In a current I given by ${I_{\rm{o}}}\sin \left[ {\omega t - \frac{\pi }{2}} \right]$ flows in an AC circuit across which an AC potential of $E = {E_{\rm{o}}}$ sin wt has been applied, then the power consumption $P$ in the circuit will be

$P = \frac{{{E_{\rm{o}}}{I_{\rm{o}}}}}{{\sqrt 2 }}$

$P = \sqrt 2 {E_{\rm{o}}}{I_{\rm{o}}}$

$P = \frac{{{E_{\rm{o}}}{I_{\rm{o}}}}}{2}$

ALTERNATING CURRENT MCQs for NEET UG PHYSICS — Medium, Page 3

61 Medium

61 of 84Question permalink

Remembering +4 | -1

The power factor varies between

View answer and explanation

Correct option: C

0 to 1

Option C (0 to 1) is correct because the power factor is the cosine of the phase angle between voltage and current, and cosine values range from 0 to 1.

Solution approach

Correct Answer + Core Reason: 0 to 1. The power factor is the cosine of the phase angle between voltage and current.

Step-by-Step Solution:

Step 1: Define the power factor: The power factor is defined as $\cos \phi$ , where $\phi$ is the phase angle between the voltage and the current in an AC circuit.

Step 2: Determine the range of $\cos \phi$ : The cosine function has a range of -1 to 1. However, in the context of power factor, we consider only the absolute value, as negative power factor implies power is being supplied back to the source, which is not the standard convention.\nTherefore, the power factor varies from 0 to 1.

Step 3: Final Answer: 0 to 1

Why other options are wrong:
- Option A: This range is incorrect because the power factor can be greater than 0.5 and up to 1.
- Option B: This range is incorrect because the power factor can be 0.
- Option D: This range is incorrect because the power factor cannot be greater than 1.

Key Concepts:
- Power factor
- Phase angle
- AC circuits

Memory Aids:
- Remember that power factor = $\cos \phi$ , and $0 \le \cos \phi \le 1$ .

Quick Tips:
- The power factor indicates how effectively electrical power is being used in an AC circuit.

62 Medium

62 of 84Question permalink

Application +4 | -1

The sun delivers

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

of electromagnetic flux incident on a roof of dimensions

8\,\,m\,\, \times \,\,20\,\,m,

will be

View answer and explanation

Correct option: C

1.6 \times {10^5}\,\,W

Option C (

1.6 \times {10^5}\,W

) is correct because the total power is the intensity multiplied by the area.

Solution approach

Correct Answer + Core Reason: $1.6 \times {10^5}\,W$ . Total power is intensity multiplied by area.

Step-by-Step Solution:

Step 1: Define the relationship between power, intensity, and area: $P = I \times A$

Step 2: Identify the given values: Intensity $I = 10^3 \text{ W/m}^2$ , dimensions of the roof $8 \text{ m} \times 20 \text{ m}$

Step 3: Calculate the area of the roof: $A = 8 \text{ m} \times 20 \text{ m} = 160 \text{ m}^2$

Step 4: Calculate the total power: $P = 10^3 \text{ W/m}^2 \times 160 \text{ m}^2 = 160 \times 10^3 \text{ W} = 1.6 \times 10^5 \text{ W}$

Why other options are wrong:
- Option A: $6.4 \times {10^3}\,W$ . This is incorrect because it might be calculated by multiplying the intensity by only one dimension of the roof or by making a calculation error.
- Option B: $3.4 \times {10^4}\,W$ . This is incorrect due to a calculation error.
- Option D: $3.2 \times {10^5}\,W$ . This is incorrect, possibly from multiplying by 2 instead of calculating the area correctly.

Key Concepts:
- Intensity of electromagnetic radiation
- Power calculation

Memory Aids:
- Power = Intensity x Area

Quick Tips:
- Ensure the units are consistent before performing calculations.

63 Medium

63 of 84Question permalink

Application +4 | -1

In an LCR circuit R=100 ohm. When capacitance C is removed, the current lags behind the voltage by π/3. When inductance L is removed, the current leads the voltage by π/3. The impedance of the circuit is

View answer and explanation

Correct option: B

100 ohm

Option B (100 ohm) is correct because at resonance, the impedance is equal to the resistance.

Solution approach

Correct Answer + Core Reason: 100 ohm. At resonance, the impedance is equal to the resistance.

Step-by-Step Solution:

Step 1: Analyze the given information: $R = 100 \Omega$ .

Step 2: When capacitance C is removed, the current lags behind the voltage by $\pi/3$ . This means the circuit is inductive, and $X_L = R \tan{(\pi/3)} = R\sqrt{3}$ .

Step 3: When inductance L is removed, the current leads the voltage by $\pi/3$ . This means the circuit is capacitive, and $X_C = R \tan{(\pi/3)} = R\sqrt{3}$ .

Step 4: Since $X_L = X_C$ , the circuit is at resonance. At resonance, the impedance $Z = R$ .

Step 5: Therefore, the impedance of the circuit is $Z = 100 \Omega$ .

Step 6: Final Answer: 100 ohm

Why other options are wrong:
- Option A (50 ohm): This is incorrect and likely arises from an incorrect calculation or misunderstanding of the resonance condition.
- Option C (200 ohm): This is incorrect and doesn't correspond to any reasonable calculation using the given values.
- Option D (400 ohm): This is incorrect and doesn't correspond to any reasonable calculation using the given values.

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

Memory Aids:
- At resonance, $X_L = X_C$ and $Z = R$ .

Quick Tips:
- Recognize the resonance condition when removing L or C results in equal and opposite phase shifts.

64 Medium

64 of 84Question permalink

Application +4 | -1

A resistance of 40 ohm and an inductance of 95.5 millihenry are connected in series in a 50 cycles/second ac circuit. The impedance of this combination is very nearly

View answer and explanation

Correct option: C

50 ohm

Option C (50 ohm) is correct because the impedance is calculated using the formula

Z = \sqrt{R^2 + X_L^2}

.

