ALTERNATING CURRENT MCQs for NEET UG PHYSICS — Easy

1 Easy

Understanding +4 | -1

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

View answer and explanation

Correct option: B

Current is ahead of e.m.f. by

\pi /2

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

\pi /2

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

2 Easy

2 of 39Question permalink

Application +4 | -1

A generator produces a voltage that is given by

V = 240\sin 120\;t

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

View answer and explanation

Correct option: C

19 Hz and 170 V

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

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

Hz and the RMS voltage is

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

V.

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

Key Concepts:
- Angular frequency
- Frequency
- RMS voltage

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

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

3 Easy

3 of 39Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: A

Infinity

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

Solution approach

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

Step-by-Step Solution:

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

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

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

Step 4: Final Answer: Infinity

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

Key Concepts:
- Inductance
- DC resistance
- Ideal choke

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

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

4 Easy

4 of 39Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: A

Eddy currents

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

Key Concepts:
- Eddy currents
- Lamination
- Transformer losses

Memory Aids:
- Lamination = Less eddy currents

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

5 Easy

5 of 39Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: B

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

Option B (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

6 Easy

6 of 39Question permalink

Remembering +4 | -1

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

\left( {{X_C}} \right)

and inductive reactance

\left( {{X_L}} \right)

is

View answer and explanation

Correct option: D

{X_C} - {X_L} = 0

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

{X_C} - {X_L} = 0

.

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

Memory Aids:
- Resonance: $X_L = X_C$

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

7 Easy

7 of 39Question permalink

Understanding +4 | -1

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

\mathrm{L}

and resistance

\mathrm{R}

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

View answer and explanation

Correct option: B

double

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

8 Easy

8 of 39Question permalink

Analyzing +4 | -1

In series

L C R

circuit voltage drop across resistance is

8 {~V}

, across inductor is

6 {~V}

and across capacitor is

12 {~V}

. Then

View answer and explanation

Correct option: D

None of these

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

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

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

9 Easy

9 of 39Question permalink

Application +4 | -1

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

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

The current will be maximum for the first time at

View answer and explanation

Correct option: D

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

Option D (

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

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

\pi/2

, so

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

, which gives

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

.

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

Key Concepts:
- Maximum value of sine function

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

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

10 Easy

10 of 39Question permalink

Application +4 | -1

An AC voltage source has an output of

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

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

View answer and explanation

Correct option: A

1.41A

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

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

and the RMS current is

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

A.

Solution approach

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

Step-by-Step Solution:

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

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

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

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

Key Concepts:
- RMS voltage
- Ohm's law

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

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

11 Easy

11 of 39Question permalink

Application +4 | -1

Write the equation of a sinusoidal voltage of rms valus

220 \mathrm{~V}

and frequency

50 \mathrm{~Hz}

View answer and explanation

Correct option: B

220 \sqrt{2} \sin (100 \pi t)

Option B is correct. The equation for a sinusoidal voltage is

V = V_0 \sin(2 \pi f t)

, where

V_0 = V_{rms} \sqrt{2}

and

f

is the frequency.

Solution approach

Correct Answer + Core Reason: $220 \sqrt{2} \sin (100 \pi t)$ . This is the correct sinusoidal voltage equation with given RMS value and frequency.

Step-by-Step Solution:

Step 1: Calculate the peak voltage: $V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2}$ V.

Step 2: Calculate the angular frequency: $\omega = 2\pi f = 2\pi (50) = 100\pi$ rad/s.

Step 3: Write the equation for the sinusoidal voltage: $V = V_0 \sin(\omega t) = 220 \sqrt{2} \sin(100\pi t)$ .

Why other options are wrong:
- Option A: The peak voltage and angular frequency are incorrect.
- Option C: The peak voltage is incorrect. It should be $220\sqrt{2}$ .
- Option D: The peak voltage and angular frequency are incorrect.

Key Concepts:
- Sinusoidal voltage equation
- RMS voltage
- Angular frequency

Memory Aids:
- Remember the relationship: $V_0 = V_{rms} \sqrt{2}$ .

Quick Tips:
- Always convert RMS voltage to peak voltage when writing the sinusoidal equation.

