ALTERNATING CURRENT MCQs for NEET UG PHYSICS — Mixed difficulty

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 143Question 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 Easy

3 of 143Question permalink

Understanding +4 | -1

An alternating e.m.f. is applied to purely capacitive circuit. The phase relation between e.m.f. and current flowing in the circuit is or
In a circuit containing capacitance only

View answer and explanation

Correct option: B

Current is ahead of e.m.f. by

\pi /2

Option B (Current is ahead of e.m.f. by

\pi /2

) is correct because in a purely capacitive circuit, the current leads the voltage by 90 degrees.

Solution approach

Correct Answer + Core Reason: Current is ahead of e.m.f. by $\pi /2$ . In a purely capacitive circuit, current leads voltage by 90 degrees.

Step-by-Step Solution:

Step 1: Recall the phase relationship between voltage and current in a capacitor.

Step 2: In a purely capacitive circuit, the current leads the voltage by $\pi/2$ radians (90 degrees).

Why other options are wrong:
- Option A: e.m.f is ahead of current by $\pi /2$ - Incorrect because voltage lags behind current in a capacitor.
- Option C: Current lags behind e.m.f. by $\pi$ - Incorrect, current leads in a capacitor, not lags, and the phase difference is $\pi/2$ , not $\pi$ .
- Option D: Current is ahead of e.m.f. by $\pi$ - Incorrect, the phase difference is $\pi/2$ , not $\pi$ .

Key Concepts:
- Phase relationship in AC circuits
- Capacitive reactance

Memory Aids:
- "ELI the ICE man": ELI (Inductor: Voltage leads Current), ICE (Capacitor: Current leads Voltage)

Quick Tips:
- Remember the ELI the ICE man mnemonic.

4 Medium

4 of 143Question 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.

5 Easy

5 of 143Question permalink

Application +4 | -1

A generator produces a voltage that is given by

V = 240\sin 120\;t

, where t is in seconds. The frequency and r.m.s. voltage are

View answer and explanation

Correct option: C

19 Hz and 170 V

Option C (19 Hz and 170 V) is correct because the frequency is

f = \frac{\omega}{2\pi} = \frac{120}{2\pi} \approx 19

Hz and the RMS voltage is

V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{240}{\sqrt{2}} \approx 170

V.

Solution approach

Correct Answer + Core Reason: 19 Hz and 170 V. The frequency and RMS voltage are calculated from the given voltage equation.

Step-by-Step Solution:

Step 1: Identify the angular frequency $\omega$ from the given equation $V = 240\sin 120t$ .

Step 2: $\omega = 120$ rad/s.

Step 3: Calculate the frequency $f$ using the formula $f = \frac{\omega}{2\pi} = \frac{120}{2\pi} \approx 19.1$ Hz.

Step 4: Calculate the RMS voltage $V_{rms}$ using the formula $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{240}{\sqrt{2}} \approx 169.7$ V.

Why other options are wrong:
- Option A: 60 Hz and 240 V - Incorrect, the frequency and RMS voltage are not calculated correctly.
- Option B: 19 Hz and 120 V - Incorrect, the RMS voltage is not calculated correctly.
- Option D: 754 Hz and 70 V - Incorrect, both frequency and RMS voltage are incorrect.

Key Concepts:
- Angular frequency
- Frequency
- RMS voltage

Memory Aids:
- $f = \frac{\omega}{2\pi}$ and $V_{rms} = \frac{V_0}{\sqrt{2}}$

Quick Tips:
- Remember the formulas for frequency and RMS voltage.

6 Medium

6 of 143Question 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.

7 Medium

7 of 143Question 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.

8 Medium

8 of 143Question 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.

9 Easy

9 of 143Question permalink

Understanding +4 | -1

In an ideal choke, ratio of its inductance L to its DC resistance R is

View answer and explanation

Correct option: A

Infinity

Option A (Infinity) is correct because an ideal choke has zero DC resistance, so the ratio of inductance to DC resistance is infinite.

Solution approach

Correct Answer + Core Reason: Infinity. An ideal choke has zero DC resistance, so the ratio of inductance to DC resistance is infinite.

