ALTERNATING CURRENT MCQs for NEET UG PHYSICS — Medium, Page 2

31 Medium

Understanding +4 | -1

A coil of resistance R and inductance L is connected to a battery of emf E volt. The final current in the coil is

View answer and explanation

Correct option: A

\frac{E}{R}

Option A is correct because after a long time, the inductor acts as a short circuit, and the final current is determined only by the resistance R and the emf E, following Ohm's law: I = E/R.

Solution approach

Correct Answer + Core Reason: $\frac{E}{R}$ . After a long time, the inductor acts as a short circuit.

Step-by-Step Solution:

Step 1: Consider the behavior of an inductor in a DC circuit after a long time.

Step 2: After a long time, the inductor acts as a short circuit, meaning its inductive reactance becomes negligible.

Step 3: The circuit then behaves as a simple resistive circuit with resistance R and emf E.

Step 4: Apply Ohm's law to find the final current: $I = \frac{E}{R}$ .

Why other options are wrong:
- Option B: $\frac{E}{L}$ : This expression does not represent the final current; it relates to the rate of change of current in the inductor.
- Option C: $\sqrt {\left( {\frac{E}{{{{R^2} + {L^2}}}}} \right)}$ : This expression is incorrect and doesn't represent the final current.
- Option D: $\sqrt {\left( {\frac{{EL}}{{{{R^2} + {L^2}}}}} \right)}$ : This expression is also incorrect and doesn't represent the final current.

Key Concepts:
- Inductor behavior in DC circuits
- Ohm's law

Memory Aids:
- Inductor acts as a short circuit in DC after a long time.

Quick Tips:
- Remember that inductors act as short circuits in DC after a long time, and capacitors act as open circuits.

32 Medium

32 of 84Question permalink

Understanding +4 | -1

The primary coil of an ideal transformer has 100 turns and secondary coil has 200 turns. Then select the correct option.

View answer and explanation

Correct option: D

The Voltage in secondary coil is two times than primary coil

Option D (The Voltage in secondary coil is two times than primary coil) is correct because the voltage is directly proportional to the number of turns in an ideal transformer.

Solution approach

Correct Answer + Core Reason: The Voltage in secondary coil is two times than primary coil. The voltage ratio is directly proportional to the turns ratio.

Step-by-Step Solution:

Step 1: Recall the transformer equation: $\frac{V_P}{V_S} = \frac{N_P}{N_S}$ .

Step 2: Given $N_P = 100$ and $N_S = 200$ , so $\frac{N_P}{N_S} = \frac{100}{200} = \frac{1}{2}$ .

Step 3: Therefore, $\frac{V_P}{V_S} = \frac{1}{2}$ , which means $V_S = 2V_P$ .

Why other options are wrong:
- Option A: The voltage in the primary coil is half that of the secondary coil.
- Option B: In an ideal transformer, power is conserved, so the power in the primary coil is equal to the power in the secondary coil.
- Option C: The current in the secondary coil is half that of the primary coil.

Key Concepts:
- Transformer equation
- Turns ratio
- Voltage ratio

Memory Aids:
- Voltage follows turns.

Quick Tips:
- Remember the direct proportionality between voltage and turns.

33 Medium

33 of 84Question permalink

Remembering +4 | -1

An A.C. source is connected to a resistive
circuit. Which of the following is true?

View answer and explanation

Correct option: C

Current and voltage are in same phase

Option C is correct. In a purely resistive AC circuit, the current and voltage are always in phase.

Solution approach

Correct Answer + Core Reason: Current and voltage are in same phase. In a purely resistive circuit, voltage and current are in phase.

Step-by-Step Solution:

Step 1: Recall the relationship between voltage and current in a resistor: $V = IR$ .

Step 2: In an AC circuit with only a resistor, the voltage and current are directly proportional at any given time.

Step 3: Therefore, the voltage and current reach their maximum and minimum values at the same time, meaning they are in phase.

Why other options are wrong:
- Option A: Current leads voltage in a capacitive circuit.
- Option B: Current lags voltage in an inductive circuit.
- Option D: The phase relationship is fixed in a purely resistive circuit.

Key Concepts:
- AC circuits with resistors
- Phase relationship between voltage and current

Memory Aids:
- In a resistor, voltage and current are always in phase.

Quick Tips:
- Remember the phase relationships for resistors, inductors, and capacitors.

34 Medium

34 of 84Question permalink

Application +4 | -1

In a circuit, the value of the alternating current is measured by hot wire ammeter as 10 ampere. Its peak value will be

View answer and explanation

Correct option: C

14.14A

Option C (14.14A) is correct because the peak value is calculated by multiplying the RMS value by

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: 14.14A. The peak value is calculated by multiplying the RMS value by $\sqrt{2}$ .

Step-by-Step Solution:

Step 1: Identify the given information: RMS current, $I_{rms} = 10$ A.

Step 2: Identify the formula to calculate the peak current: $I_{peak} = I_{rms} \times \sqrt{2}$

Step 3: Substitute the given value into the formula: $I_{peak} = 10 \times \sqrt{2}$

Step 4: Calculate the peak current: $I_{peak} = 10 \times 1.414 \approx 14.14$ A.

Step 5: Final Answer: 14.14 A

Why other options are wrong:
- Option A (10A): This is the RMS current, not the peak current.
- Option B (20A): This is incorrect. It might be obtained by multiplying the RMS current by 2, which is not the correct factor.
- Option D (7.07A): This is incorrect because it divides the RMS current by $\sqrt{2}$ instead of multiplying.

Key Concepts:
- RMS current
- Peak current
- Relationship between RMS and peak current in AC circuits: $I_{peak} = I_{rms} \times \sqrt{2}$

Memory Aids:
- Remember that peak current is always greater than RMS current.

Quick Tips:
- Always remember the relationship between RMS and peak values for sinusoidal AC signals.

35 Medium

35 of 84Question permalink

Application +4 | -1

A transformer is having 2100 turns in primary and 4200 turns in secondary. An AC source of 120 V, 10 A is connected to its primary. The secondary voltage and current are

View answer and explanation

Correct option: A

240 V,5 A

Option A (240 V,5 A) is correct because the voltage is doubled and the current is halved due to the turns ratio of the transformer.

Solution approach

Correct Answer + Core Reason: 240 V, 5 A. The secondary voltage and current are calculated using the transformer equations.