12 Easy

12 of 39Question permalink

Understanding +4 | -1

At high frequency, the capacitor offer

View answer and explanation

Correct option: B

Less reactance

Option B (Less reactance) is correct because the reactance of a capacitor is inversely proportional to the frequency:

X_C = \frac{1}{2 \pi f C}

. As frequency increases, reactance decreases.

Solution approach

Correct Answer + Core Reason:

13 Easy

13 of 39Question permalink

Remembering +4 | -1

Hot wire ammeters can be used for measuring:

View answer and explanation

Correct option: A

Both AC and DC

Option A is correct. Hot wire ammeters measure the RMS value of current, which is effective for both AC and DC.

Solution approach

Correct Answer + Core Reason: Both AC and DC. Hot wire ammeters measure RMS current, applicable to both.

Step-by-Step Solution:

Step 1: Understand the principle of a hot wire ammeter. It works based on the heating effect of current.

Step 2: The heating effect is proportional to the square of the current ( $I^2 R$ ), which is independent of the current's direction.

Step 3: Therefore, hot wire ammeters can measure both AC and DC currents.

Why other options are wrong:
- Option B: Hot wire ammeters can measure DC as well.
- Option C: Hot wire ammeters can measure AC as well.
- Option D: Hot wire ammeters can measure both AC and DC.

Key Concepts:
- Hot wire ammeter principle
- Heating effect of current
- RMS value of current

Memory Aids:
- Hot wire ammeters respond to the heating effect, which is direction-independent.

Quick Tips:
- Remember that hot wire ammeters measure the RMS value of current.

14 Easy

14 of 39Question permalink

Application +4 | -1

A series LCR circuit is subjected to an AC signal of

200 \mathrm{~V}, 50 \mathrm{~Hz}.

If the voltage across the inductor

(\mathrm{L}=10 \mathrm{~mH})

is

31.4 \mathrm{~V},

then the current in this circuit is

View answer and explanation

Correct option: C

10 \mathrm{~A}

Option C (10 A) is correct because the current is calculated using the voltage across the inductor and its inductive reactance.

Solution approach

Correct Answer + Core Reason: 10 A. The current is calculated using the voltage across the inductor and its inductive reactance.

Step-by-Step Solution:

Step 1: Identify the given information:\nVoltage across the inductor, $V_L = 31.4 \text{{ V}}$ \nInductance, $L = 10 \text{{ mH}} = 10 \times 10^{-3} \text{{ H}}$ \nFrequency, $f = 50 \text{{ Hz}}$

Step 2: Calculate the inductive reactance, $X_L$ :
$X_L = 2 \pi f L$

Step 3: Substitute the values:
$X_L = 2 \pi (50 \text{{ Hz}}) (10 \times 10^{-3} \text{{ H}})$
$X_L = 2 \times 3.14 \times 50 \times 0.01 \approx 3.14 \Omega$

Step 4: Calculate the current, $I$ :
$I = \frac{{V_L}}{{X_L}}$

Step 5: Substitute the values:
$I = \frac{{31.4 \text{{ V}}}}{{3.14 \Omega}}$
$I = 10 \text{{ A}}$

Why other options are wrong:
- Option A (68 A): This is incorrect. It likely arises from using the total voltage instead of the voltage across the inductor or miscalculating the reactance.
- Option B (63 A): This is incorrect. It likely arises from using the total voltage instead of the voltage across the inductor or miscalculating the reactance.
- Option D (10 mA): This is incorrect. It likely arises from a miscalculation or misunderstanding of the units (e.g., using mH directly without converting to H).

Key Concepts:
- Inductive Reactance: $X_L = 2 \pi f L$
- Ohm's Law for AC circuits: $V = I X$

Memory Aids:
- Remember the formula for inductive reactance: $X_L = 2 \pi f L$

Quick Tips:
- Always convert all units to SI units before plugging them into formulas.
- Remember that the voltage across the inductor is used to calculate the current through it.

15 Easy

15 of 39Question permalink

Remembering +4 | -1

A capacitor becomes a perfect insulator for

View answer and explanation

Correct option: B

direct current

Option B is correct. A capacitor blocks direct current (DC) because, after being fully charged, no further current can flow.

Solution approach

Correct Answer + Core Reason: direct current. A capacitor blocks DC after it's fully charged.

Step-by-Step Solution:

Step 1: Understand the behavior of a capacitor in a DC circuit. Initially, a capacitor allows current to flow as it charges.