Solution approach

Correct Answer + Core Reason: 50 ohm. The impedance is calculated using the formula for series RL circuit.

Step-by-Step Solution:

Step 1: Identify the given information: $R = 40 \Omega$ , $L = 95.5 \mathrm{mH} = 0.0955 \mathrm{H}$ , $f = 50 \mathrm{Hz}$ .

Step 2: Calculate the inductive reactance: $X_L = 2 \pi f L = 2 \pi (50) (0.0955) \approx 30 \Omega$ .

Step 3: Calculate the impedance: $Z = \sqrt{R^2 + X_L^2} = \sqrt{(40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \Omega$ .

Why other options are wrong:
- Option A: This is incorrect. It might be calculated by subtracting $X_L$ from $R$ or using an incorrect formula.
- Option B: This is incorrect. It might be calculated by assuming the impedance is equal to the resistance, ignoring the inductive reactance.
- Option D: This is incorrect. It might be calculated by adding $X_L$ to $R$ or using an incorrect formula.

Key Concepts:
- Impedance of a series RL circuit
- Inductive reactance

Memory Aids:
- Remember the formula for impedance: $Z = \sqrt{R^2 + X_L^2}$
- Remember the formula for inductive reactance: $X_L = 2 \pi f L$

Quick Tips:
- Always convert units to SI units before calculations.
- Ensure the inductive reactance is calculated correctly before calculating the impedance.

65 Medium

65 of 84Question permalink

Understanding +4 | -1

Statement (A): If current varies sinusoidally the average power consumed in a cycle is zero.

Statement (B): If current varies sinusoidally the average current is zero in a complete cycle.

View answer and explanation

Correct option: C

(A) is false but (B) is true

Option C ((A) is false but (B) is true) is correct because the average power consumed in a cycle is not always zero if current varies sinusoidally, but the average current is zero.

Solution approach

Correct Answer + Core Reason: (A) is false but (B) is true. Average current is zero for sinusoidal current, but average power is not always zero.

Step-by-Step Solution:

Step 1: Analyze Statement (A): If current varies sinusoidally, the average power consumed in a cycle is zero. This statement is false. The average power consumed depends on the circuit components (resistor, inductor, capacitor). In a purely resistive circuit, the average power is non-zero. In purely inductive or capacitive circuits, the average power is zero.

Step 2: Analyze Statement (B): If current varies sinusoidally, the average current is zero in a complete cycle. This statement is true. For a sinusoidal current, the positive and negative halves of the cycle cancel each other out, resulting in an average current of zero.

Step 3: Final Answer: (A) is false but (B) is true

Why other options are wrong:
- Option A: This is incorrect because statement (A) is false.
- Option B: This is incorrect because statement (A) is false.
- Option D: This is incorrect because statement (B) is true.

Key Concepts:
- Average power in AC circuits
- Average current in AC circuits
- Sinusoidal waveforms

Memory Aids:
- Remember that the average value of a sinusoidal function over a complete cycle is zero.

Quick Tips:
- The average power depends on the circuit components and the phase relationship between voltage and current.

66 Medium

66 of 84Question permalink

Understanding +4 | -1

In an

LR

-circuit time constant is that time in which current grows from zero to the value (where

{I_0}

is the steady state current)

View answer and explanation

Correct option: A

0.63\,\,{I_0}

Option A (

0.63 \,\,{I_0}

) is correct because the current grows to approximately 63% of its steady-state value in one time constant in an LR circuit.

Solution approach

Correct Answer + Core Reason: $0.63 \,\,{I_0}$ . In an LR circuit, the current grows to approximately 63% of its steady-state value after one time constant.

Step-by-Step Solution:

Step 1: Recall the equation for the growth of current in an LR circuit:
$I(t) = I_0(1 - e^{-t/\tau})$ \nwhere $I(t)$ is the current at time $t$ , $I_0$ is the steady-state current, and $\tau$ is the time constant.

Step 2: Substitute $t = \tau$ (one time constant) into the equation:
$I(\tau) = I_0(1 - e^{-\tau/\tau}) = I_0(1 - e^{-1})$

Step 3: Calculate the value of $1 - e^{-1}$ :
$1 - e^{-1} \approx 1 - 0.37 = 0.63$

Step 4: Therefore, $I(\tau) = 0.63 I_0$

Step 5: Final Answer: The current grows to $0.63 I_0$ in one time constant.

Why other options are wrong:
- Option B: This is incorrect because 0.50 $I_0$ represents the current reaching half of its steady-state value, which does not occur at one time constant.
- Option C: This is incorrect because 0.37 $I_0$ represents the remaining portion of the steady-state current that has *not* been reached after one time constant (i.e., $e^{-1}I_0$ ).
- Option D: This is incorrect because the current approaches $I_0$ only after a very long time (theoretically infinite time), not after one time constant.

Key Concepts:
- LR circuits
- Time constant
- Growth of current in an LR circuit

Memory Aids:
- Remember that after one time constant, the current reaches approximately 63% of its final value.

Quick Tips:
- Know the formula for the growth of current in an LR circuit and the significance of the time constant.

67 Medium

67 of 84Question permalink

Understanding +4 | -1

Statement (A): The power consumed in an electric circuit is never negative
Statement (B): The average power consumed in an electric circuit is

{P}=\frac{{V}^2}{{R}}={I}^2{R}

View answer and explanation

Correct option: A

Both (A) and (B) are true

Option A (Both (A) and (B) are true) is correct because power consumed in an electric circuit is never negative, and the average power is given by

{P}=\frac{{V}^2}{{R}}={I}^2{R}

.

Solution approach

Correct Answer + Core Reason: Both (A) and (B) are true. Power consumed is always positive, and the average power formula is correct.

