Alternating Current Test

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Alternating Current Test - 37

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Question 1
1 / -0

A broadcasting centre broadcasts at 300 metre band. A capacitor of capacitance 2.5$$\mu$$F is available. The value of the inductance required for resonant circuit is nearly

A
$$\displaystyle 1 \times 10^{-4} H$$

B
$$\displaystyle 1 \times 10^{-8} H$$

C
$$\displaystyle 1 \times 10^{-6} H$$

D
$$\displaystyle 1 \times 10^{-2} H$$

Solution

$$c=\lambda \nu \quad at resonance$$

$$ \nu =\dfrac { c }{ \lambda } =\dfrac { 1 }{ 2\pi \sqrt { LC } }$$

$$\sqrt { LC } =\dfrac { 300 }{ 3\times 10^8\times 2\pi } =\dfrac{ 10^{-6 }}{2\pi} $$

$$LC=\dfrac{10^{-12}}{4 \pi^2}$$

$$L=\dfrac {10^{ -12 } }{ 2.5\times 10^{ -6 }\times 4\times 9.87 } =\dfrac {10^{ -6 } }{ 98.7 } \cong {10}^{ -8 }\ H$$
Question 2
1 / -0

If in a series L-C-R ac circuit, the voltages across L, C and R are $$V_1, V_2$$ and $$V_3$$ respectively, then the voltage of the source is always

A
equal to $$\displaystyle V_1+V_2+V_3$$

B
equal to $$\displaystyle V_1-V_2+V_3$$

C
more than $$\displaystyle V_1+V_2+V_3$$

D
none of the above is true

Solution

The voltages across different elements in AC circuit add vectorially.
The voltages across inductor and capacitor are out of phase and at $$90^{\circ}$$ to that across resistor.
Hence net voltage across source $$=\sqrt{(V_1-V_2)^2+V_3^2}$$.
Question 3
1 / -0

An ac source producing emf $$\displaystyle e = E_0[cos (100 \pi t) + cos (500 \pi t)]$$ is connected in series with a capacitor and a resistor. The steady state current in the circuit is found to be $$\displaystyle i = I_1 cos (100 \pi t + \varphi_1) + I_2 cos (500 \pi t + \varphi_2)$$

A
$$\displaystyle I_1 > I_2$$

B
$$\displaystyle I_1 = I_2$$

C
$$\displaystyle I_1 < I_2$$

D
nothing can be said

Solution

We can assume two different sources connected in series with same orientation.

$${ E }_{ 1 }={ E }_{ 0 }\cos { 100\pi t } $$

$${ E }_{ 2 }={ E }_{ 0 }\cos { 500\pi t } $$

such that $$e={ E }_{ 0 }\cos { 100\pi t }+{ E }_{ 0 }\cos { 500\pi t }$$

becomes,$$e={ E }_{ 1 }+{ E }_{ 2 }$$

Now, solving using principle of superposition we get,

$$i_{ 1 }=100\pi C\cos { (100 } \pi t+\phi_1)$$

$$i_{ 2 }=500\pi C\cos { (500 } \pi t+\phi_2)$$

Final current will be,

$$i=i_{ 1 }+i_{ 2 }$$

From this we get

$$I_{ 1 }=100\pi C$$

and $$I_{ 2 }=500\pi C$$

From this we get,

$$I_{ 1 }<I_2$$.
Question 4
1 / -0

In an electric circuit the applied alternating emf is given by $$\displaystyle E = 100\sin(314t) $$ volts and current flowing $$\displaystyle I = \sin \left(314t + \dfrac{\pi}{3}\right)$$, Then, the impedance of the circuit is $$(in\ \Omega)$$

A
$$\dfrac{100}{ \sqrt{2}}$$

B
$$100$$

C
$$100 \sqrt{2}$$

D
$$None\ of\ the\ above$$

Solution

Phasor of $$E=100 \angle{(\pi /2)} $$

phasor of $$I=1 \angle{(\pi /2 + \pi /3)}$$

phasor of impedance $$=\dfrac{E}{I} = 100 \angle{(-\pi/3)} $$

impedance magnitude$$= 100\ \Omega$$
Question 5
1 / -0

An $$AC$$ voltmeter in an $$L-C-R$$ circuit reads 30 volt across resistance, 80 volt across inductance and 40 volt across capacitance. The value of applied voltage will be

A
50 Volt

B
25 Volt

C
150 Volt

D
70 Volt

Solution

$$V=\sqrt { { V }_{ R }^{ 2 }+{ ({ V }_{ L }^{ }-{ V }_{ C }^{ }) }^{ 2 } } $$

$$V=\sqrt { { 30}^{ 2 }+{ (80^{ }-40^{ }) }^{ 2 } } $$

$$V=50$$ volt
Question 6
1 / -0

A coil of 10 mH and 10 $$\Omega$$ resistance is connected in parallel to a capacitance of $$0.1 \mu F$$. The impedance of the

