Alternating Current — Page 4 | JEE Physics

Q31

A series LCR circuit has

\mathrm{L}=0.01\, \mathrm{H}, \mathrm{R}=10\, \Omega

and

\mathrm{C}=1 \mu \mathrm{F}

and it is connected to ac voltage of amplitude

\left(\mathrm{V}_{\mathrm{m}}\right) 50 \mathrm{~V}

. At frequency

60 \%

lower than resonant frequency, the amplitude of current will be approximately :

A 466 mA

B 312 mA

C 238 mA

D 196 mA

Correct Answer

Option C

Solution

\omega = 0.4{\omega _0}

...... (i)

\Rightarrow I = {V \over Z} = {{50} \over {\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}} }}

..... (ii)

\Rightarrow I = 238

mA

Q32

If

\mathrm{R}, \mathrm{X}_{\mathrm{L}}

, and

\mathrm{X}_{\mathrm{C}}

represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless :

A

\dfrac{R}{X_{L} X_{C}}

B

R X_{L} X_{C}

C

\dfrac{R}{\sqrt{X_{L} X_{C}}}

D

R \dfrac{X_{L}}{X_{C}}

Correct Answer

Option C

Solution

$R=$ Resistance

\begin{aligned} & {\left[X_{L}\right]=[R]} \\\\ & {\left[X_{C}\right]=[R]} \end{aligned}

So, $\dfrac{R}{\sqrt{X_{L} X_{C}}}$ is dimensionless.

Q33

What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?

A Both, inducting reactance and current will be doubled.

B Inductive reactance will be doubled and current will be halved.

C Both, inductive reactance and current will be halved.

D Inductive reactance will be halved and current will be doubled.

Correct Answer

Option D

Solution

{X_L} = \omega L

X{'_L} = \left( {{{{X_L}} \over 2}} \right)

$\because$

I = {V \over {{X_L}}}

&

I' = {{2V} \over {{X_L}}} = 2I

Q34

In a transformer, number of turns in the primary coil are

140

and that in the secondary coil are

280.

If current in primary coil is

4A,

then that in the secondary coil is

A

4A

B

2A

C

6A

D

10A

Correct Answer

Option B

Solution

{N_p} = 140,\,\,{N_s} = 280,\,\,{I_p} = 4A,\,\,{I_s} = ?

For a transformer

{{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}

\Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A

Q35

The power factor of

AC

circuit having resistance

(R)

and inductance

(L)

connected in series and an angular velocity

\omega

is

A

R/\omega L

B

R/{\left( {{R^2} + {\omega ^2}{L^2}} \right)^{1/2}}

C

\omega L/R

D

R/{\left( {{R^2} - {\omega ^2}{L^2}} \right)^{1/2}}

Correct Answer

Option B

Solution

The impedance triangle for resistance

\left( R \right)

and inductor

(L)

connected in series is shown in the figure. Power factor

\cos \phi = {R \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}

Q36

In an oscillating

LC

circuit the maximum charge on the capacitor is

Q

. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

A

{Q \over 2}

B

{Q \over {\sqrt 3 }}

C

{Q \over {\sqrt 2 }}

D

Q

Correct Answer

Option C

Solution

When the capacitor is completely charged, the total energy in the

L.C

circuit is with the capacitor and that energy is

E = {1 \over 2}{{{Q^2}} \over C}

When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get

{E \over 2} = {1 \over 2}{{Q{'^2}} \over C}

where

Q'

is the charge on plate of the capacitor $\therefore$

{1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}

\Rightarrow Q' = {Q \over {\sqrt 2 }}

Q37

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 form the options given below:

A Both Statement I and Statement II are false

B Statement I is true but statement II is false

C Statement I is false but statement II is true

D Both Statement I and Statement II are true

Correct Answer

Option A

Solution

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

This statement is incorrect.

Electrical resonance occurs in an AC circuit when the capacitive reactance and inductive reactance are equal, causing the impedance of the circuit to be minimum.

