Alternating Current — Page 6 | NEET Physics

Q51

The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux

\phi

linked with the primary coil is given by

\phi

=

\phi

0 + 4t, where

\phi

is webers, t is time in seconds and

\phi

0 is a constant, the output voltage across the secondary coil is

A 120 volt

B 220 volts

C 30 volts

D 90 volts.

Correct Answer

Option A

Solution

Voltage across the primary, V p =

{{d\phi } \over {dt}} = {d \over {dt}}\left( {{\phi _0} + 4t} \right) =

4 volt Also

{{{V_s}} \over {{V_p}}} = {{{N_s}} \over {{N_p}}}

Where, N s = No. of turns across primary coil = 50 N p = No. of turns across secondary coil = 1500 $\therefore$ V s =

{{1500} \over {50}} \times 4

= 120 V

Q52

A coil of inductive reactance 31

\Omega

has a resistance of 8

\Omega

. It is placed in series with a condenser of capacitative reactance 25

\Omega

. The combination is connected to an a.c. source of 110 V. The power factor of the circuit is

A 0.33

B 0.56

C 0.64

D 0.80

Correct Answer

Option D

Solution

Impedance of series LCR is Z =

\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}

=

\sqrt {{8^2} + {{\left( {31 - 25} \right)}^2}}

=

\sqrt {64 + 36}

= 10

\Omega

Power factor, cos $\phi$ =

{R \over Z} = {8 \over {10}} = 0.8

Q53

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency

f

. If L is doubled and C is changed to 4C, the frequency will be

A

f/2

B

f/4

C

8f

D

f/2

\sqrt 2

Correct Answer

Option D

Solution

We know that frequency of electrical oscillation in L.C. circuit is f =

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

$\therefore$

{{{f_1}} \over {{f_2}}} = {\left( {{{{L_2}{C_2}} \over {{L_1}{C_1}}}} \right)^{1/2}}

=

{\left( {{{2L \times 4C} \over {L \times C}}} \right)^{1/2}} = {\left( 8 \right)^{1/2}}

$\therefore$

{{{f_1}} \over {{f_2}}} = 2\sqrt 2

$\Rightarrow$

{{f_2} = {{{f_1}} \over {2\sqrt 2 }}}

$\Rightarrow$

{{f_2} = {f \over {2\sqrt 2 }}}

[As f 1 = f]

Q54

The core of a transformer is laminated because

A ratio of voltage in primary and secondary may be increased

B energy losses due to eddy currents may be minimised

C the weight of the transformer may be reduced

D rusting of the core may be prevented.

Correct Answer

Option B

Solution

The core of a transformer is laminated to minimise the energy losses due to eddy currents.

Q55

In a circuit L, C and R are connected in series with an alternating voltage source of frquency

f.

The current leads the voltage by 45 o . The value of C is

A

{1 \over {\pi f\left( {2\pi f\,L - R} \right)}}

B

{1 \over {2\pi f\left( {2\pi f\,L - R} \right)}}

C

{1 \over {\pi f\left( {2\pi f\,L + R} \right)}}

D

{1 \over {2\pi f\left( {2\pi f\,L + R} \right)}}

Correct Answer

Option D

Solution

tan $\phi$ =

{{{X_C} - {X_L}} \over R}

$\Rightarrow$

\tan \left( {{\pi \over 4}} \right) = {{{1 \over {\omega C}} - \omega L} \over R}

$\Rightarrow$ R =

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

$\Rightarrow$ R + 2 $\pi$ fL =

{1 \over {2\pi fC}}

$\Rightarrow$ C =

{1 \over {2\pi f\left( {R + 2\pi fL} \right)}}

Q56

A coil of 40 henry inductance is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is

A 5 seconds

B 1/5 seconds

C 40 seconds

D 20 seconds

Correct Answer

Option A

Solution

Time constant of LR circuit is $\tau$ = L/R. $\therefore$ $\tau$ = 40/8 = 5 sec.

Q57

For a series LCR circuit the power loss at resonance is

A

{{{V^2}} \over {\left[ {\omega L - {1 \over {\omega C}}} \right]}}

B

{I^2}L\omega

C

{I^2}R

D

{{{V^2}} \over {C\omega }}

Correct Answer

Option C

Solution

The impedance Z of a series LCR circuit is given by, Z =

\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}

At resonance, X L = X C , hence Z = R. Let, supply voltage = V R = V $\therefore$ R.M.S. current, I =

{V \over R}

$\therefore$ Power loss = VI = I 2 R

Q58

A capacitor of capacity C has reactance X. If capacitance and frequency become double then reactance will be

A 4X

B X/2

C X/4

D 2X

Correct Answer

Option C

Solution

Reactance X =

{1 \over {\omega C}} = {1 \over {2\pi fC}}

$\therefore$ X $\propto$

{1 \over {fC}}

$\therefore$

{{{X_1}} \over X} = {{fC} \over {{f_1}{C_1}}}

=

{{fC} \over {2f \times 2C}}

=

{1 \over 4}

$\Rightarrow$ X 1 =

{X \over 4}

Q59

The value of quality factor is

A

{{\omega R} \over L}

B

{1 \over {\omega RC}}

C

\sqrt {LC}

D

L/R

Correct Answer

Option B

Solution

Quality factor Q =

{{\omega L} \over R}

As

{\omega ^2} = {1 \over {LC}}

$\therefore$ Q =

{1 \over {\omega RC}}

Q60

A series LCR circuit is subjected to an ac signal of

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

. If the voltage across the inductor

(\mathrm{L}=10 \mathrm{~mH})

is

31.4 \mathrm{~V}

, then the current in this circuit is _______.

A 10 A

B 10 mA

C 68 A

D 63 A

Correct Answer

Option A

Solution

\begin{aligned} &V_L=I(\omega L)=31.4\\ &\begin{aligned} \Rightarrow \quad I & =\frac{31.4}{2 \times 3.14 \times 50 \times 10 \times 10^{-3}} \\ & =10 \mathrm{~A} \end{aligned} \end{aligned}