Step-by-Step Solution:

Step 1: Identify the given information:
$N_1 = 2100$
$N_2 = 4200$ \nV_1 = 120 V\nI_1 = 10 A

Step 2: Calculate the secondary voltage:
$\frac{{V_2}}{{V_1}} = \frac{{N_2}}{{N_1}}$ \nV_2 = V_1 * $\frac{{N_2}}{{N_1}} = 120 * \frac{{4200}}{{2100}} = 120 * 2 = 240 V$

Step 3: Calculate the secondary current:
$\frac{{I_2}}{{I_1}} = \frac{{N_1}}{{N_2}}$ \nI_2 = I_1 * $\frac{{N_1}}{{N_2}} = 10 * \frac{{2100}}{{4200}} = 10 * \frac{{1}}{{2}} = 5 A$

Step 4: Final Answer:\nV_2 = 240 V, I_2 = 5 A

Why other options are wrong:
- Option B (120 V, 10 A): This is incorrect because it assumes the transformer has no effect on voltage or current.
- Option C (240 V, 10 A): This is incorrect because it correctly calculates the secondary voltage but incorrectly assumes the current remains the same.
- Option D (120 V, 20 A): This is incorrect because it assumes the voltage remains the same and the current doubles, which is the opposite of what a step-up transformer does.

Key Concepts:
- Transformer equation: $\frac{{V_2}}{{V_1}} = \frac{{N_2}}{{N_1}} = \frac{{I_1}}{{I_2}}$

Memory Aids:
- Voltage and turns are directly proportional, while current and turns are inversely proportional.

Quick Tips:
- Remember that in an ideal transformer, power in equals power out (V1I1 = V2I2).

36 Medium

36 of 84Question permalink

Understanding +4 | -1

For a series L – C – R circuit, the phase difference between current and voltage at the condition of resonance will be

View answer and explanation

Correct option: C

Zero

Option C is correct because at resonance in a series L-C-R circuit, the inductive reactance equals the capacitive reactance, resulting in a phase difference of zero between current and voltage.

Solution approach

Correct Answer + Core Reason: Zero. At resonance, the inductive and capacitive reactances cancel each other out.

Step-by-Step Solution:

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

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

Step 3: The impedance of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$ .

Step 4: At resonance, $X_L = X_C$ , so $Z = \sqrt{R^2 + (0)^2} = R$ .

Step 5: The phase difference between current and voltage is given by $\tan \phi = \frac{X_L - X_C}{R}$ .

Step 6: At resonance, $X_L = X_C$ , so $\tan \phi = \frac{0}{R} = 0$ .

Step 7: Therefore, the phase difference $\phi = 0$ .

Why other options are wrong:
- Option A: $\pi/2$ : This is the phase difference in a purely inductive or capacitive circuit, not at resonance.
- Option B: $\pi/4$ : This is a possible phase difference, but not at resonance.
- Option D: Nothing can be said: The phase difference is zero at resonance.

Key Concepts:
- Resonance in AC circuits
- Impedance
- Phase difference

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

Quick Tips:
- Remember that at resonance, the phase difference between current and voltage is zero.

37 Medium

37 of 84Question permalink

Understanding +4 | -1

What is the average power dissipation in an ideal capacitor in an AC circuit?

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because an ideal capacitor does not dissipate energy; it only stores energy.

Solution approach

Correct Answer + Core Reason: Zero. An ideal capacitor does not dissipate power.

Step-by-Step Solution:

Step 1: Recall the behavior of an ideal capacitor in an AC circuit: An ideal capacitor stores energy in the form of an electric field. It does not dissipate energy like a resistor.

Step 2: Consider the phase relationship between voltage and current in a capacitor: In an AC circuit, the current leads the voltage by 90 degrees in a capacitor.

Step 3: Calculate the average power: The average power dissipated in a capacitor is given by $P = VI\cos \phi$ , where $\phi$ is the phase angle between the voltage and the current. In this case, $\phi = 90^\circ$ , so $P = VI\cos (90^\circ) = 0$ .

Step 4: Final Answer: Zero

Why other options are wrong:
- Option A: This value is incorrect because an ideal capacitor does not dissipate energy.
- Option B: This value is incorrect because an ideal capacitor does not dissipate energy.
- Option D: This value is incorrect because an ideal capacitor does not dissipate energy.

Key Concepts:
- Capacitors in AC circuits
- Power dissipation
- Phase relationship between voltage and current

Memory Aids:
- Remember that an ideal capacitor does not dissipate energy; it only stores energy.

Quick Tips:
- The average power dissipated in a purely capacitive circuit is zero.

38 Medium

38 of 84Question permalink

Remembering +4 | -1

The transformation ratio in the step-up transformer is

View answer and explanation

Correct option: B

Greater than one

Option B (Greater than one) is correct because a step-up transformer increases the voltage, which means the transformation ratio (secondary voltage/primary voltage) is greater than one.

Solution approach

Correct Answer + Core Reason: Greater than one. Step-up transformers increase voltage, resulting in a transformation ratio greater than 1.

Step-by-Step Solution:

Step 1: Define the transformation ratio (K) as the ratio of secondary voltage ( $V_s$ ) to primary voltage ( $V_p$ ): $K = \frac{{V_s}}{{V_p}}$

Step 2: In a step-up transformer, $V_s > V_p$ .

Step 3: Therefore, the transformation ratio $K > 1$ .

Why other options are wrong:
- Option A: A transformation ratio of one means the voltage is neither stepped up nor down.
- Option C: A transformation ratio less than one indicates a step-down transformer.
- Option D: The transformation ratio is directly related to whether it's a step-up or step-down transformer and doesn't depend on other factors in this context.

Key Concepts:
- Transformation Ratio
- Step-Up Transformer
- Step-Down Transformer

Memory Aids:
- \"Step-up\" means the ratio goes \"up\" (greater than 1).

Quick Tips:
- Remember that the transformation ratio is directly related to the voltage change in the transformer.

39 Medium

39 of 84Question permalink

Understanding +4 | -1

A choke coil has negligible resistance. The alternating potential drop across it is 220 volt and the current is 5mA. The power consumed is.

View answer and explanation

Correct option: C

zero

Option C is correct. A choke coil with negligible resistance dissipates no power because the voltage and current are 90 degrees out of phase.

Solution approach

Correct Answer + Core Reason: zero. A choke coil with negligible resistance has a 90-degree phase difference between voltage and current, resulting in zero power consumption.

Step-by-Step Solution:

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

Step 2: For a pure inductor (choke coil with negligible resistance), the phase angle $\phi = 90^\circ$ .

Step 3: Therefore, $\cos 90^\circ = 0$ , and the power consumed is $P = VI \times 0 = 0$ .

