Alternating Current Test

Result

Alternating Current Test - 64

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In an alternating current circuit, the phase difference between current l and voltage is $$\phi$$, then the Wattless component of current will be -

A
$$I\cos \phi$$

B
$$I \tan\phi$$

C
$$I \sin\phi$$

D
$$I \cos^2\phi$$

Solution
Question 2
1 / -0

In a RLC series circuit at resonance, the value of the power factor is :

A
infinity

B
zero

C
$$\dfrac{1}{\sqrt{2}}$$

D
$$1$$

Solution

At resonance, $$X_L = X_c$$
$$\implies \ Z = \sqrt{R^2 + (X_L-X_c)^2} = R$$
Power factor $$\cos\phi = \dfrac{R}{Z} = \dfrac{R}{R} = 1$$
Question 3
1 / -0

A current of 2A flow in an inductance of 10H. for generating 100V EMF in it the rate of change current will be.

A
1 A/sec

B
5 A/sec

C
10 A/sec

D
100 A/sec
Question 4
1 / -0

A source of 220V is applied in an A.C. circuit. The value of resistance is $$220 \Omega$$. Frequency & inductance are 50 Hz and 0.7 H, then wattless current is

A
$$0.5 amp$$

B
$$0.7 amp$$

C
$$1.0 amp$$

D
none

Solution

$$ \Rightarrow R = 2\Omega $$
$$ \Rightarrow {V_{rms}} = 220V$$
$$ \Rightarrow {V_ \circ } = 220\sqrt 2 $$
$$\Rightarrow {X_L} = WL = 100\pi \times 0.7$$
$$ \Rightarrow 100\dfrac{{22}}{7} \times \dfrac{7}{{10}} = 220\Omega $$
$$\therefore z = 220\sqrt 3 $$
also $$\tan \phi = \dfrac{{{V_L}}}{R} = 1$$
$$\phi = {45^ \circ }$$
$$\therefore$$ wattless current$$ = {i_ \circ }\sin \phi $$
$$ = \dfrac{{220\sqrt 3 }}{{220\sqrt 2 }}\sin {45^ \circ }$$
$$ = \dfrac{1}{{\sqrt 2 }} = 0.7A$$
so we get $$0.7A$$
Hence option$$B$$ is correct.
Question 5
1 / -0

The self inductance of a choke coil is mH. when it is connected with a 10 VDC source then the loss of power is 20 watt. When it connected with 10 volt AC source loss of power is 10 watt. The frequency of AC source will bw-

A
50 Hz

B
60 Hz

C
80 Hz

D
100 Hz

Solution

$$\begin{array}{l} When\, \, \, dc\, \, is\, \, pass\, \, through\, \, the\, \, inductor\, \, no\, \, induv\tan ce\, \, effect\, \, will\, \, be\, \, their\, \, only \\ resis\tan ce\, \, will\, \, be\, their\, \, resis\tan ce\, \, of\, \, the\, \, inductor\, \, be\, \, R \\ 20=\frac { { 100 } }{ R } \\ R=5\Omega \\ Xl=\omega \times 0.001 \\ power\, \, =\frac { { { v^{ 2 } } } }{ z } \cos \phi \\ Z=\sqrt { 25+{ \omega ^{ 2 } }\times { { 10 }^{ -6 } } } \\ \cos \phi =\frac { R }{ Z } =\frac { 5 }{ Z } \\ 10=\frac { { 100 } }{ Z } \frac { 5 }{ Z } =\frac { { 500 } }{ { { Z^{ 2 } } } } =\frac { { 500 } }{ { 25 } } +{ \omega ^{ 2 } }\times { 10^{ -6 } } \\ 25+{ \omega ^{ 2 } }\times { 10^{ -6 } }=50 \\ { \omega ^{ 2 } }=25\times { 10^{ 6 } } \\ \omega =5000 \end{array}$$
$$=50 Hz$$
Hence, Option $$A$$ is correct answer.
Question 6
1 / -0

In the phasors method for the only capacitive A.C. circuit, voltage is in +X direction then current is in direction.

A
+X

B
-X

C
+Y

D
-Y

Solution
Question 7
1 / -0

If an ac source is connected across ideal capacitor and current passing through it is dentoed by curve then instantaneous power is denoted by curve

A
$$c$$

B
$$b$$

C
$$e$$

D
$$a$$
Question 8
1 / -0

In LC oscillation resistance is $$100\Omega$$ and inductance and capacitance is $$1$$H and $$10\mu F$$. Find the half power of frequency.

A
$$266.2$$

B
$$366.2$$

C
$$166.2$$

D
$$233.2$$

Solution

$$f_n=\omega_0-\dfrac{R}{2L}=\dfrac{1}{\sqrt{LC}}-\dfrac{R}{2L}$$

$$=\dfrac{1}{\sqrt{1\times 10\times 10^{-6}}}$$ $$-\dfrac{100}{2\times 1}$$

$$=\dfrac{1000}{\sqrt{10}}-\dfrac{100}{2}$$
$$=100\left[\sqrt{10}-0.5\right]=266.27$$.
Question 9
1 / -0

If in an A.C., L-C series circuit $$X_c > X_L$$. Hence potential _________.

A
Lags behind the current by $$\pi/2$$

B
Leads the current by $$\pi$$ in phase

C
Leads the current by $$\pi/2$$ in phase

D
Lags behind the current by $$\pi$$ in phase

Solution

Potential lags by $$\pi/2$$ phase.
Question 10
1 / -0

Two inductors each of inductance L are connected in parallel . what is their equivalent inductance?

A
2 L

B
L

C
L/ 2

D
L/ 4

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Alternating Current Test - 64

Result

Alternating Current Test - 64

Score

-

Rank

-

SHARING IS CARING

Question 1

$$I\cos \phi$$

$$I \tan\phi$$

$$I \sin\phi$$

$$I \cos^2\phi$$

Solution

Question 2

infinity

zero

$$\dfrac{1}{\sqrt{2}}$$

$$1$$

Solution

Question 3

1 A/sec

5 A/sec

10 A/sec

100 A/sec

Question 4

$$0.5 amp$$

$$0.7 amp$$

$$1.0 amp$$

none

Solution

Question 5

50 Hz

60 Hz

80 Hz

100 Hz

Solution

Question 6

+X

-X

+Y

-Y

Solution

Question 7

$$c$$

$$b$$

$$e$$

$$a$$

Question 8

$$266.2$$

$$366.2$$

$$166.2$$

$$233.2$$

Solution

Question 9

Lags behind the current by $$\pi/2$$

Leads the current by $$\pi$$ in phase

Leads the current by $$\pi/2$$ in phase

Lags behind the current by $$\pi$$ in phase

Solution

Question 10

2 L

L

L/ 2

L/ 4

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.