Alternating Current Test

Result

Alternating Current Test - 31

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

A
$$\dfrac{R}{4}$$

B
$$\dfrac{R}{2}$$

C
$$R$$

D
$$\dfrac{R}{3}$$

Solution

Phase angle $$(\phi) = 45^0 = {tan}^{-1}(1)$$

$$= {tan}^{-1}(\dfrac{X_L}{R})$$

So, $$\dfrac{X_L}{R} = 1$$

$$\Rightarrow X_L = R$$
Question 2
1 / -0

When the frequency of applied emf in an LCR series circuit is less than the resonant frequency,then the nature of the circuit will be:
a) Capacitive b) resistive c) inductive

A
only a is true

B
only a and b are true

C
only b and c are true

D
a, b and c are true

Solution

$$X_C = \dfrac{1}{\omega C}$$
So, $$X_C \propto \dfrac{1}{\omega} = \dfrac{1}{2\pi f}$$
$$X_L = \omega L$$ So, $$X_L \propto \omega = 2\pi f$$
So, as frequency decreases $$X_C$$ increases So, circuit becomes capacitive circuit.
Question 3
1 / -0

Figure shows a series LCR circuit connected to a variable frequency 200V source. The source frequency which drives the circuit at resonance is

A
50 Hz

B
(50 /$$\pi$$ )Hz

C
25 Hz

D
(25 /$$\pi$$ ) Hz

Solution

For resonance frequency

$$X_L = X_C$$

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

$$2\pi f = \dfrac{1}{\sqrt{LC}}$$

$$f = \dfrac{1}{2\pi}\times \dfrac{1}{\sqrt{LC}}$$

$$= \dfrac{1}{2\pi}\times \dfrac{1}{5\times 80\times 10^{-6}}$$

$$= \dfrac{1}{2\pi}\times \dfrac{10^3}{20}$$

$$= \dfrac{25}{\pi}Hz$$
Question 4
1 / -0

A 100$$\Omega $$ resistance is connected in series with a 4H inductor. The voltage across the resistor is $$V_{R}=2sin(1000t)V$$. The voltage across the inductor is:

A
$$80sin\left ( 1000t+\dfrac{\pi }{2} \right )$$

B
$$40sin\left ( 1000t+\dfrac{\pi }{2} \right )$$

C
$$80sin\left ( 1000t-\dfrac{\pi }{2} \right )$$

D
$$40sin\left ( 1000t-\dfrac{\pi }{2} \right )$$

Solution

$$V_R = 2sin (1000t)V$$ So, $$V_{max}(R) = 2V$$

So, $$w = 1000$$

$$2\pi f = 1000$$

$$f = \dfrac{500}{\pi}Hz$$

$$X_L = wL$$

$$= 1000\times 4$$

$$= 4000 \Omega$$

$$R = 100 \Omega$$

$$i_R = \dfrac{V_R}{R} = \dfrac{2}{100} = 2\times 10^{-2}A = i_L$$
($$\because$$current is same in series)

$$V_{max}(L) = i_L\times X_L$$

$$= 4000\times 2\times 10^{-2}$$

$$= 80V$$

So, $$V_L = 80 sin (1000t + \dfrac{\pi}{2})$$

Since in inductor voltage lead by $$\dfrac{\pi}{2}$$
Question 5
1 / -0

A condenser of 10 $$\mu$$F and an inductor of 1H are connected in series with an A.C. source of frequency 50Hz. The impedance of the combination will be (take $$\pi^{2}$$ $$=$$10 ):

A
zero

B
Infinity

C
44.7$$\Omega $$

D
5.67$$\Omega $$

Solution

Impedance = $$X_L - X_C = \omega L-\dfrac{1}{\omega C}$$

$$= 2\pi f L -\dfrac{1}{2\pi f C}$$

$$= 2\pi \times 50\times 1-\dfrac{1}{2\pi \times 50\times 10\times 10^{-6}}$$

$$= \pi \left (100 - \dfrac{10^6}{2{\pi}^2\times 50\times 10} \right )$$

$$= \pi \left(100-\dfrac{10^6}{2\times 10 \times 50 \times10} \right)$$
  $$(\because {\pi}^2 = 10)$$
$$= 0$$
So, impedance $$= 0$$
Question 6
1 / -0

In an L-C-R series circuit, $$R=\sqrt{5}\Omega ,X_{L}=9\Omega ,X_{C}=7\Omega $$. If applied voltage in the circuit is 50V then impedance of the circuit in ohm will be