This typically happens in a series RLC circuit or a parallel RLC circuit.

If the circuit contains only a capacitor or an inductor, it cannot undergo electrical resonance as there is no counterpart reactance to balance the impedance.

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

This statement is also incorrect.

An AC circuit containing a pure capacitor or a pure inductor will have a power factor of 0, not a non-zero power factor.

The power factor of a capacitor is -1, and the power factor of an inductor is +1, but when only considering the reactive components, the power factor is 0.

In such a circuit, no real power is consumed, and the circuit only has reactive power.

The energy is alternately stored and released by the capacitor and inductor, but no energy is dissipated as heat or used to perform work.

As both statements are incorrect, the correct answer would be an option that states both statements are false.

Q38

Given below are two statements: Statement I : When the frequency of an a.c source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreases. Statement II : In a series LCR circuit, the value of power factor at resonance is one. In the light of given statements, choose the most appropriate answer from the options given below.

A Both Statement I and Statement II are False.

B Statement I is incorrect but Statement II is true.

C Both Statement I and Statement II are true.

D Statement I is correct but Statement II is false.

Correct Answer

Option C

Solution

Statement I is true because in a series LCR circuit, the current first increases as the frequency increases, reaching a maximum value when the circuit is at resonance.

At resonance, the inductive reactance (XL) and capacitive reactance (XC) cancel each other out, resulting in the lowest impedance (Z) and the highest current.

As the frequency continues to increase beyond resonance, the current in the circuit decreases.

Statement II is also true because, at resonance in a series LCR circuit, the inductive reactance (XL) and capacitive reactance (XC) are equal and cancel each other out.

This results in the impedance (Z) being purely resistive.

The power factor at resonance is given by the cosine of the phase angle (θ), and since the phase angle is 0° at resonance, the power factor is 1.

Thus, both statements are true, and the correct answer is Both Statement I and Statement II are true.

Q39

Given below are two statements: Statement I : Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs Statement II : Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage. In the light of the above statements, choose the correct answer from the options given below:

A Statement I is false but Statement II is true

B Both Statement I and Statement II are false

C Statement I is true but Statement II is false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

In a series LCR circuit connected to an AC source, resonance occurs at a particular frequency at which the inductive reactance is equal to the capacitive reactance, resulting in the minimum impedance of the circuit.

At this frequency, the circuit draws maximum current from the source, and thus, the maximum power is dissipated in the circuit.

Therefore, Statement I is true.

In a circuit containing only a resistor, the power dissipated is given by P = VI = I $^2$ R, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor.

The voltage and current are in phase in a purely resistive circuit, which means that the power is maximized.

Therefore, Statement II is also true.

Q40

A capacitor of capacitance

150.0 ~\mu \mathrm{F}

is connected to an alternating source of emf given by

\mathrm{E}=36 \sin (120 \pi \mathrm{t}) \mathrm{V}

. The maximum value of current in the circuit is approximately equal to :

A

\dfrac{1}{\sqrt{2}} A

B

2 \sqrt{2} A

C

\sqrt{2} A

D

2 A

Correct Answer

Option D

Solution

For a capacitor connected to an AC source, the maximum current

I_\text{max}

can be calculated using the formula:

I_\text{max} = E_\text{max} \cdot \omega C

where

E_\text{max}

is the maximum voltage, $\omega$ is the angular frequency, and

C

is the capacitance. Given the emf equation:

E = 36 \sin(120\pi t) \, \text{V}

, we can determine that

E_\text{max} = 36\, \text{V}

and

\omega = 120\pi \, \text{rad/s}

. The capacitance is given as

150.0\, \mu\text{F} = 150.0 \times 10^{-6}\, \text{F}

. Now, we can calculate the maximum current:

I_\text{max} = 36 \cdot (120\pi) \cdot (150.0 \times 10^{-6})

I_\text{max} \approx 2\, \text{A}

Thus, the correct answer is

2\, \text{A}

.