Why other options are wrong:
- Option A: This would be the power if the circuit was purely resistive.
- Option B: This is incorrect because the power consumed is zero.
- Option D: This is incorrect because the power consumed is zero.

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

Memory Aids:
- In a pure inductor, voltage leads current by 90 degrees, and no power is consumed.

Quick Tips:
- Remember that ideal inductors and capacitors do not dissipate power.

40 Medium

40 of 84Question permalink

Understanding +4 | -1

If

{E_0}

is the peak emf,

{I_0}

is the peak current and

\phi

is the phase difference between them, then the average power dissipation in the circuit is

View answer and explanation

Correct option: D

\frac{1}{2}{E_0}{I_0}\cos \phi

Option D is correct. The average power dissipation in an AC circuit is given by

\frac{1}{2} E_0 I_0 \cos \phi

, where

\phi

is the phase difference between voltage and current.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}{E_0}{I_0}\cos \phi$ . The average power dissipation depends on the phase difference.

Step-by-Step Solution:

Step 1: Recall the formula for instantaneous power: $P = EI$ .

Step 2: Express the instantaneous emf and current as $E = E_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t - \phi)$ .

Step 3: Calculate the average power: $P_{avg} = \frac{1}{T} \int_0^T E I dt = \frac{1}{T} \int_0^T E_0 \sin(\omega t) I_0 \sin(\omega t - \phi) dt = \frac{1}{2} E_0 I_0 \cos \phi$ .

Why other options are wrong:
- Option A: This is the apparent power, not the average power.
- Option B: This is the RMS power, not the average power.
- Option C: This formula uses $\sin \phi$ instead of $\cos \phi$ .

Key Concepts:
- Average power in AC circuits
- Phase difference

Memory Aids:
- Remember the formula: $P_{avg} = \frac{1}{2} E_0 I_0 \cos \phi$ .

Quick Tips:
- Always consider the phase difference when calculating average power.

41 Medium

41 of 84Question permalink

Application +4 | -1

In an ac circuit

V\,\,\,{\text{and}}\,\,\,I

are given by

V = 50\,\,\sin \,\,50\,t\,\,\,{\text{volt}}

and

I = 100\,\,\sin \left( {50t + \frac{\pi }{3}} \right)\,\,{\text{mA}}.

The power dissipated in the circuit is

View answer and explanation

Correct option: A

1.25 watt

Option A (1.25 watt) is correct because the power dissipated in an AC circuit is given by

P = V_{rms} I_{rms} \cos{\phi}

, where

\phi

is the phase difference between voltage and current.

Solution approach

Correct Answer + Core Reason: 1.25 watt. The power dissipated is calculated using RMS values and the phase difference.

Step-by-Step Solution:

Step 1: Identify the given values: $V = 50 \sin 50t \text{ V}$ , $I = 100 \sin (50t + \frac{\pi}{3}) \text{ mA} = 0.1 \sin (50t + \frac{\pi}{3}) \text{ A}$

Step 2: Determine the peak voltage and peak current: $V_0 = 50 \text{ V}$ , $I_0 = 0.1 \text{ A}$

Step 3: Calculate the RMS voltage and RMS current: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{50}{\sqrt{2}} \text{ V}$ , $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} \text{ A}$

Step 4: Determine the phase difference between voltage and current: $\phi = \frac{\pi}{3} = 60^\circ$

Step 5: Calculate the power dissipated: $P = V_{rms} I_{rms} \cos \phi = \frac{50}{\sqrt{2}} \times \frac{0.1}{\sqrt{2}} \times \cos 60^\circ$

Step 6: Simplify the expression: $P = \frac{50 \times 0.1}{2} \times \frac{1}{2} = \frac{5}{2} \times \frac{1}{2} = \frac{5}{4} = 1.25 \text{ W}$

Why other options are wrong:
- Option B: 5 watt. This is incorrect because it might be calculated without considering the phase difference or using peak values directly.
- Option C: 10 watt. This is incorrect due to a calculation error.
- Option D: 15 watt. This is incorrect due to a calculation error.

Key Concepts:
- Power in AC circuits
- RMS voltage and current
- Phase difference

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

Quick Tips:
- Remember to use RMS values when calculating power in AC circuits.
- Don't forget to include the power factor $\cos \phi$ .

42 Medium

42 of 84Question permalink

Understanding +4 | -1

The power dissipated in an AC circuit is zero if the circuit is

View answer and explanation

Correct option: C

Either purely inductive or purely capacitive

Option C (Either purely inductive or purely capacitive) is correct because in purely inductive or capacitive circuits, the voltage and current are out of phase by 90 degrees, resulting in zero average power dissipation.

Solution approach

Correct Answer + Core Reason: Either purely inductive or purely capacitive. In these circuits, the phase difference between voltage and current is 90 degrees, leading to zero power dissipation.

Step-by-Step Solution:

Step 1: Recall the formula for power dissipated in an AC circuit: $P = V_{rms} I_{rms} \cos{\phi}$ , where φ is the phase angle between voltage and current.

Step 2: In a purely resistive circuit, φ = 0, so $P = V_{rms} I_{rms}$ .

Step 3: In a purely inductive or capacitive circuit, φ = 90 degrees, so $\cos{\phi} = 0$ , and therefore $P = 0$ .

Why other options are wrong:
- Option A: In a purely resistive circuit, power is dissipated.
- Option B: In a purely inductive circuit, power dissipation is zero, but the option is not complete.
- Option D: In a purely capacitive circuit, power dissipation is zero, but the option is not complete.

Key Concepts:
- Power in AC circuits
- Phase angle
- Resistive, inductive, and capacitive circuits

Memory Aids:
- P = VI cos(phi)

Quick Tips:
- Remember that power is only dissipated in resistive components.

43 Medium

43 of 84Question permalink

Understanding +4 | -1

When the speed of a dc motor increases the armature current

View answer and explanation

Correct option: B

Decreases

Option B is correct. As the speed of a DC motor increases, the back EMF also increases, reducing the armature current.

Solution approach

Correct Answer + Core Reason: Decreases. Increased speed leads to higher back EMF, reducing armature current.

Step-by-Step Solution:

Step 1: Recall the equation for the back EMF in a DC motor: $E_b = K \phi \omega$ , where $K$ is a constant, $\phi$ is the magnetic flux, and $\omega$ is the angular speed.

Step 2: The armature current is given by $I_a = \frac{{V - E_b}}{R_a}$ , where $V$ is the applied voltage and $R_a$ is the armature resistance.

