Alternating Current Test

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Alternating Current Test - 71

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Question 1
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An inductor of inductance $$L= 200\ \text{mH}$$ and the resistors $$R_{1}$$ and $$R_{2}$$, each of which is equal to $$4\ \Omega $$, are connected to a battery of emf $$12\ \text{V}$$ through a switch $$S$$ as shown in the Figure. The switch $$S$$ is closed and after the steady state is reached, the switch $$S$$ is opened. The current in $$R_{1}$$ then is $$($$in $$A)$$

A
$$6e^{-20t}\ \text{from B to A}$$

B
$$3e^{-40t}\ \text{from A to B}$$

C
$$3e^{-40t}\ \text{from B to A}$$

D
$$6e^{-20t}\ \text{from A to B}$$

Solution

When S is closed at t = 0 s, current in $$LR_{2}$$ branch

$$i_{2}=3\left ( 1-e^{-\dfrac{R_{2}}{L}t} \right ),\dfrac{R_{2}}{L}=\dfrac{4}{2\times 10^{-1}}=20\ s^{-1}$$

$$=3(1-e^{-20t})A$$

Current i, through $$R_{1}=3A$$
When S is opened, current in R$$_{1}$$ reduces to zero just after opening.
But in $$L-R_{2}$$, it decreases exponentially

$$\therefore$$ Current through $$R_{1}=3e^{-40t}$$ A, from B to A
Question 2
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Directions For Questions

An alternating source $$V\ \text{volt},\ f\ \text{Hz}$$ is connected across an lnductance $$L($$in series wlth resistance $$R)$$ in parallel with a capacitance $$C$$ as shown above.
The admittance Y of the circuit is then defined through I= YV

where $$|\displaystyle \mathrm{Y}|=\frac{[\mathrm{R}^{2}\mathrm{X}_{\mathrm{c}}^{2}+(\mathrm{R}^{2}+\mathrm{X}_{L}^{2}-X_{L}X_{\mathrm{c}})^{2}]}{\mathrm{X}_{\mathrm{c}}(\mathrm{R}^{2}+\mathrm{X}_{L}^{2})}\frac{1}{2}$$ The current $$I$$ leads the source voltage $$V$$ by 6,

such that $$\displaystyle \tan\partial=\frac{(\mathrm{R}^{2}+X_{L}-X_{L}\mathrm{X}_{\mathrm{c}})}{\mathrm{R}\mathrm{X}_{\mathrm{c}}}$$ with $$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$$ and $$\displaystyle X_{C}=\dfrac{1}{\omega \mathrm{C}}$$

...view full instructions

The frequency at which the impedance of the circuit becomes maximum is

A
$$\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}+\dfrac{R^{2}}{L^{2}}}$$

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

C
$$\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$$

D
$$\dfrac{1}{2\pi }\dfrac{R}{L}$$

Solution

Magnitude of admittance $$Y=\dfrac{1}{Z}$$ is given by

$$|\displaystyle

\mathrm{Y}|=\dfrac{[\mathrm{R}^{2}\mathrm{X}_{\mathrm{c}}^{2}+(\mathrm{R}^{2}+\mathrm{X}_{L}^{2}-X_{L}X_{\mathrm{c}})^{2}]}{\mathrm{X}_{\mathrm{c}}(\mathrm{R}^{2}+\mathrm{X}_{L}^{2})}\dfrac{1}{2}$$

Now, $$|Z|$$ is maximum or $$|Y|$$ is minimum at $$\omega =\omega _{0}$$ such that

$$R^{2}+X_{L}^{2}-X_{L}X_{C}=0$$
where $$X_L = w_oL$$ and $$X_C = \dfrac{1}{w_o C}$$
Or $$R^{2}+\omega _{0}^{2}L^{2}=\omega _{0}L\times \dfrac{1}{\omega _{0}C}=\dfrac{L}{C}$$

or $$\omega _{0}^{2}=\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}$$

or $$\omega _{0}=\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$$

Hence resonance frequency $$f_{0}=\dfrac{w_o}{2\pi}=\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$$
Question 3
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Directions For Questions

In the circuit shown, the switch S is in position A carrying certain current i. Now, at time t $$=$$ 0 s, when the current i $$=$$ 0.5 A, the switch S is moved from A to B. The phase difference between the source voltage e $$=$$ 200 sin $$\omega $$tV and the current is 45$$^{o}$$

...view full instructions

The reading shown in AC Voltmeter V when S is moved to B is:

A
120 V

B
100 V

C
140 V

D
160 V

Solution

Phase difference $$=\phi =45$$

$$cos \phi = \dfrac{R}{\sqrt {X_L^2+R^2}} = \dfrac{1}{\sqrt 2}$$

Solving, $$ X_L=R$$

$$\implies X_L=100\Omega$$

Impedance is $$Z = \sqrt {R^2+X_L^2} = 100\sqrt 2$$

Current in the circuit:

$$I_o=\dfrac{e}{Z} = \dfrac{200}{(\sqrt{2})100}=\sqrt{2}A$$

$$I_{rms}=1A$$

AC voltmeter shows RMS reading. Hence,

$$V_{rms}=I_{rms} R =(100\Omega)(1A)$$

$$=100V$$
Question 4
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In the circuit diagram shown, $$X_{C}=100\Omega , X_{L}=200 \Omega$$ & $$R=100 \Omega .$$ The effective current through the source is:

