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

31 Easy

Application +4 | -1

A transformer with turns ratio

\frac{N_1}{N_2}=\frac{50}{1}

is connected to a 120 volt AC supply. If primary and secondary circuit resistance are

1.5 \mathrm{~k} \Omega

and

1 ~\Omega

respectively then find out power output.

View answer and explanation

Correct option: A

5.76 \mathrm{~W}

Option A (

5.76 \mathrm{~W}

) is correct because it is the power output calculated using the transformer turns ratio and the given resistances and voltage.

Solution approach

Correct Answer + Core Reason: $5.76 \mathrm{~W}$ . The power output is calculated using the turns ratio and resistances.

Step-by-Step Solution:

Step 1: The turns ratio is given as $\frac{N_1}{N_2} = \frac{50}{1}$ . The primary voltage is $V_1 = 120 \, V$ .

Step 2: The secondary voltage $V_2$ is given by $\frac{V_1}{V_2} = \frac{N_1}{N_2}$ , so $V_2 = V_1 \times \frac{N_2}{N_1} = 120 \times \frac{1}{50} = 2.4 \, V$ .

Step 3: The secondary current $I_2$ is given by $I_2 = \frac{V_2}{R_2}$ , where $R_2 = 1 \, \Omega$ is the secondary resistance. So, $I_2 = \frac{2.4}{1} = 2.4 \, A$ .

Step 4: The power output $P_{out}$ is given by $P_{out} = I_2^2 R_2 = (2.4)^2 \times 1 = 5.76 \, W$ .

Why other options are wrong:
- Option B: This is incorrect due to an error in the calculation of the secondary current or power.
- Option C: This is incorrect due to an error in the calculation of the secondary current or power.
- Option D: This is incorrect due to an error in the calculation of the secondary current or power.

Key Concepts:
- Transformer
- Turns ratio
- Voltage and current relationship in a transformer
- Power calculation

Memory Aids:
- Remember the formulas: $\frac{V_1}{V_2} = \frac{N_1}{N_2}$ and $P = I^2 R$

Quick Tips:
- Calculate the secondary voltage and current correctly.
- Use the correct resistance value for power calculation.

32 Easy

32 of 39Question permalink

Application +4 | -1

An electric heater rated 220 V and 550 W is connected to A.C mains. The current drown by it is

View answer and explanation

Correct option: B

2.5 A

Option B (2.5 A) is correct because it is calculated using the formula I = P/V.

Solution approach

Correct Answer + Core Reason: 2.5 A. The current is calculated using the power and voltage ratings of the heater.

Step-by-Step Solution:

Step 1: Identify the given values: Power (P) = 550 W, Voltage (V) = 220 V.

Step 2: Use the formula relating power, voltage, and current: $P = V \times I$ .

Step 3: Rearrange the formula to solve for current: $I = \frac{P}{V}$ .

Step 4: Substitute the given values: $I = \frac{550 \text{{ W}}}{220 \text{{ V}}}$ .

Step 5: Calculate the current: $I = 2.5 \text{{ A}}$ .

Why other options are wrong:
- Option A: This is incorrect. It might be obtained by dividing voltage by power or by making a calculation error.
- Option C: This is incorrect. It might be obtained by dividing power by voltage squared or by making a calculation error.
- Option D: This is incorrect. It might be obtained by dividing voltage squared by power or by making a calculation error.

Key Concepts:
- Electrical power
- Voltage
- Current

Memory Aids:
- P = VI (Power equals Voltage times Current)

Quick Tips:
- Always check the units before performing calculations.
- Remember the relationship between power, voltage, and current.

33 Easy

33 of 39Question permalink

Understanding +4 | -1

The instantaneous values of current and emf in an ac circuit are

I = 1/\sqrt 2

\sin 314\;t\;amp

and

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

respectively. The phase difference between E and I will be

View answer and explanation

Correct option: A

- \pi /6\;rad

Option A (

-\pi/6 \;rad

) is correct because the phase difference is directly observed from the given equations of emf and current.

Solution approach

Correct Answer + Core Reason: $- \pi /6\\;rad$ . The phase difference is the difference between the phase angles of the emf and current.

Step-by-Step Solution:

Step 1: Identify the given equations for current and emf:
$I = \frac{{1}}{{\sqrt 2 }} \sin 314t \\;amp$
$E = \sqrt 2 \sin (314t - \frac{\pi}{6}) \\;V$

Step 2: Determine the phase angle for the current and emf.\nThe phase angle for the current is $0$ rad.\nThe phase angle for the emf is $-\frac{\pi}{6}$ rad.

Step 3: Calculate the phase difference between E and I.\nPhase difference = Phase angle of E - Phase angle of I\nPhase difference = $(-\frac{\pi}{6}) - 0 = -\frac{\pi}{6}$ rad

Step 4: Final Answer: The phase difference is $-\frac{\pi}{6}$ rad.