Step 2: Once the capacitor is fully charged, the voltage across it equals the source voltage, and the current stops flowing.

Step 3: Therefore, a capacitor acts as an open circuit (perfect insulator) for DC after being fully charged.

Why other options are wrong:
- Option A: Capacitors allow alternating current (AC) to pass through due to continuous charging and discharging.
- Option C: Capacitors block DC but allow AC.
- Option D: The capacitor acts as an insulator for DC.

Key Concepts:
- Capacitor behavior in DC circuits
- Capacitor charging and discharging

Memory Aids:
- Capacitors "block DC, pass AC".

Quick Tips:
- Remember that capacitors store charge and their behavior differs in AC and DC circuits.

16 Easy

16 of 39Question permalink

Application +4 | -1

A transformer is used to light 140 W, 24 V lamp from 240 V AC mains. The current in the mains is 0.7 A. The efficiency of transformer is nearest to

View answer and explanation

Correct option: B

0.8

Option B (0.8) is correct because the efficiency is the ratio of output power to input power, which is approximately 0.8.

Solution approach

Correct Answer + Core Reason: 0.8. Efficiency is the ratio of output power to input power.

Step-by-Step Solution:

Step 1: Identify the given information.\nOutput power (lamp) = 140 W\nInput voltage = 240 V\nInput current = 0.7 A

Step 2: Calculate the input power.\nInput Power = Voltage × Current
$P_{in} = V_{in} \times I_{in}$

Step 3: Substitute the given values.
$P_{in} = 240 \times 0.7 = 168 \text{{ W}}$

Step 4: Calculate the efficiency.\nEfficiency = (Output Power / Input Power) × 100%
$\eta = \frac{{P_{out}}}{{P_{in}}} = \frac{{140}}{{168}}$

Step 5: Calculate the final efficiency.
$\eta = 0.833 \approx 0.8$

Step 6: Express as a decimal.\nEfficiency = 0.8

Why other options are wrong:
- Option A (0.9): This is incorrect as it overestimates the efficiency. The calculation shows the efficiency is closer to 0.8.
- Option C (0.7): This is incorrect as it underestimates the efficiency. The calculation shows the efficiency is closer to 0.8.
- Option D (0.6): This is incorrect as it significantly underestimates the efficiency. The calculation shows the efficiency is closer to 0.8.

Key Concepts:
- Transformer efficiency
- Power calculation (P = VI)

Memory Aids:
- Efficiency = Output / Input

Quick Tips:
- Always ensure the units are consistent before calculating efficiency.
- Efficiency is always less than or equal to 1.

17 Easy

17 of 39Question permalink

Understanding +4 | -1

In an ac circuit with voltage V and current I, the power dissipated is

View answer and explanation

Correct option: D

Depends on the phases between V and I

Option D (Depends on the phases between V and I) is correct because the power dissipated in an AC circuit depends on the phase difference between voltage and current, as given by

P = VI\cos\phi

.

Solution approach

Correct Answer + Core Reason: Depends on the phases between V and I. The power dissipated depends on the phase difference between voltage and current.

Step-by-Step Solution:

Step 1: Recall the formula for power in an AC circuit.

Step 2: The power dissipated in an AC circuit is given by $P = VI\cos\phi$ , where $\phi$ is the phase difference between voltage and current.

Step 3: Therefore, the power depends on the phase difference.

Why other options are wrong:
- Option A: $VI$ - Incorrect because it doesn't account for the phase difference.
- Option B: $\frac{{1}}{{2}}VI$ - Incorrect because it only applies in specific cases and doesn't account for the general phase difference.
- Option C: $\frac{{1}}{{\sqrt 2 }}VI$ - Incorrect because it only applies in specific cases and doesn't account for the general phase difference.

Key Concepts:
- AC power
- Power factor

Memory Aids:
- Power factor is $\cos\phi$ .

Quick Tips:
- Remember the power formula $P = VI\cos\phi$ .

18 Easy

18 of 39Question permalink

Remembering +4 | -1

Transformer works on the principle of

View answer and explanation

Correct option: A

Mutual induction between two coils

Option A (Mutual induction between two coils) is correct because a transformer's operation is based on the principle of mutual induction between its primary and secondary coils.

