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1

AIEEE 2005

MCQ (Single Correct Answer)
A coil of inductance $$300$$ $$mH$$ and resistance $$2\,\Omega $$ is connected to a source of voltage $$2$$ $$V$$. The current reaches half of its steady state value in
A
$$0.1$$ $$s$$
B
$$0.05$$ $$s$$
C
$$0.3$$ $$s$$
D
$$0.15$$ $$s$$

Explanation

KEY CONCEPT : The charging of inductance given

by, $$i = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right)$$

$${{{i_0}} \over 2} = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right) \Rightarrow {e^{ - {{Rt} \over L}}} = {1 \over 2}$$

Taking log on both the sides,

$$ - {{Rt} \over L} = \log 1 - \log 2$$

$$ \Rightarrow t = {L \over R}\log 2 = {{300 \times {{10}^{ - 3}}} \over 2} \times 0.69$$

$$ \Rightarrow t - 0.1\,\sec .$$
2

AIEEE 2005

MCQ (Single Correct Answer)
The self inductance of the motor of an electric fan is $$10$$ $$H$$. In order to impart maximum power at $$50$$ $$Hz$$, it should be connected to a capacitance of
A
$$8\mu F$$
B
$$4\mu F$$
C
$$2\mu F$$
D
$$1\mu F$$

Explanation

For maximum power, $${X_L} = X{}_C,$$ which yields

$$C = {1 \over {{{\left( {2\pi n} \right)}^2}L}} = {1 \over {4{\pi ^2} \times 50 \times 50 \times 10}}$$

$$\therefore$$ $$C = 0.1 \times {10^{ - 5}}F = 1\mu F$$
3

AIEEE 2004

MCQ (Single Correct Answer)
In a $$LCR$$ circuit capacitance is changed from $$C$$ to $$2$$ $$C$$. For the resonant frequency to remain unchaged, the inductance should be changed from $$L$$ to
A
$$L/2$$
B
$$2L$$
C
$$4L$$
D
$$L/4$$

Explanation

For resonant frequency to remain same $$LC$$ should be const. $$LC=$$ const

$$ \Rightarrow LC = L' \times 2C \Rightarrow L' = {L \over 2}$$
4

AIEEE 2004

MCQ (Single Correct Answer)
A metal conductor of length $$1$$ $$m$$ rotates vertically about one of its ends at angular velocity $$5$$ radians per second. If the horizontal component of earth's magnetic field is $$0.2 \times {10^{ - 4}}T,$$ then the $$e.m.f.$$ developed between the two ends of the conductor is
A
$$5mV$$
B
$$50\mu V$$
C
$$5\mu V$$
D
$$50mV$$

Explanation

$$\ell = 1m,\,\,\omega = 5\,rad/s,\,\,B = 0.2 \times {10^{ - 4}}T$$

$$\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V$$

Questions Asked from Alternating Current and Electromagnetic Induction

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