Why other options are wrong:
- Option B: This is incorrect because it considers the phase difference to be $-\pi/3$ which is not directly observed from the given equations.
- Option C: This is incorrect because it takes the phase difference as positive $\pi/6$ instead of negative.
- Option D: This is incorrect because it considers the phase difference to be $\pi/3$ which is not directly observed from the given equations.

Key Concepts:
- AC circuits
- Phase difference
- Instantaneous values of current and emf

Memory Aids:
- Remember that the phase difference is the angle by which one sinusoidal quantity leads or lags another.

Quick Tips:
- Directly compare the arguments of the sine functions to find the phase difference.

34 Easy

34 of 39Question permalink

Understanding +4 | -1

A transformer of frequency

60 \mathrm{~Hz}

and

120 \mathrm{~V}

input has

8: 1

turn ratio. The frequency of the output is

View answer and explanation

Correct option: D

60 \mathrm{~Hz}

Option D (60 Hz) is correct because the frequency of the output voltage in a transformer is the same as the frequency of the input voltage.

Solution approach

Correct Answer + Core Reason: $60 \mathrm{~Hz}$ . The frequency remains unchanged in a transformer.

Step-by-Step Solution:

Step 1: Understand the function of a transformer: It changes the voltage level but not the frequency.

Step 2: The frequency of the input AC signal is the same as the frequency of the output AC signal.

Step 3: Therefore, if the input frequency is 60 Hz, the output frequency is also 60 Hz.

Why other options are wrong:
- Option A: The frequency is not divided by the turns ratio.
- Option B: The frequency is not multiplied by the turns ratio.
- Option C: The frequency does not change in a transformer.

Key Concepts:
- Transformer operation
- Frequency invariance

Memory Aids:
- Transformer = Same frequency

Quick Tips:
- The frequency is always constant across the transformer.

35 Easy

35 of 39Question permalink

Application +4 | -1

The ratio of turns in primary and secondary coils of a transformer is 1 : 20. The ratio of currents in primary and secondary coils will be

View answer and explanation

Correct option: B

0.834027778

Answer A (0.055555556) is incorrect. The correct answer is the inverse of the turns ratio, which is 20. The provided answer is the inverse of 18.

Solution approach

Correct Answer + Core Reason: 20. The ratio of currents is the inverse of the turns ratio.

Step-by-Step Solution:

Step 1: Identify the given information:
$\frac{{N_1}}{{N_2}} = \frac{{1}}{{20}}$

Step 2: Identify the formula to use:
$\frac{{I_1}}{{I_2}} = \frac{{N_2}}{{N_1}}$

Step 3: Substitute the values:
$\frac{{I_1}}{{I_2}} = \frac{{20}}{{1}} = 20$

Step 4: Final Answer:
$\frac{{I_1}}{{I_2}} = 20$

Why other options are wrong:
- Option A (0.055555556): This is equal to 1/18, which is incorrect.
- Option B (0.834027778): This appears to be a random number and doesn't relate to the turns ratio.
- Option C (0.319444444): This also appears to be a random number and doesn't relate to the turns ratio.
- Option D (16.66736111): This is approximately 50/3, which is also incorrect.

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

Memory Aids:
- The current ratio is the inverse of the turns ratio.

Quick Tips:
- Remember that voltage and turns are directly proportional, while current and turns are inversely proportional.

36 Easy

36 of 39Question permalink

Application +4 | -1

A series LCR circuit with

R=20^{\prime} \Omega, L=1.5 \mathrm{H}

and

\mathrm{C}=35

\mu \mathrm{F}

is connected to a variable-frequency

200 \mathrm{~V}

ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power in

\mathrm{Kw}

transferred to the circuit in one complete cycle?

View answer and explanation

Correct option: A

2

Answer A (2) is correct because at resonance, the impedance is equal to the resistance, and the average power is calculated using

P = V^2/R

.

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the given information:\nR = 20 $\Omega$ \nL = 1.5 H\nC = 35 $\mu$ F\nV = 200 V

Step 2: At resonance, the impedance of the circuit is equal to the resistance (R).

Step 3: Calculate the average power transferred to the circuit:
$P = \frac{{V^2}}{{R}}$

Step 4: Substitute the values:\nP = $\frac{{(200)^2}}{{20}} = \frac{{40000}}{{20}} = 2000 W = 2 kW$

Step 5: Final Answer:\nP = 2 kW

Why other options are wrong:
- Option B (0): This is incorrect because at resonance, the current is maximum, and power is dissipated in the resistor.
- Option C (0): This is incorrect for the same reason as option B.
- Option D (0): This is incorrect for the same reason as option B.

Key Concepts:
- Resonance in LCR circuits
- Average power in AC circuits: $P = V_{rms}I_{rms} \cos \phi$
- At resonance, $\cos \phi = 1$

Memory Aids:
- At resonance, impedance is minimum and equal to resistance.

Quick Tips:
- Remember that at resonance, the inductive and capacitive reactances cancel each other out.

37 Easy

37 of 39Question permalink

Understanding +4 | -1

In an AC series circuit, the instantaneous current is maximum when the instantaneous voltage is maximum. The circuit element connected to the source will be

View answer and explanation

Correct option: C

Pure resistor

Option C is correct because in a pure resistor circuit, the instantaneous current and voltage are in phase, meaning they reach their maximum values at the same time.

