Alternating Current — Page 8 | NEET Physics

Q71

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)

Q72

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

Q73

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.

Q74

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.

Q75

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}

Q76

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}

Q77

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.

Q78

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.

Q79

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

Q80

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