Alternating Current — Page 4 | NEET Physics

Q31

Which of the following combinations should be selected for better tuning of an L-C-R circuit used for combination ?

A R = 20

\Omega

, L = 1.5 H, C = 35

\mu

F

B R = 25

\Omega

, L = 2.5 H, C = 45

\mu

F

C R = 15

\Omega

, L = 3.5 H, C = 30

\mu

F

D R = 25

\Omega

, L = 1.5 H, C = 45

\mu

F

Correct Answer

Option C

Solution

Quality factor of an L-C-R circuit is given by, Q =

{1 \over R}\sqrt {{L \over C}}

So, Q 1 =

{1 \over {20}}\sqrt {{{1.5} \over {35 \times {{10}^{ - 6}}}}}

= 10.35 Q 2 =

{1 \over {25}}\sqrt {{{2.5} \over {45 \times {{10}^{ - 6}}}}}

= 9.43 Q 3 =

{1 \over {15}}\sqrt {{{3.5} \over {30 \times {{10}^{ - 6}}}}}

= 22.77 Q 4 =

{1 \over {25}}\sqrt {{{1.5} \over {45 \times {{10}^{ - 6}}}}}

= 7.30 As Q 3 is maximum of Q 1 , Q 2 , Q 3 , and Q 4 .

Hence, option (c) should be selected for better tuning of an L-C-R circuit.

Q32

A small signal voltage V(t) = V 0 sin

\omega

t is applied across an ideal capacitor C

A Current

I(t)

is in phase with voltage

V(t)

.

B Current

I(t)

leads voltage V(t) by 180 o

C Current

I(t)

, legs voltage

V(t)

by 90 o .

D Over a full cycle the capacitor C foes not consume any energy from the voltage source.

Correct Answer

Option D

Solution

In ideal capacitor, I leads from voltage by 90 o and capacitor does not consume energy.

Q33

An inductor 20 mH, a capacitor 50

\mu F

and a resistor 40

\Omega

are connected in series across a source of emf V = 10 sin 340t. The power loss in A.C. circuit is

A 0.76 W

B 0.89 W

C 0.51 W

D 0.67 W

Correct Answer

Option C

Solution

X L = $\omega$ L = 340 $\times$ 20 $\times$ 10 -3 = 6.8

\Omega

X C =

{1 \over {340 \times 50 \times {{10}^{ - 6}}}}

= 58.82

\Omega

Average power in impedance Z =

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

=

\sqrt {{{40}^2} + {{\left( {58.82 - 6.8} \right)}^2}}

= 65.62

\Omega

Power loss in A.C. circuit, = V rms I rms cos $\phi$ =

{1 \over 2}

V 0 I 0 cos $\phi$ =

{1 \over 2} \times 10 \times {{10} \over {65.62}} \times {{40} \over {65.62}}

= 0.46 W

Q34

A series R-C circuit is connected to an alternating voltage source. Consider two situations : (a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is

I

and voltage across capacitor is

V

then

A

{i_a} > {i_b}

B

{V_a} = {V_b}

C

{V_a} < {V_b}

D

{V_a} > {V_b}

Correct Answer

Option D

Solution

For series R – C circuit, capacitive reactance, Z=

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

Current through resistor, i = Current in the circuit =

{V \over {{Z_c}}} = {V \over {\sqrt {{R^2} + {{\left( {{1 \over {C\omega }}} \right)}^2}} }}

Voltage across capacitor, V = iX C =

{V \over {\sqrt {{R^2} + {{\left( {{1 \over {C\omega }}} \right)}^2}} }} \times {1 \over {C\omega }}

=

{V \over {\sqrt {{R^2}{\omega ^2}{C^2} + 1} }}

When mica is introduced capacitance will increase, hence voltage across capacitor gets decrease.