Solution approach

Correct Answer + Core Reason: Pure resistor. In a resistor, voltage and current are in phase.

Step-by-Step Solution:

Step 1: Recall the phase relationship between voltage and current in different circuit elements.

Step 2: In a pure resistor, the voltage and current are in phase. This means that they reach their maximum and minimum values at the same time.

Step 3: In a pure inductor, the current lags behind the voltage by 90 degrees ( $\pi/2$ ).

Step 4: In a pure capacitor, the current leads the voltage by 90 degrees ( $\pi/2$ ).

Step 5: Since the instantaneous current is maximum when the instantaneous voltage is maximum, the circuit element must be a pure resistor.

Why other options are wrong:
- Option A: Pure inductor: In an inductor, the current lags behind the voltage, so they are not maximum at the same time.
- Option B: Pure capacitor: In a capacitor, the current leads the voltage, so they are not maximum at the same time.
- Option D: Combination of capacitor and an inductor: The phase relationship in a combination of capacitor and inductor depends on the values of L and C and the frequency. It's not guaranteed that voltage and current will be in phase.

Key Concepts:
- Phase relationship between voltage and current in AC circuits
- Resistor
- Inductor
- Capacitor

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 relationships in resistors, inductors, and capacitors.

38 Easy

38 of 39Question permalink

Application +4 | -1

If an 8 Ω resistance and 6 Ω reactance are present in an ac series circuit then the impedance of the circuit will be

View answer and explanation

Correct option: C

10 ohm

Option C (10 ohm) is correct because impedance is the vector sum of resistance and reactance in an AC circuit.

Solution approach

Correct Answer + Core Reason: 10 ohm. Impedance is the vector sum of resistance and reactance.

Step-by-Step Solution:

Step 1: Identify the given values: Resistance $R = 8 \Omega$ , Reactance $X = 6 \Omega$ .

Step 2: Calculate the impedance using the formula: $Z = \sqrt{{R^2 + X^2}}$

Step 3: Substitute the values: $Z = \sqrt{{(8 \Omega)^2 + (6 \Omega)^2}}$

Step 4: Calculate the impedance: $Z = \sqrt{{64 + 36}} = \sqrt{{100}} = 10 \Omega$

Step 5: Final Answer: 10 ohm

Why other options are wrong:
- Option A (20 ohm): This is incorrect as it is obtained by simply adding the resistance and reactance, which is not the correct way to calculate the impedance.
- Option B (5 ohm): This is incorrect and doesn't correspond to any reasonable calculation using the given values.
- **Option D ( $14\sqrt 2 \;ohm$ ): This is incorrect and doesn't correspond to any reasonable calculation using the given values.

Key Concepts:
- AC circuit analysis
- Impedance
- Resistance
- Reactance

Memory Aids:
- Impedance is like the "total resistance" in an AC circuit.

Quick Tips:**
- Remember that impedance is the vector sum of resistance and reactance, not the simple arithmetic sum.

39 Easy

39 of 39Question permalink

Application +4 | -1

An alternating current of frequency ‘f’ is flowing in a circuit containing a resistance R and a choke L in series. The impedance of this circuit is

View answer and explanation

Correct option: B

\sqrt {{R^2} + 4{\pi ^2}{f^2}{L^2}}

Option B (

\sqrt {{R^2} + 4{\pi ^2}{f^2}{L^2}}

) is correct because the impedance of a series RL circuit is given by

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

, where

X_L = 2\pi fL

.

Solution approach

Correct Answer + Core Reason: $\sqrt {{R^2} + 4{\pi ^2}{f^2}{L^2}}$ . The impedance of a series RL circuit is $Z = \sqrt{R^2 + X_L^2}$ , where $X_L = 2\pi fL$ .

Step-by-Step Solution:

Step 1: Recall the formula for impedance in a series RL circuit.

Step 2: The impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$ , where $X_L$ is the inductive reactance.

Step 3: The inductive reactance is given by $X_L = 2\pi fL$ .

Step 4: Substitute $X_L$ into the impedance formula: $Z = \sqrt{R^2 + (2\pi fL)^2} = \sqrt{R^2 + 4\pi^2 f^2 L^2}$ .

Why other options are wrong:
- Option A: $R + 2\pi fL$ - Incorrect, this is a simple addition and doesn't account for the phase relationship.
- Option C: $\sqrt {{R^2} + {L^2}}$ - Incorrect, this doesn't include the frequency or the $2\pi$ factor.
- Option D: $\sqrt {{R^2} + 2\pi fL}$ - Incorrect, the inductive reactance is squared in the impedance formula.

Key Concepts:
- Impedance in AC circuits
- Inductive reactance

Memory Aids:
- $Z = \sqrt{R^2 + X_L^2}$ and $X_L = 2\pi fL$

Quick Tips:
- Remember the impedance formula for series RL circuits.

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