Alternating Current — Page 5 | NEET Physics

Q41

A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be

A 2.0 A

B 4.0 A

C 8.0 A

D

{{20} \over {\sqrt {13} }}A

Correct Answer

Option B

Solution

Current flowing in the coil is I =

{{200} \over Z} = {{200} \over {\sqrt {{R^2} + {{\left( {\omega L} \right)}^2}} }} = {{200} \over {\sqrt {{{\left( {30} \right)}^2} + {{\left( {40} \right)}^2}} }}

= 4 A

Q42

The r.m.s. value of potential difference V shown in the figure is

A

{{{V_0}} \over {\sqrt 3 }}

B V 0

C

{{{V_0}} \over {\sqrt 2 }}

D

{{{V_0}} \over 2}

Correct Answer

Option C

Solution

V rms =

\sqrt {{{{T \over 2}{{\left( {{V_0}} \right)}^2} + 0} \over T}} = {{{V_0}} \over {\sqrt 2 }}

Q43

An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3

\Omega

, the phase difference between the applied voltage and the current in the circuit is

A

\pi /6

B

\pi /4

C

\pi /2

D zero

Correct Answer

Option B

Solution

tan $\phi$ =

{{{X_L}} \over R}

=

{3 \over 3}

= 1 $\Rightarrow$ $\phi$ = 45 o or

{\pi \over 4}

Q44

In the ac circuit an alternating voltage e =

200\sqrt 2

sin100t volts is connected to a capacitor of capacity 1

\mu

F. The r.m.s. value of the current in the circuit is

A 10 mA

B 100 mA

C 200 mA

D 20 mA

Correct Answer

Option D

Solution

e 0 =

200\sqrt 2

V, $\omega$ = 100 rad s -1 The capacitive reactance is X C =

{1 \over {\omega C}} = {1 \over {100 \times 1 \times {{10}^{ - 6}}}}

The r.m.s. value of the current in the circuit is i r.m.s =

{{{V_{rms}}} \over {{X_C}}} = {{{{{e_0}} \over {\sqrt 2 }}} \over {{1 \over {\omega C}}}}

=

{{{{200\sqrt 2 } \over {\sqrt 2 }}} \over {{1 \over {100 \times {{10}^{ - 6}}}}}}

= 200 × 100 × 10 –6 A = 2 × 10 –2 A = 20 mA

Q45

A condenser of capacity C is charged to a potential difference of V 1 . The plates of th condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V 2 is

A

{\left( {{{C{{\left( {{V_1} - {V_2}} \right)}^2}} \over L}} \right)^{{1 \over 2}}}

B

{{C\left( {V_1^2 - V_2^2} \right)} \over L}

C

{{C\left( {V_1^2 + V_2^2} \right)} \over L}

D

{\left( {{{C\left( {V_1^2 - V_2^2} \right)} \over L}} \right)^{{1 \over 2}}}

Correct Answer

Option D

Solution

q = q 0 cos $\omega$ t $\Rightarrow$ cos $\omega$ t =

{q \over {{q_0}}} = {{C{V_2}} \over {C{V_1}}} = {{{V_2}} \over {{V_1}}}

Current through the inductor I =

{{dq} \over {dt}} = {d \over {dt}}\left( {{q_0}\cos \omega t} \right)

= - q 0 $\omega$ sin $\omega$ t $\therefore$ |I| =

{C{V_1}{1 \over {\sqrt {LC} }}{{\left[ {1 - {{\cos }^2}\omega t} \right]}^{1/2}}}

=

{V_1}\sqrt {{C \over L}} {\left[ {1 - \left( {{{{V_2}} \over {{V_1}}}} \right)} \right]^{1/2}}

=

{\left( {{{C\left( {V_1^2 - V_2^2} \right)} \over L}} \right)^{{1 \over 2}}}

Q46

A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drwn by the primary windings of the transformer is

A 3.6 ampere

B 2.8 ampere

C 2.5 ampere

D 5.0 ampere

Correct Answer

Option D

Solution

Efficiency of the transformer, $\eta$ = Output power Input power $\Rightarrow$

{{80} \over {100}} = {{{V_s}{I_s}} \over {{V_p}{I_p}}}

$\Rightarrow$ I p =

{{440 \times 2} \over {{{80} \over {100}} \times 220}}

= 5 A

Q47

Power dissipated in an LCR series circuit connected to an A.C. source of emf

\varepsilon

is

A

{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - {1 \over {\omega }}} \right)}^2}} } \over R}

B

{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} } \over R}

C

{{{\varepsilon ^2}R} \over {\sqrt {{R^2} + {{\left( {L - {1 \over {C\omega }}} \right)}^2}} }}

D

{{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} \right]}}

Correct Answer

Option D

Solution

Power dissipated, P = E rms I rms cos $\phi$ cos $\phi$ =

{R \over Z}

But I rms =

{{{E_{rms}}} \over Z}

$\therefore$ P =

E_{rms}^2.{R \over {{Z^2}}}

Also we know, Z =

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

So, P =

E_{rms}^2.{R \over {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}}

=

{{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} \right]}}

Q48

In an a.c. circuit the e.m.f. (

\varepsilon

) and the current (i) at any instant are given respectively by

\varepsilon

= E 0 sin

\omega

t,

i

=

I

0 sin(

\omega

t

-

\phi

) The average power in the circuit over one cycle of a.c. is

A

{{{E_0}{I_0}} \over 2}\cos \phi

B

{{E_0}{I_0}}

C

{{{E_0}{I_0}} \over 2}

D

{{{E_0}{I_0}} \over 2}\sin \phi

Correct Answer

Option A

Solution

Average power =

{{{E_0}{I_0}} \over 2}\cos \phi

Q49

A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 amp, the efficiency of the transformer is approximately

A 50%

B 90%

C 10%

D 30%

Correct Answer

Option B

Solution

Efficiency of the transformer =

{{{P_{output}}} \over {{P_{input}}}} \times 100

=

{{100} \over {220 \times 0.5}} \times 100

= 90 %

Q50

What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10

\mu

F and

\omega

= 1000 s

-

1 ?

A 1 mH

B cannot be calculated unless R is known

C 10 mH

D 100 mH

Correct Answer

Option D

Solution

Condition for which current is maximum in a series LCR circuit is,

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

$\Rightarrow$ 1000 =

{1 \over {\sqrt {L\left( {10 \times {{10}^6}} \right)} }}

$\Rightarrow$ L = 100 mH