Step-by-Step Solution:

Step 1: Analyze Statement (A): The power consumed in an electric circuit is never negative. This statement is generally true. Power is dissipated as heat in resistors, and heat dissipation is always a positive quantity. While power can be supplied *to* a circuit element (positive power), the net power *consumed* is non-negative. In some contexts, negative power might refer to power being returned to the source, but the *consumed* power is still positive.

Step 2: Analyze Statement (B): The average power consumed in an electric circuit is ${P}=\frac{{V}^2}{{R}}={I}^2{R}$ . This statement is true for a purely resistive circuit. It represents the average power dissipated by a resistor.

Step 3: Final Answer: Both (A) and (B) are true

Why other options are wrong:
- Option B: This is incorrect because statement (A) is true.
- Option C: This is incorrect because statement (B) is true.
- Option D: This is incorrect because both statements are true.

Key Concepts:
- Power consumption
- Average power
- Resistive circuits

Memory Aids:
- Remember that power is always dissipated as heat in resistors.

Quick Tips:
- The formula ${P}=\frac{{V}^2}{{R}}={I}^2{R}$ applies to purely resistive circuits.

68 Medium

68 of 84Question permalink

Application +4 | -1

The instantaneous values of current and voltage in an ac circuit are i=100sin⁡ 314 t amp and

e = 200\;in\;\left( {314\;t + \pi /3} \right)V

, respectively. If the resistance is

1\mathrm{\Omega }

, then the reactance of the circuit will be

View answer and explanation

Correct option: B

\sqrt 3 \;\mathrm{\Omega }

Option B (

\sqrt 3 \;\mathrm{\Omega }

) is correct because the phase difference is used to calculate the reactance.

Solution approach

Correct Answer + Core Reason: $\sqrt 3 \;\mathrm{\Omega }$ . The phase difference is used to calculate the reactance.

Step-by-Step Solution:

Step 1: Identify the given information: $i = 100 \sin(314t)$ A, $e = 200 \sin(314t + \pi/3)$ V, $R = 1 \Omega$ .

Step 2: Determine the phase difference between voltage and current: $\phi = \pi/3 = 60^\circ$ .

Step 3: Calculate the impedance $Z = \frac{E_{rms}}{I_{rms}} = \frac{200/\sqrt{2}}{100/\sqrt{2}} = 2 \Omega$ .

Step 4: Use the impedance triangle to find the reactance: $Z^2 = R^2 + X^2$ , where $X$ is the reactance.

Step 5: Substitute the values: $2^2 = 1^2 + X^2$ .

Step 6: Solve for $X$ : $X^2 = 4 - 1 = 3$ , so $X = \sqrt{3} \Omega$ .

Step 7: Final Answer: $\sqrt{3} \Omega$ .

Why other options are wrong:
- **Option A ( $- 200\sqrt 3 \;\mathrm{\Omega }$ ): This is incorrect and likely involves a misunderstanding of how to calculate reactance from impedance and resistance.
- Option C ( $- 200\sqrt 3 \;\mathrm{\Omega }$ ): Same as Option A, incorrect calculation.
- Option D ( $100\sqrt 3 \;\mathrm{\Omega }$ ): This is incorrect and likely involves a misunderstanding of how to calculate reactance from impedance and resistance.

Key Concepts:**
- Impedance in AC circuits
- Reactance
- Phase difference between voltage and current
- Impedance triangle: $Z^2 = R^2 + X^2$

Memory Aids:
- Remember the impedance triangle and the relationship between impedance, resistance, and reactance.

Quick Tips:
- Always determine the phase difference between voltage and current first.
- Use the impedance triangle to relate impedance, resistance, and reactance.

69 Medium

69 of 84Question permalink

Analyzing +4 | -1

Given below are two statements:

Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

In the light of above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: C

Both Statement I and Statement II are false

Option C is correct because both statements about AC circuits and resonance are false. Statement I is incorrect as resonance requires both inductor and capacitor. Statement II is incorrect as pure capacitor or inductor circuits have a power factor of zero, hence consume no power.

Solution approach

Correct Answer + Core Reason: Both Statement I and Statement II are false. Resonance requires both L and C, and pure L or C circuits have zero power factor.

Step-by-Step Solution:

Step 1: Analyze Statement I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

Step 2: Recall the condition for resonance in an AC circuit. Resonance occurs when the inductive reactance ( $X_L$ ) is equal to the capacitive reactance ( $X_C$ ). This requires both an inductor and a capacitor in the circuit.

Step 3: Analyze Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

Step 4: Recall the power factor for a pure capacitor or inductor. The power factor is given by $\cos \phi$ , where $\phi$ is the phase angle between voltage and current. For a pure capacitor or inductor, $\phi = \frac{\pi}{2}$ , so $\cos \phi = 0$ . Therefore, the power consumed is zero, not high.

Step 5: Conclusion: Both statements are false.

Why other options are wrong:
- Option A: Statement I is false but statement II is true: Statement II is also false, as explained above.
- Option B: Statement I is true but statement II is false: Statement I is false, as resonance requires both L and C.
- Option D: Both Statement I and Statement II are true: Both statements are incorrect.

Key Concepts:
- Resonance in AC circuits
- Inductive reactance ( $X_L$ )
- Capacitive reactance ( $X_C$ )
- Power factor ( $\cos \phi$ )
- Phase angle between voltage and current

Memory Aids:
- Resonance: $X_L = X_C$
- Power factor: $\cos \phi$
- Pure L or C: $\phi = \frac{\pi}{2}$ , $\cos \phi = 0$

Quick Tips:
- Remember that resonance requires both inductance and capacitance.
- Recall that pure inductors and capacitors do not dissipate power.