A
2 A

B
$$2 \sqrt{2} A$$

C
0.5 A

D
$$\sqrt{0.4} A$$

Solution

$$I_{R}=\dfrac{V}{R}=\dfrac{200}{100}=2A$$

$${I}'=\dfrac{V}{X_{L}-X_{C}}=\dfrac{200}{100}=2A$$

$$I=\sqrt{I{_{R}}^{2}+{I}'^{2}}=2\sqrt{2} Amp.$$
Question 5
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In a series L, C circuit , which of the following represents variation of magnitude of reactance (X) with frequency (f) ?

A

B

C

D

Solution

$$X=\left ( \omega L- \dfrac{1}{\omega C} \right ).$$

This is of the form $$y=x-\dfrac{1}{x}$$ which gives us the graph as shown.

For the values of frequency between 0 and 1 we take its absolute value and take the mirror image of the part which is below the x axis.
Question 6
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Directions For Questions

In the $$L-C$$ circuit shown, $$C=1\mu F$$. With capacitor charged to $$100V$$, Switch $$S$$ is suddenly closed at time $$t=0$$. The circuit then oscillates at $${10}^{3}Hz$$.

...view full instructions

Calculate $$\omega$$ and $$T$$.

A
$$6.28\times{10}^{-3}rad/s$$; $${10}^{-3}s$$

B
$$6.28\times{10}^{3}rad/s$$; $${10}^{-3}s$$

C
$$7.28\times{10}^{3}rad/s$$; $${10}^{-3}s$$

D
$$7.28\times{10}^{-3}rad/s$$; $${10}^{-3}s$$

Solution

Oscillation frequency $$f = 10^3 \ Hz$$
Angular frequency $$w = 2\pi f = 2\pi\times 10^3 = 6.28\times 10^3 \ rad/s$$
Time period $$T = \dfrac{1}{f} = \dfrac{1}{10^3} = 10^{-3} \ s^{-1}$$
Question 7
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Switch S is closed at t = 0. After sufficiently long time an iron rod is inserted into the inductor L. Then, the bulb :

A
glows more brightly

B
gets dimmer

C
glows with the same brightness

D
gets momentarily dimmer and then glows more brightly

Solution

When the iron rod is inserted into the inductor, the inductance of the coil increases.
As a result potential difference across the inductor increases, potential difference across the resistor( bulb ) decreases, so the bulb becomes dimmer.
Question 8
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The instantaneous potential difference between points A and B is (Phase angle is $$37^0$$)

A
$$8 \sin (50\pi t+37\dfrac{\pi }{180})$$

B
$$8 \sin (50\pi t-37\dfrac{\pi }{180})$$

C
$$10 \sin (50\pi t )$$

D
$$10 \cos (50\pi t )$$

Solution

Impedance $$z=\sqrt{(8-2)^{2}+(8)^{2}}= 10\Omega $$

current lags voltage by $$37^{\circ}$$,then

$$i=\dfrac{10}{10} \sin (50\pi t-37^{\circ} )$$

$$V_{AB}=i\times R=8\sin (50\pi t-37^{\circ} )$$
Question 9
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Identify the graph which correctly represents the variation of capacitive reactance $$X_C$$ with frequency.

A

B

C

D

Solution

$$X_C=\displaystyle \frac{1}{\omega C}=\frac{1}{2\pi fC}$$
or $$X_C\propto \displaystyle\frac{1}{f}$$
i.e., $$X_C$$ versus $$f$$ graph is a rectangular hyperbola.
Question 10
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The frequency of oscillation of current in the inductance is

A
$$\dfrac {1}{3\sqrt {LC}}$$

B
$$\dfrac {1}{6\pi \sqrt {LC}}$$

C
$$\dfrac {1}{\sqrt {LC}}$$

D
$$\dfrac {1}{3\pi \sqrt {LC}}$$

Solution

Step 1: Find equivalent inductance
Here, both inductor are connected in series
So, Equivalent inductance is given by
$$\mathrm{L}_{\mathrm{eq}}=\mathrm{L}+2\mathrm{~ L}=3L$$

$$\textbf{Step 2: Find equivalent Capacitance}$$
Here, both capacitors are connected in parallel
So, Equivalent capacitance is given by
$$\mathrm{C}_{\mathrm{eq}}=\mathrm{C}+2\mathrm{C}=3C$$

$$\textbf{Step 3: Find the frequency of oscillation.}$$
Hence, Frequency of oscillation is
$$\Rightarrow\mathrm{f}=\frac{1}{2\pi\sqrt{\mathrm{L}_{\mathrm{eq}}\mathrm{C}_{\mathrm{eq}}}}$$
$$\Rightarrow\frac{1}{6\pi\sqrt{\mathrm{LC}}}$$
Therefore, The correct option is 'B' is $$\frac{1}{6 \pi \sqrt{L C}}$$.