
- Option D: Both assertion and reason are not false; the reason is true.

Key Concepts:
- L-R circuits
- Phase difference between voltage and current in inductive and resistive circuits.

Memory Aids:
- ELI the ICE man (Voltage leads current in Inductor, Current leads voltage in Capacitor).

Quick Tips:
- Remember the phase relationships in L, C, and R circuits.

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Application +4 | -1

In order to obtain a time constant of 10 s in a R-C circuit containing a resistance of

{10^3}\mathrm{\Omega }

, the capacity of the condenser should be

View answer and explanation

Correct option: D

10000 μF

Option D (10000 μF) is correct because the capacitance is calculated using the time constant formula, resulting in 10000 μF.

Solution approach

Correct Answer + Core Reason: 10000 μF. The capacitance is calculated using the formula for the time constant in an RC circuit.

Step-by-Step Solution:

Step 1: Identify the given information:\nTime constant, $\tau = 10 \text{ s}$ \nResistance, $R = 10^3 \Omega = 1000 \Omega$

Step 2: Recall the formula for the time constant in an RC circuit:
$\tau = RC$

Step 3: Solve for the capacitance, $C$ :
$C = \frac{\tau}{R}$

Step 4: Substitute the given values:
$C = \frac{10 \text{ s}}{1000 \Omega} = 0.01 \text{ F}$

Step 5: Convert the capacitance to microfarads:
$C = 0.01 \text{ F} = 0.01 \times 10^6 \mu \text{F} = 10000 \mu \text{F}$

Why other options are wrong:
- Option A: This option gives a capacitance of 10 μF, which is incorrect. It might result from a miscalculation or misunderstanding of the units.
- Option B: This option gives a capacitance of 100 μF, which is incorrect. It might result from a miscalculation or misunderstanding of the units.
- Option C: This option gives a capacitance of 1000 μF, which is incorrect. It might result from a miscalculation or misunderstanding of the units.

Key Concepts:
- Time constant in an RC circuit: $\tau = RC$

Memory Aids:
- Remember the formula: "Tau equals RC"

Quick Tips:
- Always ensure the units are consistent before performing calculations.
- Convert the final answer to the required units.

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Analyzing +4 | -1

Voltage and current in an ac circuit are given by

V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right)

and

I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)

View answer and explanation

Correct option: C

Current leads the voltage by

60^{\circ}

Option C (Current leads the voltage by

60^{\circ}

) is correct because the phase angle of the current is

\frac{\pi}{6}

and the phase angle of the voltage is

-\frac{\pi}{6}

, so the current leads the voltage by

\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{3} = 60^{\circ}

.

Solution approach

Correct Answer + Core Reason: Current leads the voltage by $60^{\circ}$ . The phase angle difference indicates the current leads the voltage.

Step-by-Step Solution:

Step 1: Identify the given information: $V = 5 \sin(100\pi t - \frac{\pi}{6})$ and $I = 4 \sin(100\pi t + \frac{\pi}{6})$ .

Step 2: Determine the phase angle of the voltage: $\phi_V = -\frac{\pi}{6}$ .

Step 3: Determine the phase angle of the current: $\phi_I = \frac{\pi}{6}$ .

Step 4: Calculate the phase difference between the current and voltage: $\Delta \phi = \phi_I - \phi_V = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ .

Step 5: Convert the phase difference to degrees: $\frac{\pi}{3} = \frac{180^{\circ}}{3} = 60^{\circ}$ .

Step 6: Since the phase difference is positive, the current leads the voltage by $60^{\circ}$ .

Step 7: Final Answer: Current leads the voltage by $60^{\circ}$ .

Why other options are wrong:
- Option A: Voltage leads the current by $30^{\circ}$ . Incorrect, the current leads the voltage.
- Option B: Current leads the voltage by $30^{\circ}$ . Incorrect, the current leads the voltage by 60 degrees.
- Option D: Voltage leads the current by $60^{\circ}$ . Incorrect, the current leads the voltage.

Key Concepts:
- Phase difference between voltage and current
- AC circuits

Memory Aids:
- Remember that a positive phase difference means the current leads the voltage.

Quick Tips:
- Always determine the phase angles of both voltage and current before calculating the phase difference.

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Remembering +4 | -1

Core of transformer is made up of

View answer and explanation

Correct option: A

soft iron

Option A (soft iron) is correct because soft iron has high permeability and low hysteresis loss, making it suitable for transformer cores.

Solution approach

Correct Answer + Core Reason: soft iron. Soft iron is used due to its high permeability and low hysteresis losses.

Step-by-Step Solution:

Step 1: Identify the requirements for a transformer core material.

Step 2: The core material should have high permeability to concentrate the magnetic flux and low hysteresis loss to minimize energy loss during magnetization and demagnetization cycles.

Step 3: Soft iron possesses these properties.

Step 4: Therefore, soft iron is the ideal material for the core of a transformer.

Why other options are wrong:
- Option B: Steel has higher hysteresis losses compared to soft iron.
- Option C: Iron is a general term, but soft iron specifies the desired properties.
- Option D: Alnico is a permanent magnet material and not suitable for transformer cores.

Key Concepts:
- Transformer core materials
- Permeability
- Hysteresis loss

Memory Aids:
- Soft iron = Soft core = Transformer core

Quick Tips:
- Remember that transformer cores need to be easily magnetized and demagnetized.

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Understanding +4 | -1

Mark the correct statement.

View answer and explanation

Correct option: A

Ideal inductor does not dissipate power in an AC circuit

Option A is correct because an ideal inductor stores energy but does not dissipate it in an AC circuit.

Solution approach

Correct Answer + Core Reason: Ideal inductor does not dissipate power in an AC circuit. An ideal inductor only stores energy in its magnetic field and returns it to the circuit, resulting in zero average power dissipation.

Step-by-Step Solution:

Step 1: Recall the properties of an ideal inductor in an AC circuit.

Step 2: An ideal inductor has zero resistance. Therefore, it does not dissipate energy as heat.

Step 3: The power dissipated in an inductor is given by $P = I_{rms} V_{rms} \cos{\phi}$ , where $\phi$ is the phase angle between voltage and current. For an ideal inductor, $\phi = \frac{\pi}{2}$ , so $\cos{\phi} = 0$ , and $P = 0$ .

Why other options are wrong:
- Option B: An ideal inductor does not dissipate any power, so it cannot dissipate maximum power.
- Option C: In an inductor, current lags behind the voltage by $\frac{\pi}{2}$ , not $\pi$ .
- Option D: In an inductor, current lags behind the voltage, it does not lead.

Key Concepts:
- AC circuits
- Inductors
- Power dissipation
- Phase angle

Memory Aids:
- ELI the ICE man: Voltage (E) leads Current (I) in an Inductor (L), Current (I) leads Voltage (E) in a Capacitor (C).

Quick Tips:
- Remember that ideal inductors and capacitors do not dissipate power; they only store energy.

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