Alternating Current Test

Result

Alternating Current Test - 85

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 R-C series A.C. circuit, current is by $$\delta $$ than voltage.

A
$$ahead,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{1}{{\omega CR}}} \right)$$

B
$$behind,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{1}{{\omega CR}}} \right)$$

C
$$ahead,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{{\omega C}}{R}} \right)$$

D
$$behind,Where\;\delta = {\tan ^{ - 1}}\left( {\dfrac{{\omega C}}{R}} \right)$$
Question 2
1 / -0

An A.C. circuit consists of a resistance and a choke in series. The resistance is of $$220$$ $$\Omega$$ and choke is of $$0.7$$ henry. The power absorbed from $$220$$ volts and $$50$$ Hz, source connected with the circuit, is?

A
$$120.08$$ watt

B
$$109.97$$ watt

C
$$100.08$$ watt

D
$$98.08$$ watt

Solution
Question 3
1 / -0

In the series LCR circuit, the voltmeter and ammeter readings are respectively.

A
$$V=250V, I=4A$$

B
$$V=150V, I=2A$$

C
$$V=1000V, I=5A$$

D
$$V=100V, I=2A$$

Solution

Ref. image 1

As, current leads the voltage across capacitor by $$\pi/2$$, while current lags from voltage in inductor by, $$\pi/2$$. Also current and voltage are in same phase across resistance

Ref. image 2

as sum of net voltage = Source voltage
Sum of $$V_L , V_C$$ & $$V_R = \sqrt{V_R^2 + (V_L - V_C)^2}$$

as $$V_L = 200 V, \, V_C = 200 V, V_R = V$$

$$\Rightarrow \sqrt{V^2 + (200 - 200)^2} = 100 volt$$

$$\Rightarrow V = 100 volt$$

Also $$V_L = iX_L, V_C = i X_C \, \left\{\begin{matrix} X_L = inductive \, reactance\\ X_C = capacitive \, reactance\end{matrix}\right.$$

Also $$V_L = V_C = 200$$

$$\Rightarrow X_L = X_C$$

Now, net impedance of the circuit

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

as $$X_L = X_C$$

$$\Rightarrow Z = R $$ and $$R = 50 \Omega$$ given

$$\Rightarrow Z = 50 \Omega$$

Npw, current in the circuit

$$i = \dfrac{source \, voltage}{impedance} = \dfrac{100}{Z}$$

$$\Rightarrow i = \dfrac{100}{50} = 2A$$

Hence, value of $$V = 100\, volt \,\, A = 2A$$

Option D is correct.
Question 4
1 / -0

An alternating emf is applied across a parallel combination of a resistance $$R,$$ capacitance $$C$$ and an inductance $$L.$$ If $$I _ { R } , I _ { L } , I _ { C }$$ are the currents through $$R ,$$ $$L$$ and $$C ,$$ respectively, then the diagram which correctly represents, the phase relationship among $$I _ { R } , I _ { L } , I _ { C }$$ and source emf $$E,$$ is given by

A

B

C

D
Question 5
1 / -0

In an A.C circuit, a resistance of R ohm is connected in series with an indutance L.If phase angle between voltage and current be$${ 45 }^{ 0 }$$, the value of inductive reacatance will be

A
R/4

B
R/2

C
R

D
R/5

Solution
Question 6
1 / -0

In the series LCR circuit (Fig. 7.33) , the voltmeter and ammeter readings are :

A
$$ V = 100 \,V , I = 2A $$

B
$$ V = 100 \,V , I = 5A $$

C
$$ V = 1000 \,V , I = 2A $$

D
$$ V = 300 \,V , I = 1A $$

Solution

Here , $$ V_L = V_C $$ . They are in opposite phase . Hence , they will cancel each other . Now the resultant potential difference is equal to the applied difference $$ = 100\,V $$

$$ Z = R $$ $$ ( \therefore \, X_L = X_C ) $$

$$ \therefore $$ $$ I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{V_{rms}}{R} = \dfrac{100}{50} = 2\,A $$
Question 7
1 / -0

In a circuit, the phase of the current is lagging behind $$\pi /3$$ angle with phase of voltage. Element present in circuit are:

A
$$R$$ and $$C$$

B
$$R$$ and $$L$$

C
$$L$$ and $$C$$

D
only $$L$$

Solution
Question 8
1 / -0

The alternating current is given by the equation $$i = 10\sin \left ( 100 \pi t + \dfrac{\pi}{6} \right ).$$ The current attains its first maximum at t is :

A
$$\dfrac{1}{600}s$$

B
$$\dfrac{1}{50}s$$

C
$$\dfrac{1}{100}s$$

D
$$\dfrac{1}{300}s$$

Solution

Given, $$i = 10\sin \left(100\pi t + \dfrac{\pi}{6}\right)$$

At $$t = 0$$ $$i = 10\sin \dfrac{\pi}{6} = 5$$

$$\dfrac{di}{dt} = 10 \cos \left(100\pi t + \dfrac{\pi}{6} \right) (100 \pi)$$

$$\left[ \dfrac{di}{dt}\right]_{t=0} = 10\cos \left(\dfrac{\pi}{6}\right) (100) \pi >0$$

$$\dfrac{di}{dt}$$ at $$t = 0$$ is +ve, so sin function is increasing at $$t = 0$$. therefore next maxima will be $$i = 10$$

$$10 = 10\sin \left(100\pi t + \dfrac{\pi}{6}\right)$$

$$1 = \sin \left(100\pi t + \dfrac{\pi}{6}\right)$$

$$\dfrac{\pi}{2} = 100\pi t + \dfrac{\pi}{6}$$

$$100\pi t = \dfrac{2\pi}{ 6} = \dfrac{\pi}{3}$$

$$t = \dfrac{1}{300}s$$.
Question 9
1 / -0

How can you decrease current in an alternating current circuit without any loss of power?

A
By using pure inductor

B
By using pure resistance

C
By using resistance and inductor

D
By using resistance and capacitor

Solution

By using pure inductor
Question 10
1 / -0

In a radio receiver, the short wave and medium wave station are tuned by using the same capacitor but coils of difference inductance $$L_s$$ and $$L_m$$ respectively then

A
$$L_s > L_m$$

B
$$L_s < L_m$$

C
$$L_s = L_m$$

D
$$None\ of\ these$$

Solution

As $$v=\dfrac {c}{\lambda}\Rightarrow v_m=\dfrac {c}{\lambda_m}$$ and $$v_s =\dfrac {c}{\lambda_s}$$
$$\because \lambda_m > \lambda_s \Rightarrow v_m < v_s$$
Also $$v_m=\dfrac {1}{2\pi \sqrt {L_m C}}$$ and $$v_s =\dfrac {1}{2\pi \sqrt {L_s C}}$$
$$\Rightarrow \dfrac {v_m}{v_s}=\sqrt {\dfrac {L_s}{L_m}}\Rightarrow L_s < L_m$$