Alternating Current Test

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Alternating Current Test - 27

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Question 1
1 / -0

If resonant frequency is $$f$$ and capacitance become $$4$$ times, then resonant frequency will be :

A
$$\dfrac {f}{2}$$

B
$$2f$$

C
$$f$$

D
$$\dfrac {f}{4}$$

Solution

Frequency, $$f = \dfrac {1}{2\pi \sqrt {LC}}$$
When capacitance $$= 4C$$, then
$$f' = \dfrac {1}{2\pi \sqrt {4LC}} = \dfrac {f}{2}$$.
Question 2
1 / -0

When the frequency of the AC voltage applied to series LCR circuit is gradually increased from a low value, the impedance of the circuit:

A
monotonically increases.

B
first increases and then decreases.

C
first decreases and then increases.

D
monotonically decreases.

Solution

Impedance of series LCR circuit is given by:
$${ Z }^{ 2 }={ \left( \omega L-\cfrac { 1 }{ \omega C } \right) }^{ 2 }+{ R }^{ 2 }$$

Drawing the above curve for impedance v/s frequency, we get the plot as shown in the attached figure. Hence, impedance first decreases and then increases.
Question 3
1 / -0

If A.C. voltage is applied to a pure capacitor, then voltage across the capacitor _________.

A
Leads the current by phase angle $$\left(\displaystyle\frac{\pi}{2}\right)$$ rad.

B
Leads the current by phase angle $$(\pi)$$ rad.

C
Lags behind the current by phase angle $$\left(\displaystyle\frac{\pi}{2}\right)$$ rad.

D
Lags behind the current by phase angle $$(\pi)$$ rad.

Solution

In a capacitor, as the voltage applied is increased, there is current going into the capacitor. Now, as the applied voltage reaches the positive peak, the charging (in flow of current) stops. As the voltage starts to reduce, current flows out of the capacitor and continues until the voltage reaches the negative peak and discharging stops.
If we trace this out on a graph we will see that the current is leading the voltage by 90°, reaching zero current at the peaks of the voltage and maximum current as the voltage is going through zero. Hence, the voltage acrross the capacitor lags behind current by the phase angle $$(\frac{\pi}{2})$$ rad.
Question 4
1 / -0

The reactance of inductance at $$10^{4}Hz$$ is $$10^{4}\Omega$$. Its reactance at $$2\times 10^{4}Hz$$ will be :

A
$$10^{4}\Omega$$

B
$$2\times 10^{4}\Omega$$

C
$$3\times 10^{4}\Omega$$

D
$$4\times 10^{4}\Omega$$

Solution

$$X_{L} = wL = 2\pi fL$$
$$X_{L}\propto f$$
If frequency is double, then reactance will be double.
Question 5
1 / -0

Power dissipated in pure inductance will be

A
$$\dfrac {LI^{2}}{2}$$

B
$$2LI^{2}$$

C
$$\dfrac {LI^{2}}{4}$$

D
Zero

Solution

Power is dissipated only in the Resistor , Capacitor and Inductor only store energy. So Inductance power dissipated is Zero.
Question 6
1 / -0

When the frequency of AC is doubled, the impedance of an LCR series circuit:

A
is halved

B
is doubled

C
increases

D
decreases

Solution
Question 7
1 / -0

If $$V=100 \sin 100t$$ volt, and $$I=100 \sin(100t+\dfrac {\pi}{6})A$$. then find the watt less power in watt?

A
$$10^{4}$$

B
$$10^{3}$$

C
$$10^{2}$$

D
$$2.5 \times 10^{3}{\sqrt{3}}$$

Solution

$$P= V_{rms} \times I_{rms} \times \cos \phi$$

$$\quad= \large\frac{V_0I_0}{\sqrt{2}\times \sqrt{2}}\times \cos \dfrac{\pi}{6}$$

$$\quad= \large\frac{100 \times 100}{2} \times \frac{\sqrt{3}}{2}=2.5\times 10^3\sqrt{3}W$$
Question 8
1 / -0

In AC circuit having only capacitor, the current _____.

A
Leads the voltage by $$\dfrac {\pi}{2}$$ in phase

B
Lags behind the voltage by $$\dfrac {\pi}{2}$$ in phase

C
Leads the voltage by $$\pi$$ in phase

D
Lags behind the voltage by $$\pi$$ in phase

Solution

For a purely capacitive circuit current leads the voltage by $$\pi/2$$ in phase.
Question 9
1 / -0

'Z' is not.

A
Atomic number

B
Impedance

C
Zeta potential

D
Partition function

Solution

Z- Atomic number we study in modern physics
Z- impedance we study in alternating current
Z-zeta potential
Question 10
1 / -0

In a pure capacitive circuit if the frequency of ac source is doubled, then its capacitive reactance will be

A
remains same

B
doubled

C
halved

D
zero

Solution

Capacitive reactance, $$X_c = \dfrac{1}{2 \pi \nu C}$$

We can see that $$ X_c \propto \dfrac{1}{\nu}$$,

if $$\nu$$ is doubled, $$X_c$$ will be halved

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 27

Result

Alternating Current Test - 27

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Rank

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SHARING IS CARING

Question 1

$$\dfrac {f}{2}$$

$$2f$$

$$f$$

$$\dfrac {f}{4}$$

Solution

Question 2

monotonically increases.

first increases and then decreases.

first decreases and then increases.

monotonically decreases.

Solution

Question 3

Leads the current by phase angle $$\left(\displaystyle\frac{\pi}{2}\right)$$ rad.

Leads the current by phase angle $$(\pi)$$ rad.

Lags behind the current by phase angle $$\left(\displaystyle\frac{\pi}{2}\right)$$ rad.

Lags behind the current by phase angle $$(\pi)$$ rad.

Solution

Question 4

$$10^{4}\Omega$$

$$2\times 10^{4}\Omega$$

$$3\times 10^{4}\Omega$$

$$4\times 10^{4}\Omega$$

Solution

Question 5

$$\dfrac {LI^{2}}{2}$$

$$2LI^{2}$$

$$\dfrac {LI^{2}}{4}$$

Zero

Solution

Question 6

is halved

is doubled

increases

decreases

Solution

Question 7

$$10^{4}$$

$$10^{3}$$

$$10^{2}$$

$$2.5 \times 10^{3}{\sqrt{3}}$$

Solution

Question 8

Leads the voltage by $$\dfrac {\pi}{2}$$ in phase

Lags behind the voltage by $$\dfrac {\pi}{2}$$ in phase

Leads the voltage by $$\pi$$ in phase

Lags behind the voltage by $$\pi$$ in phase

Solution

Question 9

Atomic number

Impedance

Zeta potential

Partition function

Solution

Question 10

remains same

doubled

halved

zero

Solution

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