Step-by-Step Solution:

Step 1: Understand the properties of an ideal choke. An ideal choke has inductance (L) but zero DC resistance (R).

Step 2: The ratio of inductance to DC resistance is L/R.

Step 3: Since R = 0 for an ideal choke, the ratio L/R = L/0 = Infinity.

Step 4: Final Answer: Infinity

Why other options are wrong:
- Option B: Zero. This would be the case if the inductance were zero, which is not true for a choke.
- Option C: Unity. This would mean the inductance and DC resistance are equal, which is not true for an ideal choke.
- Option D: hundred. This is an arbitrary value and not related to the properties of an ideal choke.

Key Concepts:
- Inductance
- DC resistance
- Ideal choke

Memory Aids:
- Remember that an ideal choke has zero DC resistance.

Quick Tips:
- Always remember the properties of ideal components.

10 Medium

10 of 143Question 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.

11 Medium

11 of 143Question 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.

12 Medium

12 of 143Question 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.

13 Medium

13 of 143Question 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.

14 Medium

14 of 143Question 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.

15 Easy

15 of 143Question permalink

Understanding +4 | -1

The core of a transformer is laminated to reduce energy losses due to

View answer and explanation

Correct option: A

Eddy currents

Option A (Eddy currents) is correct because laminating the core increases the resistance to eddy current flow, thus reducing energy losses.

Solution approach

Correct Answer + Core Reason: Eddy currents. Lamination reduces eddy current losses in the transformer core.

Step-by-Step Solution:

Step 1: Define eddy currents: Circulating currents induced in the core of a transformer due to the changing magnetic field.

Step 2: These currents dissipate energy as heat, leading to losses.

Step 3: Lamination increases the resistance to these currents by breaking the core into thin, insulated sheets.

Step 4: This reduces the magnitude of eddy currents and, consequently, the energy losses.

Why other options are wrong:
- Option B: Hysteresis losses are reduced by using soft iron, not lamination.
- Option C: Resistance in the winding causes copper losses, which are separate from core losses.
- Option D: Eddy currents are the reason for lamination.

Key Concepts:
- Eddy currents
- Lamination
- Transformer losses

Memory Aids:
- Lamination = Less eddy currents

Quick Tips:
- Lamination is specifically for reducing eddy current losses in the core.

16 Medium

16 of 143Question 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.

17 Easy

17 of 143Question permalink

Understanding +4 | -1

Power delivered by the source of the circuit becomes maximum, when

View answer and explanation

Correct option: B

wL = \frac{1}{{wc}}

Option B (

wL = \frac{1}{{wc}}

) is correct because maximum power is delivered at resonance, where the inductive reactance equals the capacitive reactance.

Solution approach

Correct Answer + Core Reason: $wL = \frac{1}{{wc}}$ . Maximum power is delivered at resonance.

Step-by-Step Solution:

Step 1: Define the condition for maximum power delivery: Maximum power is delivered by the source when the impedance of the circuit is minimized. This occurs at resonance in an RLC circuit.

Step 2: State the resonance condition: At resonance, the inductive reactance ( $X_L$ ) is equal to the capacitive reactance ( $X_C$ ).
$X_L = X_C$
$wL = \frac{1}{{wC}}$

Step 3: Final Answer: $wL = \frac{1}{{wc}}$

Why other options are wrong:
- Option A: This option is incorrect as it does not represent the resonance condition.
- Option C: This option is incorrect as it does not represent the resonance condition.
- Option D: This option is incorrect as it does not represent the resonance condition.

Key Concepts:
- Resonance in RLC circuits
- Inductive reactance
- Capacitive reactance
- Impedance

Memory Aids:
- Remember that at resonance, $X_L = X_C$ .

Quick Tips:
- Maximum power transfer occurs when the source impedance matches the load impedance (in AC circuits, this means minimizing the difference between inductive and capacitive reactances).

18 Medium

18 of 143Question 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.