Solution approach

Correct Answer + Core Reason: Mutual induction between two coils. A transformer transfers energy between circuits through mutual induction.

Step-by-Step Solution:

Step 1: A transformer consists of two or more coils that are electrically isolated but magnetically linked.

Step 2: When an alternating current flows through the primary coil, it creates a changing magnetic field.

Step 3: This changing magnetic field induces a voltage in the secondary coil due to mutual induction.

Step 4: The voltage induced in the secondary coil is proportional to the rate of change of magnetic flux and the number of turns in the secondary coil.

Why other options are wrong:
- Option B: Self induction refers to the induction within a single coil, not between two coils as in a transformer.
- Option C: While inductors store energy, the primary operating principle of a transformer is mutual induction, not energy storage.
- Option D: The correct answer is one of the options provided, so this is incorrect.

Key Concepts:
- Mutual Induction
- Faraday's Law of Electromagnetic Induction
- Transformer Operation

Memory Aids:
- Transformers use \"mutual\" induction because they have \"multiple\" coils.

Quick Tips:
- Remember that transformers require AC current to operate due to the changing magnetic field needed for induction.

19 Easy

19 of 39Question permalink

Application +4 | -1

The current flowing through an ac circuit is given by

I = 5\sin (120\pi t)A.

How long will the current take to reach the peak value starting from zero?

View answer and explanation

Correct option: D

\frac{1}{{240}}\,s

Option D is correct. The time to reach the peak value is one-quarter of the period, calculated as

\frac{1}{4f} = \frac{1}{240}

s.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{240}}\,s$ . Time to peak is one-quarter of the period.

Step-by-Step Solution:

Step 1: Identify the angular frequency: $\omega = 120\pi$ rad/s.

Step 2: Calculate the frequency: $f = \frac{\omega }{{2\pi }} = \frac{{120\pi }}{{2\pi }} = 60$ Hz.

Step 3: Calculate the period: $T = \frac{1}{f} = \frac{1}{{60}}$ s.

Step 4: The time to reach the peak value is one-quarter of the period: $t = \frac{T}{4} = \frac{1}{{4 \times 60}} = \frac{1}{{240}}$ s.

Why other options are wrong:
- Option A: This is the period, not the time to reach the peak.
- Option B: This is incorrect. The time is much smaller.
- Option C: This is half the period, not the time to reach the peak.

Key Concepts:
- Angular frequency
- Frequency and period relationship

Memory Aids:
- Time to peak = T/4.

Quick Tips:
- Remember the relationship between frequency, period, and angular frequency.

20 Easy

20 of 39Question permalink

Application +4 | -1

An ac voltage is applied to a resistance R and inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is

View answer and explanation

Correct option: C

π/4

Option C (π/4) is correct because the phase difference is given by the arctangent of the ratio of inductive reactance to resistance, which are equal in this case.

Solution approach

Correct Answer + Core Reason: π/4. The phase difference is the arctangent of the ratio of reactance to resistance.

Step-by-Step Solution:

Step 1: Identify the given values: Resistance $R = 3 \Omega$ , Inductive Reactance $X_L = 3 \Omega$ .

Step 2: Calculate the phase difference using the formula: $\phi = \arctan{\frac{{X_L}}{{R}}}$

Step 3: Substitute the values: $\phi = \arctan{\frac{{3 \Omega}}{{3 \Omega}}}$

Step 4: Calculate the phase difference: $\phi = \arctan{1} = \frac{{\pi}}{{4}}$

Step 5: Final Answer: π/4

Why other options are wrong:
- Option A (Zero): This is incorrect because it would only be true if the circuit was purely resistive.
- Option B (π/6): This is incorrect and doesn't correspond to the correct phase difference when resistance and inductive reactance are equal.
- Option D (π/2): This is incorrect because it would only be true if the circuit was purely inductive.

