Alternating Current Test

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Alternating Current Test - 77

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Weekly Quiz Competition

Question 1
1 / -0

A pure inductor $$I$$, a cpacitory $$C$$ and a resistance $$R$$ are connected across a battery of emf $$E$$ and internal resistance $$r$$ as shown in figure. Switch $${S}_{w}$$ is closed at $$t=0$$, select the correct alternative(s).

A
current through resistance $$R$$ is zero all the time

B
current through resistance $$R$$ is zero at $$t=0$$ and $$t\rightarrow \infty$$

C
maximum charge stored in the capacitor is $$CE$$

D
maximum energy stored in the inductor is equal to the maximum energy stored in the capacitor
Question 2
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An $$LCR$$ series circuit with $$R=100\Omega$$ is connected to a $$200\ V,50\ Hz$$ a.c source. When only the capacitance is removed, the current leads the voltage by $$60^{o}$$. When only the inductance is removed, the current leads the voltage by $$60^{o}$$. The current in the circuit is :

A
$$2A$$

B
$$1A$$

C
$$\dfrac {\sqrt {3}}{2}A$$

D
$$\dfrac {2}{\sqrt {3}}A$$

Solution

Given,
$$R=100\Omega$$
$$V=200V$$
$$f=50Hz$$
$$\phi=60^0$$
When a capacitance is removed, the current leads the voltage by $$\phi$$
$$tan\phi=\dfrac{X_L}{R}$$
$$X_L=R tan60=\sqrt{3}R$$. . . .(1)
When the inductance is removed current leads the voltage by $$\phi$$
$$tan\phi=\dfrac{X_C}{R}$$
$$X_C=\sqrt{3}R$$. . . . . .(2)
From equation (1) and (2), we get
$$X_C=X_L$$ (this is the condition of resonance)
And at the resonance, $$Z=R$$
The current in the circuit is
$$I=\dfrac{V}{R}$$
$$I=\dfrac{200}{100}$$
$$I=2A$$
The correct option is A.
Question 3
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A condenser of capacity $$ 20 \mu F$$ is first charged and then discharged through a $$10 mH $$ inductance. Neglecting the resistance of the coil, the frequency of the resulting vibrations will be

A
$$356$$ cycle/s

B
$$35.6$$ cycle/s

C
$$356 \times 10^3 $$ cycle/s

D
$$3.56$$ cycle/s

Solution

The frequency of the resulting vibrations is given as,

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

$$f = \dfrac{1}{{2\pi \sqrt {10 \times {{10}^{ - 3}} \times 20 \times {{10}^{ - 6}}} }}$$

$$f = 356\;{\rm{cycle/s}}$$
Question 4
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In an LCR circuit, the resonating frequency is $$500$$ $$kH_z$$. If the value of $$L$$ is doubled and value of $$C$$ is decreased to $$\cfrac{1}{8}$$ times of its initial values, then the new resonating frequency in $$kH_z$$ will be

A
$$250$$

B
$$500$$

C
$$1000$$

D
$$2000$$

Solution

$$\textbf{Step 1 - Formula of resonant frequency,}$$
$$f = \dfrac {1}{2\pi \sqrt {LC}}$$

Given,
$$L_{2} = 2L_{1}$$

$$C_{2} = \dfrac {C_{1}}{8}$$

$$f_{1} = 500\ kH_{2}$$

$$\textbf{Step 2 - Comparing two resonant frequency,}$$

$$\dfrac {f_{1}}{f_{2}} = \dfrac {\dfrac {1}{2\pi \sqrt {L_{1}C_{1}}}}{\dfrac {1}{2\pi \sqrt {L_{2}C_{2}}}} = \sqrt {\dfrac {L_{2}C_{2}}{L_{1}C_{1}}}$$

$$\dfrac {f_{1}}{f_{2}} = \sqrt {\dfrac {1}{4}} = \dfrac {1}{2}$$

$$f_{2} = 2f_{1} = 2\times 500\ kH_{2}$$

$$f_{2} = 1000\ kH_{2}$$
Question 5
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A $$0.21 - H $$ inductor and a $$88- \Omega $$ resistor are connected in series to a $$220-V , 50-Hz $$ AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. Use $$ \pi = 22/7 $$

