Alternating Current — Page 10 | NEET Physics

Q91

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k

\Omega

, an inductor of inductive reactance XL = 250

\pi

\Omega

and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take

\pi

2 = 10)

A 4

\mu

F

B 25

\mu

F

C 400

\mu

F

D 40

\mu

F

Correct Answer

Option A

Solution

From maximum average power XL = XC 250 $\pi$ =

{1 \over {2\pi (50)C}}

$\Rightarrow$ C = 4 $\times$ 10 $-$ 6

Q92

A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5

\Omega

, L=25 mH and C=1000

\mu

F. The total impedance, and phase difference between the voltage across the source and the current will respectively be :

A 10

\Omega

and tan

-

1

\left( {{5 \over 3}} \right)

B

7\,\Omega

and 45o

C

10\,\Omega

and tan

-

1

\left( {{8 \over 3}} \right)

D

7\,\Omega

and tan

-

1

\left( {{5 \over 3}} \right)

Correct Answer

Option B

Solution

It is given that e0 = 283 V; $\omega$ = 320. The inductor reactance is XL = 320 $\times$ 25 $\times$ 10 $-$ 3 = 8

\Omega

The capacitor reactance is

{X_C} = {1 \over {\omega C}} = {1 \over {320 \times 1000 \times {{10}^{ - 6}}}} = {{1000} \over {320}} = 3.1\,\Omega

It is given that R = 5

\Omega

. Therefore, the total impedance is

Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {50} = 7\,\Omega

and the phase difference between the voltage across the source and the current is

\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ

Q93

A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches

\left( {{1 \over n}} \right)

times of its maximum value, is :

A

{L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)

B

{L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)

C

{L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)

D

{L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)

Correct Answer

Option B

Solution

P.E. in inductor,

U = {1 \over 2}L{I^2}

U \propto {I^2}

{U \over {{U_0}}} = {\left( {{I \over {{I_0}}}} \right)^2}

{1 \over n} = {\left( {{I \over {{I_0}}}} \right)^2}

I = {{{I_0}} \over {\sqrt n }}

We know,

I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

{{{I_0}} \over {\sqrt n }} = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

$\Rightarrow$

{e^{ - {{Rt} \over L}}}

= 1 -

{1 \over {\sqrt n }}

taking ln & solving we get,

t = {L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)

Q94

The phase difference between the alternating current and

emf

is

{\pi \over 2}.

Which of the following cannot be the constituent of the circuit?

A

R,L

B

C

alone

C

L

alone

D

L, C

Correct Answer

Option A

Solution

The phase difference between the alternating current and emf in an AC circuit depends on the components in the circuit: In a purely resistive ( $R$ ) circuit, the current and emf are in phase, meaning the phase difference is $0$ .

In a purely inductive ( $L$ ) circuit, the current lags behind the emf by $\dfrac{\pi}{2}$ , meaning the phase difference is $\dfrac{\pi}{2}$ .

In a purely capacitive ( $C$ ) circuit, the current leads the emf by $\dfrac{\pi}{2}$ , again meaning the phase difference is $\dfrac{\pi}{2}$ .

In an $L$ - $R$ or $L$ - $C$ circuit, the phase difference depends on the relative values of $L$ , $R$ , and $C$ and can be anywhere between $0$ and $\dfrac{\pi}{2}$ .

Therefore, if the phase difference between the alternating current and emf is $\dfrac{\pi}{2}$ , then the circuit cannot contain only a resistor ( $R$ ) since that would give a phase difference of $0$ .

So, the answer is Option A: $R,L$ .

The phase difference would not be $\dfrac{\pi}{2}$ if the circuit contains both a resistor and an inductor.

Q95

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:

A L

\leftrightarrow

k, C

\leftrightarrow

b, R

\leftrightarrow

m

B L

\leftrightarrow

m, C

\leftrightarrow

k, R

\leftrightarrow

b

C L

\leftrightarrow

m, C

\leftrightarrow

{1 \over k}

, R

\leftrightarrow

b

D L

\leftrightarrow

{1 \over b}

, C

\leftrightarrow

{1 \over m}

, R

\leftrightarrow

{1 \over k}

Correct Answer

Option C

Solution

For spring mass damped oscillator ma = - kx - bv $\Rightarrow$ ma + kx + bv = 0 $\Rightarrow$

m{{{d^2}x} \over {d{t^2}}}

+ b

{{dx} \over {dt}}

+ kx = 0 ....(1) For LCR circuit L

{{di} \over {dt}}

+ iR +

{q \over C}

= 0 $\Rightarrow$ L

{{{d^2}q} \over {d{t^2}}}

+ R

{{dq} \over {dt}}

+

{q \over C}

= 0 .....(2) Comparing (1) and (2), we get L $\leftrightarrow$ m, C $\leftrightarrow$

{1 \over k}

, R $\leftrightarrow$ b