Alternating Current — Page 6 | JEE Physics

Q51

In an a.c. circuit, voltage and current are given by:

V=100 \sin (100 t) V

and

I=100 \sin \left(100 t+\dfrac{\pi}{3}\right) \mathrm{mA}

respectively. The average power dissipated in one cycle is:

A 5 W

B 25 W

C 2.5 W

D 10 W

Correct Answer

Option C

Solution

\begin{aligned} & P_{\text{avg }}=V_{\text{rms }} I_{r m s} \cos (\Delta \phi) \\ & =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \left(\frac{\pi}{3}\right) \\ & =\frac{10^4}{2} \times \frac{1}{2} \times 10^{-3} \\ & =\frac{10}{4}=2.5 \mathrm{~W} \end{aligned}

Q52

An alternating voltage

V(t)=220 \sin 100 \pi t

volt is applied to a purely resistive load of

50 \Omega

. The time taken for the current to rise from half of the peak value to the peak value is:

A 7.2 ms

B 3.3 ms

C 5 ms

D 2.2 ms

Correct Answer

Option B

Solution

Rising half to peak

\begin{aligned} & \mathrm{t}=\mathrm{T} / 6 \\ & \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms} \end{aligned}

Q53

In LC circuit the inductance L = 40 mH and capacitance C = 100

\mu

F. If a voltage V(t) = 10sin(314t) is applied to the circuit, the current in the circuit is given as :

A 0.52 cos 314 t

B 5.2 cos 314 t

C 0.52 sin 314 t

D 10 cos 314 t

Correct Answer

Option A

Solution

Z = xC – xL =

{1 \over {\omega C}} - \omega L

=

{1 \over {314 \times 100 \times {{10}^{ - 6}}}} - 314 \times 40 \times {10^{ - 3}}

= 19.28

\Omega

As Vm = ImZ $\Rightarrow$ 10 = Im $\times$ 19.28 $\Rightarrow$ Im =

{{10} \over {19.28}}

= 0.52 A $\therefore$ I = 0.52 sin(314t +

{\pi \over 2}

) = 0.52 cos(314t)

Q54

An AC current is given by I = I1 sin

\omega

t + I2 cos

\omega

t. A hot wire ammeter will give a reading :

A

{{{I_1} + {I_2}} \over {\sqrt 2 }}

B

\sqrt {{{I_1^2 - I_2^2} \over 2}}

C

\sqrt {{{I_1^2 + I_2^2} \over 2}}

D

{{{I_1} + {I_2}} \over {2\sqrt 2 }}

Correct Answer

Option C

Solution

{I_{RMS}} = \sqrt {{{\int {{I^2}dt} } \over {\int {dt} }}}

I_{RMS}^2 = \int\limits_0^T {{{{{({I_1}\sin \omega t + {I_2}\cos \omega t)}^2}dt} \over T}}

= {1 \over T}\int\limits_0^T {(I_1^2{{\sin }^2}\omega t + I_2^2{{\cos }^2}\omega t + 2{I_1}{I_2}\sin \omega t\cos \omega t)dt}

= {{I_1^2} \over 2} + {{I_2^2} \over 2} + 0

{I_{RMS}} = \sqrt {{{I_1^2 + I_2^2} \over 2}}

Q55

A circuit element

\mathrm{X}

when connected to an a.c. supply of peak voltage

100 \mathrm{~V}

gives a peak current of

5 \mathrm{~A}

which is in phase with the voltage. A second element

\mathrm{Y}

when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by

\dfrac{\pi}{2}

. If

\mathrm{X}

and

\mathrm{Y}

are connected in series to the same supply, what will be the rms value of the current in ampere?

A

\dfrac{10}{\sqrt{2}}

B

\dfrac{5}{\sqrt{2}}

C

5 \sqrt{2}

D

\dfrac{5}{2}

Correct Answer

Option D

Solution

Element X should be resistive with, $R=\dfrac{100}{5}=20 \Omega$ Element Y should be inductive with,

X_{L}=\frac{100}{5}=20 \Omega

When X and Y are connector in series,

\begin{aligned} &Z=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \Omega \\\\ &I=\frac{100}{Z}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} \\\\ &i_{\mathrm{rms}}=\frac{1}{\sqrt{2}} I \\\\ &=\frac{5}{2} \end{aligned}

Q56

In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes

x

times its initial resonant frequency

\omega_0

. The value of

x

is :

A 1/4

B 1/16

C 4

D 16

Correct Answer

Option A

Solution

The resonance frequency of LC oscillations circuit is

\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\\\ & \mathrm{L} \rightarrow 2 \mathrm{~L} \\\\ & \mathrm{C} \rightarrow 8 \mathrm{C} \\\\ & \omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} \\\\ & \omega=\frac{\omega_0}{4} \end{aligned}

So $\mathrm{x}=\dfrac{1}{4}$

Q57

In a

LCR

circuit capacitance is changed from

C

to

2

C

. For the resonant frequency to remain unchaged, the inductance should be changed from

L

to

A

L/2

B

2L

C

4L

D

L/4

Correct Answer

Option A

Solution

For resonant frequency to remain same

LC

should be const.

LC=

const

\Rightarrow LC = L' \times 2C \Rightarrow L' = {L \over 2}

Q58

The self inductance of the motor of an electric fan is

10

H

. In order to impart maximum power at

50

Hz

, it should be connected to a capacitance of

A

8\mu F

B

4\mu F

C

2\mu F

D

1\mu F

Correct Answer

Option D

Solution

For maximum power,

{X_L} = X{}_C,

which yields

C = {1 \over {{{\left( {2\pi n} \right)}^2}L}} = {1 \over {4{\pi ^2} \times 50 \times 50 \times 10}}

$\therefore$

C = 0.1 \times {10^{ - 5}}F = 1\mu F

Q59

In a series resonant

LCR

circuit, the voltage across

R

is

100

volts and

R = 1\,k\Omega

with

C = 2\mu F.

The resonant frequency

\omega

is

200

rad/s

. At resonance the voltage across

L

is

A

2.5 \times {10^{ - 2}}V

B

40

V

C

250

V

D

4 \times {10^{ - 3}}V

Correct Answer

Option C

Solution

Across resistor,

I = {V \over R} = {{100} \over {1000}} = 0.1A

At resonance,

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

= {1 \over {200 \times 2 \times {{10}^{ - 6}}}} = 2500

Voltage across

L

is

I{X_L} = 0.1 \times 2500 = 250V

Q60

In an

AC

generator, a coil with

N

turns, all of the same area

A

and total resistance

R,

rotates with frequency

\omega

in a magnetic field

B.

The maximum value of

emf

generated in the coil is

A

N.A.B.R.

\omega

B

N.A.B

C

N.A.B.R.

D

N.A.B.

\omega

Correct Answer

Option D

Solution

e = - {{d\phi } \over {dt}} = - {{d\left( {N\overrightarrow B .\overrightarrow A } \right)} \over {dt}}

= - N{d \over {dt}}\left( {BA\,\cos \,\omega t} \right) = NBA\omega \sin \,\omega t

\Rightarrow {e_{\max }} = NBA\omega