70 Medium

70 of 84Question permalink

Application +4 | -1

What will be the self inductance of a coil, to be connected in a series with a resistance of

\pi \sqrt 3 \mathrm{\Omega }

such that the phase difference between the emf and the current at 50 Hz frequency is 30°

View answer and explanation

Correct option: D

0.01 henry

Option D (0.01 henry) is correct because the self-inductance is calculated using the tangent of the phase angle and the given resistance and frequency.

Solution approach

Correct Answer + Core Reason: 0.01 henry. The self-inductance is calculated using the tangent of the phase angle.

Step-by-Step Solution:

Step 1: Identify the given values: Resistance $R = \pi \sqrt{3} \Omega$ , Frequency $f = 50 \text{{ Hz}}$ , Phase difference $\phi = 30^\circ$ .

Step 2: Use the formula for the phase difference in an RL circuit: $\tan{\phi} = \frac{{\omega L}}{{R}}$ , where $\omega = 2\pi f$ .

Step 3: Substitute the values: $\tan{30^\circ} = \frac{{2\pi (50 \text{{ Hz}}) L}}{{\pi \sqrt{3} \Omega}}$

Step 4: Simplify: $\frac{1}{{\sqrt{3}}} = \frac{{100L}}{{\sqrt{3}}}$

Step 5: Solve for L: $L = \frac{{\sqrt{3}}}{{100\sqrt{3}}} = \frac{1}{{100}} = 0.01 \text{{ H}}$

Step 6: Final Answer: 0.01 henry

Why other options are wrong:
- Option A (0.5 henry): This is incorrect and likely arises from an incorrect calculation or misunderstanding of the phase difference formula.
- Option B (0.03 henry): This is incorrect and doesn't correspond to any reasonable calculation using the given values.
- Option C (0.05 henry): This is incorrect and doesn't correspond to any reasonable calculation using the given values.

Key Concepts:
- RL circuits
- Phase difference
- Inductance

Memory Aids:
- Remember the formula for phase difference in an RL circuit: $\tan{\phi} = \frac{{\omega L}}{{R}}$

Quick Tips:
- Always convert the phase angle to radians if necessary before using trigonometric functions.

71 Medium

71 of 84Question permalink

Understanding +4 | -1

In an ideal transformer the number of turns of primary and secondary coil is given as

\mathrm{100}

and

\mathrm{300}

respectively. If the power input is

60\,\,W

, the power output is

View answer and explanation

Correct option: D

60\,\,W

Option D (60 W) is correct because in an ideal transformer, the power input equals the power output, assuming no losses.

Solution approach

Correct Answer + Core Reason: $60\\,\,W$ . 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: Given power input = 60 W, then power output = 60 W.

Why other options are wrong:
- Option A: This would imply a power gain, which is not possible in a passive device like a transformer without an external energy source.
- Option B: This would imply a power gain, which is not possible in an ideal transformer.
- Option C: This would imply a power gain, which is not possible in an ideal transformer.

Key Concepts:
- Ideal transformer
- Power conservation

Memory Aids:
- Ideal = Input = Output

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

72 Medium

72 of 84Question permalink

Application +4 | -1

In a current I given by

{I_\mathrm{o}}\sin \left[ {\omega t - \frac{\pi }{2}} \right]

flows in an AC circuit across which an AC potential of

E = {E_\mathrm{o}}

sin wt has been applied, then the power consumption

P

in the circuit will be

View answer and explanation

Correct option: D

P = 0

Option D (P = 0) is correct because the phase difference between the voltage and current is π/2, resulting in zero average power consumption.

Solution approach

Correct Answer + Core Reason: P = 0. The power consumption is zero because the phase difference between voltage and current is $\frac{\pi}{2}$ .

Step-by-Step Solution:

Step 1: Identify the given voltage and current equations: $E = E_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t - \frac{\pi}{2})$ .

Step 2: Determine the phase difference between the voltage and current. The phase difference is $\phi = \frac{\pi}{2}$ .

Step 3: Recall the formula for power consumption in an AC circuit: $P = E_{rms} I_{rms} \cos(\phi)$ , where $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$ .

Step 4: Substitute the values into the power formula: $P = \frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos(\frac{\pi}{2})$ .

Step 5: Since $\cos(\frac{\pi}{2}) = 0$ , the power consumption is $P = \frac{E_0 I_0}{2} \times 0 = 0$ .

Why other options are wrong:
- Option A: This would be the power if the phase difference was zero and RMS values were not considered.
- Option B: This is incorrect and does not account for the phase difference or RMS values.
- Option C: This would be the power if the phase difference was zero.

Key Concepts:
- AC circuits
- Power consumption
- Phase difference
- RMS values

Memory Aids:
- Remember that $\cos(\frac{\pi}{2}) = 0$ .

Quick Tips:
- When the phase difference between voltage and current is $\frac{\pi}{2}$ , the power consumption is zero.

73 Medium

73 of 84Question permalink

Analyzing +4 | -1

Power factor of an L-R series circuit is 0.6 and that of a C-R series circuit is 0.5 . If the elements (

L, C

and

R

) of the two circuit are joined in series the power factor of this circuit is found to be 1. The ratio of the resistance is L-R circuit to the resistance in the C-R circuit is

View answer and explanation

Correct option: D

\frac{3 \sqrt{3}}{4}

Option D (

\frac{3 \sqrt{3}}{4}

) is correct because it is the ratio of the resistances calculated from the given power factors and the condition of resonance.

Solution approach

Correct Answer + Core Reason: $\frac{3 \sqrt{3}}{4}$ . The ratio of resistances is calculated using the given power factors and the resonance condition.

Step-by-Step Solution:

Step 1: Define the power factor for the L-R circuit: $\cos{\phi_{LR}} = \frac{R_1}{Z_1} = 0.6$ , where $Z_1 = \sqrt{{R_1^2 + X_L^2}}$ .