Step 3: As the speed increases, $E_b$ increases, which reduces the armature current $I_a$ .

Why other options are wrong:
- Option A: The armature current decreases, not increases.
- Option C: The armature current changes with speed.
- Option D: The armature current generally decreases as speed increases.

Key Concepts:
- Back EMF in a DC motor
- Armature current

Memory Aids:
- Higher speed, higher back EMF, lower armature current.

Quick Tips:
- Remember that back EMF opposes the applied voltage.

44 Medium

44 of 84Question permalink

Application +4 | -1

A current of 10 A in the primary coil of a circuit is reduced to zero. If the coefficient f mutual inductance is 3H and emf induced in secondary coil is 30 kV, time taken for the change of current is

View answer and explanation

Correct option: C

{10^{ - 3}}s

Option C (

{10^{ - 3}}s

) is correct because the time taken for the change of current is calculated using the formula

E = -M \frac{dI}{dt}

, resulting in

{10^{ - 3}}s

.

Solution approach

Correct Answer + Core Reason: ${10^{ - 3}}s$ . The time taken for the change of current is calculated using the formula $E = -M \frac{dI}{dt}$ .

Step-by-Step Solution:

Step 1: Identify the given information:\nMutual inductance, $M = 3 \text{ H}$ \nChange in current, $dI = 10 \text{ A} - 0 \text{ A} = 10 \text{ A}$ \nInduced emf, $E = 30 \text{ kV} = 30 \times 10^3 \text{ V}$

Step 2: Recall the formula for induced emf due to mutual inductance:
$E = -M \frac{dI}{dt}$

Step 3: Solve for the time taken, $dt$ :
$dt = -\frac{M \cdot dI}{E}$ \nSince we are only interested in the magnitude, we can ignore the negative sign.
$dt = \frac{3 \text{ H} \times 10 \text{ A}}{30 \times 10^3 \text{ V}} = \frac{30}{30000} \text{ s} = 10^{-3} \text{ s}$

Why other options are wrong:
- Option A: This option gives a time of ${10^3}s$ , which is incorrect. It might result from a miscalculation or misunderstanding of the units.
- Option B: This option gives a time of ${10^2}s$ , which is incorrect. It might result from a miscalculation or misunderstanding of the units.
- Option D: This option gives a time of ${10^{ - 2}}s$ , which is incorrect. It might result from a miscalculation or misunderstanding of the units.

Key Concepts:
- Mutual inductance
- Induced emf: $E = -M \frac{dI}{dt}$

Memory Aids:
- Remember the formula: "E equals M di by dt"

Quick Tips:
- Always ensure the units are consistent before performing calculations.
- Pay attention to the change in current and the induced emf.

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45 of 84Question permalink

Analyzing +4 | -1

Current in an ac circuit is given by

I=3 \sin \omega t+4

\cos \omega t

then

View answer and explanation

Correct option: C

if voltage applied is

V=V_m \sin \omega t

then the circuit must be containing resistance and capacitance

Option C (if voltage applied is

V=V_m \sin \omega t

then the circuit must be containing resistance and capacitance) is correct because the current has both sine and cosine components, indicating a phase shift that can only be achieved with both resistance and capacitance.

Solution approach

Correct Answer + Core Reason: if voltage applied is $V=V_m \sin \omega t$ then the circuit must be containing resistance and capacitance. The current has both sine and cosine components, indicating a phase shift that can only be achieved with both resistance and capacitance.

Step-by-Step Solution:

Step 1: Analyze the given current: $I = 3 \sin \omega t + 4 \cos \omega t$ .

Step 2: Rewrite the current in the form $I = I_m \sin(\omega t + \phi)$ , where $I_m = \sqrt{3^2 + 4^2} = 5$ and $\phi = \arctan(4/3)$ .

Step 3: The RMS value of the current is $I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{5}{\sqrt{2}}$ . This makes options A and B incorrect.

Step 4: If the voltage is $V = V_m \sin \omega t$ , and the current is $I = 5 \sin(\omega t + \phi)$ , then there is a phase difference between the voltage and current.

Step 5: A circuit containing only resistance will have voltage and current in phase. A circuit containing only inductance will have the current lagging the voltage by 90 degrees. A circuit containing only capacitance will have the current leading the voltage by 90 degrees.

Step 6: Since the current has both sine and cosine components, it indicates that the circuit must contain both resistance and either inductance or capacitance. Since the phase angle is positive, the current leads the voltage, indicating the presence of capacitance.

Step 7: Final Answer: if voltage applied is $V=V_m \sin \omega t$ then the circuit must be containing resistance and capacitance.

Why other options are wrong:
- Option A: rms value of current is $5 {~A}$ . Incorrect, the RMS value is $\frac{5}{\sqrt{2}}$ .
- Option B: rms value of current is $5 \sqrt{2} {~A}$ . Incorrect, the RMS value is $\frac{5}{\sqrt{2}}$ .
- Option D: if voltage applied is $V=V_m \sin \omega t$ , the circuit may contain resistance and inductance. Incorrect, the current leads the voltage, indicating capacitance.

Key Concepts:
- RMS value of current
- Phase difference between voltage and current
- AC circuits with R, L, and C

Memory Aids:
- Remember that in a capacitive circuit, current leads voltage.
- Remember that in an inductive circuit, current lags voltage.

Quick Tips:
- Always rewrite the current equation in the form $I = I_m \sin(\omega t + \phi)$ to determine the phase difference.

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Remembering +4 | -1

Transformer is based upon the principle of

View answer and explanation

Correct option: B

mutual induction

Option B (mutual induction) is correct because a transformer works by transferring energy from the primary to the secondary coil through mutual induction.

Solution approach

Correct Answer + Core Reason: mutual induction. Transformers operate based on the principle of mutual induction between two coils.

Step-by-Step Solution:

Step 1: Define mutual induction: The phenomenon where a changing current in one coil induces an electromotive force (EMF) in a nearby coil.

Step 2: In a transformer, the primary coil carries an alternating current, creating a changing magnetic flux.

Step 3: This changing flux links with the secondary coil, inducing an EMF in it, thus transferring energy.

Why other options are wrong:
- Option A: Self-induction occurs within a single coil, not between two coils as in a transformer.
- Option C: Eddy currents are a consequence of electromagnetic induction but not the primary principle of operation.
- Option D: Mutual induction is the correct principle.

Key Concepts:
- Mutual induction
- Electromagnetic induction
- Transformer operation

Memory Aids:
- Transformer = Transfer = Mutual induction

Quick Tips:
- Remember that transformers rely on the interaction between two coils.

