Alternating Current — Page 9 | NEET Physics

Q81

For a series LCR circuit with R = 100

\Omega

, L = 0.5 mH and C = 0.1 pF connected across 220V

-

50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :

A 0

^\circ

, resistive circuit

B

\approx

90

^\circ

, predominantly inductive circuit

C 0

^\circ

, resonance circuit

D

\approx

90

^\circ

, predominantly capacitive circuit

Correct Answer

Option D

Solution

R = 100

\Omega

{X_L} = \omega L = 50\pi \times {10^{ - 3}}

{X_C} = {1 \over {\omega C}} = {{{{10}^{11}}} \over {100\pi }}

{X_C} > > {X_L}

&

\left| {{X_C} - {X_L}} \right| > > R

Q82

A resistor

'R'

and

2\mu F

capacitor in series is connected through a switch to

200

V

direct supply. Across the capacitor is a neon bulb that lights up at

120

V.

Calculate the value of

R

to make the bulb light up

5

s

after the switch has been closed.

\left( {{{\log }_{10}}2.5 = 0.4} \right)

A

1.7 \times {10^5}\,\Omega

B

2.7 \times {10^6}\,\Omega

C

3.3 \times {10^7}\,\Omega

D

1.3 \times {10^4}\,\Omega

Correct Answer

Option B

Solution

We have,

V = {V_0}\left( {1 - {e^{ - t/RC}}} \right)

\Rightarrow 120 - 200\left( {1 - {e^{ - t/RC}}} \right)

\Rightarrow t = RC\,in\,\left( {2.5} \right)

\Rightarrow R = 2.71 \times {10^6}\Omega

Q83

Match with

List - I		List - II
(a)	$\omega L > {1 \over {\omega C}}$	(i)	Current is in phase with emf
(b)	$\omega L = {1 \over {\omega C}}$	(ii)	Current lags behind the applied emf
(c)	$\omega L < {1 \over {\omega C}}$	(iii)	Maximum current occurs
(d)	Resonant frequency	(iv)	Current leads the emf

A a(ii), b(i), c(iv), d(iii)

B a(ii), b(i), c(iii), d(iv)

C a(iii), b(i), c(iv), d(ii)

D a(iv), b(iii), c(ii), d(i)

Correct Answer

Option A

Solution

\omega L = {1 \over {\omega C}},{X_L} = {X_C}

So current in phase with EMF At resonance, current have maximum value.

Q84

AC voltage V(t) = 20 sin

\omega

t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take

\varepsilon

0 = 8.85

\times

10

-

12 F/m]

A 55.58

\mu

A

B 21.14

\mu

A

C 27.79

\mu

A

D 83.37

\mu

A

Correct Answer

Option C

Solution

Given, AC voltage, V(t) = 20 sin $\omega$ t volt.

Frequency, f = 50Hz Separation between the plates, d = 2 mm = 2 $\times$ 10 $-$ 3 m Area, A = 1 m2 As,

C = {{{\varepsilon _0}A} \over d}

where,

{{\varepsilon _0}}

= absolute electrical permittivity of free space = 8.854 $\times$ 10 $-$ 12 N $-$ 1 kg2m $-$ 2

C = {{{\varepsilon _0} \times 1} \over {2 \times {{10}^{ - 3}}}}

.... (i) Capacitive reactance

({X_C}) = {1 \over {\omega C}}

.... (ii) From Eqs. (i) and (ii), we get

{X_C} = {{2 \times {{10}^{ - 3}}} \over {2 \times 50\pi \times {\varepsilon _0}}}

( $\because$ $\omega$ = 2 $\pi$ f)

= {{2 \times {{10}^{ - 3}}} \over {25 \times 4\pi {\varepsilon _0}}}

\Rightarrow {X_C} = {{2 \times {{10}^{ - 3}}} \over {25}} \times 9 \times {10^9}

\Rightarrow {X_C} = {{18} \over {25}} \times {10^6}\,\Omega

By using Ohm's law, As,

{I_0} = {{{V_0}} \over {{X_C}}} = {{20 \times 25} \over {18}} \times {10^{ - 6}} = 27.78 \times {10^{ - 6}}

$\Rightarrow$ I0 = 27.78 $\mu$ A $\therefore$ The amplitude of the oscillating displacement current for applied AC voltage will be approximately 27.79 $\mu$ A.