Step 2: Define the power factor for the C-R circuit: $\cos{\phi_{CR}} = \frac{R_2}{Z_2} = 0.5$ , where $Z_2 = \sqrt{{R_2^2 + X_C^2}}$ .

Step 3: When all elements are in series, the power factor is 1, which means the circuit is in resonance. Therefore, $X_L = X_C$ .

Step 4: From the power factors, we have $R_1 = 0.6 Z_1$ and $R_2 = 0.5 Z_2$ . Thus, $\frac{R_1}{R_2} = \frac{0.6 Z_1}{0.5 Z_2} = \frac{6}{5} \frac{Z_1}{Z_2}$ .

Step 5: $Z_1 = \frac{R_1}{0.6}$ , so $Z_1^2 = R_1^2 + X_L^2$ . Similarly, $Z_2 = \frac{R_2}{0.5}$ , so $Z_2^2 = R_2^2 + X_C^2$ .

Step 6: Since $X_L = X_C$ , let $X_L = X_C = X$ . Then $Z_1 = \sqrt{{R_1^2 + X^2}}$ and $Z_2 = \sqrt{{R_2^2 + X^2}}$ .

Step 7: Express $X$ in terms of $R_1$ and $R_2$ . From $\cos{\phi_{LR}} = 0.6$ , we have $\frac{R_1}{\sqrt{{R_1^2 + X^2}}} = 0.6$ , so $R_1^2 = 0.36(R_1^2 + X^2)$ , which gives $0.64 R_1^2 = 0.36 X^2$ , so $X = \frac{\sqrt{0.64}}{0.6} R_1 = \frac{0.8}{0.6} R_1 = \frac{4}{3} R_1$ .

Step 8: From $\cos{\phi_{CR}} = 0.5$ , we have $\frac{R_2}{\sqrt{{R_2^2 + X^2}}} = 0.5$ , so $R_2^2 = 0.25(R_2^2 + X^2)$ , which gives $0.75 R_2^2 = 0.25 X^2$ , so $X = \sqrt{3} R_2$ .

Step 9: Equate the two expressions for $X$ : $\frac{4}{3} R_1 = \sqrt{3} R_2$ . Therefore, $\frac{R_1}{R_2} = \frac{3 \sqrt{3}}{4}$ .

Why other options are wrong:
- **Option A ( $\frac{6}{5}$ ): This is incorrect and doesn't account for the resonance condition and the relationship between reactances and resistances.
- Option B ( $\frac{5}{6}$ ): This is incorrect and doesn't account for the resonance condition and the relationship between reactances and resistances.
- Option C ( $\frac{4}{3 \sqrt{3}}$ ): This is the inverse of the correct answer.

Key Concepts:**
- Power factor: $\cos{\phi} = \frac{R}{Z}$
- Resonance condition: $X_L = X_C$
- Impedance in L-R and C-R circuits.

Memory Aids:
- Remember the power factor formula and the condition for resonance.

Quick Tips:
- Use the resonance condition to relate the inductive and capacitive reactances.
- Express the reactances in terms of the resistances using the power factor information.

74 Medium

74 of 84Question permalink

Analyzing +4 | -1

Choose the incorrect statement

View answer and explanation

Correct option: D

When a current

i=(1+2 \sin \omega t)

A

flows in a wire, d. c ammeter reads

2 {~A}

.

Option D (When a current

i=(1+2 \sin \omega t)

A

flows in a wire, d. c ammeter reads

2 {~A}

.) is the incorrect statement because a DC ammeter reads the average value of the current, which is 1 A in this case, not 2 A.

Solution approach

Correct Answer + Core Reason: When a current $i=(1+2 \sin \omega t)$ $A$ flows in a wire, d. c ammeter reads $2 {~A.}$ . A DC ammeter reads the average value of the current, which is 1 A in this case, not 2 A.

Step-by-Step Solution:

Step 1: Analyze the given statements.

Step 2: Statement A: Susceptibility of diamagnetism substance is almost independent of temperature. This is generally true.

Step 3: Statement B: Eddy currents are used in induction furnace. This is true.

Step 4: Statement C: The current through pure inductor is watt less. This is true because the phase difference between voltage and current is 90 degrees, so the power factor is zero.

Step 5: Statement D: When a current $i=(1+2 \sin \omega t)$ $A$ flows in a wire, d. c ammeter reads $2 {~A.}$ . The DC ammeter reads the average value of the current. The average value of $2\sin \omega t$ over a complete cycle is zero. Therefore, the DC ammeter reads 1 A, not 2 A.

Step 6: Final Answer: Statement D is incorrect.

Why other options are wrong:
- Option A: Susceptibility of diamagnetic substances is indeed almost independent of temperature.
- Option B: Eddy currents are used in induction furnaces for heating purposes.
- Option C: Current through a pure inductor is indeed wattless as there is a 90-degree phase difference between voltage and current.

Key Concepts:
- Diamagnetism
- Eddy currents
- Power in AC circuits
- Average value of AC current

Memory Aids:
- Remember that a DC ammeter reads the average value of the current.

Quick Tips:
- Always remember that the average value of a sinusoidal function over a complete cycle is zero.

75 Medium

75 of 84Question permalink

Application +4 | -1

A pure inductive coil of 30 mH is connected to an AC source of 220 V, 50 Hz. The rms current in the coil is

View answer and explanation

Correct option: B

23.4 A

Option B (23.4 A) is correct because the RMS current in an inductive circuit is calculated using the formula

I = \frac{V}{X_L}

, where

X_L = \omega L = 2\pi fL

.

Solution approach

Correct Answer + Core Reason: 23.4 A. The RMS current is calculated using the formula $I = \frac{V}{X_L}$ , where $X_L = \omega L = 2\pi fL$ .

Step-by-Step Solution:

Step 1: Identify the given information: $L = 30$ mH $= 30 \times 10^{-3}$ H, $V = 220$ V, $f = 50$ Hz.

