Alternating Current Test

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Alternating Current Test - 62

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Question 1
1 / -0

If $$L$$ denotes the inductance of an inductor through which a current $$I$$ is flowing, then the dimensional formula of $$LI^2$$ is

A
$$[MLT^{-2}]$$

B
$$[ML^2T^{-2}]$$

C
$$[M^2L^2T^{-2}]$$

D
not expressible in terms of $$M.L.T.$$

Solution
Question 2
1 / -0

An alternating emf of angular frequency $$\omega $$ is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency :

A
$$\dfrac{\omega}{4 }$$

B
$$\dfrac{\omega}{2} $$

C
$$\omega $$

D
$$2\omega $$

Solution

The instantaneous values of emf and current in inductive circuit are given by

$$[E={{E}_{0}}\sin \omega t]$$ and $$[i={{i}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)]$$ respectively.

So, $$[{{P}_{inst}}=Ei={{E}_{0}}\sin \omega t\times {{i}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)] $$

$$[={{E}_{0}}{{i}_{0}}\sin \omega t\left( \sin \omega t\cos \frac{\pi }{2}-\cos \omega t\sin \frac{\pi }{2} \right)] $$

$$[={{E}_{0}}{{i}_{0}}\sin \omega t\ \cos \omega t]$$

$$ [=\frac{1}{2}{{E}_{0}}{{i}_{0}}\sin 2\omega t]$$

$$ [(\sin 2\omega t=2\sin \omega t\ \cos \omega t)]$$

Hence, angular frequency of instantaneous power is $$[2\omega ].$$
Question 3
1 / -0

The rms speed of molecules of a gas is 200 m/s at $$27 ^ { \circ } C$$ and 1 atmosphere pressure.The rms speed at $$127 ^ { \circ } C$$ and double pressure is

A
$$\sqrt { \dfrac { 800 } { 3 } m }$$

B
$$\dfrac { 100 \sqrt { 2 } } { 3 } m / s$$

C
$$\dfrac { 400 } { \sqrt { 3 } } m / s$$

D
$$\dfrac { 800 } { 3 } m / s$$

Solution
Question 4
1 / -0

Let $$\ell, r, c$$ and $$v$$ represent inductance, resistance, capacitance and voltage, respectively. The dimension of $$\dfrac{\ell}{rcv}$$ is $$SI$$ units will be:

A
$$[LTA]$$

B
$$[LA^{-2}]$$

C
$$[A^{-1}]$$

D
$$[LT^2]$$

Solution

$$\left[\dfrac{\ell}{r}\right] = T$$

$$[CV] = AT$$

So, $$\left[\dfrac{\ell}{rCV}\right] = \dfrac{T}{AT} = A^{-1}$$
Question 5
1 / -0

For LCR series resonant circuit_______

A
impedence is maximum and current is minimum

B
impedence is minimum and current is maximum

C
impedence is minimum and current is minimum

D
impedence is maximum and current is maximum

Solution
Question 6
1 / -0

In an A.C. circuit a capacitor of $$1 \mu F$$ value is connected to a source of frequency 1000 rad/sec. The value of capacitive reactance will be

A
$$10 \Omega$$

B
$$100 \Omega$$

C
$$1000 \Omega$$

D
$$10,000 \Omega$$

Solution

$$\begin{array}{l} c=1\mu f \\ w=1000\, \, rad/\sec \\ { X_{ c } }=\frac { 1 }{ { cw } } =\frac { 1 }{ { 1\times { { 10 }^{ -6 } }\times 1000 } } =1000\Omega \end{array}$$
$$ \therefore$$ captive reactance $$=1000\Omega $$
Hence,
Option $$C$$ is correct answer.
Question 7
1 / -0

In the given figure the impedance of the circuit will be

A
$$120ohm$$

B
$$50ohm$$

C
$$60ohm$$

D
$$90ohm$$

Solution
Question 8
1 / -0

When 100 volt DC source is applied across a coil, a current of 1 A flows through it. When 100 V AC source of 50 Hz is applied to the same coil, only 0.5 A current flows. Calculate the inductance of the coil.

A
$$(\pi/\sqrt{3})H$$

B
$$(\sqrt{3}/\pi)H$$

C
$$(2/\pi)H$$

D
None of these
Question 9
1 / -0

In the only capacitive circuit, A.C voltage is

A
leading by $$\pi$$

B
lagging behind by $$\pi$$

C
lagging behind by $$\pi/2$$

D
leading by $$\pi/2$$

Solution
Question 10
1 / -0

if value of R is changed then

A
voltage across L remains same

B
voltage across C remains sam

C
voltage across L-C combination remains same w

D
voltage across L-C combination changes

Solution

The values of $$R$$ is changed then voltage across $$L-C$$ combination changes.
Hence, Option $$D$$ is correct.