47 Medium

47 of 84Question permalink

Application +4 | -1

The instantaneous current and voltage of an AC circuit are given by

i=10 \sin (314 t) A

and

V=100 \sin (314 t) V

What is the power dissipation in the circuit?

View answer and explanation

Correct option: B

500 \mathrm{~W}

Option B (500 W) is correct because the power dissipation is calculated as P = Vrms * Irms, where Vrms and Irms are the root mean square values of voltage and current.

Solution approach

Correct Answer + Core Reason: $500 \mathrm{~W}$ . The power dissipation is calculated using the RMS values of voltage and current.

Step-by-Step Solution:

Step 1: Identify the given values: $i=10 \sin (314 t) A$ and $V=100 \sin (314 t) V$ .

Step 2: Determine the peak current and voltage: $I_0 = 10 \text{{ A}}$ and $V_0 = 100 \text{{ V}}$ .

Step 3: Calculate the RMS values of current and voltage: $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \text{{ A}}$ and $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \text{{ V}}$ .

Step 4: Since the voltage and current are in phase, the power dissipation is given by $P = V_{rms} I_{rms}$ .

Step 5: Substitute the values: $P = \frac{100}{\sqrt{2}} \times \frac{10}{\sqrt{2}} = \frac{1000}{2} = 500 \text{{ W}}$ .

Why other options are wrong:
- Option A: This is incorrect and might be obtained by not considering the RMS values.
- Option C: This is incorrect and might be obtained by making a calculation error.
- Option D: This is incorrect and might be obtained by making a calculation error.

Key Concepts:
- AC circuits
- Power dissipation
- RMS values
- In-phase voltage and current

Memory Aids:
- Remember the formula for power dissipation in an AC circuit when voltage and current are in phase: $P = V_{rms} I_{rms}$ .

Quick Tips:
- Always convert units to SI units before performing calculations.
- Remember to use RMS values when calculating power in AC circuits.

48 Medium

48 of 84Question permalink

Understanding +4 | -1

The maximum current through a pure capacitor of capacitance

\mathrm{C}

is

\mathrm{I}_0

. The maximum emf is

\varepsilon_0

. If the frequency is

50 \mathrm{~Hz}

, the average power dissipated through the capacitor is

View answer and explanation

Correct option: B

Zero

Option B (Zero) is correct because a pure capacitor does not dissipate power; it only stores energy.

Solution approach

Correct Answer + Core Reason: Zero. A pure capacitor only stores energy and does not dissipate power.

Step-by-Step Solution:

Step 1: Recall the properties of a pure capacitor in an AC circuit.

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

Why other options are wrong:
- Option A: This would be the average power if the phase difference was zero, which is not the case for a capacitor.
- Option C: This is incorrect because the power dissipated in a pure capacitor is zero due to the phase difference between voltage and current.
- Option D: This expression does not correctly represent the average power dissipated in a capacitor, which is zero.

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

Memory Aids:
- ELI the ICE man: 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.

49 Medium

49 of 84Question permalink

Application +4 | -1

What is the self-inductance of a coil which produces 5 V when the current changes from 3 A to 2 A in one millisecond?

View answer and explanation

Correct option: B

5 mH

Option B (5 mH) is correct because the self-inductance is calculated using the formula

E = -L \frac{dI}{dt}

, resulting in 5 mH.

Solution approach

Correct Answer + Core Reason: 5 mH. The self-inductance is calculated using the formula $E = -L \frac{dI}{dt}$ .

Step-by-Step Solution:

Step 1: Identify the given information:\nInduced voltage, $E = 5 \text{ V}$ \nChange in current, $dI = 3 \text{ A} - 2 \text{ A} = 1 \text{ A}$ \nChange in time, $dt = 1 \text{ ms} = 1 \times 10^{-3} \text{ s}$

Step 2: Recall the formula for induced voltage in an inductor:
$E = -L \frac{dI}{dt}$

Step 3: Solve for the self-inductance, $L$ :
$L = -\frac{E}{dI/dt} = -\frac{E \cdot dt}{dI}$ \nSince we are only interested in the magnitude, we can ignore the negative sign.
$L = \frac{5 \text{ V} \times 1 \times 10^{-3} \text{ s}}{1 \text{ A}} = 5 \times 10^{-3} \text{ H} = 5 \text{ mH}$

Why other options are wrong:
- Option A: This option gives a self-inductance of 5000 H, which is incorrect. It might result from a miscalculation or misunderstanding of the units.
- Option C: This option gives a self-inductance of 50 H, which is incorrect. It might result from a miscalculation or misunderstanding of the units.
- Option D: This option gives a self-inductance of 5 H, which is incorrect. It might result from a miscalculation or misunderstanding of the units.

Key Concepts:
- Self-inductance
- Induced voltage in an inductor: $E = -L \frac{dI}{dt}$

Memory Aids:
- Remember the formula: "E equals L di by dt"

Quick Tips:
- Always ensure the units are consistent before performing calculations.
- Pay attention to the change in current and the change in time.

50 Medium

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Understanding +4 | -1

A bulb and a capacitor are in series with an ac source. On increasing frequency how will glow of the bulb change

View answer and explanation

Correct option: B

The glow increases

Option B is correct. Increasing the frequency decreases the capacitive reactance, allowing more current to flow and increasing the bulb's glow.

Solution approach

Correct Answer + Core Reason: The glow increases. Increasing frequency reduces capacitive reactance, increasing current.

Step-by-Step Solution:

Step 1: Recall the formula for capacitive reactance: $X_C = \frac{1}{{{\omega C}}} = \frac{1}{{{{2\pi fC}}}}$ .

Step 2: As the frequency $f$ increases, the capacitive reactance $X_C$ decreases.

Step 3: With lower reactance, the current in the circuit increases: $I = \frac{V}{Z}$ , where $Z$ is the impedance.

Step 4: An increase in current leads to a brighter glow of the bulb.

Why other options are wrong:
- Option A: The glow increases, not decreases.
- Option C: The glow changes with frequency.
- Option D: The bulb's glow increases, it does not quench.

Key Concepts:
- Capacitive reactance
- Frequency dependence of reactance
- Current in AC circuits

Memory Aids:
- Remember the formula: $X_C = \frac{1}{{{\omega C}}}$ .

Quick Tips:
- Higher frequency means lower capacitive reactance and higher current.

