Alternating Current Test

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Alternating Current Test - 40

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Question 1
1 / -0

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

A
$$10mH$$

B
$$12mH$$

C
$$16mH$$

D
$$20mH$$

Solution

The applied voltage leads the current by $$45^o$$
i.e., $$\dfrac{X_L}{R}=\tan 45^o$$
$$ X_L=R$$
$$\Rightarrow 2\times3.14\times 1000 L=100$$
$$\Rightarrow 6280L=100$$
$$\Rightarrow L=\dfrac{25}{1570}=0.01592\cong 16 mH$$
Question 2
1 / -0

In a L-C-R circuit, as the frequency of an alternating current increases the impedance of the circuit

A
increases continuously.

B
decreases continuously.

C
remains constant.

D
None of these.

Solution

Impedance first decreases then increases. At resonance frequency $$Z$$ is minimum.
Question 3
1 / -0

The instantaneous voltage through a device of impedence $$20\Omega$$ is $$e=80\sin { 100\pi t } $$. The effective value of the current is

A
$$3A$$

B
$$2.828A$$

C
$$1.732A$$

D
$$4A$$

Solution

Given equation, $$e=80\sin { \left(100 \pi t \right) } .......(i)$$
Standard equation of instantaneous voltage given by $$E={e}_{m}\sin { \left( \omega t \right) } ..........(ii)$$
Compare $$(i)$$ and $$(ii)$$, we get $${e}_{m}=80V$$
where $${e}_{m}$$ is the voltage amplitude.
Current amplitude $${I}_{m}=\cfrac{{e}_{m}}{Z}$$ where $$Z=$$ impedence
$$=\dfrac{80}{20}=4A$$

$${ I }_{ r.m.s }=\cfrac { 4 }{ \sqrt { 2 } } =\cfrac { 4\sqrt { 2 } }{ 2 } =2\sqrt { 2 } =2.828A$$
Question 4
1 / -0

The impedance of a series $$L-C-R$$ circuit in an $$AC$$ circuit is

A
$$\displaystyle \sqrt{R+(X_L-X_C)}$$

B
$$\displaystyle \sqrt{R^2+(X^2_L-X^2_C)}$$

C
$$R$$

D
None of these

Solution

$$Z=\sqrt{R^2+(X_L-X_C)^2}$$
So, Option $$ D $$ is the correct Answer
Question 5
1 / -0

In the $$AC$$ network shown in figure, the rms current flowing through the inductor and capacitor are $$0.6A$$ and $$0.8A$$ respectively. Then the current coming out of the source is

A
$$1.0A$$

B
$$1.4A$$

C
$$0.2A$$

D
None of the above

Solution

$$I_C$$ is $$90^o$$ ahead of the applied voltage and $$I_L$$ lags behind
the applied voltage by $$90^o$$. So, there is a phase difference of
$$180^o$$ between $$I_L$$ and $$I_C$$.
$$\therefore \displaystyle I=I_C-I_L=0.2A$$
Question 6
1 / -0

For watt-less power in an $$AC$$ circuit the phase angle between the current and voltage is

A
$$0^o$$

B
$$90^o$$

C
$$45^o$$

D
Not possible

Solution

Watt-less power in an AC circuit is basically power supposed to be generated by inductive and capacitive reactance and since they are not resistor they generate any heat , and these power wasted is called watt-less power and its phase angle is always $$90^o$$ as it has only capacitor and inductor.
Question 7
1 / -0

In a parallel $$L-C-R$$ circuit as shown in figure if $$I_R, I_L, I_C$$ and $$I$$ represents the rms values of current flowing through resistor, inductor, capacitor and the source, then choose the appropriate correct answer