Q35

A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z'. the power drawn will be

A

P\left( {{R \over Z}} \right)

B P

C

P{\left( {{R \over Z}} \right)^2}

D

P\sqrt {{R \over Z}}

Correct Answer

Option C

Solution

For pure resistor circuit, power P =

{{V_{rms}^2} \over R}

$\Rightarrow$

{V_{rms}^2}

= PR For L-R series circuit, power P' = V rms I rms cos $\phi$ =

{V_{rms}} \times {{{V_{rms}}} \over Z} \times {R \over Z}

=

{{V_{rms}^2R} \over {{Z^2}}}

=

{{PR \times R} \over {{Z^2}}}

$\therefore$ P' =

P{\left( {{R \over Z}} \right)^2}

Q36

A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are

A 300 V, 15 A

B 450 V, 15 A

C 450 V, 13.5 A

D 600 V, 15 A

Correct Answer

Option B

Solution

Now P p = V p I p or I p = P p /V p = 3000/200 = 15 A Also, V s = Ps /I s = 2700/6 = 450 V

Q37

The primary of a transformer when connected to a dc battery of 10 Volt draws a current of 1 mA. The number of turns of the primary and secondary windings are 50 and 100 respectively. The voltage in the secondary and the current drawn by the circuit in the secondary are respectively

A 20 V and 2.0 mA

B 10 V and 0.5 mA

C Zero volt and therefore no current

D 20 V and 0.5 mA

Correct Answer

Option C

Solution

Transformer cannot work on dc. $\therefore$ V s = 0 and I s = 0

Q38

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when

A a capacitance of reactance X C = X L is included in the same citcuit.

B an iron rod is inserted in the coil.

C frequency of the AC source is decreased.

D number of turns in the coil is reduced.

Correct Answer

Option B

Solution

If a bulb B and AC source are connected in series with self-inductance coil, then in such case the brightness of the bulb decreases when an iron rod is inserted in the coil which further increases impedance of the circuit.

Q39

The instantaneous values of alternating current and voltage in a circuit are given as

i = {1 \over {\sqrt 2 }}

sin (100

\pi

t) ampere

e = {1 \over {\sqrt 2 }}\sin \left( {100\pi t + {\pi \over 3}} \right)

Volt The average power in watts consumed in the circuit is

A

{1 \over 4}

B

{{\sqrt 3 } \over 4}

C

{1 \over 2}

D

{1 \over 8}

Correct Answer

Option D

Solution

i rms =

{{{i_0}} \over {\sqrt 2 }} = {{{1 \over {\sqrt 2 }}} \over {\sqrt 2 }} = {1 \over 2}

A e rms =

{{{e_0}} \over {\sqrt 2 }} = {{{1 \over {\sqrt 2 }}} \over {\sqrt 2 }} = {1 \over 2}

V Average power consumed in the circuit, P = i rms e rms cos $\phi$ =

\left( {{1 \over 2}} \right)\left( {{1 \over 2}} \right)\cos {\pi \over 3}

=

{1 \over 8}

W

Q40

In an electrical circuit R, L, C and a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is If instead, C is removed from the circuit, the phase difference is again. The power factor of the circuit is

A

{1 \over 2}

B

{1 \over {\sqrt 2 }}

C 1

D

{{\sqrt 3 } \over 2}

Correct Answer

Option C

Solution

When L is removed, the phase difference between the voltage and current is tan $\phi$ 1 =

{{{X_C}} \over R}

$\Rightarrow$

\tan {\pi \over 3} = {{{X_C}} \over R}

$\Rightarrow$ X C =

\sqrt 3 R

When C is removed, the phase difference between the voltage and current is tan $\phi$ 2 =

{{{X_L}} \over R}

$\Rightarrow$

\tan {\pi \over 3} = {{{X_L}} \over R}

$\Rightarrow$ X L =

\sqrt 3 R

As X L = X C , the series LCR circuit is in resonance. Net impedence, Z =

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

= R Power factor, cos $\phi$ =

{R \over Z} = {R \over R} = 1