51 Medium

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Understanding +4 | -1

Same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the emf of AC is increased, the effect on the value of the current

View answer and explanation

Correct option: D

Decreases in the first circuit and increases in the other

Option D is correct because increasing the frequency decreases the current in an inductive circuit (due to increased inductive reactance) and increases the current in a capacitive circuit (due to decreased capacitive reactance).

Solution approach

Correct Answer + Core Reason: Decreases in the first circuit and increases in the other. Inductive reactance increases with frequency, while capacitive reactance decreases.

Step-by-Step Solution:

Step 1: Consider the effect of frequency on inductive and capacitive reactance.

Step 2: Inductive reactance is given by $X_L = \omega L = 2 \pi f L$ , where f is the frequency and L is the inductance. As frequency increases, inductive reactance increases.

Step 3: Capacitive reactance is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$ , where f is the frequency and C is the capacitance. As frequency increases, capacitive reactance decreases.

Step 4: In the first circuit (only inductance), as frequency increases, $X_L$ increases, and the current decreases (since $I = \frac{V}{X_L}$ ).

Step 5: In the second circuit (only capacitance), as frequency increases, $X_C$ decreases, and the current increases (since $I = \frac{V}{X_C}$ ).

Why other options are wrong:
- Option A: Increases in the first circuit and decreases in the other: This is incorrect because current decreases in the inductive circuit and increases in the capacitive circuit.
- Option B: Increases in both the circuits: This is incorrect because current decreases in the inductive circuit.
- Option C: Decreases in both the circuits: This is incorrect because current increases in the capacitive circuit.

Key Concepts:
- Inductive reactance
- Capacitive reactance
- Frequency

Memory Aids:
- $X_L = \omega L$
- $X_C = \frac{1}{\omega C}$

Quick Tips:
- Remember how frequency affects inductive and capacitive reactance.

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Understanding +4 | -1

A resistor R, an inductor L and a capacitor C are connected in series to an oscillator of frequency

n

, if the resonant frequency is

{n_r}

, then the current lags behind voltage, when

View answer and explanation

Correct option: D

n > {n_r}

Option D (

n > {n_r}

) is correct because when the frequency is greater than the resonant frequency, the inductive reactance dominates, causing the current to lag behind the voltage.

Solution approach

Correct Answer + Core Reason: $n > {n_r}$ . When the frequency is greater than the resonant frequency, the circuit is inductive, and the current lags the voltage.

Step-by-Step Solution:

Step 1: Recall the behavior of an RLC series circuit at different frequencies relative to the resonant frequency.

Step 2: At resonance ( $n = n_r$ ), the impedance is minimum, and the current and voltage are in phase.

Step 3: When $n < n_r$ , the capacitive reactance dominates, and the current leads the voltage.

Step 4: When $n > n_r$ , the inductive reactance dominates, and the current lags the voltage.

Why other options are wrong:
- Option A: At $n = 0$ , the capacitor acts as an open circuit, so the concept of lagging/leading is not applicable.
- Option B: When $n < n_r$ , the current leads the voltage.
- Option C: At $n = n_r$ , the current and voltage are in phase.

Key Concepts:
- Resonance in RLC circuits
- Inductive reactance
- Capacitive reactance
- Phase relationship between current and voltage

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

Quick Tips:
- Use ELI the ICE man to remember the phase relationships.

53 Medium

53 of 84Question permalink

Analyzing +4 | -1

Assertion (A)If the frequency of the applied

AC

is doubled, then the power factor of a series

R-L

circuit decreases.

Reason (R) Power factor of series R-L circuit s given' by

\cos \theta=\frac{2 {R}}{\sqrt{{R}^2+\omega^2 {~L}^2}}

View answer and explanation

Correct option: C

If Assertion is true but Reason is false.

Option C is correct because the assertion is true, but the reason is false. Increasing the frequency in an R-L circuit decreases the power factor. The correct formula for the power factor is

\cos \theta = \frac{{R}}{{\sqrt{{R^2 + (\omega L)^2}}}}

Solution approach

Correct Answer + Core Reason: If Assertion is true but Reason is false. The assertion is correct, but the given formula for the power factor is incorrect.

Step-by-Step Solution:

Step 1: Analyze the Assertion: If the frequency of the applied AC is doubled, then the power factor of a series R-L circuit decreases.

Step 2: Recall the formula for the power factor in a series R-L circuit: $\cos \theta = \frac{{R}}{{\sqrt{{R^2 + X_L^2}}}} = \frac{{R}}{{\sqrt{{R^2 + (\omega L)^2}}}}$ , where $\omega = 2 \pi f$ is the angular frequency and f is the frequency.

Step 3: If the frequency (f) is doubled, then $\omega$ is also doubled. This increases the inductive reactance ( $X_L = \omega L$ ). As $X_L$ increases, the denominator in the power factor equation increases, and therefore the power factor ( $\cos \theta$ ) decreases. So, the assertion is true.

Step 4: Analyze the Reason: Power factor of series R-L circuit is given by $\cos \theta=\frac{{2 {R}}}{{\sqrt{{R}^2+\omega^2 {~L}^2}}}$ .

Step 5: Compare the given reason with the correct formula for the power factor. The correct formula is $\cos \theta = \frac{{R}}{{\sqrt{{R^2 + (\omega L)^2}}}}$ . The reason provides an incorrect formula (with a 2R in the numerator). So, the reason is false.

Step 6: Conclusion: The assertion is true, but the reason is false.

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

Key Concepts:
- Power factor in AC circuits
- Series R-L circuit
- Inductive reactance ( $X_L = \omega L$ )
- Angular frequency ( $\omega = 2 \pi f$ )

Memory Aids:
- Power factor: $\cos \theta = \frac{{R}}{Z}$ , where Z is the impedance.
- Impedance in R-L circuit: $Z = \sqrt{{R^2 + X_L^2}}$

Quick Tips:
- Remember the correct formula for the power factor in a series R-L circuit.
- Understand how frequency affects inductive reactance and the power factor.

54 Medium

54 of 84Question permalink

Analyzing +4 | -1

A series R-C circuit is connected to AC Voltage source. Consider two cases: (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current

{I_R}

through the resistor and voltage

{V_c}

across the capacitor are compared in the two cases. Which of the following is/are true?

View answer and explanation

Correct option: B

I_R^A < I_R^B

Option B (

I_R^A < I_R^B

) is correct because introducing a dielectric increases capacitance, leading to a lower impedance and thus a higher current.

Solution approach

Correct Answer + Core Reason: $I_R^A < I_R^B$ . Introducing a dielectric increases capacitance, decreasing impedance and increasing current.