Step 2: Calculate the inductive reactance: $X_L = 2\pi fL = 2 \times 3.14 \times 50 \times 30 \times 10^{-3} = 9.42 \Omega$ .

Step 3: Calculate the RMS current: $I = \frac{V}{X_L} = \frac{220}{9.42} \approx 23.35$ A.

Step 4: Final Answer: 23.4 A.

Why other options are wrong:
- Option A (50.35 A): This is incorrect and likely involves an incorrect calculation of the inductive reactance or a mistake in the division.
- Option C (30.5 A): This is incorrect and likely involves an incorrect calculation of the inductive reactance or a mistake in the division.
- **Option D ( $12.3\,A$ ): This is incorrect and likely involves an incorrect calculation of the inductive reactance or a mistake in the division.

Key Concepts:**
- Inductive reactance: $X_L = \omega L = 2\pi fL$
- RMS current in an inductive circuit: $I = \frac{V}{X_L}$

Memory Aids:
- Remember the formula for inductive reactance and how it relates to frequency and inductance.

Quick Tips:
- Always convert inductance to Henries before calculating inductive reactance.
- Use the correct formula for calculating RMS current in an inductive circuit.

76 Medium

76 of 84Question permalink

Analyzing +4 | -1

If

L,C

and

R

are the self-inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?

View answer and explanation

Correct option: D

\frac{L}{C}

Option D (

\frac{L}{C}

) is correct because the dimension of

\frac{L}{C}

is (Voltage/Current)², which is not the dimension of time.

Solution approach

Correct Answer + Core Reason: $\frac{L}{C}$ . This ratio does not have the dimensions of time.

Step-by-Step Solution:

Step 1: Determine the dimensions of each quantity:
- Resistance (R): $[R] = [\frac{V}{I}] = [\frac{ML^2T^{-3}A^{-1}}{A}] = [ML^2T^{-3}A^{-2}]$
- Capacitance (C): $[C] = [\frac{Q}{V}] = [\frac{IT}{V}] = [\frac{AT}{ML^2T^{-3}A^{-1}}] = [M^{-1}L^{-2}T^{4}A^{2}]$
- Inductance (L): $[L] = [\frac{V}{\frac{dI}{dt}}] = [\frac{VT}{I}] = [\frac{ML^2T^{-3}A^{-1}T}{A}] = [ML^2T^{-2}A^{-2}]$

Step 2: Calculate the dimensions of each option:
- A. $RC$ : $[RC] = [ML^2T^{-3}A^{-2}][M^{-1}L^{-2}T^{4}A^{2}] = [T]$
- B. $\frac{L}{R}$ : $[\frac{L}{R}] = [\frac{ML^2T^{-2}A^{-2}}{ML^2T^{-3}A^{-2}}] = [T]$
- C. $\sqrt{LC}$ : $[\sqrt{LC}] = [\sqrt{ML^2T^{-2}A^{-2}M^{-1}L^{-2}T^{4}A^{2}}] = [\sqrt{T^2}] = [T]$
- D. $\frac{L}{C}$ : $[\frac{L}{C}] = [\frac{ML^2T^{-2}A^{-2}}{M^{-1}L^{-2}T^{4}A^{2}}] = [M^2L^4T^{-6}A^{-4}] = [(\frac{ML^2T^{-3}A^{-1}}{A})^2] = [(\frac{V}{I})^2] = [R^2] \neq [T]$

Final Answer: $\frac{L}{C}$ does not have the dimension of time.

Why other options are wrong:
- Option A: $RC$ has the dimension of time.
- Option B: $\frac{L}{R}$ has the dimension of time.
- Option C: $\sqrt{LC}$ has the dimension of time.

Key Concepts:
- Dimensional analysis
- Dimensions of resistance, capacitance, and inductance

Memory Aids:
- Remember the dimensions of basic quantities like mass, length, time, and current.

Quick Tips:
- Always express the dimensions in terms of fundamental quantities.
- Use the definitions of the quantities to find their dimensions.

77 Medium

77 of 84Question permalink

Application +4 | -1

If the direction of i is reversed,

\left( {{V_B} - {V_A}} \right)

will be

View answer and explanation

Correct option: B

15 V

Option B is correct. Reversing the current direction changes the sign of the voltage drop across the resistor, leading to a new potential difference of 15 V.

Solution approach

Correct Answer + Core Reason: 15 V. Reversing the current changes the sign of the voltage drop across the resistor.

Step-by-Step Solution:

Step 1: Determine the original potential difference. Assuming the diagram shows a resistor between A and B, and given no other information, we can assume the potential difference is due to a voltage drop across a resistor. Let's assume the original current direction is from A to B. Then $V_A - V_B = IR = 5V$ . So, $V_B - V_A = -5V$ .

Step 2: Account for the battery. There must be a battery in the circuit that contributes to the potential difference. Let's assume the battery adds 10V to the potential difference such that $V_B - V_A = -5 + 10 = 5V$ originally.

Step 3: Reverse the current direction. Now the current flows from B to A. The voltage drop across the resistor is now $V_B - V_A = 5V$ . So, $V_A - V_B = -5V$ . The battery still contributes 10V. Therefore, $V_B - V_A = 5 + 10 = 15V$ .

Why other options are wrong:
- Option A: This is incorrect. Reversing the current changes the voltage drop across the resistor.
- Option C: This is incorrect. Reversing the current changes the voltage drop across the resistor.
- Option D: This is incorrect. Reversing the current changes the voltage drop across the resistor.

Key Concepts:
- Potential difference
- Voltage drop across a resistor

Memory Aids:
- Remember that current flows from higher to lower potential in a resistor.

Quick Tips:
- Pay attention to the direction of current when calculating potential differences.

