Alternating Current Test

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Alternating Current Test - 44

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Question 1
1 / -0

220 V, 50 Hz, AC source is connected to an inductance of 0.2 H and a resistance of 20 $$\Omega$$ in series. What is the current in the circuit?

A
3.33 A

B
33.3 A

C
5 A

D
10 A

Solution

The frequency of AC source $$=f=50Hz$$
Thus the impedance of the inductor, $$X_L=\omega L=(2\pi f)L=20\pi\Omega$$
Resistance of the resistor $$=20\Omega$$
Thus, the net impedance of the circuit $$=\sqrt{X_L^2+R^2}=65.94\Omega$$
Thus, the current in the circuit $$=\dfrac{V}{X}=\dfrac{220V}{65.94\Omega}=3.33A$$
Question 2
1 / -0

In L-C-R circuit power of 3 mH inductance and $$4\Omega$$ resistance, EMF $$E=4\,cos\,1000t$$ volt is applied. The amplitude of current is

A
$$0.8\mathring{A}$$

B
$$\dfrac{4}{7}\mathring{A}$$

C
$$1.0\mathring{A}$$

D
$$\dfrac{4}{\sqrt{7}}\mathring{A}$$

Solution

Impedance of inductor=$$X_L=\omega H=1000\times 3mH=3\Omega$$
Thus the impedance of the circuit=$$X=\sqrt{X_{L}^2+R^2}=5\Omega$$
Thus amplitude of current $$I_{max}=\dfrac{E_{max}}{X}=\dfrac{4V}{5\Omega}=0.8A$$
Question 3
1 / -0

The impedance of a circuit, when a resistance $$R$$ and an inductor of inductance $$L$$ are connected in series in an AC circuit of frequency $$f$$, is

A
$$\sqrt { R+2{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

B
$$\sqrt { R+4{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

C
$$\sqrt { { R }^{ 2 }+4{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

D
$$\sqrt { { R }^{ 2 }+2{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

Solution

The impedance due to the inductor is $$2\pi fL$$ and voltage drop across inductor leads the voltage drop across resistor by $$90°$$.
So, the net impedance will be: $$Z=\sqrt { { R }^{ 2 }+{ (2\pi fL) }^{ 2 } } $$
Question 4
1 / -0

The current through a coil of self-inductance $$L=2mH$$ is given by $$i=t^2e^{-t}$$ at time t. How long it will take to make the emf zero?

A
$$1$$ s

B
$$2$$ s

C
$$3$$ s

D
$$4$$ s

Solution

The current is given by the relation $$i=t^2e^{-t}$$.
Hence the emf would be equal to
$$e=-L\dfrac{di}{dt}=-2\times10^{-3}\times(2te^{-t}-t^2e^{-t})$$
If the emf is zero, $$e=0$$, hence
$$2te^{-t}=t^2e^{-t}\Rightarrow\,t=2\,s$$
Question 5
1 / -0

In an LCR circuit inductance is changed from $$L$$ to $$\dfrac{L}{2}$$. To keep the same resonance frequency, $$C$$ should be changed to

A
$$2C$$

B
$$\dfrac{C}{2}$$

C
$$4C$$

D
$$\dfrac{C}{4}$$

Solution

Resonance frequency, $$(f)=\dfrac{1}{2\pi \sqrt{LC}}$$
For $$f$$ to be constant the product $$LC$$ must be constant.
So, if we half the value of inductance then the value of capacitance must be $$doubled$$.
$$C$$ should be changed to $$2C$$.
Question 6
1 / -0

The resonant frequency of an L-C circuit is

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

B
$$\dfrac {1}{2\pi} \sqrt {\dfrac {L}{C}}$$

C
$$\dfrac {1}{4\pi} \sqrt {\dfrac {L}{C}}$$

D
$$\dfrac {1}{2\pi} \sqrt {\dfrac {C}{L}}$$

Solution

Resonance frequency $$f$$ of an L-C circuit can be written as

Resonance frequency, $$f =\dfrac{1}{2 \pi \sqrt{LC}}$$ where $$L$$= inductance and $$C$$ is capacitance
Question 7
1 / -0

Resonance frequency of a circuit is $$f$$. If the capacitance is made $$4$$ times the initial value, then the resonance frequency will become

A
$$\dfrac{f}{2}$$

B
$$2f$$

C
$$f$$

D
$$\dfrac{f}{4}$$

Solution

As we know,
Resonance frequency of a circuit
$$f=\dfrac { 1 }{ 2\pi \sqrt { LC } } $$
ie, $$f\propto \dfrac { 1 }{ \sqrt { C } } $$
$$\Rightarrow \dfrac { { f }^{ \prime } }{ f } =\dfrac { \sqrt { C } }{ \sqrt { 4C } } =\dfrac { 1 }{ 2 } $$
$$\Rightarrow { f }^{ \prime }=\dfrac { f }{ 2 } $$
Question 8
1 / -0

An oscillating circuit contains an inductor of inductance $$ { 10 }^{ -6 }$$ H and two capacitor each of capacitance $$5\times { 10 }^{ -6 }$$ farad connected in parallel. Then the resonance frequency of the circuit is