Step-by-Step Solution:

Step 1: Analyze the circuit in case A (without dielectric). The impedance $Z_A$ is given by $Z_A = \sqrt{{R^2 + X_C^2}}$ , where $X_C = \frac{1}{{\omega C}}$ . The current $I_R^A = \frac{V}{Z_A}$ .

Step 2: Analyze the circuit in case B (with dielectric). The capacitance becomes $C' = kC = 4C$ . The new capacitive reactance is $X_C' = \frac{1}{{\omega C'}} = \frac{1}{{4\omega C}} = \frac{X_C}{4}$ . The impedance $Z_B = \sqrt{{R^2 + (X_C')^2}} = \sqrt{{R^2 + (\frac{X_C}{4})^2}}$ . The current $I_R^B = \frac{V}{Z_B}$ .

Step 3: Compare the currents. Since $X_C' < X_C$ , it follows that $Z_B < Z_A$ . Therefore, $I_R^B > I_R^A$ .

Step 4: Compare the voltages. $V_C^A = I_R^A X_C$ and $V_C^B = I_R^B X_C'$ . Since $I_R^B > I_R^A$ and $X_C' < X_C$ , it's not immediately clear which voltage is larger. However, since the current increases more significantly than the reactance decreases, $V_C^B$ will be smaller than $V_C^A$ .

Why other options are wrong:
- **Option A ( $I_R^A > I_R^B$ ): This is incorrect because the current increases when the dielectric is introduced.
- Option C ( $V_C^A > V_C^B$ ): This is correct, but the question asks for all true statements, and option B is also true.
- Option D ( $V_C^A < V_C^B$ ): This is incorrect because the voltage across the capacitor decreases when the dielectric is introduced.

Key Concepts:**
- Impedance in R-C circuit: $Z = \sqrt{{R^2 + X_C^2}}$
- Capacitive Reactance: $X_C = \frac{1}{{\omega C}}$
- Effect of dielectric on capacitance: $C' = kC$

Memory Aids:
- Dielectric increases capacitance, which decreases impedance, leading to increased current.

Quick Tips:
- Remember that introducing a dielectric material increases the capacitance of the capacitor.
- Understand how changes in capacitance affect the impedance and current in an AC circuit.

55 Medium

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Remembering +4 | -1

In step-up transformer, relation between number of turns in primary

\left( {{N_p}} \right)

and number of turns is secondary

\left( {{N_s}} \right)

coils is

View answer and explanation

Correct option: A

{N_s}

is greater than

{N_p}

Option A (

{N_s}

is greater than

{N_p}

) is correct because in a step-up transformer, the number of turns in the secondary coil is greater than the number of turns in the primary coil to increase the voltage.

Solution approach

Correct Answer + Core Reason: ${N_s}$ is greater than ${N_p}$ . In a step-up transformer, the secondary coil has more turns than the primary coil.

Step-by-Step Solution:

Step 1: Recall the relationship between the number of turns and voltage in a transformer: $\frac{{V_s}}{{V_p}} = \frac{{N_s}}{{N_p}}$

Step 2: In a step-up transformer, $V_s > V_p$ .

Step 3: Therefore, $N_s > N_p$ .

Why other options are wrong:
- Option B: If ${N_p}$ is greater than ${N_s}$ , it would be a step-down transformer.
- Option C: If ${N_s}$ is equal to ${N_p}$ , the voltage would remain the same (no step-up or step-down).
- Option D: ${N_p} = 2{N_s}$ would represent a step-down transformer, not a step-up transformer.

Key Concepts:
- Transformer Equation
- Step-Up Transformer
- Number of Turns

Memory Aids:
- \"Step-up\" means the \"secondary\" has more turns.

Quick Tips:
- Remember the direct relationship between the number of turns and the voltage in a transformer.

56 Medium

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Understanding +4 | -1

Which of the following quantities remains constant in a step-down transformer ?

View answer and explanation

Correct option: C

Power

Option C (Power) is correct because, ideally, a transformer conserves power; the power in the primary coil equals the power in the secondary coil.

Solution approach

Correct Answer + Core Reason: Power. Ideally, a transformer conserves power.

Step-by-Step Solution:

Step 1: Recall the principle of a transformer.

Step 2: A transformer transfers electrical energy from one circuit to another through electromagnetic induction.

Step 3: Ideally, the power in the primary coil equals the power in the secondary coil. $P_{in} = P_{out}$

Why other options are wrong:
- Option A: Current - Incorrect, current changes in a step-down transformer.
- Option B: Voltage - Incorrect, voltage changes in a step-down transformer.
- Option D: None of these - Incorrect, power remains constant.

Key Concepts:
- Transformer principle
- Conservation of power

Memory Aids:
- Ideal transformer: Power in = Power out

Quick Tips:
- Remember that transformers transfer power, ideally without loss.

57 Medium

57 of 84Question permalink

Application +4 | -1

In an L – C – R circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

View answer and explanation

Correct option: C

L/2

Option C (L/2) is correct because the resonant frequency is inversely proportional to the square root of the product of inductance and capacitance.

Solution approach

Correct Answer + Core Reason: L/2. The resonant frequency is inversely proportional to the square root of LC.

Step-by-Step Solution:

Step 1: Recall the formula for resonant frequency: $f = \frac{1}{{2\pi \sqrt{{LC}}}}$

Step 2: We want the resonant frequency to remain unchanged, so $f_1 = f_2$ .

Step 3: Let the initial values be $L_1 = L$ and $C_1 = C$ . Let the final values be $L_2$ and $C_2 = 2C$ .

Step 4: Set up the equation: $\frac{1}{{2\pi \sqrt{{L_1C_1}}}} = \frac{1}{{2\pi \sqrt{{L_2C_2}}}}$

Step 5: Simplify: $\sqrt{{L_1C_1}} = \sqrt{{L_2C_2}}$

Step 6: Substitute the given values: $\sqrt{{LC}} = \sqrt{{L_2(2C)}}$

Step 7: Square both sides: $LC = 2CL_2$

Step 8: Solve for $L_2$ : $L_2 = \frac{{LC}}{{2C}} = \frac{L}{2}$

Step 9: Final Answer: L/2

Why other options are wrong:
- Option A (4L): This is incorrect and would significantly decrease the resonant frequency.
- Option B (2L): This is incorrect and would decrease the resonant frequency.
- Option D (L/4): This is incorrect and would increase the resonant frequency.

