Alternating Current Test

Result

Alternating Current Test - 66

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

Which one of the following represents capacitive reatance versus angular frequency graph?

A

B

C

D

Solution
Question 2
1 / -0

If $$ E_0 $$ represents the peak value of the voltage in an ac circuit, the r.m.s. value of the voltage will be[

A
$$ \dfrac{E_0}{\pi} $$

B
$$ \dfrac{E_0}{2} $$

C
$$ \dfrac{E_0}{\sqrt{\pi}} $$

D
$$ \dfrac{E_0}{\sqrt{2}} $$

Solution

If $$E_0$$ represents the peak value of the voltage in an ac circuit then ,
we know,
$$E_{rms}=\dfrac {E_0}{\sqrt 2}$$
Question 3
1 / -0

The resistance of a coil for dc is in ohms. In ac, the resistance

A
will remains same

B
will increase

C
will decrease

D
will be zero

Solution

The coil having inductance $$L$$ besides the resistance $$R$$. Hence for ac it’s effective resistance $$ \sqrt{R^2 + X^{2}_{L}} $$ will be larger than it’s resistance $$R$$ for dc.
Question 4
1 / -0

An alternating current of frequency ' f ' is flowing in a circuit containing a resistance R and a choke L in series. The impedance of this circuit is

A
$$ R + 2 \pi \, fL $$

B
$$ \sqrt{R^2 + 4 \, \pi^2 f^2 L^2} $$

C
$$ \sqrt{R^2 + L^2} $$

D
$$ \sqrt{R^2 + 2 \pi \, fL }$$

Solution

As we know,
$$ Z = \sqrt{R^2 + X_{L}^{2}} , X_L = \omega L $$ and $$ \omega = 2 \pi f $$
$$ \therefore \, Z = \sqrt{R^2 + 4 \pi^2 f^2 L^2} $$
Question 5
1 / -0

Power delivered by the source of the circuit becomes maximum, when

A
$$ \omega L = \omega C $$

B
$$ \omega L = \dfrac{1}{\omega C} $$

C
$$ \omega L = - \left ( \dfrac{1}{\omega C} \right )^2 $$

D
$$ \omega L = \sqrt{\omega C} $$

Solution

As we know,
power delivered by the source of the circuit becomes maximum when, load resistance equals to source resistance.

we know, in L-C-R circuit,

load resistance is inductive reactance and source resistance is reactance of capacitor.

e.g., $$X_L=X_C$$

or, $$\omega L=\dfrac{1}{\omega C}$$

hence, Power delivered by the source of the circuit becomes maximum, when , $$\omega L=\dfrac{1}{\omega C}$$
Question 6
1 / -0

An alternating e.m.f. is applied to purely capacitive circuit. The phase relation between e.m.f. and current flowing in the circuit is or In a circuit containing capacitance only

A
e.m.f is ahead of current by $$ \dfrac{\pi}{2} $$

B
Current is ahead of e.m.f by $$ \dfrac{\pi}{2} $$

C
Current lags behind e.m.f by $$ \pi $$

D
Current is ahead of e.m.f by $$ \pi $$

Solution

As we know,
For purely capacitive circuit $$ e = e_0 \, sin \omega t $$
$$ i = i_0 \, sin \left ( \omega t + \dfrac{\pi}{2} \right ) $$ i.e., current is ahead of emf by $$ \dfrac{\pi}{2} $$
Question 7
1 / -0

A series ac circuit consist of an inductor and a capacitor. The inductance and capacitance is respectively $$1$$ henry and $$ 25\mu F.$$ If the current is maximum in circuit then angular frequency will be

A
$$ 200 $$

B
$$ 100 $$

C
$$ 50 $$

D
$$ 200 / 2\pi $$

Solution

The current in the LC circuit becomes maximum when resonance occurs. So

$$ \omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{1 \times 25 \times 10^{-6}}} = \dfrac{1000}{5} = 200 $$ rad/sec
Question 8
1 / -0

In a series LCR circuit, operated with an ac of angular frequency $$ \omega $$ ,the total impedance is

A
$$ [R^2 + (L \omega - C \omega)^2]^{1/2} $$

B
$$ \left [ R^2 + \left (L \omega - \dfrac{1}{C \omega} \right )^2 \right ]^{1/2} $$

C
$$ \left [R^2 + \left (L \omega - \dfrac{1}{C \omega} \right )^2 \right ]^{-1/2} $$

D
$$ \left [(R \omega)^2 + \left (L \omega - \dfrac{1}{C \omega} \right )^2 \right ]^{1/2} $$

Solution

As we know,
$$X_L=L\omega \ and \ X_C=\dfrac1{\omega C}$$
now,
From the above figure,
$$Z=\sqrt{R^2+(X_L-X_C)^2}$$

$$Z=[R^2+(L\omega -\dfrac1{\omega C})^2]^{\dfrac12}$$
Question 9
1 / -0

When $$ 100 $$ volt dc is applied across a coil, a current of $$1$$ amp flows through it. When $$ 100$$ volt ac at $$ 50$$ cycle $$s^{-1} $$ is applied to the same coil, only $$0.5$$ ampere current flows. The impedance of the coil is

A
$$ 100 \Omega $$

B
$$ 200 \Omega $$

C
$$ 300 \Omega $$

D
$$ 400 \Omega $$

Solution

When dc is supplied $$ R = \dfrac{V}{i} = \dfrac{100}{1} = 100 \, \Omega $$

When ac is supplied $$ Z = \dfrac{V}{i} = \dfrac{100}{0.5} = 200 \, \Omega $$
Question 10
1 / -0

If resistance of $$ 100 \,\Omega $$ inductance of $$ 0.5 $$ henry and capacitance of $$ 10 \times 10^{-6} F $$ are connected in series through $$ 50 $$ Hz ac supply, then impedance is

A
$$ 1.876 $$

B
$$ 18.76 $$

C
$$ 189.72 $$

D
$$ 101.3 $$

Solution

As we know,
$$ Z = \sqrt{R^2 + (X_L - X_C)^2} $$

$$ = \sqrt{100^2 + \left ( 0.5 \times 100 \pi - \dfrac{1}{10 \times 10^{-6} \times 100 \pi} \right )^2} = 189.72 \, \Omega $$