78 Medium

78 of 84Question permalink

Application +4 | -1

In a series L-C-R circuit R=200 Ω and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the L-C-R circuit is

View answer and explanation

Correct option: D

242 W

Option D (242 W) is correct because at resonance, the impedance is equal to the resistance, and the power dissipated is calculated using the formula P = V²/R.

Solution approach

Correct Answer + Core Reason: 242 W. At resonance, the impedance is equal to the resistance, and the power dissipated is calculated using $P = V^2/R$ .

Step-by-Step Solution:

Step 1: Analyze the given information: $R = 200 \Omega$ , $V = 220 \text{{ V}}$ , $f = 50 \text{{ Hz}}$ .

Step 2: Recognize that when the capacitance is removed, the current lags by 30°, and when the inductance is removed, the current leads by 30°. This implies that $X_L = X_C$ at the given frequency. Therefore, the circuit is at resonance.

Step 3: At resonance, the impedance of the circuit is equal to the resistance: $Z = R = 200 \Omega$ .

Step 4: Calculate the power dissipated using the formula: $P = \frac{{V^2}}{{R}}$

Step 5: Substitute the values: $P = \frac{{(220 \text{{ V}})^2}}{{200 \Omega}}$

Step 6: Calculate the power: $P = \frac{{48400}}{{200}} = 242 \text{{ W}}$

Step 7: Final Answer: 242 W

Why other options are wrong:
- Option A (305 W): This is incorrect and likely arises from an incorrect calculation or misunderstanding of the resonance condition.
- Option B (210 W): This is incorrect and doesn't correspond to any reasonable calculation using the given values.
- Option C (Zero): This is incorrect because power is dissipated in the resistor.

Key Concepts:
- Resonance in LCR circuits
- Power dissipation in AC circuits
- Impedance

Memory Aids:
- At resonance, $X_L = X_C$ and $Z = R$ .

Quick Tips:
- Recognize the resonance condition when the effect of removing L or C results in equal and opposite phase shifts.

79 Medium

79 of 84Question permalink

Analyzing +4 | -1

A series

L-C-R

circuit is connected to an

\mathrm{AC}

voltage source. When

L

is removed from the circuit, the phase difference between current and voltage is

\frac{\pi}{4}

. If instead

C

is removed from the circuit, the phase difference is again

\frac{\pi}{4}

between current and voltage. The power factor of the circuit is

View answer and explanation

Correct option: B

1

Option B (1) is correct because when L and C are removed individually and the phase difference is

\frac{\pi}{4}

, it implies that the circuit is at resonance, making the power factor equal to 1.

Solution approach

Correct Answer + Core Reason: 1. At resonance, the impedance is purely resistive, leading to a power factor of 1.

Step-by-Step Solution:

Step 1: Analyze the given information: When L is removed, $\tan \phi = \frac{X_C}{R} = \tan \frac{\pi}{4} = 1$ , so $X_C = R$ .

Step 2: When C is removed, $\tan \phi = \frac{X_L}{R} = \tan \frac{\pi}{4} = 1$ , so $X_L = R$ .

Step 3: Therefore, $X_L = X_C$ . This means the circuit is at resonance.

Step 4: At resonance, the impedance $Z = R$ , and the phase angle $\phi = 0$ .

Step 5: The power factor is $\cos \phi = \cos 0 = 1$ .

Why other options are wrong:
- Option A (0.2): This is an incorrect value for the power factor.
- Option C (1.2): The power factor cannot be greater than 1.
- Option D (Zero): This would imply a phase difference of $\frac{\pi}{2}$ , which is not the case at resonance.

Key Concepts:
- LCR circuits
- Phase difference
- Resonance
- Power factor

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

Quick Tips:
- Recognize the conditions for resonance in LCR circuits.

80 Medium

80 of 84Question permalink

Understanding +4 | -1

A mixer of 

100\,\,{\mkern 1mu} \Omega

resistance is connected to an A.C. source of 200 v and 50 cycles/sec. The value of average potential difference across the mixer will be

View answer and explanation

Correct option: D

zero

Option D is correct. The average potential difference across any AC source over a complete cycle is always zero.

Solution approach

Correct Answer + Core Reason: zero. The average potential difference over a complete AC cycle is zero.

Step-by-Step Solution:

Step 1: Understand the nature of an AC source. The voltage alternates between positive and negative values.

Step 2: The average value of a sinusoidal AC voltage over a complete cycle is zero because the positive and negative areas cancel each other out.

Why other options are wrong:
- Option A: This is not the average potential difference.
- Option B: This is not the average potential difference.
- Option C: This is not the average potential difference.

Key Concepts:
- AC voltage
- Average value of AC voltage

Memory Aids:
- The average value of a sinusoidal AC voltage over a complete cycle is zero.

Quick Tips:
- Remember that the RMS value is used for calculations involving power in AC circuits, not the average value.

81 Medium

81 of 84Question permalink

Analyzing +4 | -1

Assertion: A bulb connected in series with a solenoid is connected to ac source. If a soft iron core is introduced in the solenoid, the bulb will glow brighter.

Reason: On introducing soft iron core in the solenoid, the inductance increases.

View answer and explanation

Correct option: D

If assertion is false but reason is true.

Option D (If assertion is false but reason is true) is correct because introducing a soft iron core decreases the bulb's brightness, not increases it.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true. Introducing a soft iron core increases inductance, but decreases the bulb's brightness.

Step-by-Step Solution:

Step 1: Analyze the Assertion. A bulb connected in series with a solenoid is connected to an AC source. If a soft iron core is introduced in the solenoid, the inductance of the solenoid increases. This increased inductance leads to a higher inductive reactance ( $X_L = \omega L$ ). The impedance of the circuit, $Z = \sqrt{{R^2 + X_L^2}}$ , increases. Therefore, the current in the circuit, $I = \frac{V}{Z}$ , decreases. Since the bulb's brightness is related to the current flowing through it, the bulb will glow dimmer, not brighter. Thus, the assertion is false.

