Alternating Current Test

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Alternating Current Test - 19

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Question 1
1 / -0

The instantaneous value of emf and current in an A.C. circuit are; $$E=1.414sin\left ( 100\pi t-\frac{\pi }{4} \right )$$ , $$I=0.707sin(100\pi t)$$ . The impedance of the circuit will be

A
$$1\Omega $$

B
$$2\Omega $$

C
$$\sqrt{2}\Omega $$

D
$$\dfrac{1}{2}\Omega $$

Solution

Given : $$I_{max} = 0.707 \ A$$ and $$E_{max} = 1.414 \ $$ volts
Impedance $$Z = \dfrac{E_{max}}{I_{max}}$$
$$Z= \dfrac{1.414}{0.707}$$
$$Z= 2 \Omega$$
Question 2
1 / -0

The capacitive reactance of 50 $$\mu$$F capacitance at a frequency of $$2 \times 10^{3}$$Hz will be ____ $$\Omega $$

A
$$\frac{2}{\pi }$$

B
$$\frac{3}{\pi }$$

C
$$\frac{4}{\pi }$$

D
$$\frac{5}{\pi }$$

Solution

Capacitive reactance = $$\frac{1}{\omega C}$$

$$= \frac{1}{2\pi f C}$$

$$= \frac{1}{2\pi 2\times 10^3 \times 50\times 10^{-6}}$$

$$= \frac{5}{\pi} \Omega$$
Question 3
1 / -0

The phase angle between current and voltage in a purely inductive circuit is :

A
$$zero$$

B
$$\pi $$

C
$$\pi $$/4

D
$$\pi $$/2

Solution

In the above image waveform of current and voltage in puerly inductive circuit with time is shown.
It is clear from the image that current lags voltage by $$90^o$$.
Hence phase angle between current and voltage in purely inductive circuit is $$\pi /2$$
Question 4
1 / -0

The instantaneous value of emf and current in an A.C. circuit are; $$E=1.414sin\left ( 100\pi t-\frac{\pi }{4} \right )$$ , $$I=0.707sin(100\pi t)$$ . The admittance of the circuit will be _______ mho.

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

B
$$\sqrt{2}$$

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

D
$$2$$

Solution

Admittance =$$\dfrac{1}{impedance}= \dfrac{I_{max}}{V_{max}}$$
$$= \dfrac{0.707}{1.414}$$
$$= \dfrac{1}{2}$$mho
Question 5
1 / -0

In a circuit, the frequency is $$f=\dfrac{1000}{2\pi }$$Hz and the inductance is 2 henry, then the reactance will be

A
200$$\Omega $$

B
200$$ \mu$$ $$\Omega $$

C
2000$$\Omega $$

D
2000$$ \mu$$ $$\Omega $$

Solution

Given : $$f = \dfrac{1000}{2\pi}$$ $$L = 2 \ H$$
Thus $$w = 2\pi f = 2\pi\times \dfrac{1000}{2\pi} = 1000$$
Reactance $$X_L =\omega L$$
$$X_L = 1000\times 2= 2000\Omega$$
Question 6
1 / -0

The instantaneous value of emf and current in an A.C. circuit are: $$E=1.414sin\left ( 100\pi t-\frac{\pi }{4} \right )$$, $$I=0.707sin(100\pi t)$$ .The reactance of the circuit
will be

A
$$\sqrt{2}$$ $$\frac{1}{2}\Omega$$

B
2$$\Omega$$

C
$$\frac{1}{2}\Omega$$

D
$$\frac{1}{\sqrt{2}}\Omega$$

Solution

reactance of the circuit = $$\frac{E_{max}}{I_{max}}$$
$$= \frac{1.414}{0.707}$$
$$= 2 \Omega$$
Question 7
1 / -0

The instantaneous value of emf and current in an
A.C. circuit are; $$E=1.414sin\left ( 100\pi t-\dfrac{\pi }{4} \right )$$ ,
$$I=0.707sin(100\pi t)$$ . RMS value of current will be

A
1A

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

C
$$\sqrt{2}A$$

D
$$\dfrac{1}{2}A$$

Solution

Given : $$I_{max} = 0.707 \ A$$
Rms value of current $$I_{rms} = \dfrac{I_{max}}{\sqrt{2}}$$
$$I_{rms}= \dfrac{.707}{\sqrt{2}}$$
$$= \dfrac{1}{2}A$$
Question 8
1 / -0

The instantaneous value of emf and current in an A.C. circuit are; $$E=1.414sin\left ( 100\pi t-\dfrac{\pi }{4} \right )$$ , $$I=0.707sin(100\pi t)$$. The RMS value of emf will be

A
$$2\sqrt{2}V$$

B
$$1 V$$

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

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

Solution

$$E_{rms} = \dfrac{E_{max}}{\sqrt{2}}$$

$$= \dfrac{1.414}{\sqrt{2}}$$$$= 1 V$$
Question 9
1 / -0

In an AC circuit containing only capacitance, the current :

A
leads the voltage by 180

B
lags the voltage by 90

C
leads the voltage by 90

D
remains in phase with the voltage

Solution

Current leads by $$90^{\circ}$$
Question 10
1 / -0

An inductive coil has a resistance of 100 $$\Omega $$ . When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by $$45^{o}$$. The inductance of the coil is

A
$$\dfrac{1}{100\pi }$$H

B
$$\dfrac{1}{20\pi }$$H

C
$$\dfrac{1}{40\pi }$$H

D
$$\dfrac{1}{60\pi }$$H

Solution

$$X_L = \omega L$$

$$L = \dfrac{X_L}{\omega }$$ $$= \dfrac{100}{2\pi f}$$ $$= \dfrac{100}{2\pi 1000}$$ $$= \dfrac{1}{20\pi}H$$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 19

Result

Alternating Current Test - 19

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SHARING IS CARING

Question 1

$$1\Omega $$

$$2\Omega $$

$$\sqrt{2}\Omega $$

$$\dfrac{1}{2}\Omega $$

Solution

Question 2

$$\frac{2}{\pi }$$

$$\frac{3}{\pi }$$

$$\frac{4}{\pi }$$

$$\frac{5}{\pi }$$

Solution

Question 3

$$zero$$

$$\pi $$

$$\pi $$/4

$$\pi $$/2

Solution

Question 4

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

$$\sqrt{2}$$

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

$$2$$

Solution

Question 5

200$$\Omega $$

200$$ \mu$$ $$\Omega $$

2000$$\Omega $$

2000$$ \mu$$ $$\Omega $$

Solution

Question 6

$$\sqrt{2}$$ $$\frac{1}{2}\Omega$$

2$$\Omega$$

$$\frac{1}{2}\Omega$$

$$\frac{1}{\sqrt{2}}\Omega$$

Solution

Question 7

1A

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

$$\sqrt{2}A$$

$$\dfrac{1}{2}A$$

Solution

Question 8

$$2\sqrt{2}V$$

$$1 V$$

$$\dfrac{1}{2}V$$

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

Solution

Question 9

leads the voltage by 180

lags the voltage by 90

leads the voltage by 90

remains in phase with the voltage

Solution

Question 10

$$\dfrac{1}{100\pi }$$H

$$\dfrac{1}{20\pi }$$H

$$\dfrac{1}{40\pi }$$H

$$\dfrac{1}{60\pi }$$H

Solution

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