Key Concepts:
- AC circuit analysis
- Phase difference
- Inductive reactance
- Resistance

Memory Aids:
- $\arctan{1} = \frac{{\pi}}{{4}}$

Quick Tips:
- Remember the formula for phase difference in an RL circuit: $\phi = \arctan{\frac{{X_L}}{{R}}}$

21 Easy

21 of 39Question permalink

Application +4 | -1

A series LCR circuit with inductance 10 H, capacitance

10\,\mu F

resistance

50\,\,\Omega

is connected to an ac source of voltage, V = 200 sin(100 t) volt. If the resonant frequency of the LCR circuit is

{v_ \circ },

and the frequency of the ac source is

v

, then :

View answer and explanation

Correct option: C

{v_ \circ } = v = \frac{{50}}{\pi }Hz

Option C (

{v_ \circ } = v = \frac{{50}}{\pi }Hz

) is correct because both the resonant frequency and the frequency of the AC source are calculated to be

\frac{{50}}{\pi }Hz

.

Solution approach

Correct Answer + Core Reason: ${v_ \circ } = v = \frac{{50}}{\pi }Hz$ . The resonant frequency and source frequency are equal and calculated using the given values.

Step-by-Step Solution:

Step 1: Identify the given information:\nInductance, $L = 10 \text{ H}$ \nCapacitance, $C = 10 \mu \text{F} = 10 \times 10^{-6} \text{ F}$ \nResistance, $R = 50 \Omega$ \nVoltage, $V = 200 \sin(100t)$

Step 2: Determine the angular frequency of the AC source:\nFrom the voltage equation, $V = 200 \sin(\omega t)$ , we have $\omega = 100 \text{ rad/s}$ .\nThe frequency of the AC source is $v = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} \text{ Hz}$ .

Step 3: Calculate the resonant frequency of the LCR circuit:\nThe resonant angular frequency is given by $\omega_0 = \frac{1}{\sqrt{LC}}$ .\nSo, $\omega_0 = \frac{1}{\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-4}}} = \frac{1}{10^{-2}} = 100 \text{ rad/s}$ .\nThe resonant frequency is ${v_ \circ } = \frac{\omega_0}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} \text{ Hz}$ .

Step 4: Compare the source frequency and resonant frequency:\nWe have $v = \frac{50}{\pi} \text{ Hz}$ and ${v_ \circ } = \frac{50}{\pi} \text{ Hz}$ . Therefore, ${v_ \circ } = v = \frac{50}{\pi} \text{ Hz}$ .

Why other options are wrong:
- Option A: This option incorrectly states the frequency of the AC source as 100 Hz, while it is $\frac{50}{\pi} \text{ Hz}$ .
- Option B: This option incorrectly states both frequencies as 50 Hz, which is not the calculated value.
- Option D: This option mixes up the values of the source frequency and resonant frequency.

Key Concepts:
- Resonant frequency in LCR circuits: $\omega_0 = \frac{1}{\sqrt{LC}}$
- Relationship between angular frequency and frequency: $\omega = 2\pi v$

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

Quick Tips:
- Always convert units to SI units before calculations.
- Carefully identify the angular frequency from the voltage equation.

22 Easy

22 of 39Question permalink

Application +4 | -1

A step-up transformer has transformation ratio of

\mathrm{3 : 2}

. What is the voltage in secondary if voltage in primary is

30\,\,V

View answer and explanation

Correct option: A

45\,\,V

Option A (

45\,V

) is correct because the voltage in the secondary coil is calculated using the transformation ratio of the transformer.

Solution approach

Correct Answer + Core Reason: $45\,V$ . The secondary voltage is calculated using the turns ratio.

Step-by-Step Solution:

Step 1: The transformation ratio is given as $\frac{N_1}{N_2} = \frac{3}{2}$ , where $N_1$ is the number of turns in the primary coil and $N_2$ is the number of turns in the secondary coil.

Step 2: The relationship between the voltages and the number of turns in a transformer is given by $\frac{V_1}{V_2} = \frac{N_1}{N_2}$ , where $V_1$ is the voltage in the primary coil and $V_2$ is the voltage in the secondary coil.

Step 3: We are given $V_1 = 30 \, V$ . We need to find $V_2$ .

Step 4: Substituting the given values into the equation, we get $\frac{30}{V_2} = \frac{3}{2}$ .

Step 5: Solving for $V_2$ , we get $V_2 = \frac{30 \times 2}{3} = 20 \times 1.5 = 45 \, V$ .

Why other options are wrong:
- Option B: This is incorrect because it divides the primary voltage by the transformation ratio instead of multiplying.
- Option C: This is incorrect because it multiplies the primary voltage by the transformation ratio squared.
- Option D: This is incorrect because it performs an incorrect calculation with the given values.