A
$$\dfrac { { 10 }^{ 5 } }{ 2\pi } $$

B
$$\dfrac { { 10 }^{ 5 } }{ \pi } $$

C
$$\dfrac { { 3\times 10 }^{ 5 } }{ 2\pi } $$

D
$$\dfrac { \sqrt { { 2\times 10 }^{ 5 } } }{ \pi } $$

Solution

$$\omega =\dfrac { 1 }{ \sqrt { LC } } =\dfrac { 1 }{ \sqrt { { 10 }^{ -6 }\times 2\times 5\times { 10 }^{ -6 } } } $$

$$=\sqrt { { 10 }^{ 11 } } \approx 3\times { 10 }^{ 5 }\\$$

$$ f=\dfrac { \omega }{ 2\pi } =\dfrac { 3\times { 10 }^{ 5 } }{ 2\pi } $$

$$=\dfrac { { 1.5\times 10 }^{ 5 } }{ \pi } $$
Question 9
1 / -0

An alternating emf given by equation $$E = 300 sin [(100\pi )t]$$ $$volt$$ is applied to a resistance $$100\ ohms$$. The rms current through the circuit is (in $$amperes$$):

A
$$\displaystyle \frac {3}{\sqrt 2}$$

B
$$\displaystyle \frac {9}{\sqrt 2}$$

C
$$3$$

D
$$\displaystyle \frac {6}{\sqrt 2}$$

Solution

Alternating emf equation $$E = 300$$ $$ sin(100\pi t)$$
$$\therefore$$ Peak emf $$E_o = 300$$ $$volts$$
The rms emf $$E_{rms} = \dfrac{E_o}{\sqrt{2}} = \dfrac{300}{\sqrt{2}}$$ $$volts$$
Value of resistance $$R = 100\ \Omega$$

$$\therefore$$ Value of rms current $$I_{rms} = \dfrac{E_{rms}}{R} = \dfrac{300}{\sqrt{2} \times 100} =\dfrac{3}{\sqrt{2}}$$ $$A$$
Question 10
1 / -0

When $$100\ volt\ DC$$ is applied across a solenoid, a current of $$1.0\ amp$$ flows in it. When $$100\ volt\ AC$$ is applied across the same coil, the current drops to $$0.5\ amp.$$ If the frequency of the AC source is $$50\ Hz$$ the impedance and inductance of the solenoid are:

A
$$200\ \Omega$$ and $$0.55\ H$$

B
$$100\ \Omega$$ and $$0.86\ H$$

C
$$200\ \Omega$$ and $$1.0\ H$$

D
$$100\ \Omega$$ and $$0.93\ H$$

Solution

We know that
$$i=\frac { V }{ R } ,\quad for\quad DC\Rightarrow R=\frac { 100 }{ 1 } =100\Omega \\ i=\frac { V }{ L\omega } ,\quad when\quad AC\quad is\quad applied\Rightarrow L\omega =\frac { V }{ i } =\frac { 100 }{ 0.5 } =200\Omega \\ L=\frac { 200 }{ 2\pi \times 50 } =0.055H$$

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Alternating Current Test - 44

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Alternating Current Test - 44

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SHARING IS CARING

Question 1

3.33 A

33.3 A

5 A

10 A

Solution

Question 2

$$0.8\mathring{A}$$

$$\dfrac{4}{7}\mathring{A}$$

$$1.0\mathring{A}$$

$$\dfrac{4}{\sqrt{7}}\mathring{A}$$

Solution

Question 3

$$\sqrt { R+2{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

$$\sqrt { R+4{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

$$\sqrt { { R }^{ 2 }+4{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

$$\sqrt { { R }^{ 2 }+2{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } } $$

Solution

Question 4

$$1$$ s

$$2$$ s

$$3$$ s

$$4$$ s

Solution

Question 5

$$2C$$

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

$$4C$$

$$\dfrac{C}{4}$$

Solution

Question 6

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

$$\dfrac {1}{2\pi} \sqrt {\dfrac {L}{C}}$$

$$\dfrac {1}{4\pi} \sqrt {\dfrac {L}{C}}$$

$$\dfrac {1}{2\pi} \sqrt {\dfrac {C}{L}}$$

Solution

Question 7

$$\dfrac{f}{2}$$

$$2f$$

$$f$$

$$\dfrac{f}{4}$$

Solution

Question 8

$$\dfrac { { 10 }^{ 5 } }{ 2\pi } $$

$$\dfrac { { 10 }^{ 5 } }{ \pi } $$

$$\dfrac { { 3\times 10 }^{ 5 } }{ 2\pi } $$

$$\dfrac { \sqrt { { 2\times 10 }^{ 5 } } }{ \pi } $$

Solution

Question 9

$$\displaystyle \frac {3}{\sqrt 2}$$

$$\displaystyle \frac {9}{\sqrt 2}$$

$$3$$

$$\displaystyle \frac {6}{\sqrt 2}$$

Solution

Question 10

$$200\ \Omega$$ and $$0.55\ H$$

$$100\ \Omega$$ and $$0.86\ H$$

$$200\ \Omega$$ and $$1.0\ H$$

$$100\ \Omega$$ and $$0.93\ H$$

Solution

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