Alternating Current Test

Result

Alternating Current Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

In the given circuit, the potential difference across the $$6 \mu F$$ capacitor in steady state is

A
$$1\ V$$

B
$$6\ V$$

C
$$3\ V$$

D
$$4\ V$$
Question 2
1 / -0

In general in an alternating current circuit for a complete cycle

A
The average value of current is zero

B
The average value of square of the current is zero

C
Average power dissipation is zero

D
All of these
Question 3
1 / -0

The natural frequency of an LC circuit is 125 kHZz. when the capacitor is totally filled with a dielectric material, the natural frequency decreases by 25 kHz. Dielectric constant of the material is nearly.

A
3.33

B
2.12

C
1.56

D
1.91

Solution
Question 4
1 / -0

The impedance of coaxial cable, when its inductance is 0.40 $$\mu H$$ and capacitance is 1$$\times { 10 }^{ -11 }$$ F can be

A
$$2\times { 10 }^{ 2 }{ \Omega }$$

B
100$$\Omega $$

C
$$3\times { 10 }^{ 3 }{ \Omega }$$

D
$$3\times { 10 }^{ -2 }{ \Omega }$$

Solution

Usiing $$Z=\sqrt {\dfrac {L}{C}}$$, we get $$Z=\sqrt {\dfrac {0.40\times 10^{-6}}{10^{-11}}}=2\times 10^2 \Omega$$
Question 5
1 / -0

A capacitor is connected to an A.C generator.The ratio of reactance and impedance of capacitor is-

A
1

B
Less than 1

C
greater than 1

D
zero

Solution
Question 6
1 / -0

When induced emf in inductor coil is $$50%$$ of its maximum value then stored energy in inductor coil in the given circuit will be-

A
$$2.5 \,mJ$$

B
$$5 \,mJ$$

C
$$15 \,mJ$$

D
$$20 \,J$$

Solution

For max. induced emf across inductor will be at $$t = 0$$, when it will behave as O.C.
$$V = 2v = V_{max}$$ ($$\coprod$$el circuit Voltage will be same)
$$50%$$ of its max. That means $$1$$ volt.

When voltage of $$1$$ volt appears across inductor
Then, $$2 = 1 \times i + 1$$
$$[I=1 Amp]$$

Energy stored in inductor $$= \dfrac{1}{2} Li^2$$

$$= \dfrac{1}{2}\times 5 \times 1^2 \times 10^{-3}$$
$$E=2.5 mJ$$
Question 7
1 / -0

Even if a physical quantity depends upon three quantites,out of which two are dimensionally same, then the formula cannot be derived by the method of dimensions.This statement.

A
may be true

B
may be false

C
must be true

D
must be false

Solution
Question 8
1 / -0

In an LCR circuit $$ R=100 \Omega $$ when capacitance C is removed, the current lags behind the voltage by $$ \pi / 3 $$. when inductance L is removed, the current leads the voltage by $$ \pi / 3 $$. the imoedance of the circuit is

A
50 ohm

B
100 ohm

C
200 ohm

D
400 ohm

Solution

When $$C$$ is removed circuit becomes $$RL$$ circuit hence
$$ tan \dfrac{\pi}{3} = \dfrac{X_L}{R} $$ ....(i)

When $$L$$ is removed circuit becomes $$RC$$ circuit hence
$$ tan \dfrac{\pi}{3} = \dfrac{X_C}{R} $$ ......(ii)
From equation (i) and (ii) we obtain $$ X_L = X_C $$. This is the condition of resonance and in resonance $$ Z = R = 100 \, \Omega $$
Question 9
1 / -0

An inductive circuit contains resistance of 10 ohms and an inductance of 20 H. If an A.c voltage of 120 volt and frequency 60 Hz is applied to this circuit , the current would be nearly

A
$$0.16 amp$$

B
$$0.3 amp$$

C
$$0.48 amp$$

D
$$0.80 amp$$

Solution

$$Z=\sqrt { { R }^{ 2 }+{ (\omega L) }^{ 2 } } =\sqrt { { 10 }^{ 2 }+{ (120\pi \times 2) }^{ 2 } } =376\Omega \\ I=\cfrac { V }{ Z } =\cfrac { 120 }{ 376 } =0.3A$$
Question 10
1 / -0

The alternating current in a circular is described by the graph as shown in figure. The rms current obtained from the graph would be

A
1.4 A

B
2.2 A

C
1.6 A

D
2.6 A

Solution

$${i_{rms}} = \sqrt {\frac{{1 \times \frac{T}{2} + \frac{T}{2} \times 4}}{T}} $$
$$ = \sqrt {\frac{{5T}}{{25}}} = \sqrt {2.5} = 1.58A$$
Hence,
option $$(C)$$ is correct answer.