Alternating Current — Page 8 | JEE Physics

Q71

Match with .

List - I		List - II
(a)	Rectifier	(i)	Used either for stepping up or stepping down the a.c. voltage
(b)	Stabilizer	(ii)	Used to convert a.c. voltage into d.c. voltage
(c)	Transformer	(iii)	Used to remove any ripple in the rectified output voltage
(d)	Filter	(iv)	Used for constant output voltage even when the input voltage or load current change

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

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

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

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

Correct Answer

Option A

Solution

(a) Rectifier : used to convert a a.c. voltage into d.c. voltage. (b) Stabilizer : used for constant output voltage even when the input voltage or load current change (c) Transformer : used either for stepping up or stepping down the a.c. voltage. (d) Filter : used to remove any ripple in the rectified output voltage.

Q72

An alternating current is given by the equation i = i1 sin

\omega

t + i2 cos

\omega

t. The rms current will be :

A

{1 \over {\sqrt 2 }}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}

B

{1 \over {\sqrt 2 }}({i_1} + {i_2})

C

{1 \over {\sqrt 2 }}{({i_1} + {i_2})^2}

D

{1 \over 2}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}

Correct Answer

Option A

Solution

{I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos \theta }

{I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos 90^\circ }

{I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}(0)} = \sqrt {I_1^2 + I_2^2}

We know that,

{I_{rms}} = {{{I_0}} \over {\sqrt 2 }}

So,

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

Q73

Match with

List - I		List - II
(a)	Phase difference between current and voltage in a purely resistive AC circuit	(i)	${\pi \over 2}$ ; current leads voltage
(b)	Phase difference between current and voltage in a pure inductive AC circuit	(ii)	zero
(c)	Phase difference between current and voltage in a pure capacitive AC circuit	(iii)	${\pi \over 2}$ ; current lags voltage
(d)	Phase difference between current and voltage in an LCR series circuit	(iv)	${\tan ^{ - 1}}\left( {{{{X_C} - {X_L}} \over R}} \right)$ Choose the most appropriate answer from the options given below :

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

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

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

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

Correct Answer

Option A

Solution

(a) phase difference b/w current & voltage in a purely resistive AC circuit is zero (b) phase difference b/w current & voltage in a pure inductive AC circuit is

{\pi \over 2}

; current lags voltage. (c) phase difference b/w current & voltage in a pure capacitive AC circuit is

{\pi \over 2}

; current lead voltage. (d) phase difference b/w current & voltage in an LCR series circuit is =

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

Q74

In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :

A The bandwidth of resonance circuit will increase.

B The resonance frequency will increase.

C The quality factor will increase.

D The quality factor and the resonance frequency will remain constant.

Correct Answer

Option A

Solution

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

$\Rightarrow$ 2 $\pi$ f =

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

$\Rightarrow$ f =

{1 \over {2\pi \sqrt {LC} }}

f does not depends on resistance(R). Quality factor,

Q = {{\omega L} \over R}

$\Rightarrow$

Q \propto {1 \over R}

So if R increase then Q will decrease. Also,

Q = {{\omega L} \over R} = {\omega \over {\Delta \beta}}

where

\Delta \beta

= bandwidth

\Delta \beta = {R \over L}

So if R increase then

\Delta \beta

will increase too.

Q75

In an

a.c.

circuit the voltage applied is

E = {E_0}\,\sin \,\omega t.

The resulting current in the circuit is

I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).

The power consumption in the circuit is given by

A

P = \sqrt 2 {E_0}{I_0}

B

P = {{{E_0}{I_0}} \over {\sqrt 2 }}

C

P=zero

D

P = {{{E_0}{I_0}} \over 2}

Correct Answer

Option C

Solution

KEY CONCEPT : We know that power consumed in a.c. circuit is given by,

P = {E_{rms}}{I_{rms}}\cos \phi

Here,

E = {E_0}\sin \omega t

I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right)

which implies that the phase difference,

\phi = {\pi \over 2}

$\therefore$

P = {E_{rms}}.{I_{rms}}.\cos {\pi \over 2} = 0

\left( {\,\,} \right.

as

\left. {\,\,\cos {\pi \over 2} = 0\,\,} \right)

Q76

A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

A halved

B same

C Zero

D double

Correct Answer

Option D

Solution

To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $R$ .