A
$$\displaystyle I=I_R+I_L+I_C$$

B
$$\displaystyle I=I_R+I_L-I_C$$

C
$$I_L$$ or $$I_C$$ may be greater than $$I$$

D
None of the above

Solution

Voltage $$V$$ is same across all the elements.
currents, $$I_R=V/R;I_L=V/\omega L;I_C=\omega C\times I$$, where $$\omega$$ is frequency of current source $$I$$
now total current is $$I$$ is the vector sum of three currents, and cannot be simply added like vectors. Thus option A and B are incorrect.
The vector sum $$I$$ can be lower than any of its $$I_R,I_L,I_C$$. So, option C is correct.
Question 8
1 / -0

In a series $$L-C-R$$ circuit, current in the circuit is $$11A$$ when the applied voltage is $$220V$$. Voltage across the capacitor is $$200V$$. If value of resistor $$20\Omega$$, then the voltage across the unknown inductor is

A
Zero

B
$$200V$$

C
$$20V$$

D
None of these

Solution

The given LCR circuit is in resonance. Inductive reactance magnitude $$X_L$$ increases as frequency increases while capacitive reactance magnitude $$X_C$$ decreases with the increase in frequency. At one particular frequency, these two reactances are equal in magnitude but opposite in sign; that frequency is called the resonant frequency $$f_O$$ for the given circuit.
Hence, at resonance:
$$X_L = X_C$$
So the voltage across rthe unknown inductor is $$200V$$ as same as the voltage across the capacitor.
Question 9
1 / -0

In an $$AC$$ circuit, the impedance is $$\sqrt 3$$ times the reactance, then the phase angle is

A
$$60^o$$

B
$$30^o$$

C
zero

D
none of these

Solution

$$\displaystyle \sin \phi=\frac{X}{Z}$$$$\displaystyle =\frac{1}{\sqrt 3}$$
$$\displaystyle \therefore \phi=\sin^{-1}\left(\frac{1}{\sqrt 3}\right)$$
Question 10
1 / -0

A circuit contains resistance $$R$$ and an inductance $$L$$ in series. An alternating voltage $$V=V_0\sin \omega t$$ is applied across it. The currents in $$R$$ and $$L$$ respectively will be

A
$$I_R=I_0\cos \omega t, I_L=I_0\cos \omega t$$

B
$$I_R=-I_0\sin \omega t, I_L=I_0\cos \omega t$$

C
$$I_R=I_0\sin \omega t, I_L=-I_0\cos \omega t$$

D
None of the above.

Solution

The effective impedance is $$Z= \sqrt{R^2 + (\omega ^2 L^2}$$
$$I = \frac{V_0}{Z}\sin( \omega t + \phi)$$ where $$\tan\phi = \frac{\omega L}{R}$$

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Alternating Current Test - 40

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Alternating Current Test - 40

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Question 1

$$10mH$$

$$12mH$$

$$16mH$$

$$20mH$$

Solution

Question 2

increases continuously.

decreases continuously.

remains constant.

None of these.

Solution

Question 3

$$3A$$

$$2.828A$$

$$1.732A$$

$$4A$$

Solution

Question 4

$$\displaystyle \sqrt{R+(X_L-X_C)}$$

$$\displaystyle \sqrt{R^2+(X^2_L-X^2_C)}$$

$$R$$

None of these

Solution

Question 5

$$1.0A$$

$$1.4A$$

$$0.2A$$

None of the above

Solution

Question 6

$$0^o$$

$$90^o$$

$$45^o$$

Not possible

Solution

Question 7

$$\displaystyle I=I_R+I_L+I_C$$

$$\displaystyle I=I_R+I_L-I_C$$

$$I_L$$ or $$I_C$$ may be greater than $$I$$

None of the above

Solution

Question 8

Zero

$$200V$$

$$20V$$

None of these

Solution

Question 9

$$60^o$$

$$30^o$$

zero

none of these

Solution

Question 10

$$I_R=I_0\cos \omega t, I_L=I_0\cos \omega t$$

$$I_R=-I_0\sin \omega t, I_L=I_0\cos \omega t$$

$$I_R=I_0\sin \omega t, I_L=-I_0\cos \omega t$$

None of the above.

Solution

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