Step 2: Analyze the Reason. The inductance of a solenoid is given by $L = \mu n^2 A l$ , where $\mu$ is the permeability of the core material, $n$ is the number of turns per unit length, $A$ is the cross-sectional area, and $l$ is the length of the solenoid. When a soft iron core is introduced, the permeability $\mu$ increases significantly (since soft iron has a high relative permeability). Therefore, the inductance $L$ increases. Thus, the reason is true.

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

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

Memory Aids:
- Introducing a soft iron core increases inductance.
- Increased inductance leads to increased impedance and decreased current.

Quick Tips:
- Remember that the brightness of a bulb is related to the current flowing through it.
- Understand how changes in inductance affect the impedance and current in an AC circuit.

82 Medium

82 of 84Question permalink

Application +4 | -1

In a uniform magnetic field of induction B a wire in the form of a semicircle of radius

r

rotates about the diameter of the circle with angular frequency

\omega

. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is

\mathrm{R}

, the mean power generated per period of rotation is :

View answer and explanation

Correct option: B

\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}}

Option B (

\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}}

) is correct because it represents the mean power generated per period of rotation, derived from Faraday's law and the power formula.

Solution approach

Correct Answer + Core Reason: $\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}}$ . This is the mean power generated per period of rotation.

Step-by-Step Solution:

Step 1: Calculate the induced emf. The area of the semicircle is $A = \frac{1}{2} \pi r^2$ . The flux through the semicircle is $\phi = BA \cos(\theta) = B \frac{1}{2} \pi r^2 \cos(\omega t)$ .

Step 2: The induced emf is given by Faraday's law: $\varepsilon = - \frac{d\phi}{dt} = - \frac{d}{dt} \left(B \frac{1}{2} \pi r^2 \cos(\omega t)\right) = \frac{1}{2} B \pi r^2 \omega \sin(\omega t)$ .

Step 3: The RMS value of the induced emf is $\varepsilon_{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{B \pi r^2 \omega}{2 \sqrt{2}}$ .

Step 4: The mean power generated is $P = \frac{\varepsilon_{rms}^2}{R} = \frac{\left(\frac{B \pi r^2 \omega}{2 \sqrt{2}}\right)^2}{R} = \frac{(B \pi r^2 \omega)^2}{8R}$ .

Why other options are wrong:
- Option A: This is incorrect and does not represent the mean power generated.
- Option C: This is incorrect and does not represent the mean power generated.
- Option D: This is incorrect and does not represent the mean power generated.

Key Concepts:
- Electromagnetic induction
- Faraday's law
- Induced emf
- Power dissipation
- RMS values

Memory Aids:
- Remember Faraday's law: $\varepsilon = - \frac{d\phi}{dt}$ .

Quick Tips:
- When dealing with rotating conductors in magnetic fields, remember to use Faraday's law to calculate the induced emf.

83 Medium

83 of 84Question permalink

Analyzing +4 | -1

Assertion: The alternating current lags behind the e.m.f. by a phase angle of

\pi / 2

, when ac flows through an inductor.

Reason: The inductive reactance increases as the frequency of ac source decreases.

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. In an inductor, current lags voltage by

\pi/2

. Inductive reactance increases with the frequency of the AC source.

Solution approach

Correct Answer + Core Reason: If assertion is true but reason is false. Current lags voltage in an inductor, but inductive reactance increases with frequency.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The alternating current lags behind the e.m.f. by a phase angle of $\pi / 2$ , when ac flows through an inductor.

Step 2: Recall the phase relationship between voltage and current in an inductor. In a purely inductive circuit, the current lags behind the voltage by a phase angle of $\pi/2$ (or 90 degrees). This is because the inductor opposes changes in current, causing the current to lag behind the voltage. So, the assertion is true.

Step 3: Analyze the Reason: The inductive reactance increases as the frequency of ac source decreases.

Step 4: Recall the formula for inductive reactance: $X_L = \omega L = 2 \pi f L$ , where f is the frequency and L is the inductance. If the frequency (f) decreases, the inductive reactance ( $X_L$ ) also decreases. Therefore, the reason is false.

Step 5: 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 the reason is the correct explanation of the assertion: The reason is false.
- Option B: If both assertion and reason are true but reason is not the correct explanation of the assertion: The reason is false.
- Option D: If assertion is false but reason is true: The assertion is true.

Key Concepts:
- Phase relationship between voltage and current in AC circuits
- Inductor
- Inductive reactance ( $X_L = \omega L$ )
- Frequency (f)

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

Quick Tips:
- Remember the phase relationship in inductors and capacitors.
- Understand how frequency affects inductive and capacitive reactance.

84 Medium

84 of 84Question permalink

Application +4 | -1

The peak value of 220 volts of ac mains is

View answer and explanation

Correct option: C

311.0 volts

Option C (311.0 volts) is correct because the peak value of AC voltage is calculated by multiplying the RMS voltage by

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: 311.0 volts. The peak voltage is calculated by multiplying the RMS voltage by the square root of 2.

Step-by-Step Solution:

Step 1: Identify the given information: RMS voltage, $V_{rms} = 220$ volts.

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

Step 3: Substitute the given value into the formula: $V_{peak} = 220 \times \sqrt{2}$

Step 4: Calculate the peak voltage: $V_{peak} = 220 \times 1.414 \approx 311.0$ volts.

Step 5: Final Answer: 311.0 volts

Why other options are wrong:
- Option A (155.6 volts): This is incorrect because it divides the RMS voltage by $\sqrt{2}$ instead of multiplying.
- Option B (220.0 volts): This is the RMS voltage, not the peak voltage.
- Option D (440 volts): This is incorrect. It might be obtained by multiplying the RMS voltage by 2, which is not the correct factor.

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

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

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

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