Key Concepts:
- Resonance in LCR circuits
- Resonant frequency

Memory Aids:
- Remember the formula for resonant frequency: $f = \frac{1}{{2\pi \sqrt{{LC}}}}$

Quick Tips:
- When dealing with resonant frequency problems, remember the inverse square root relationship between frequency and L and C.

58 Medium

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Remembering +4 | -1

In a series

L C R

circuit, at resonance, the bandwidth frequency can be represented as

View answer and explanation

Correct option: D

\frac{R}{2 \pi L}

Option D (

\frac{R}{2 \pi L}

) is correct because the bandwidth frequency in a series LCR circuit at resonance is given by the formula

\Delta f = \frac{R}{2 \pi L}

.

Solution approach

Correct Answer + Core Reason: $\frac{R}{2 \pi L}$ . The bandwidth frequency in a series LCR circuit at resonance is given by the formula $\Delta f = \frac{R}{2 \pi L}$ .

Step-by-Step Solution:

Step 1: Recall the formula for bandwidth frequency ( $\Delta f$ ) in a series LCR circuit at resonance: $\Delta f = \frac{R}{2 \pi L}$ , where R is the resistance and L is the inductance.

Step 2: This formula directly represents the bandwidth frequency.

Step 3: Final Answer: $\frac{R}{2 \pi L}$

Why other options are wrong:
- Option A: $\frac{R}{L C}$ . This is incorrect and does not represent the bandwidth frequency.
- Option B: $\frac{R}{C}$ . This is incorrect and does not represent the bandwidth frequency.
- Option C: $\frac{R L}{2 \pi}$ . This is incorrect and does not represent the bandwidth frequency.

Key Concepts:
- Resonance in LCR circuits
- Bandwidth frequency

Memory Aids:
- Remember the formula for bandwidth frequency: $\Delta f = \frac{R}{2 \pi L}$

Quick Tips:
- Always remember the formula for bandwidth frequency in a series LCR circuit at resonance.

59 Medium

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Understanding +4 | -1

The power is transmitted from a power house on high voltage ac because

View answer and explanation

Correct option: B

It is more economical due to less power wastage

Option B (It is more economical due to less power wastage) is correct because transmitting power at high voltage reduces current, thus reducing

I^2R

losses in the transmission lines.

Solution approach

Correct Answer + Core Reason: It is more economical due to less power wastage. High voltage reduces current, reducing $I^2R$ losses.

Step-by-Step Solution:

Step 1: Recall the relationship between power, voltage, and current: $P = VI$ .

Step 2: To transmit a certain power $P$ , if the voltage $V$ is high, the current $I$ will be low.

Step 3: The power loss in the transmission lines is given by $P_{loss} = I^2R$ , where $R$ is the resistance of the lines.

Step 4: Since the current is low at high voltage, the power loss is also low, making it more economical.

Why other options are wrong:
- Option A: Electric current travels faster at higher volts - Incorrect, the speed of current is not significantly affected by voltage.
- Option C: It is difficult to generate power at low voltage - Incorrect, power can be generated at various voltages.
- Option D: Changes of stealing transmission lines are minimized - Incorrect, this is not the primary reason.

Key Concepts:
- Power transmission
- Power loss in transmission lines

Memory Aids:
- $P = VI$ and $P_{loss} = I^2R$

Quick Tips:
- Remember that high voltage transmission reduces current and thus reduces power loss.

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Analyzing +4 | -1

An ac source of angular frequency

\omega

is fed across a resistor

r

and a capacitor

C

in series. The current in the circuit measured is

l

. If now the frequency of source is changed to

\omega / 3

(but maintaining the same voltage), the current in the circuit is found to reduce by a factor of three. The power factor of the circuit at the original frequency

\omega

is

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because the problem states the current reduces by a factor of 3 when the frequency is reduced to

\omega/3

, implying resonance at the original frequency.

Solution approach

Correct Answer + Core Reason: Zero. The current reduces by a factor of 3 when the frequency is changed to $\omega/3$ , implying resonance at $\omega$ , hence power factor is zero.

Step-by-Step Solution:

Step 1: Define the initial impedance $Z_1$ at frequency $\omega$ : $Z_1 = \sqrt{{r^2 + (X_C)^2}} = \sqrt{{r^2 + (\frac{1}{{\omega C}})^2}}$ . The current is $I = \frac{V}{Z_1}$ .

Step 2: Define the impedance $Z_2$ at frequency $\frac{\omega}{3}$ : $Z_2 = \sqrt{{r^2 + (X_C')^2}} = \sqrt{{r^2 + (\frac{1}{{\frac{\omega}{3} C}})^2}} = \sqrt{{r^2 + (\frac{3}{{\omega C}})^2}}$ . The current is $\frac{I}{3} = \frac{V}{Z_2}$ .

Step 3: Relate the two currents: $\frac{I}{3} = \frac{V}{Z_2} = \frac{1}{3} \frac{V}{Z_1}$ . Therefore, $Z_2 = 3Z_1$ .

Step 4: Substitute the expressions for $Z_1$ and $Z_2$ : $\sqrt{{r^2 + (\frac{9}{{\omega^2 C^2}})}} = 3 \sqrt{{r^2 + (\frac{1}{{\omega^2 C^2}})}}$ .

Step 5: Square both sides: $r^2 + \frac{9}{{\omega^2 C^2}} = 9(r^2 + \frac{1}{{\omega^2 C^2}})$ .

Step 6: Simplify: $r^2 + \frac{9}{{\omega^2 C^2}} = 9r^2 + \frac{9}{{\omega^2 C^2}}$ . This implies $8r^2 = 0$ , so $r = 0$ .

Step 7: If $r = 0$ , the circuit is purely capacitive. The power factor is $\cos{\phi} = \cos{90^\circ} = 0$ .

Why other options are wrong:
- Option A (1): This would be the power factor for a purely resistive circuit.
- Option B (1 / 2): This would be the power factor for some R-C circuits but not under the given conditions.
- Option D (1 / 3): This would be the power factor for some R-C circuits but not under the given conditions.

Key Concepts:
- Impedance in an R-C circuit: $Z = \sqrt{{R^2 + X_C^2}}$
- Capacitive Reactance: $X_C = \frac{1}{{\omega C}}$
- Power factor: $\cos{\phi} = \frac{R}{Z}$

Memory Aids:
- Power factor is 1 for purely resistive circuits and 0 for purely capacitive or inductive circuits.

Quick Tips:
- Analyze how the change in frequency affects the impedance and current in the circuit.
- Remember that the power factor is the cosine of the phase angle between voltage and current.

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