At resonance, the impedance $Z$ of the series LCR circuit is equal to the resistance $R$ , and thus:

Z = R

The amplitude of the current $I_0$ at resonance is given by Ohm's law:

I_0 = \frac{V_0}{R}

where $V_0$ is the amplitude of the voltage supplied.

Now, if the resistance $R$ is halved while keeping the voltage amplitude $V_0$ , capacitance $C$ , and inductance $L$ the same, the new resistance becomes:

R_{new} = \frac{R}{2}

The new current amplitude $I_0'$ at resonance is given by the modified Ohm's law:

I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 \times I_0

Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is: Option D: double

Q77

In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t i = 20 sin

\left( {30t - {\pi \over 4}} \right)

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively

A 50, 0

B 50, 10

C

{{1000} \over {\sqrt 2 }},10

D

{{50} \over {\sqrt 2 }}

Correct Answer

Option C

Solution

Wattless current, here $\phi$ is the angle between i and e. Average power, Pav = Vrms Irms cos $\phi$ =

{{100} \over {\sqrt 2 }} \times {{20} \over {\sqrt 2 }}

cos

{\pi \over 4}

=

{{1000} \over {\sqrt 2 }}

watt.

Q78

A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is

I_0

. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

A

\dfrac{\mathrm{I}_0}{2}

B

\dfrac{\mathrm{I}_0}{\sqrt{2}}

C

2 \mathrm{I}_0

D

\mathrm{I_0}

Correct Answer

Option A

Solution

At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel each other out, leaving only the resistance.

Therefore, the amplitude of the current is given by:

I_0 = \frac{E}{R}

Now, if the resistance is doubled, the new resistance becomes $2R$ . The current amplitude at resonance then becomes:

I = \frac{E}{2R} = \frac{1}{2}\left(\frac{E}{R}\right) = \frac{I_0}{2}

Thus, the amplitude of current at resonance when the resistance is doubled is $\dfrac{I_0}{2}$ .

The correct option is A.

Q79

A fully charged capacitor

C

with initial charge

{q_0}

is connected to a coil of self inductance

L

at

t=0.

The time at which the energy is stored equally between the electric and the magnetic fields is :

A

{\pi \over 4}\sqrt {LC}

B

2\pi \sqrt {LC}

C

\sqrt {LC}

D

\pi \sqrt {LC}

Correct Answer

Option A

Solution

Energy stored in magnetic field

= {1 \over 2}L{i^2}

Energy stored in electric field

= {1 \over 2}{{{q^2}} \over C}

$\therefore$

{1 \over 2}L{i^2} = {1 \over 2}{{{q^2}} \over C}

Also

q = {q_0}\,\cos \,\omega t

and

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

On solving

t = {\pi \over 4}\sqrt {LC}

Q80

Primary side of a transformer is connected to

230 \mathrm{~V}, 50 \mathrm{~Hz}

supply. Turns ratio of primary to secondary winding is

10: 1

. Load resistance connected to secondary side is

46 \Omega

. The power consumed in it is :

A 11.5 W

B 12.5 W

C 10.0 W

D 12.0 W

Correct Answer

Option A

Solution

\begin{aligned} & \frac{V_1}{V_2}=\frac{N_1}{N_2} \\ & \frac{230}{V_2}=\frac{10}{1} \\ & V_2=23 \mathrm{~V} \end{aligned}

Power consumed

=\frac{\mathrm{V}_2^2}{\mathrm{R}}

=\frac{23 \times 23}{46}=11.5 \mathrm{~W}