Key Concepts:
- Transformer
- Transformation ratio
- Voltage relationship in a transformer

Memory Aids:
- Remember the formula: $\frac{V_1}{V_2} = \frac{N_1}{N_2}$

Quick Tips:
- Identify the primary and secondary sides correctly.
- Use the correct transformation ratio.

23 Easy

23 of 39Question permalink

Understanding +4 | -1

The current in series LCR circuit will be maximum when ω is

View answer and explanation

Correct option: D

\sqrt {1/LC}

Option D (

\sqrt {1/LC}

) is correct because the current in a series LCR circuit is maximum at resonance, which occurs when the angular frequency equals

\sqrt {1/LC}

.

Solution approach

Correct Answer + Core Reason: $\sqrt {1/LC}$ . The current is maximum at resonance, where the impedance is minimized.

Step-by-Step Solution:

Step 1: Understand the concept of resonance in an LCR circuit.

Step 2: Recall the formula for resonant angular frequency ( $\omega$ ) in a series LCR circuit: $\omega = \frac{1}{\sqrt{LC}}$

Step 3: At resonance, the inductive reactance ( $X_L = \omega L$ ) equals the capacitive reactance ( $X_C = \frac{1}{\omega C}$ ). This minimizes the impedance of the circuit, leading to maximum current.

Step 4: Therefore, the current is maximum when $\omega = \frac{1}{\sqrt{LC}}$

Why other options are wrong:
- Option A: As large as possible: While increasing frequency can increase current up to a point, it's not the condition for *maximum* current. Beyond resonance, the impedance increases again.
- Option B: Equal to natural frequency of LCR system: This is a correct statement, but it's not in the correct mathematical form. The natural frequency *is* $\frac{1}{\sqrt{LC}}$ , but the option doesn't provide this formula.
- Option C: $\sqrt {LC}$ : This is the inverse of the correct resonant frequency. Using this value would lead to a very low current due to high impedance.

Key Concepts:
- Resonance in LCR circuits
- Impedance
- Inductive Reactance ( $X_L$ )
- Capacitive Reactance ( $X_C$ )
- Resonant Frequency: $\omega = \frac{1}{\sqrt{LC}}$

Memory Aids:
- LC circuits resonate at a specific frequency.
- Resonance = Maximum Current

Quick Tips:
- Remember the formula for resonant frequency. It's fundamental to LCR circuit analysis.
- Understand the relationship between impedance, reactance, and frequency in LCR circuits.

24 Easy

24 of 39Question permalink

Application +4 | -1

Power dissipated in an LCR series circuit connected to an a.c. source of

emf\;E

is

View answer and explanation

Correct option: A

{E^2}R/\left[ {{R^2} + {{\left( {L\omega - \frac{1}{{C\omega }}} \right)}^2}} \right]

Option A (

{E^2}R/ {{R^2} + {{\left( {L\omega - \frac{{1}}{{C\omega }}} \right)}^2}} )

is correct because it represents the power dissipated in an LCR series circuit, derived from the formula

P = \frac{{E^2}}{Z} \cos \phi

, where Z is the impedance and

\cos \phi

is the power factor.

Solution approach

Correct Answer + Core Reason: ${E^2}R/ {{R^2} + {{\left( {L\omega - \frac{{1}}{{C\omega }}} \right)}^2}}$

25 Easy

25 of 39Question permalink

Application +4 | -1

In a L-C-R series circuit, the potential difference between the terminals of the inductance is 60 V, between the terminals of the capacitor is 30 V and that across the resistance is 40 V. Then, supply voltage will be equal to

View answer and explanation

Correct option: A

50 V

Option A (50 V) is correct because the supply voltage is the vector sum of the voltage across the resistor and the net voltage across the inductor and capacitor.

Solution approach

Correct Answer + Core Reason: 50 V. The supply voltage is the vector sum of the voltages.

Step-by-Step Solution:

Step 1: Identify the given voltages: $V_L = 60 \text{{ V}}$ , $V_C = 30 \text{{ V}}$ , $V_R = 40 \text{{ V}}$ .

Step 2: Calculate the net reactive voltage: $V_{LC} = V_L - V_C = 60 \text{{ V}} - 30 \text{{ V}} = 30 \text{{ V}}$ .