A
$$\displaystyle 10^2 \Omega$$

B
$$\displaystyle 10^4 \Omega$$

C
$$\displaystyle 10^6 \Omega$$

D
$$\displaystyle 10^8 \Omega$$
Question 7
1 / -0

The reactance of a condenser of capacity 50 $$\mu$$F for an $$AC$$ of frequency $$2 \times 10^3$$ Hertz will be

A
5 ohm

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

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

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

Solution

Given : $$\nu = 2000$$ Hz , $$C = 50\mu F$$

Capacitive reactance $$X_c = \dfrac{1}{2\pi \nu C}$$

$$ = \dfrac{1}{2\pi (2000)\times 50\times 10^{-6}} $$

$$= \dfrac{5}{\pi}$$ $$\Omega$$
Question 8
1 / -0

The capacitive reactance of a condenser of capacity 25$$\mu$$F for an AC of frequency 4000 Hz will be

A
$$\displaystyle \frac{5}{\pi} \Omega$$

B
$$\displaystyle \frac{10}{\pi} \Omega$$

C
$$\displaystyle 5 \pi \Omega$$

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

Solution

$${ X }_{ C }=\dfrac { 1 }{ \omega C } $$

$${ X }_{ C }=\dfrac { 1 }{2\pi f\times C } $$

$${ X }_{ C }=\dfrac { 1 }{2\pi \times 4000\times 25\times { 10 }^{ -6 } } $$

$${ X }_{ C }=\dfrac { 5 }{ \pi } $$.
Question 9
1 / -0

In the figure two identical bulbs, each with filament resistance $$100\ \Omega$$ are connected to a resistor $$R = 100\ \Omega$$, and an inductor $$\displaystyle (X_L = 100 \Omega)$$ as shown in the Figure. Then, which bulb glows more

A
$$\displaystyle B_1$$

B
$$\displaystyle B_2$$

C
both glow equally

D
cannot be predicted

Solution

Impedance in branch containing bulb1

$$Z_1=200\Omega$$

impedance in branch containing bulb 2

$${ Z }_{ 2 }=\sqrt { { R }^{ 2 }+{ { X }_{ L } }^{ 2 } } $$

$${ Z }_{ 2 }=\sqrt { { 100 }^{ 2 }+{ 100 }^{ 2 } } $$

$${ Z }_{ 2 }=100\sqrt { 2 } $$

Since,

$$Z_1>Z_2$$

$$B_2$$ will glow more than $$B_1$$.
Question 10
1 / -0

In an ac circuit, the resistance of R $$\Omega$$ is connected in series with an inductance $$I$$. If phase angle between voltage and current be $$\displaystyle 45^{\circ}$$, the value of inductive reactance will be:

A
$$R/2$$

B
$$\displaystyle R/\sqrt 2$$

C
$$R$$

D
Insufficient data

Solution

$$ \dfrac { R }{ \sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } } =cos(45)=\dfrac { 1 }{ \sqrt { 2 } } \\ \Rightarrow 2{ R }^{ 2 }={ R }^{ 2 }+{ X }_{ L }^{ 2 }\quad \\ \Rightarrow \left| { X }_{ L } \right| =R $$

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Alternating Current Test - 37

Result

Alternating Current Test - 37

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SHARING IS CARING

Question 1

$$\displaystyle 1 \times 10^{-4} H$$

$$\displaystyle 1 \times 10^{-8} H$$

$$\displaystyle 1 \times 10^{-6} H$$

$$\displaystyle 1 \times 10^{-2} H$$

Solution

Question 2

equal to $$\displaystyle V_1+V_2+V_3$$

equal to $$\displaystyle V_1-V_2+V_3$$

more than $$\displaystyle V_1+V_2+V_3$$

none of the above is true

Solution

Question 3

$$\displaystyle I_1 > I_2$$

$$\displaystyle I_1 = I_2$$

$$\displaystyle I_1 < I_2$$

nothing can be said

Solution

Question 4

$$\dfrac{100}{ \sqrt{2}}$$

$$100$$

$$100 \sqrt{2}$$

$$None\ of\ the\ above$$

Solution

Question 5

50 Volt

25 Volt

150 Volt

70 Volt

Solution

Question 6

$$\displaystyle 10^2 \Omega$$

$$\displaystyle 10^4 \Omega$$

$$\displaystyle 10^6 \Omega$$

$$\displaystyle 10^8 \Omega$$

Question 7

5 ohm

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

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

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

Solution

Question 8

$$\displaystyle \frac{5}{\pi} \Omega$$

$$\displaystyle \frac{10}{\pi} \Omega$$

$$\displaystyle 5 \pi \Omega$$

$$\displaystyle \frac{\pi}{5} \Omega$$

Solution

Question 9

$$\displaystyle B_1$$

$$\displaystyle B_2$$

both glow equally

cannot be predicted

Solution

Question 10

$$R/2$$

$$\displaystyle R/\sqrt 2$$

$$R$$

Insufficient data

Solution

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