Alternating Current Test

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Alternating Current Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

An inductance coil of $$1$$ $$H$$ and a condenser of capacity $$1$$ $$pF$$ produce resonance. The resonant frequency will be

A
$$\displaystyle \dfrac{10^6}{\pi} Hz$$

B
$$\displaystyle 27 \pi \times 10^6 Hz$$

C
$$\displaystyle \dfrac{2\pi}{10^6} Hz$$

D
$$\displaystyle \dfrac{10^6}{2\pi} Hz$$

Solution

If $$f$$ is resonant frequency then

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

$$L = 1H$$

$$C = 1pF = {10}^{-12}$$

$$f = \dfrac{1}{2\pi\sqrt{ 1 \times {10}^{-12}}}$$

$$f = \dfrac{{10}^{6}}{2\pi} Hz$$
Question 2
1 / -0

A resistance $$R$$ and a capacitor $$C$$ are joined to a source of $$AC$$ of constant e.m.f and variable frequency. The potential difference across $$C$$ is $$V$$. If the frequency of $$AC$$ is gradually increased, $$V$$ will

A
increase

B
decrease

C
remain constant

D
first increase and then decrease

Solution

In complex plane, $$ V = \dfrac { {V}_{AC} }{ 1+ jRC\omega } $$
Therefore as $$\omega ($$frequency$$)$$ increases,$$ V $$ decreases.
Question 3
1 / -0

The natural frequency of a $$L-C$$ circuit is

A
$$\displaystyle \frac{1}{2\pi \sqrt{LC}}$$

B
$$\displaystyle \frac{1}{2\pi} \sqrt{\frac{C}{L}}$$

C
$$\displaystyle \frac{1}{2\pi} \sqrt{\frac{L}{C}}$$

D
$$\displaystyle \sqrt{LC}$$

Solution

When circuit is at natural frequency,

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

$$f=\dfrac { 1 }{ 2\pi \sqrt { LC } } $$
Question 4
1 / -0

If the values of $$L, C$$ and $$R$$ in a series $$L-C-R$$ circuit are 10 mH, 100 $$\mu F$$ and $$100 \Omega$$ respectively then the value of resonant frequency will be

A
$$\displaystyle \frac{10^3}{2 \pi} Hz$$

B
$$\displaystyle 2\times 10^3 Hz$$

C
$$\displaystyle 2\times \frac{10^3}{pi}Hz$$

D
$$\displaystyle 10^3 Hz$$

Solution

Resonant frequency $$= \dfrac { 1 }{ 2\pi \sqrt { LC } } $$

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

$$ =\dfrac { { 10 }^{ 3 } }{ 2\pi } Hz$$
Question 5
1 / -0

In an $$L-C-R$$ circuit the value of $$X_L,X_C$$ and $$R$$ are 300 $$\Omega$$, 200 $$\Omega$$ and 100 $$\Omega$$ respectively.The total impedance of the circuit will be

A
600 $$\Omega$$

B
200 $$\Omega$$

C
141 $$\Omega$$

D
310 $$\Omega$$

Solution

Total impedance $$= R + j({X}_{L} -{X}_{C} ) $$

$$= 100 +j100 $$

$$=100\sqrt { 2 } \measuredangle 45 $$

$$=141 \measuredangle 45 $$
Question 6
1 / -0

The resultant reactance in an $$L-C-R$$ circuit is

A
$$\displaystyle X_L+X_C$$

B
$$\displaystyle X_L-X_C$$

C
$$\displaystyle \sqrt{X^2_L+X^2_C}$$

D
$$\displaystyle \sqrt{X^2_L-X^2_C}$$

Solution

Reactance is the nonresistive component of impedance in an AC circuit, arising from the effect of inductance or capacitance or both and causing the current to be out of phase with the electromotive force causing it.
Therefore, reactance of the $$L-C-R$$ circuit is $$ {X}_{L} - {X}_{C} $$.
Question 7
1 / -0

The angular frequency of an AC source is 10 radian/sec. The reactance of $$1 \mu F$$ capacitor will be