Step 3: Calculate the supply voltage using the Pythagorean theorem: $V = \sqrt{{V_R^2 + V_{LC}^2}}$

Step 4: Substitute the values: $V = \sqrt{{(40 \text{{ V}})^2 + (30 \text{{ V}})^2}}$

Step 5: Calculate the supply voltage: $V = \sqrt{{1600 + 900}} = \sqrt{{2500}} = 50 \text{{ V}}$

Step 6: Final Answer: 50 V

Why other options are wrong:
- Option B (70 V): This is incorrect as it might be obtained by simply adding the voltage across the resistor to the voltage difference across the inductor and capacitor, without considering the phase difference.
- Option C (130 V): This is incorrect as it is obtained by directly adding all the voltages, which is not the correct way to calculate the total voltage in an AC circuit.
- Option D (10 V): This is incorrect and doesn't correspond to any reasonable calculation using the given values.

Key Concepts:
- AC circuit analysis
- Phasor diagrams
- Voltage in series LCR circuit

Memory Aids:
- Remember to use the vector sum of voltages in AC circuits.

Quick Tips:
- Always subtract the capacitive reactance from the inductive reactance (or vice versa) before calculating the total impedance.

26 Easy

26 of 39Question permalink

Understanding +4 | -1

In an ideal transformer, the turns ratio is

\frac{N_P}{N_S}=\frac{1}{2}.

The ratio

V_S: V_P

is equal to (the symbols carry their usual meaning):

View answer and explanation

Correct option: A

2: 1

Option A (2: 1) is correct because the voltage ratio is the inverse of the turns ratio, so

V_S/V_P = N_S/N_P = 2/1

.

Solution approach

Correct Answer + Core Reason: $2: 1$ . The voltage ratio is directly proportional to the turns ratio.

Step-by-Step Solution:

Step 1: Recall the relationship between turns ratio and voltage ratio in an ideal transformer: $\frac{N_P}{N_S} = \frac{V_P}{V_S}$ .

Step 2: Given $\frac{N_P}{N_S} = \frac{1}{2}$ , we need to find $\frac{V_S}{V_P}$ .

Step 3: Since $\frac{V_P}{V_S} = \frac{N_P}{N_S} = \frac{1}{2}$ , then $\frac{V_S}{V_P} = \frac{2}{1}$ .

Why other options are wrong:
- Option B: This would be true if the turns ratio was 1:1.
- Option C: This is the square of the turns ratio, which relates to power in some cases, but not directly to the voltage ratio.
- Option D: This is the inverse of the correct answer.

Key Concepts:
- Transformer turns ratio
- Voltage ratio

Memory Aids:
- Voltage follows turns (directly proportional)

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

27 Easy

27 of 39Question permalink

Application +4 | -1

The current flowing in a step down transformer 220 V to 22 V having impedance 220 Ω is

View answer and explanation

Correct option: B

0.1A

Option B (0.1A) is correct because it is calculated using Ohm's law on the secondary side of the transformer.

Solution approach

Correct Answer + Core Reason: 0.1A. The current is calculated using Ohm's law on the secondary side.

Step-by-Step Solution:

Step 1: Identify the given information: Secondary voltage $V_s = 22 \text{{ V}}$ , Impedance $Z = 220 \Omega$ .

Step 2: Apply Ohm's Law: $I = \frac{{V_s}}{{Z}}$

Step 3: Substitute the values: $I = \frac{{22 \text{{ V}}}}{{220 \Omega}}$

Step 4: Calculate the current: $I = 0.1 \text{{ A}}$

Step 5: Final Answer: 0.1 A

Why other options are wrong:
- Option A (0.041666667): This value is incorrect and likely arises from an incorrect calculation or misunderstanding of the transformer's operation.
- Option C (2 mA): This value is too small and likely results from dividing the voltage by a much larger impedance value than the given one.
- Option D (0.1 mA): This value is also too small and suggests a similar error to Option C, using an incorrect impedance value.

Key Concepts:
- Ohm's Law: $V = IR$
- Transformers

Memory Aids:
- V = IR (Voltage = Current x Resistance)

Quick Tips:
- Always use the correct voltage and impedance values for the secondary side when calculating the secondary current.