A
$$ 2 A, tan^{-1} 3/4 $$

B
$$ 14.4 A, tan^{-1} 7/8 $$

C
$$ 14.4 A, tan^{-1} 8/7 $$

D
$$ 3.28 A, tan^{-1} 2/11 $$

Solution

The impedance of the circuit is given as,

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

$$Z = \sqrt {{{\left( {88} \right)}^2} + {{\left( {2 \times \pi \times 50 \times 0.21} \right)}^2}} $$

$$Z = 110\;\Omega $$

The current in the circuit is given as,

$$I = \dfrac{V}{Z}$$

$$I = \dfrac{{220}}{{110}}$$

$$ = 2\;{\rm{A}}$$

The phase angle is given as,

$${\rm{\theta }} = {\tan ^{ - 1}}\left( {\dfrac{{{X_L}}}{R}} \right)$$

$${\rm{\theta }} = {\tan ^{ - 1}}\left( {\dfrac{{2 \times \pi \times 50 \times 0.21}}{{88}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$$

Thus, the current in the circuit is $$2\;{\rm{A}}$$ and the phase angle is $${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$$.
Question 6
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A coil has resistance $$30 \ ohm $$ and inductive reactance $$ 20 \ ohm$$ at 50 Hz frequency. If an ac source of 200 V,100 Hz is connected across the coil, the current in the coil will be

A
$$ 20\sqrt 13 A$$

B
2.0 A

C
4.0 A

D
none

Solution
Question 7
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In L-C oscillation if frequency of oscillation of charge is $$f$$, then frequency of oscillation of magnetic energy is

A
$$f$$

B
$$2f$$

C
$$\cfrac{f}{2}$$

D
$$4f$$

Solution
Question 8
1 / -0

An inductor of inductance $$2.0$$ $$H$$ and a resistor of resistance $$10$$ $$\Omega $$ are connected senes to a battery of EMF $$20$$ $$V$$ in a circuit as shown.The key $${ K }_{ 1 }$$ been kept closed for a long time. Then at $$t$$ = $$0$$ , $${ K }_{ 1 }$$i s opened and key $${ K }_{ 2 }$$ is closed simultaneously, The rate of decrease of current in the circuit at $$t$$ = $$1.0 s$$ wili be ($${ e }^{ 5 }$$ = $$150$$)

A
$$\dfrac { 1 }{ 15 } $$ $$A/s$$

B
$$\dfrac { 2 }{ 15 } $$ $$A/s$$

C
$$\dfrac { 1 }{ 5 } $$ $$A/s$$

D
$$\dfrac { 4 }{ 15 } $$ $$A/s$$

Solution
Question 9
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A capacitor of capacitance $$C$$ has initial charge $${ Q }_{ 0 }$$ and connected to an inductor of inductance $$L$$ as shown. At $$t=0$$ switch $$S$$ is closed. The current through the inductor when energy in the capacitor is three tirnes the energy of inductor is

A
$$\dfrac { { Q }_{ 0 } }{ 2\sqrt { LC } }$$

B
$$\dfrac { { Q }_{ 0 } }{ \sqrt { LC } }$$

C
$$\dfrac { 2{ Q }_{ 0 } }{ \sqrt { LC } } $$

D
$$\dfrac { 4{ Q }_{ 0 } }{ \sqrt { LC } } $$

Solution
Question 10
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For the circuit shown in the figure the current through the inductor is $$1.6\ A$$ , while the current through the condenser is $$0.4\ A,$$ then the current $$l$$ drawn from the source is :

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

B
$$1.65\ A$$

C
$$1.2\ A$$

D
$$2.0\ A$$

Solution

In capacitor current leads voltage by $$90$$ degree while in inductor current lags voltage by $$90$$ degrees. So thus inductor current lags the capacitor current by $$180$$ degrees.

Thus current drawn from the source is equal to difference of inductor current and capacitor current.

Thus current drawn from the source

$$I = {I_L} - {I_C}$$

$$I = 1.6 - 0.4$$

$$I = 1.2\;{\rm{A}}$$

Thus current drawn from the source is $$1.2\;{\rm{A}}$$.