A
$$\displaystyle 10^4 \Omega$$

B
$$\displaystyle 10^2 \Omega$$

C
$$\displaystyle 10^1 \Omega$$

D
$$\displaystyle 10^5 \Omega$$

Solution

Reactance $$= \dfrac { 1 }{ C\omega } \Omega $$

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

$$={ 10 }^{ 5 }\Omega $$
Question 8
1 / -0

In the following circuit the values of $$L,C,R$$ and $$E_0$$ are 0.01 H, $$\displaystyle 10^{-5} F, 25 \Omega$$ and 220 volt respectively. The value of current flowing in the circuit at $$f=0$$ and $$f=\infty$$ will respectively be

A
8 $$A$$ and 0 $$A$$

B
0 $$A$$ and 0 $$A$$

C
8 $$A$$ and 8 $$A$$

D
0 $$A$$ and 8 $$A$$

Solution

$$I= \dfrac { 220 }{ 25+j\left(.01\times 2\pi f\quad -\quad \dfrac { 1 }{ { 10 }^{ -5 }2\pi f } \right) } $$

$$I = 0$$ at $$f= 0$$

$$I=0$$ at $$f= \infty $$
Question 9
1 / -0

The values of $$X_L, X_C$$ and $$R$$ in an AC circuit are 8 $$\Omega$$, 6$$\Omega$$ and 10 $$\Omega$$ respectively. The total impedance of the circuit

A
$$\displaystyle 10.2 \Omega$$

B
$$\displaystyle 12.2 \Omega$$

C
$$\displaystyle 10 \Omega$$

D
$$\displaystyle 24.4 \Omega$$

Solution

Total impedance $$= R +j({X}_{L} - {X}_{C} )$$

$$ =10 +j(8-6) = 10+j2 $$

$$= 10.2\quad \measuredangle { tan }^{ -1 }\left(\dfrac { 2 }{ 10 } \right) $$
Question 10
1 / -0

A solenoid of length 10 cm, diameter 1 cm, number of turns 500 with relative permeability of the core 2000, is connected to an ac source of frequency 50 Hz. Then, the reactance is

A
zero

B
55 $$\Omega$$

C
105 $$\Omega$$

D
155 $$\Omega$$

Solution

Inductance of the solenoid $$L =\dfrac { \mu { N }^{ 2 }A }{ l } = 0.493 H $$

reactance of this solenoid $$=L\omega=L 2\pi f=155\Omega $$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 36

Result

Alternating Current Test - 36

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SHARING IS CARING

Question 1

$$\displaystyle \dfrac{10^6}{\pi} Hz$$

$$\displaystyle 27 \pi \times 10^6 Hz$$

$$\displaystyle \dfrac{2\pi}{10^6} Hz$$

$$\displaystyle \dfrac{10^6}{2\pi} Hz$$

Solution

Question 2

increase

decrease

remain constant

first increase and then decrease

Solution

Question 3

$$\displaystyle \frac{1}{2\pi \sqrt{LC}}$$

$$\displaystyle \frac{1}{2\pi} \sqrt{\frac{C}{L}}$$

$$\displaystyle \frac{1}{2\pi} \sqrt{\frac{L}{C}}$$

$$\displaystyle \sqrt{LC}$$

Solution

Question 4

$$\displaystyle \frac{10^3}{2 \pi} Hz$$

$$\displaystyle 2\times 10^3 Hz$$

$$\displaystyle 2\times \frac{10^3}{pi}Hz$$

$$\displaystyle 10^3 Hz$$

Solution

Question 5

600 $$\Omega$$

200 $$\Omega$$

141 $$\Omega$$

310 $$\Omega$$

Solution

Question 6

$$\displaystyle X_L+X_C$$

$$\displaystyle X_L-X_C$$

$$\displaystyle \sqrt{X^2_L+X^2_C}$$

$$\displaystyle \sqrt{X^2_L-X^2_C}$$

Solution

Question 7

$$\displaystyle 10^4 \Omega$$

$$\displaystyle 10^2 \Omega$$

$$\displaystyle 10^1 \Omega$$

$$\displaystyle 10^5 \Omega$$

Solution

Question 8

8 $$A$$ and 0 $$A$$

0 $$A$$ and 0 $$A$$

8 $$A$$ and 8 $$A$$

0 $$A$$ and 8 $$A$$

Solution

Question 9

$$\displaystyle 10.2 \Omega$$

$$\displaystyle 12.2 \Omega$$

$$\displaystyle 10 \Omega$$

$$\displaystyle 24.4 \Omega$$

Solution

Question 10

zero

55 $$\Omega$$

105 $$\Omega$$

155 $$\Omega$$

Solution

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