28 Easy

28 of 39Question permalink

Remembering +4 | -1

The peak voltage of the ac source is equal to:

View answer and explanation

Correct option: D

\sqrt 2

times the rms value of the ac source

Option D (

\sqrt 2

times the rms value of the ac source) is correct because the peak voltage is related to the RMS voltage by the factor of

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: $\sqrt 2$ times the rms value of the ac source. The peak voltage is related to the RMS voltage by the factor of $\sqrt{2}$ .

Step-by-Step Solution:

Step 1: Recall the relationship between peak voltage ( $V_p$ ) and RMS voltage ( $V_{rms}$ ): $V_p = \sqrt{2} V_{rms}$ .

Step 2: This formula directly states that the peak voltage is $\sqrt{2}$ times the RMS voltage.

Step 3: Final Answer: $\sqrt 2$ times the rms value of the ac source

Why other options are wrong:
- Option A: $1/\sqrt 2$ times the rms value of the ac source. This is the inverse relationship; RMS voltage is $1/\sqrt 2$ times the peak voltage.
- Option B: The value of voltage supplied to the circuit. This is ambiguous and doesn't specify whether it's RMS or peak.
- Option C: The rms value of the ac source. This is the RMS voltage itself, not the peak voltage.

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.

29 Easy

29 of 39Question permalink

Application +4 | -1

In an AC circuit, V and I are given by

V = 150\sin \left( {150\mathrm{t}} \right)\mathrm{\;volt\;and\;}I = 150\sin \left( {150\mathrm{t} + \frac{\mathrm{\pi }}{3}} \right)\mathrm{amp}

. The power dissipated in the circuit is

View answer and explanation

Correct option: B

5625 W

Option B (5625 W) is correct because the power dissipated is given by

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

, where

\phi = \pi/3

, so

P = (150/\sqrt{2})(150/\sqrt{2})\cos(\pi/3) = 5625

W.

Solution approach

Correct Answer + Core Reason: 5625 W. The power dissipated is calculated using the RMS values of voltage and current and the phase difference.

Step-by-Step Solution:

Step 1: Identify the peak voltage and current: $V_0 = 150$ V and $I_0 = 150$ A.

Step 2: Calculate the RMS voltage and current: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{150}{\sqrt{2}}$ V and $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{150}{\sqrt{2}}$ A.

Step 3: Identify the phase difference $\phi = \frac{{\pi }}{{3}}$ .

Step 4: Calculate the power dissipated using the formula $P = V_{rms}I_{rms}\cos\phi = \frac{150}{\sqrt{2}} \cdot \frac{150}{\sqrt{2}} \cdot \cos \frac{{\pi }}{{3}} = \frac{{150 \cdot 150}}{2} \cdot \frac{1}{2} = \frac{{22500}}{4} = 5625$ W.

Why other options are wrong:
- Option A: Zero - Incorrect, the phase difference is not $\pi/2$ .
- Option C: 150 W - Incorrect, this doesn't account for RMS values and the phase difference.
- Option D: 106 W - Incorrect, this is not the correct calculation.

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

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

Quick Tips:
- Remember to use RMS values when calculating power in AC circuits.

30 Easy

30 of 39Question permalink

Understanding +4 | -1

A bulb is connected first with dc and then ac of same voltage it will shine brightly with

View answer and explanation

Correct option: D

Equally with both

Option D (Equally with both) is correct because the bulb's brightness depends on the RMS voltage, which is the same for both AC and DC in this scenario.

Solution approach

Correct Answer + Core Reason: Equally with both. The brightness depends on the RMS voltage, which is the same for both AC and DC.

Step-by-Step Solution:

Step 1: Understand the concept of RMS voltage.

Step 2: RMS voltage is the effective voltage that produces the same heating effect in a resistor as a DC voltage of the same magnitude.

Step 3: Since the AC and DC voltages are the same, the brightness will be the same.

Why other options are wrong:
- Option A: AC - Incorrect because the voltage is the same for both AC and DC.
- Option B: DC - Incorrect because the voltage is the same for both AC and DC.
- Option C: Brightness will be in ratio 1/1.4 - Incorrect because the voltage is the same for both AC and DC.

Key Concepts:
- RMS voltage
- Power dissipation in a resistor

Memory Aids:
- RMS voltage is the effective voltage.

Quick Tips:
- Remember that brightness is related to power, which is related to the square of the voltage.

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