Alternating Current Test

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Alternating Current Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The phase difference between alternating emf and current in a purely capacitive circuit will be

A
zero

B
$$\displaystyle \pi$$

C
$$\displaystyle -\frac{\pi}{2}$$

D
$$\displaystyle \frac{\pi}{2}$$

Solution

In a purely capacitive circuit, current leads the voltage by $$\pi /2 $$.
Question 2
1 / -0

The square root of the product of inductance and capacitance has dimensions of

A
length

B
mass

C
time

D
dimensionless

Solution

$$\displaystyle f = \dfrac{1}{\sqrt{LC}} $$

$$ or \ \ \ \sqrt{LC} = T$$
Question 3
1 / -0

The resonant frequency in an anti resonant circuit is :

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

B
$$\displaystyle \frac{1}{2\pi} \sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$$

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

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

Solution

For resonance

$${ X }_{ L }={ X }_{ C }$$

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

$${ \omega }^{ 2 }=\dfrac { 1 }{ LC } $$

$$\omega =\dfrac { 1 }{ \sqrt { LC } } $$

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

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

In an A.C. circuit, the potential difference across an inductance and a resistance joined in series are respectively $$16 V$$ and $$20 V$$. The total potential difference across the circuit is

A
$$20.0V$$

B
$$25.6 V$$

C
$$31.4 V$$

D
$$53.5 V$$

Solution

Total potential = $$\sqrt{V_R^2+V_L^2}$$

$$= \sqrt{(16)^2+(20)^2}$$

$$= \sqrt{256+400}$$

$$= \sqrt{656}$$

$$= 25.6V$$
Question 5
1 / -0

An LCR series circuit contains $$L = 8 H$$, C$$=$$0.5 $$\mu $$ F and R $$=$$ 100$$\Omega $$ .The resonant frequency of the circuit is

A
$$\dfrac{100}{\pi }Hz$$

B
$$\dfrac{500}{\pi }Hz$$

C
$$\dfrac{250}{\pi }Hz$$

D
$$\dfrac{125}{\pi }Hz$$

Solution

For resonant frequency reactance should be zero.
So, $$X_L-X_C = 0$$
$$X_L = X_C$$
$$\omega L = \dfrac{1}{\omega C}$$
$$\implies \ \omega = \sqrt{\dfrac{1}{2C}}$$
Or $$2\pi f = \dfrac{1}{\sqrt{2C}}$$
Thus resonant frequency $$f = \dfrac{1}{2\pi}\times \dfrac{1}{\sqrt{2C}}$$
$$f= \dfrac{1}{2\pi}\times \dfrac{1}{\sqrt{8\times 0.5\times 10^{-6}}}$$
$$f= \dfrac{1}{2\pi}\times \dfrac{10^3}{2}$$
$$f= \dfrac{250}{\pi}Hz$$
Question 6
1 / -0

In the circuit shown, a 30 V d.c source gives a current 2.0A as recorded in the ammeter A and 30V a.c. source of frequency 100Hz gives a current 1.2A. The inductive reactance is

A
10ohm

B
20 ohm

C
$$5\sqrt{34}$$ ohm

D
40 ohm

Solution

In DC current

$$V = iR$$

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

$$R = \dfrac{30}{2} = 15\Omega$$

In a.c. current

$$V = i Impedance$$

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

$$Impedance = \dfrac{30}{1.2} = 25\omega$$

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

$$(25)^2 = R^2+(X_L)^2$$

$$X_L = \sqrt{(25)^2-R^2}$$

$$= \sqrt{625-225}$$

$$= \sqrt{400}$$

$$= 20\Omega$$
Question 7
1 / -0

If the values of inductance and frequency in an AC circuit are 2 henry and $$\displaystyle \dfrac{10^3}{2\pi}$$ Hz respectively then the value of inductive reactance will be

A
$$\displaystyle \dfrac{ 2\times 10^3}{\pi} \Omega$$

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

C
$$\displaystyle 10^3 \Omega$$

D
$$\displaystyle 2\times 10^3 \Omega$$

Solution

Inductive reactance $$= L\omega =2\times 2\pi \dfrac { { 10 }^{ 3 } }{ 2\pi } =2\times { 10 }^{ 3 }\Omega $$
Question 8
1 / -0

.In the given circuit, the phase difference between voltages across R and C is

A
zero

B
$$\pi /2$$

C
$$\pi $$

D
$$3\pi /2$$

Solution

Voltage phase difference between R and C is $$\dfrac{\pi}{2}$$ ,
R and L is $$\dfrac{\pi}{2}$$ and, L and C is $$\pi$$ for a LCR circuit such that voltage in L leads by $$\dfrac{\pi}{2}$$ from voltage in R and voltage in R leads by $$\dfrac{\pi}{2}$$ from voltage in C.
Question 9
1 / -0

An experimentalist has a coil of 3 mH. He wants to make a circuit whose frequency is 106 Hz. The capacity of condenser used will be

A
$$752.22 \mu F$$

B
$$852.22 \mu F$$

C
$$950.22 \mu F$$

D
zero

Solution

Circuit has to work properly at resonance therfore,

$$\omega =\dfrac { 1 }{ \sqrt { LC } } $$

$$C=\dfrac { 1 }{ { \omega }^{ 2 }L } $$

$$C=\dfrac { 1 }{ 4\pi ^{ 2 }f^2L } $$

$$C=\dfrac { 1 }{ 4\pi ^{ 2 }\times 106^2\times 3\times { 10 }^{ -3 } } $$

$$C=752.22\mu F$$
Question 10
1 / -0

The phase difference between the applied emf and the line current in an anti resonant circuit at resonance is

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

B
$$\displaystyle \pi$$ radian

C
$$\displaystyle \frac{3\pi}{2}$$ radian

D
zero

Solution

We know that phase difference is given by:

$$\phi =\arctan { (\dfrac { ({ X }_{ L }-{ X }_{ c }) }{ R } ) } $$

at resonance

$${ X }_{ L }-{ X }_{ c }=0$$

Therefore $$\phi =0$$.

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Alternating Current Test - 21

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Alternating Current Test - 21

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Question 1

zero

$$\displaystyle \pi$$

$$\displaystyle -\frac{\pi}{2}$$

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

Solution

Question 2

length

mass

time

dimensionless

Solution

Question 3

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

$$\displaystyle \frac{1}{2\pi} \sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$$

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

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

Solution

Question 4

$$20.0V$$

$$25.6 V$$

$$31.4 V$$

$$53.5 V$$

Solution

Question 5

$$\dfrac{100}{\pi }Hz$$

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

$$\dfrac{250}{\pi }Hz$$

$$\dfrac{125}{\pi }Hz$$

Solution

Question 6

10ohm

20 ohm

$$5\sqrt{34}$$ ohm

40 ohm

Solution

Question 7

$$\displaystyle \dfrac{ 2\times 10^3}{\pi} \Omega$$

$$\displaystyle 2\times 10^2 \Omega$$

$$\displaystyle 10^3 \Omega$$

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

Solution

Question 8

zero

$$\pi /2$$

$$\pi $$

$$3\pi /2$$

Solution

Question 9

$$752.22 \mu F$$

$$852.22 \mu F$$

$$950.22 \mu F$$

zero

Solution

Question 10

$$\displaystyle \frac{\pi}{2}$$ radian

$$\displaystyle \pi$$ radian

$$\displaystyle \frac{3\pi}{2}$$ radian

zero

Solution

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