A
$$2$$

B
$$3$$

C
$$2\sqrt{5}$$

D
$$3\sqrt{5}$$

Solution

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

$$= \sqrt{5+(9-7)^2}$$

$$= \sqrt{5+(2)^2}$$

$$= \sqrt{5+4}$$

$$= \sqrt{9}$$

$$= 3\Omega$$
Question 7
1 / -0

A coil of inductance 0.1H is connected to 50V, 100Hz generator and current is found to be 0.5A. The potential difference across resistance of the coil is

A
15V

B
20V

C
25V

D
39V

Solution

$$Z= \dfrac{V}{i}$$

$$= \dfrac{50}{0.5}$$

$$= 100\Omega$$

$$Z^2 = X_L^+R^2$$

$$(100)^2 = (2\pi fL)^2+R^2$$

$$(100)^2-[2\pi 100(0.1)]^2 = R^2$$

$$6052 = R^2$$

$$R = 78\Omega$$

$$V_R = iR$$

$$= 0.5\times 78$$

$$= 39V$$
Question 8
1 / -0

In an LR circuit, R $$=$$ 10 $$\Omega $$ and L $$=$$ 2H. If an alternating voltage of 120V and 60Hz is connected in this circuit, then the value of current flowing in it will be _______ A (nearly)

A
0.32

B
0.16

C
0.48

D
0.8

Solution

Impedance(z) = $$\sqrt{R^2+X_L^2}$$

$$= \sqrt{(10)^2+(2\pi fL)^2}$$

$$= \sqrt{(10)^2+(2\pi \times 60\times 2)^2}$$

$$= \sqrt{100+{\pi}^2\times 57600}$$

$$= \sqrt{100+568480}$$

$$= 754.04$$

$$i = \dfrac{V}{z}$$

$$= \dfrac{120}{754.04}$$

$$= 0.16A$$
Question 9
1 / -0

A 220V, 50Hz a.c. generator is connected to an inductor and a 50$$\Omega $$ resistance in series. The current in the circuit is 1.0A. The P.D. across the inductor is:

A
102.2 V

B
186.4 V

C
214.24 V

D
302 V

Solution

Current = $$\dfrac{V}{Z}$$

$$1 = \dfrac{220}{Z}$$

$$\Rightarrow Z = 220\Omega$$

$$Z = \sqrt{X_L^2+R^2}$$

$$(220)^2 = X_L^2+(50)^2$$

$$X_L^2 = 45,900$$

$$X_L = 214.24$$

$$V_L = iX_L$$

$$= (1)214.24$$

$$= 214.24V$$
Question 10
1 / -0

A circuit operating at $$\dfrac{360}{2\pi }$$ Hz $$\Omega $$ contains a 1 F $$\mu $$ capacitor and a 20 resistor. The inductor must be added in series to make the phase angle for the circuit zero is

A
7.7 H

B
10 H

C
3.5 H

D
15 H

Solution

To make phase angle zero, circuit should be purely resistive, so, reactance should be zero.
Hence, $$X_L = X_C$$
$$wL = \dfrac{1}{\omega C}$$
$$L = \dfrac{1}{\omega^2 C}$$
$$L= \dfrac{1}{(2\pi f)^2C}$$
$$L= \dfrac{1}{(2\pi \dfrac{360}{2\pi})^2\times 10^{-6}}$$
$$L= \dfrac{10^6}{129600}$$
$$L= 7.7H$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Alternating Current Test - 31

Result

Alternating Current Test - 31

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac{R}{4}$$

$$\dfrac{R}{2}$$

$$R$$

$$\dfrac{R}{3}$$

Solution

Question 2

only a is true

only a and b are true

only b and c are true

a, b and c are true

Solution

Question 3

50 Hz

(50 /$$\pi$$ )Hz

25 Hz

(25 /$$\pi$$ ) Hz

Solution

Question 4

$$80sin\left ( 1000t+\dfrac{\pi }{2} \right )$$

$$40sin\left ( 1000t+\dfrac{\pi }{2} \right )$$

$$80sin\left ( 1000t-\dfrac{\pi }{2} \right )$$

$$40sin\left ( 1000t-\dfrac{\pi }{2} \right )$$

Solution

Question 5

zero

Infinity

44.7$$\Omega $$

5.67$$\Omega $$

Solution

Question 6

$$2$$

$$3$$

$$2\sqrt{5}$$

$$3\sqrt{5}$$

Solution

Question 7

15V

20V

25V

39V

Solution

Question 8

0.32

0.16

0.48

0.8

Solution

Question 9

102.2 V

186.4 V

214.24 V

302 V

Solution

Question 10

7.7 H

10 H

3.5 H

15 H

Solution

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.