Alternating Current — Page 9 | JEE Physics

Q81

The core of any transformer is laminated so as to

A reduce the energy loss due to eddy currents

B make it light weight

C make it robust and strong

D increase the secondary voltage

Correct Answer

Option A

Solution

Laminated core provide less area of cross-section for the current to flow.

Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.

Q82

An ac current is represented as

i=5 \sqrt{2}+10 \cos \left(650 \pi t+\dfrac{\pi}{6}\right) A m p

The r.m.s value of the current is

A 10 Amp

B

5 \sqrt{2} \mathrm{~Amp}

C 100 Amp

D 50 Amp

Correct Answer

Option A

Solution

To find the root mean square (RMS) value of the given alternating current, follow these steps: The current is represented as: $i = 5\sqrt{2} + 10 \cos \left(650\pi t + \dfrac{\pi}{6}\right) \, \text{Amp}$ Here, the time-independent DC component is $5\sqrt{2}$ and the AC component is $10 \cos \left(650\pi t + \dfrac{\pi}{6}\right)$ .

Calculate the square of the current, $i^2$ : $i^2 = \left(5\sqrt{2}\right)^2 + \left(10 \cos \left(650\pi t + \dfrac{\pi}{6}\right)\right)^2 + 2 \times 5\sqrt{2} \times 10 \cos \left(650\pi t + \dfrac{\pi}{6}\right)$ Simplifying, we have: $i^2 = 50 + 100 \cos^2 \left(650\pi t + \dfrac{\pi}{6}\right) + 100\sqrt{2} \cos \left(650\pi t + \dfrac{\pi}{6}\right)$ Find the average value $\langle i^2 \rangle$ : The average value of $\cos$ terms over a period is zero, simplifying our equation to: $\langle i^2 \rangle = 50 + \dfrac{100}{2} + 0$ This simplifies to: $\langle i^2 \rangle = 50 + 50 = 100$ Calculate the RMS current: The RMS value is the square root of the mean of the squares of the current: $\langle i \rangle = \sqrt{100} = 10 \, \text{Amp}$ Thus, the RMS value of the current is 10 Amps.

Q83

A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10A, then the input voltage and current in the primary coil are :

A 220 V and 20 A

B 220 V and 10A

C 440 V and 5A

D 440 V and 20 A

Correct Answer

Option C

Solution

Given NP = 300, Ns = 150, P0 = 2200W Is = 10 A P0 = V0I0 $\Rightarrow$ 2200 = V0 × 10 $\Rightarrow$ V0 = 220 V $\because$

{{{V_i}} \over {{V_0}}} = {{{N_P}} \over {{N_S}}} \Rightarrow {V_i} = 2 \times 220 = 440\,V

Also, P0 = ViIi

\Rightarrow {I_i} = {{2200} \over {440}} = 5A

Q84

A 10

\Omega

resistance is connected across 220V

-

50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :

A 2.5 ms

B 1.5 ms

C 3.0 ms

D 4.5 ms

Correct Answer

Option A

Solution

$\Rightarrow$ i = i0sin $\omega$ t when i = i0 i0 = i0sin $\omega$ t1 $\Rightarrow$ $\omega$ t1 =

{\pi \over 2}

..... (i) When i =

{{{i_1}} \over {\sqrt 2 }}

{{{i_1}} \over {\sqrt 2 }}

= i0sin $\omega$ t2 $\Rightarrow$ $\omega$ t2 =

{\pi \over 4}

...... (ii) Time taken by current from maximum value to rms value

\Rightarrow ({t_1} - {t_2}) = {\pi \over {2\omega }} - {\pi \over {4\omega }} = {\pi \over {4\omega }} = {\pi \over {4 \times 2\pi f}}

= {1 \over {8 \times 50}}

{1 \over {400}}

sec = 2.5 ms

Q85

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

Q86

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

Q87

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.

Q88

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.

Q89

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}

Q90

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}