Q85

For an RLC circuit driven with voltage of amplitude vm and frequency

{\omega _0}

=

{1 \over {\sqrt {LC} }}

the current exhibits resonance. The quality factor, Q is given by :

A

{{CR} \over {{\omega _0}}}

B

{{{\omega _0}L} \over R}

C

{{{\omega _0}R} \over L}

D

{R \over {\left( {{\omega _0}C} \right)}}

Correct Answer

Option B

Solution

Quality factor (Q) =

{{Angular\,\,{\mathop{\rm Re}\nolimits} sonance} \over {Bandwith}}

=

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

=

{{{\omega _0}} \over {{R \over L}}}

=

{{{\omega _0}L} \over R}

Q86

A series L.R circuit connected with an ac source

E=(25 \sin 1000 t) V

has a power factor of

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

. If the source of emf is changed to

\mathrm{E}=(20 \sin 2000 \mathrm{t}) \mathrm{V}

, the new power factor of the circuit will be :

A

\dfrac{1}{\sqrt{3}}

B

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

C

\dfrac{1}{\sqrt{5}}

D

\dfrac{1}{\sqrt{7}}

Correct Answer

Option C

Solution

\begin{aligned} & E=25 \sin (1000 t) \\\\ & \cos \theta=\frac{1}{\sqrt{2}} \end{aligned}

LR circuit

\begin{aligned} & \text{ Initially } \frac{R}{\omega_1 L}=\frac{1}{\tan \theta}=\frac{1}{\tan 45^{\circ}}=1 \\\\ & X_L=\omega_1 L \\\\ & \omega_2=2 \omega_1, \text{ given } \\\\ & \tan \theta^{\prime}=\frac{\omega_2 L}{R}=\frac{2 \omega_1 L}{R} \\\\ & \tan \theta^{\prime}=2 \\\\ & \cos \theta^{\prime}=\frac{1}{\sqrt{5}} \end{aligned}

Q87

A circuit connected to an ac source of emf e = e0sin(100t) with t in seconds, gives a phase difference of

\pi

/4 between the emf e and current i. Which of the following circuits will exhibit this ?

A RC circuit with R = 1 k

\Omega

and C = 1μF

B RL circuit with R = 1k

\Omega

and L = 1mH

C RC circuit with R = 1k

\Omega

and C = 10 μF

D RL circuit with R = 1 k

\Omega

and L = 10 mH

Correct Answer

Option C

Solution

Given phase difference =

{\pi \over 4}

and $\omega$ = 100 rad/s $\Rightarrow$ Reactance (X) = Resistance (R) Now by checking option. Option (A) R = 1000

\Omega

and Xc =

{1 \over {{{10}^{ - 6}} \times 100}} = {10^4}\Omega

Option (B) R = 103

\Omega

and XL =

{10^{ - 3}} \times 100 = 10^{-1} \Omega

Option (C) R = 103

\Omega

and Xc =

{1 \over {{10 \times {10}^{ - 6}} \times 100}} = {10^3}\Omega

Option (D) R = 103

\Omega

and XL =

10 \times {10^{ - 3}} \times 100 = 1\Omega

Q88

An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5

\Omega

resistor. The ratio of the currents at time t =

\infty

and at t = 40 s is close to : (Take e2 = 7.389)

A 1.06

B 0.84

C 1.15

D 1.46

Correct Answer

Option A

Solution

i = i0(1 -

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

) i $\infty$ = i0(1 -

{e^{ - \infty }}

) = i0 $\therefore$

{{{i_\infty }} \over {{i_{40s}}}}

=

{{{i_0}} \over {{i_0}\left( {1 - {e^{ - {{5 \times 40} \over {10 \times {{10}^{ - 3}}}}}}} \right)}}

=

{1 \over {\left( {1 - {e^{ - 2000}}} \right)}}

$\approx$ 1 Then most appropriate option is 1.06

Q89

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :

A 0.044 H

B 0.065 H

C 80 H

D 0.08 H

Correct Answer

Option B

Solution

From the circuit, we have

R = {{80} \over {10}} = 8\,\Omega

10 = {{220} \over {\sqrt {{R^2} + X_L^2} }} \Rightarrow \sqrt {64 + X_L^2} = 22

\Rightarrow X_L^2 = 484 - 64 = 420

\Rightarrow {X_L} = \sqrt {420} = 20.5\,\Omega

\Rightarrow \omega L = 20.5

L = {{20.5} \over {2\pi \times 50}} = {{20.5} \over {314}} = 0.065\,H

Q90

An AC voltage

V=20 \sin 200 \pi t

is applied to a series LCR circuit which drives a current

I=10 \sin \left(200 \pi t+\dfrac{\pi}{3}\right)

. The average power dissipated is:

A 21.6 W

B 200 W

C 173.2 W

D 50 W

Correct Answer

Option D

Solution

\begin{aligned} & =\mathrm{IV} \cos \phi \\ & =\frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ} \\ & =50 \mathrm{~W} \end{aligned}