19 Easy

19 of 143Question permalink

Remembering +4 | -1

In L – C – R series circuit the resonance condition in terms of capacitive reactance

\left( {{X_C}} \right)

and inductive reactance

\left( {{X_L}} \right)

is

View answer and explanation

Correct option: D

{X_C} - {X_L} = 0

Option D is correct because the resonance condition in an L-C-R series circuit is when the capacitive reactance equals the inductive reactance, i.e.,

{X_C} - {X_L} = 0

.

Solution approach

Correct Answer + Core Reason: ${X_C} - {X_L} = 0$ . Resonance occurs when capacitive and inductive reactances are equal.

Step-by-Step Solution:

Step 1: Recall the condition for resonance in an L-C-R series circuit.

Step 2: Resonance occurs when the inductive reactance ( $X_L$ ) is equal to the capacitive reactance ( $X_C$ ).

Step 3: Express the resonance condition mathematically: $X_L = X_C$ .

Step 4: Rearrange the equation to get: $X_C - X_L = 0$ .

Step 5: Conclusion: The resonance condition is ${X_C} - {X_L} = 0$ .

Why other options are wrong:
- Option A: ${X_C} + {X_L} = 0$ : This is incorrect because it implies that both reactances are zero, which is not the resonance condition.
- Option B: ${X_C} = 0$ : This is incorrect because it only considers the capacitive reactance and ignores the inductive reactance.
- Option C: ${X_L} = 0$ : This is incorrect because it only considers the inductive reactance and ignores the capacitive reactance.

Key Concepts:
- Resonance in AC circuits
- Inductive reactance ( $X_L$ )
- Capacitive reactance ( $X_C$ )

Memory Aids:
- Resonance: $X_L = X_C$

Quick Tips:
- Remember that resonance occurs when the inductive and capacitive reactances are equal.

20 Easy

20 of 143Question permalink

Understanding +4 | -1

A LCR circuit is at resonance for a capacitor C, inductance

\mathrm{L}

and resistance

\mathrm{R}

. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

View answer and explanation

Correct option: B

double

Option B is correct because at resonance, the current amplitude is inversely proportional to the resistance. Halving the resistance doubles the current amplitude.

Solution approach

Correct Answer + Core Reason: double. At resonance, current amplitude is inversely proportional to resistance.

Step-by-Step Solution:

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

Step 2: At resonance, the impedance of the circuit is equal to the resistance (R). The inductive reactance ( $X_L$ ) and capacitive reactance ( $X_C$ ) cancel each other out.

Step 3: The current amplitude at resonance is given by $I = \frac{V}{R}$ , where V is the voltage amplitude and R is the resistance.

Step 4: If the resistance is halved (R' = R/2), the new current amplitude will be $I' = \frac{V}{R'} = \frac{V}{R/2} = 2 \frac{V}{R} = 2I$ .

Step 5: Therefore, the current amplitude at resonance will be doubled.

Why other options are wrong:
- Option A: Zero: The current will not be zero if there is a voltage source and a resistance.
- Option C: same: The current will change if the resistance changes.
- Option D: halved: The current will double, not halve, if the resistance is halved.

Key Concepts:
- Resonance in AC circuits
- Impedance
- Current amplitude

Memory Aids:
- At resonance, $Z = R$
- $I = \frac{V}{R}$

Quick Tips:
- Remember that at resonance, the impedance is equal to the resistance.
- Understand the relationship between current, voltage, and resistance.

21 Medium

21 of 143Question 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.

22 Medium

22 of 143Question permalink

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.

23 Medium

23 of 143Question permalink

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.

24 Easy

24 of 143Question permalink

Analyzing +4 | -1

In series

L C R

circuit voltage drop across resistance is

8 {~V}

, across inductor is

6 {~V}

and across capacitor is

12 {~V}

. Then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the source voltage, power factor, and voltage drop relationships are not as described in the other options for the given LCR circuit.

Solution approach

Correct Answer + Core Reason: None of these. The voltage of the source, power factor, and voltage drops across elements are not as described in the other options.

Step-by-Step Solution:

Step 1: Calculate the source voltage ( $V$ ) using the formula for series LCR circuit:
$V = \sqrt{{V_R^2 + (V_L - V_C)^2}}$ \nwhere $V_R$ is the voltage across the resistor, $V_L$ is the voltage across the inductor, and $V_C$ is the voltage across the capacitor.

