Alternating Current Questions — JEE Physics

Q1

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}

Q2

The current flowing through an ac circuit is given by I = 5 sin(120

\pi

t)A How long will the current take to reach the peak value starting from zero?

A

{1 \over {60}}

s

B 60 s

C

{1 \over {120}}

s

D

{1 \over {240}}

s

Correct Answer

Option D

Solution

\omega = 120\pi

\Rightarrow T = {1 \over {60}}\sec

The current will take its peak value in

{T \over 4}

time So

t = {T \over 4}

= {1 \over {240}}s

Q3

The equation of current in a purely inductive circuit is

5 \sin \left(49\, \pi t-30^{\circ}\right)

. If the inductance is

30 \,\mathrm{mH}

then the equation for the voltage across the inductor, will be :

\left\{\right.

Let

\left.\pi=\dfrac{22}{7}\right\}

A

1.47 \sin \left(49 \pi t-30^{\circ}\right)

B

1.47 \sin \left(49 \pi t+60^{\circ}\right)

C

23.1 \sin \left(49 \pi t-30^{\circ}\right)

D

23.1 \sin \left(49 \pi t+60^{\circ}\right)

Correct Answer

Option D

Solution

V(t) = I\omega L\sin (49\pi t - 30^\circ + 90^\circ )

= 5 \times 49\pi \times {{30} \over {1000}}\sin (49\pi t + 60^\circ )

= 23.1\sin (49\pi t + 60^\circ )

Q4

A transformer operating at primary voltage

8 \,\mathrm{kV}

and secondary voltage

160 \mathrm{~V}

serves a load of

80 \mathrm{~kW}

. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

A

800 \,\Omega

and

1.06 \,\Omega

B

10 \,\Omega

and

500 \,\Omega

C

800 \,\Omega

and

0.32 \,\Omega

D

1.06 \,\Omega

and

500 \,\Omega

Correct Answer

Option C

Solution

{V_1}{i_1} = {V_2}{i_2} = 80

kW

\Rightarrow {i_1} = 10\,A

and

{i_2} = {{80 \times 1000} \over {160}} = 500\,A

\Rightarrow {R_1} = {{{V_1}} \over {{i_1}}} = 800\,\Omega

and

{R_2} = {{160} \over {500}} = 0.32\,\Omega

Q5

An alternating emf

\mathrm{E}=440 \sin 100 \pi \mathrm{t}

is applied to a circuit containing an inductance of

\dfrac{\sqrt{2}}{\pi} \mathrm{H}

. If an a.c. ammeter is connected in the circuit, its reading will be :

A 4.4 A

B 1.55 A

C 2.2 A

D 3.11 A

Correct Answer

Option C

Solution

I = {V \over {\omega L}}

= {{440} \over {100\pi \times {{\sqrt 2 } \over \pi }}} = {{44} \over {10\sqrt 2 }}

\Rightarrow {I_{rms}} = {I \over {\sqrt 2 }} = {{44} \over {20}} = 2.2\,A

Q6

A coil of inductance 1 H and resistance

100 \,\Omega

is connected to a battery of 6 V. Determine approximately : (a) The time elapsed before the current acquires half of its steady - state value. (b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given

\ln 2=0.693, \mathrm{e}^{-3 / 2}=0.25

)

A t = 10 ms; U = 2 mJ

B t = 10 ms; U = 1 mJ

C t = 7 ms; U = 1 mJ

D t = 7 ms; U = 2 mJ

Correct Answer

Option C

Solution

i(t) = {V \over R}(1 - {e^{ - Rt/L}})

...... (1)

{L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms

...... (2)

{V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})

\Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms

U = {1 \over 2}L{i^2} = {1 \over 2}{[1 - {e^{ - 15/10}}]^2}{\left[ {{6 \over {100}}} \right]^2}

= {1 \over 2}{[1 - 0.25]^2} \times 36 \times {10^{ - 4}}

= 1\,mJ

Q7

An alternating voltage source

\mathrm{V}=260 \sin (628 \mathrm{t}

) is connected across a pure inductor of

5 \mathrm{mH}

Inductive reactance in the circuit is :

A

6.28 \Omega

B

0.318 \Omega

C

0.5 \Omega

D

3.14 \Omega

Correct Answer

Option D

Solution

$\omega$ = 628 rad/s $X_{L}=L \omega$

\begin{aligned} & =5 \mathrm{mH} \times 628 \\\\ & =3.14 \Omega \end{aligned}

Q8

In a series LR circuit with

\mathrm{X_L=R}

, power factor P1. If a capacitor of capacitance C with

\mathrm{X_C=X_L}

is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :

A 1 :

\sqrt2

B 1 : 3

C 1 : 2

D 1 : 1

Correct Answer

Option A

Solution

{X_L} = R

\Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}

Now,

{X_L} = {X_C} = R

\Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1

\Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}

Q9

In an

LCR

series

a.c.

circuit, the voltage across each of the components,

L,C

and

R

is

50V

. The voltage across the

L.C

combination will be :

A

100V

B

50\sqrt 2

C

50

V

D

0

V

(zero)

Correct Answer

Option D

Solution

Since the phase difference between

L

&

C

is

\pi ,

$\therefore$ net voltage difference across

LC=50-50=0

Q10

Alternating current can not be measured by

D.C.

ammeter because

A Average value of current for complete cycle is zero

B

A.C.

Changes direction

C

A.C.

can not pass through

D.C.

Ammeter

D

D.C.

Ammeter will get damaged.

Correct Answer

Option A

Solution

D.C.

ammeter measure average current in

AC

current, average current is zero for complete cycle. Hence reading will be zero.