Alternating Current Test

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Alternating Current Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

For a series LCR circuit the power loss at resonance is : -

A
$\dfrac{V^{2}}{\left [ \omega L-\dfrac{1}{\omega C} \right ]}$

B
$I^{2}L\omega$

C
$I^{2}R$

D
$\dfrac{V^{2}}{C\omega }$

Solution
Question 2
1 / -0

The natural frequency of the circuit shown in fig. is

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

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

C
$\dfrac {2}{\sqrt {LC}}$

D
none of these

Solution

$L_{eq}=L+L=2L$

$C_{eq}=C/2$

$\therefore natural frequency = \dfrac{1}{\sqrt {L_{eq} C_{eq} }}$
Question 3
1 / -0

The capacitance in an oscillatory LC circuit is increased by 1%. The change in inductance required to restore its frequency of oscillation is to

A
decrease it by 0.5%

B
increase it by 1%

C
decrease it by 1%

D
decrease it by 2%

Solution

$\omega_r=constant$

$\Rightarrow LC=constant$

$\Rightarrow LdC+CdL=0$

$\Rightarrow \dfrac {dL}{L}=-\dfrac {dC}{C}=-1 %$
Question 4
1 / -0

In alternating current

A
The direction of current is always positive

B
The direction of current is always negative

C
The direction of current changes constantly

D
The direction of current is either positive or negative

Solution

C

Electric charge in alternating current (AC) changes direction periodically. The voltage in AC circuits also periodically reverses because the current changes direction.
Question 5
1 / -0

The frequency of A.C mains in India is

A
$30 Hz$

B
$50Hz$

C
$60Hz$

D
$120Hz$

Solution

In India, the AC mains supply is referred to as single-phase alternating current and corresponds to a voltage of 230 V at a frequency of $50 Hz$ , similar to most European countries. Whereas in the USA, AC mains supply uses $60 Hz$
Question 6
1 / -0

An inductor, a resistor and a capacitor are joined in series with an AC source. As the frequency of the source is slightly increased from a very low value, the reactance

A
of the inductor increases

B
of the resistor increases

C
of the capacitor increases

D
of the circuit increases

Solution

A

Resonance of inductor, $X_L = 2 \pi f L$

Here f is the frequency of Ac, L is inductance of inductor. Reactance is directly proportional to frequency. Therefore, with increase in frequency reactance also increases.

When current passes through inductor, magnetic field develop across inductor, magnetic field cut by inductor and emf is induced. This emf is called back emf which opposes the applied emf. Back emf increase with increase in frequency of applied emf. Therefore opposition, i.e. reactance also increases with increase in frequency.
Question 7
1 / -0

The circuit shown in Fig. acts as a

A
tuned filter

B
low pass filter

C
high pass filter

D
rectifier

Solution

The circuit shown in figure has capacitor and inductor in parallel so their will be current flowing continuously with energy being transformed into electrical in capacitor and magnetic in inductor, hence it is a tuned filter.
Question 8
1 / -0

A steady potential difference of 100 V produces heat at a constant rate in a resistor. The alternating voltage which will produce half the heating effect in the same resister will be

A
100 V

B
50 V

C
70.7 V

D
141.4 V

Solution

The power supplied to the resistor by DC source $=\dfrac{V_{dc}^2}{R}$
Energy given by AC source $=\int_0^T \dfrac{V_0^2}{R}dt$

Hence, $\int_0^T \dfrac{V_0^2sin^2\omega t}{R}dt=\dfrac{1}{2}\times \dfrac{V_{dc}^2T}{R}$

$\implies V_0=V_{dc}=100V$
Question 9
1 / -0

An amplitude modulated (AM) radio operates at $550 kHz$ to $1650 kHz$ . If $L$ is fixed and $C$ is varied for tuning then minimum and maximum value of $C$ is

A
C, 3C

B
C, 6C

C
C, 9C

D
C, 12C

Solution

$\displaystyle \frac{f_{max}}{f_{min}} = 3$

$\therefore \dfrac{\sqrt{LC_{max}}}{\sqrt{LC_{min}}} = 3$

$\dfrac{C_{max}}{C_{min}}= 9$

$C_{min}= C$

$\therefore C_{max}=9C$
Question 10
1 / -0

When $4\ V$ DC is connected across an inductor, current is $0.2\ A$ . When AC of $4\ V$ is applied the current is $0.1\ A$ . Then self inductance of the coil is:
[Given $\omega = 1000\ rads^{-1}$ ]

A
$20\ mH$

B
$40\ mH$

C
$20 \sqrt{3}\ mH$

D
none of these

Solution

$\displaystyle Z = \frac{4}{0.1} = 40\ \Omega = \sqrt{R^2 + (L\omega)^2}$

and $\displaystyle R = \frac{4}{0.2} = 20\ \Omega$

$\displaystyle \therefore\ L\omega = 20\sqrt3 \Omega$

or $\displaystyle L = 20\sqrt3\ mH$

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Alternating Current Test - 34

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Alternating Current Test - 34

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Question 1

V2[ωL−1ωC]\dfrac{V^{2}}{\left [ \omega L-\dfrac{1}{\omega C} \right ]}[ωL−ωC1​]V2​

I2LωI^{2}L\omega I2Lω

I2RI^{2}RI2R

V2Cω\dfrac{V^{2}}{C\omega }CωV2​

Solution

Question 2

1LC\dfrac {1}{\sqrt {LC}}LC​1​

12LC\dfrac {1}{2\sqrt {LC}}2LC​1​

2LC\dfrac {2}{\sqrt {LC}}LC​2​

none of these

Solution

Question 3

decrease it by 0.5%

increase it by 1%

decrease it by 1%

decrease it by 2%

Solution

Question 4

The direction of current is always positive

The direction of current is always negative

The direction of current changes constantly

The direction of current is either positive or negative

Solution

Question 5

30Hz30 Hz30Hz

50Hz50Hz50Hz

60Hz60Hz60Hz

120Hz120Hz120Hz

Solution

Question 6

of the inductor increases

of the resistor increases

of the capacitor increases

of the circuit increases

Solution

Question 7

tuned filter

low pass filter

high pass filter

rectifier

Solution

Question 8

100 V

50 V

70.7 V

141.4 V

Solution

Question 9

C, 3C

C, 6C

C, 9C

C, 12C

Solution

Question 10

20 mH20\ mH20 mH

40 mH40\ mH40 mH

203 mH20 \sqrt{3}\ mH203​ mH

none of these

Solution

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$\dfrac{V^{2}}{\left [ \omega L-\dfrac{1}{\omega C} \right ]}$

$I^{2}L\omega$

$I^{2}R$

$\dfrac{V^{2}}{C\omega }$

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

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

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

$30 Hz$

$50Hz$

$60Hz$

$120Hz$

$20\ mH$

$40\ mH$

$20 \sqrt{3}\ mH$