Step 2: Substitute the given values: $V_R = 8 \text{{ V}}$ , $V_L = 6 \text{{ V}}$ , and $V_C = 12 \text{{ V}}$ :
$V = \sqrt{{8^2 + (6 - 12)^2}}$

Step 3: Calculate the source voltage:
$V = \sqrt{{64 + (-6)^2}} = \sqrt{{64 + 36}} = \sqrt{{100}} = 10 \text{{ V}}$

Step 4: Determine the phase relationship between voltage and current. Since $V_C > V_L$ , the circuit is capacitive, and the current leads the voltage.

Step 5: Calculate the power factor ( $\cos \phi$ ) using the formula:
$\cos \phi = \frac{{R}}{Z} = \frac{{V_R}}{V} = \frac{{8}}{10} = 0.8 = \frac{{4}}{5}$

Step 6: Analyze the voltage drops. In a series LCR circuit, the voltage drop across individual elements can be greater than the source voltage due to resonance effects.

Step 7: Conclusion: The voltage of the source is 10 V, and the current leads the voltage. The power factor is 4/5. Therefore, none of the given options are correct.

Why other options are wrong:
- Option A: Voltage of the source will be leading current in the circuit. This is incorrect because since $V_C > V_L$ , the current leads the voltage, not the other way around.
- Option B: Voltage drop across each element will be less than the applied voltage. This is incorrect because the voltage across the capacitor (12 V) is greater than the source voltage (10 V).
- Option C: Power factor of circuit will be $4 / 3$ . This is incorrect because the power factor is calculated as $4/5$ , not $4/3$ .
- Option D: None of these. This is the correct answer because the other options are demonstrably false.

Key Concepts:
- Series LCR circuit
- Source voltage calculation
- Phase relationship between voltage and current
- Power factor

Memory Aids:
- "ELI the ICE man" (ELI: Voltage leads current in an inductive circuit; ICE: Current leads voltage in a capacitive circuit)
- Power factor: $\cos \phi = \frac{{R}}{Z}$

Quick Tips:
- Remember the formula for source voltage in a series LCR circuit.
- Determine whether the circuit is inductive or capacitive based on the relative values of $V_L$ and $V_C$ .
- The power factor is always between 0 and 1.

25 Easy

25 of 143Question permalink

Application +4 | -1

The instantaneous value of current in an A.C. circuit is

I = 2\sin \left( {100\;\pi \;t + \pi /3} \right)A

The current will be maximum for the first time at

View answer and explanation

Correct option: D

t = \frac{1}{{600}}s

Option D (

t = \frac{1}{{600}}s

) is correct because the current is maximum when the argument of the sine function is

\pi/2

, so

100\pi t + \pi/3 = \pi/2

, which gives

t = \frac{1}{{600}}s

.

Solution approach

Correct Answer + Core Reason: $t = \frac{1}{{600}}s$ . The current is maximum when the argument of the sine function is $\pi/2$ .

Step-by-Step Solution:

Step 1: The current is given by $I = 2\sin \left( {100\pi t + \frac{{\pi }}{{3}}} \right)A$ .

Step 2: The current will be maximum when $\sin \left( {100\pi t + \frac{{\pi }}{{3}}} \right) = 1$ .

Step 3: This occurs when $100\pi t + \frac{{\pi }}{{3}} = \frac{{\pi }}{{2}}$ .

Step 4: Solve for $t$ : $100\pi t = \frac{{\pi }}{{2}} - \frac{{\pi }}{{3}} = \frac{{\pi }}{{6}}$ .

Step 5: $t = \frac{{\pi }}{{6}} \cdot \frac{1}{{100\pi }} = \frac{1}{{600}}s$ .

Why other options are wrong:
- Option A: $t = \frac{1}{{100}}s$ - Incorrect, this does not satisfy the condition for maximum current.
- Option B: $t = \frac{1}{{200}}s$ - Incorrect, this does not satisfy the condition for maximum current.
- Option C: $t = \frac{1}{{400}}s$ - Incorrect, this does not satisfy the condition for maximum current.

