Alternating Current Test

Result

Alternating Current Test - 52

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 the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be

A
0 V, 8 A

B
150 V, A 8

C
150 V, 3 A

D
0 V, 3 A

Solution

Solution:-
Given:- $$X_C=X_L$$, V=240V, R=30$$\Omega$$
Since X_L=X_C, So,
Z=$$\sqrt{R^2+{\left(X_L-X_C\right)}^2}$$=R
So, potential drop across C $$\&$$ L is zero and current is $$\rightarrow$$
i=$$\cfrac{V}{Z}=\cfrac{V}{R}=\cfrac{240}{30}=8A$$
i=8A
Question 2
1 / -0

A $$5cm$$ long solenoid having $$10$$ ohm resistance and $$5mH$$ inductance is joined to a $$10V$$ battery. At steady state, the current through the solenoid (in ampere) will be

A
$$5$$

B
$$2$$

C
$$1$$

D
zero

Solution

At steady state inductor behave like short circuit.
$$ i = \cfrac{v}{r} = \cfrac{10}{10} = 1A$$
Question 3
1 / -0

Consider the $$R-L-C$$ circuit given below. The circuit is driven by a $$50\ Hz\ AC$$ source with peak voltage $$220\ V$$. If $$R = 400\Omega, C = 200\mu F$$ and $$L = 6H$$, the maximum current in the circuit is closest to

A
$$0.120\ A$$

B
$$0.55\ A$$

C
$$1.2\ A$$

D
$$5.5\ A$$

Solution

At maximum current $$X_{c} = X_{L}$$
so Z = R
so $$I = \dfrac{v}{R}$$
so I = 0.55 A
Question 4
1 / -0

Consider $$L, C, R$$ circuit as shown in figure, with a.c. source of peak value $$V$$ and angular frequency $$\omega$$. Then the peak value of current through the ac source.

A
$$\dfrac {V}{\sqrt {\dfrac {1}{R^{2}} + \left (\omega L - \dfrac {1}{\omega C}\right )^{2}}}$$

B
$$V \sqrt {\dfrac {1}{R^{2}} + \left (\omega C - \dfrac {1}{\omega L}\right )^{2}}$$

C
$$\dfrac {V}{\sqrt {R^{2} + \left (\omega L - \dfrac {1}{\omega C}\right )^{2}}}$$

D
$$\dfrac {VR\omega C}{\sqrt {\omega^{2}C^{2} + R(\omega^{2} C^{2} - 1)^{2}}}$$

Solution

$$X_c=\cfrac 1 {j\omega c}$$
$$X_L=j\omega L$$
And $$R$$ are in parallel,
Therefore, equivalent admittance of the circuit is
$$Y=\cfrac { 1 }{ { X }_{ c } } +\cfrac { 1 }{ { X }_{ L } } +\cfrac { 1 }{ R } .$$
$$Y=\cfrac { 1 }{ R } +j(c\omega -\cfrac { 1 }{ L\omega } )$$
$$\left| Y \right| =\sqrt { \cfrac { 1 }{ { R }^{ 2 } } +({ \omega c-\cfrac { 1 }{ \omega L } ) }^{ 2 } } $$
Current $$I=VY=V\sqrt { { \cfrac { 1 }{ { R }^{ 2 } } +({ \omega c-\cfrac { 1 }{ \omega L } ) }^{ 2 } } } $$
Question 5
1 / -0

Two coils have mutual inductance $$0.005H$$. The current changes in the first coil according to equation $$I={ I }_{ 0 }\sin { \omega t } $$, where $${ I }_{ 0 }=10A$$ and $$\omega =100\pi rad\quad { s }^{ -1 }$$. The maximum value of emf in the second coil is

A
$$5$$

B
$$5\pi $$

C
$$0.5\pi $$

D
$$\pi$$

Solution

Current is given as $$I = I_o \ \sin wt = 10\sin (100\pi t)$$
Rate of change of current $$\dfrac{dI}{dt} = I_o w \cos wt = 1000\pi \cos (100\pi t)$$
Thus $$\dfrac{dI}{dt}\bigg|_{max} = 1000\pi $$
Given : $$L = 0.005 \ H$$
Maximum emf $$E_{max} = L\dfrac{dI}{dt}\bigg|_{max} = 0.005\times 1000\pi = 5\pi$$
Question 6
1 / -0

In the series LCR circuit shown the impedance is:

A
$$200 \Omega$$

B
$$100\Omega$$

C
$$300\Omega$$

D
$$500\Omega$$

Solution

Hence , $$L = 1 H , C = 20 \mu F = 20 \times 10^{-6}F$$
R$$ = 300 \Omega, \upsilon = \dfrac{50}{\pi} Hz$$
The inductive reactance is
$$X_L = 2 \pi \upsilon L \, = \, 2 \times \pi \times \dfrac{50}{\pi} \times 1 = 100 \Omega$$
The capacitive reactance is
$$X_C = \dfrac{1}{2 \pi \upsilon C} = \dfrac{1}{2 \times \pi \times \dfrac{50}{\pi} \times 20 \times 10^{-6}} = 500 \Omega$$
The impedance of the series LCR cicuit is
$$ Z = \sqrt{R^2 + (X_C -X_L)^2}\,\, = \,\,\sqrt{(300)^2 + (500-100)^2} $$

$$Z = \sqrt{(300)^2 + (400)^2} = 500\, \Omega $$
Question 7
1 / -0

The reciprocal of the impedance of an electric current is?

A
Admittance

B
Susceptance

C
Conductance

D
Reactance

Solution

As conductance is the complement of resistance, there is also a complementary expression of reactance or impedance, called susceptance
Question 8
1 / -0

A capacitor 'C' is connected across a D.C. source, the reactance of capacitor will be _________.

A
ZERO

B
HIGH

C
LOW

D
INFINITE

Solution

Frequency of a source voltage $$f = 0$$
Reactance of capacitance $$X_C = \dfrac{1}{2\pi f C}$$
$$\implies \ X_C = \infty$$
Correct answer is option D.
Question 9
1 / -0

A 5 $$\mu$$F capacitor is connected to a 200 V, 100 Hz ac source. The capacitive reactance is:

A
$$212 \Omega$$

B
$$312 \Omega$$

C
$$318 \Omega$$

D
$$412 \Omega$$

Solution

Given: The capacitance of the capacitor is $$5\mu F$$.
The voltage supplied is $$200\ V$$.
The frequency of the voltage supplied is $$100\ Hz$$.

To find: The capacitive reactance of the capacitor.

The reactance of the capacitor is given by:
$$X_C=\cfrac{1}{\omega_C}=\cfrac{1}{2\pi f_C}$$
$$\Rightarrow\cfrac{1}{2\pi\times100\times5}\\ \quad=318\ \Omega$$
Question 10
1 / -0

Phase difference between voltage and current in a capacitor in an ac circuit is then

A
$$\pi$$

B
$$\pi/2$$

C
$$0$$

D
$$\pi/3$$

Solution

In a capacitive ac circuits , the voltage lags
behind the current in phase by $$\pi /2$$ radian.