Alternating Current — Page 3 | JEE Physics

Q21

Given below are two statements : Statement I : In an LCR series circuit, current is maximum at resonance. Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source. In the light of the above statements, choose the correct from the options given below :

A Statement I is true but Statement II is false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are true

D Both Statement I and Statement II are false

Correct Answer

Option C

Solution

Statement I : True Statement II : True Current in purely resistive circuit is equal to current in LCR circuit with same resistance at resonance, otherwise more.

Q22

A capacitor of capacitance

100 \mu \mathrm{F}

is charged to a potential of

12 \mathrm{~V}

and connected to a

6.4 \mathrm{~mH}

inductor to produce oscillations. The maximum current in the circuit would be :

A 2.0 A

B 3.2 A

C 1.5 A

D 1.2 A

Correct Answer

Option C

Solution

By energy conservation

\begin{aligned} & \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\text{max }}^2 \\ & \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \\ & =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12 \\ & =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A} \end{aligned}

Q23

A 0.07 H inductor and a 12

\Omega

resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take

\pi

as

{{22} \over 7}

]

A 8.8 A and

{\tan ^{ - 1}}\left( {{{11} \over 6}} \right)

B 88 A and

{\tan ^{ - 1}}\left( {{{11} \over 6}} \right)

C 0.88 A and

{\tan ^{ - 1}}\left( {{{11} \over 6}} \right)

D 8.8 A and

{\tan ^{ - 1}}\left( {{{6} \over 11}} \right)

Correct Answer

Option A

Solution

\phi = {\tan ^{ - 1}}\left( {{{{X_L}} \over R}} \right)

{X_L} = \omega L

{X_L} = 2 \times {{22} \over 7} \times 50 \times 0.07 = 22\Omega

\phi = {\tan ^{ - 1}}\left( {{{22} \over {12}}} \right)

R = 12\Omega

\phi = {\tan ^{ - 1}}\left( {{{11} \over 6}} \right)

Z = \sqrt {X_L^2 + {R^2}} = 25.059

I = {V \over Z} = {{220} \over {25.059}} = 8.77A

Q24

A 100

\Omega

resistance, a 0.1

\mu

F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.

A 0.70 H

B 70.3 mH

C 7.03

\times

10

-

5 H

D 70.3 H

Correct Answer

Option D

Solution

C = 0.1 $\mu$ F = 10 $-$ 7 F Resonant frequency = 60 Hz.

{\omega _0} = {1 \over {\sqrt {LC} }}

2\pi {f_0} = {1 \over {\sqrt {LC} }} \Rightarrow L = {1 \over {4{\pi ^2}f_0^2C}}

by putting values

L \simeq 70.3

Hz.

Q25

In an ac circuit, an inductor, a capacitor and a resistor are connected in series with XL = R = XC. Impedance of this circuit is :

A 2R2

B Zero

C R

D R

\sqrt 2

Correct Answer

Option C

Solution

Z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}} = R

$\because$ XL = XC Option (c)

Q26

If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?

A RC

B

{L \over R}

C

\sqrt{LC}

D

{L \over C}

Correct Answer

Option D

Solution

U = {1 \over 2}L{i^2} = {1 \over 2}C{V^2}

So,

\left[ {{L \over C}} \right] = {{{V^2}} \over {{i^2}}} = {R^2}

is not the dimension of time.

Q27

A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25

\mu

F and R = 100

\Omega

. The phase difference (

\Phi

) between the applied voltage and resultant current will be :

A tan

-

1(0.17)

B tan

-

1(9.46)

C tan

-

1(0.30)

D tan

-

1(13.33)

Correct Answer

Option A

Solution

{X_L} = 3000 \times 10 \times {10^{ - 3}} = 30\,\Omega

{X_C} = {1 \over {3000 \times 25}} \times {10^6} = {{40} \over 3}\,\Omega

So

{X_L} - {X_C} = 30 - {{40} \over 3} = {{50} \over 3}\,\Omega

\tan \theta = {{{X_L} - {X_C}} \over R} = {{50/3} \over {100}} = {1 \over 6}

So

\theta = {\tan ^{ - 1}}(0.17)

Q28

In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be : (f1 = original resonant frequency, Q1 = original quality factor)

A

{f_2} = {{{f_1}} \over 2}

and

{Q_2} = {Q_1}

B

{f_2} = {f_1}

and

{Q_2} = {{{Q_1}} \over {{Q_2}}}

C

{f_2} = 2{f_1}

and

{Q_2} = {Q_1}

D

{f_2} = {f_1}

and

{Q_2} = 2{Q_1}

Correct Answer

Option A

Solution

We know, Quality factor (Q factor)

{Q_1} = {{{w_1}} \over {\Delta w}}

= {1 \over {\sqrt {LC} }} \times {L \over R}

= {1 \over R}\sqrt {{L \over C}}

Now, when

L' = 2L

and

C' = 2C

then

{Q_2} = {1 \over R}\sqrt {{{2L} \over {2C}}} = {1 \over R}\sqrt {{L \over C}} = {Q_1}

$\therefore$ Q2 remains same as Q1. Also, as

{w_1} = {1 \over {\sqrt {LC} }}

\Rightarrow 2\pi {f_1} = {1 \over {\sqrt {LC} }}

\Rightarrow {f_1} = {1 \over {2\pi \sqrt {LC} }}

$\therefore$ When

L' = 2L

and

C' = 2C

then new resonating frequency

{f_2} = {1 \over {2\pi \sqrt {2L \times 2C} }} = {1 \over {2\pi \times 2\sqrt {LC} }} = {1 \over 2} \times {f_1}

Q29

To increase the resonant frequency in series LCR circuit,

A source frequency should be increased.

B another resistance should be added in series with the first resistance.

C another capacitor should be added in series with the first capacitor.

D the source frequency should be decreased.

Correct Answer

Option C

Solution

Resonant frequency

= {1 \over {\sqrt {LC} }} = {\omega _0}

$\Rightarrow$ If we decrease C, $\omega$ 0 would increase $\Rightarrow$ Another capacitor should be added in series.

Q30

A direct current of

4 \mathrm{~A}

and an alternating current of peak value

4 \mathrm{~A}

flow through resistance of

3\, \Omega

and

2\,\Omega

respectively. The ratio of heat produced in the two resistances in same interval of time will be :

A 3 : 2

B 3 : 1

C 3 : 4

D 4 : 3

Correct Answer

Option B

Solution

Ratio =

{{i_1^2{R_1}} \over {{{\left( {{{{i_2}} \over {\sqrt 2 }}} \right)}^2}{R_2}}} = {{{4^2} \times 3} \over {{{\left( {{4 \over {\sqrt 2 }}} \right)}^2} \times 2}}

$\Rightarrow$ Ratio = 3 : 1