Key Concepts:
- Maximum value of sine function

Memory Aids:
- $\sin(\theta) = 1$ when $\theta = \pi/2$

Quick Tips:
- Remember that the sine function reaches its maximum value of 1 at $\pi/2$ .

26 Easy

26 of 143Question permalink

Application +4 | -1

An AC voltage source has an output of

\mathrm{\Delta }V = \left( {200\mathrm{V}} \right)\sin 2\pi ft.

, This source is connected to a 100 Ω resistor. RMS current in the resistance is

View answer and explanation

Correct option: A

1.41A

Option A (1.41A) is correct because the RMS voltage is

V_{rms} = \frac{200}{\sqrt{2}}

and the RMS current is

I_{rms} = \frac{V_{rms}}{R} = \frac{200}{\sqrt{2} \cdot 100} = \sqrt{2} \approx 1.41

A.

Solution approach

Correct Answer + Core Reason: 1.41A. The RMS current is calculated from the RMS voltage and resistance.

Step-by-Step Solution:

Step 1: Identify the peak voltage: $V_0 = 200$ V.

Step 2: Calculate the RMS voltage: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}}$ V.

Step 3: Calculate the RMS current using Ohm's law: $I_{rms} = \frac{V_{rms}}{R} = \frac{{200/\sqrt{2}}}{{100}} = \frac{2}{{\sqrt{2}}} = \sqrt{2} \approx 1.41$ A.

Why other options are wrong:
- Option B: 2.41A - Incorrect, this is not the correct calculation.
- Option C: 3.41A - Incorrect, this is not the correct calculation.
- Option D: 0.71A - Incorrect, this is half of the correct answer, likely from dividing by $\sqrt{2}$ twice.

Key Concepts:
- RMS voltage
- Ohm's law

Memory Aids:
- $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $V = IR$

Quick Tips:
- Remember to use RMS values when applying Ohm's law to AC circuits.

27 Medium

27 of 143Question permalink

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.

28 Hard

28 of 143Question permalink

Understanding +4 | -1

The sum of the instantaneous current values over one complete cycle is zero, and the average current is zero. This means average power consumed is

View answer and explanation

Correct option: D

\frac{1}{2} \mathrm{i}_{\mathrm{m}}^2 \mathrm{R}

Option D (

\frac{1}{2} \mathrm{i}_{\mathrm{m}}^2 \mathrm{R}

) is correct because the average power consumed in an AC circuit is related to the RMS current, which is

\frac{i_m}{\sqrt{2}}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2} \mathrm{i}_{\mathrm{m}}^2 \mathrm{R}$ . The average power is calculated using the RMS current value.

Step-by-Step Solution:

Step 1: Recall the formula for average power in an AC circuit: $P_{avg} = I_{rms}^2 R$

Step 2: Recall the relationship between RMS current and peak current: $I_{rms} = \frac{I_m}{\sqrt{2}}$

Step 3: Substitute the expression for $I_{rms}$ into the power formula: $P_{avg} = \left( \frac{I_m}{\sqrt{2}} \right)^2 R$

Step 4: Simplify the expression: $P_{avg} = \frac{I_m^2}{2} R = \frac{1}{2} I_m^2 R$

Step 5: Replace $I_m$ with $i_m$ to match the notation in the options: $P_{avg} = \frac{1}{2} i_m^2 R$

Why other options are wrong:
- Option A: $\mathrm{i}_{\mathrm{m}}^2 \mathrm{R}$ This is incorrect because it uses the peak current squared directly, without accounting for the RMS value.
- Option B: Zero. This is incorrect because even though the average current is zero, the power dissipated is not zero due to the squared current term.
- Option C: $\frac{1}{2}$ . This is incorrect because it doesn't include the current and resistance terms in the power calculation.

Key Concepts:
- Average power in AC circuits
- RMS current
- Relationship between RMS and peak current

Memory Aids:
- Remember that average power involves the square of the RMS current.

Quick Tips:
- Always use RMS values when calculating average power in AC circuits.

29 Medium

29 of 143Question permalink

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.

30 Medium

30 of 143Question permalink

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.

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