Alternating Current Test

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Alternating Current Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

In case of $$AC$$ circuits, impedance is given by the ratio of
I. Peak values of voltage and current
II. rms values of voltage and current
III. Instantaneous values of voltage and current.

A
I only

B
II only

C
I and II

D
I, II and III

Solution
Question 2
1 / -0

A loop is formed by two parallel conduction connected by a solenoid with inductance $$ L = 10^3 H $$ and a conducting rod of mass $$m=1 gm$$ which can freely slide (without friction) over a pair of conductors. The conductors are located in a horizontal plane in a uniform magnetic field $$B = 4 T $$ in the direction shown. The distance between conductors is equal to $$ t = 1 m. $$ At the moment $$ t = 0 $$ , the rod is imparted an initial velocity $$V_0$$ to the right. Determine angular frequency (in rad/s) of oscillation of rod.

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution
Question 3
1 / -0

A coil of $$10^{-2}\ H$$ inductance carries a current $$I=2\sin (100t)A$$. When current is half of its peak value then at that instant the induced emf in coil :-

A
$$1V$$

B
$$\sqrt{2}\ V$$

C
$$\sqrt{3}\ V$$

D
$$2V$$

Solution

$$I=2\sin (100t)$$

$$\dfrac{I_{max}}{2}=2\sin (100t)$$

$$1=2\sin (100t)$$

$$\dfrac{dI}{dt}=2\times 100 \times \cos (100t)$$

$$=200 \times \cos \left ( \dfrac{\pi}{6} \right )$$

$$=200 \times \dfrac{\sqrt 3}{2}=100\sqrt 3$$

$$\epsilon =L \times \dfrac{di}{dt}$$

$$=10^{-2} \times 100\sqrt 3$$

$$=\sqrt 3V$$
Question 4
1 / -0

Current in resistance is $$A$$, then

A
$$V_{s}=15\ V$$

B
impedance of network is $$15\ Omega$$

C
power factor of given circuit is $$(0.46)$$ lagging (current is lagging)

D
NONE of the above

Solution
Question 5
1 / -0

If $$\omega_t$$ and $$\omega_2$$ are half power frequencies of a series $$LCR$$ circuit, then resonant frequency $$\omega$$, can be expressed as :

A
$$\omega_r = \dfrac{\omega_1+\omega_2}{2}$$

B
$$\omega_r=\sqrt{\omega_1\omega_2}$$

C
$$\omega_r= \sqrt{\dfrac{\omega_1^2+\omega_2^2}{2}}$$

D
$$\omega_r = \dfrac{\omega_1\omega_2}{\omega_1+\omega_2}$$

Solution
Question 6
1 / -0

In the given circuit, the $$AC$$ source has $$\omega=100\ rad/s$$. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

A
The current through the circuit, $$I$$ is $$0.3\ A$$

B
The current through the circuit, $$I$$ is $$0.3\sqrt {2}\ A$$

C
The voltage across $$100\Omega $$ resistor $$10\sqrt {2}V$$

D
The voltage across $$50\Omega $$ resistor $$10\ V$$

Solution

Given that,

Resistance $${{R}_{C}}=100\Omega $$

$$\omega =100\,rad/s$$

$${{X}_{L}}=50\,H$$

$${{R}_{L}}=50\Omega $$

$${{X}_{C}}=100\mu F$$

Now, the impedance is

$$ {{z}_{1}}=\sqrt{{{R}^{2}}+X_{C}^{2}} $$

$$ {{Z}_{1}}=\sqrt{{{\left( 100 \right)}^{2}}+{{\left( 100 \right)}^{2}}} $$

$$ {{Z}_{1}}=100\sqrt{2} $$

Now the current is

$${{I}_{1}}=\dfrac{20}{100\sqrt{2}}\,A$$

Now, the impedance is

$$ {{Z}_{L}}=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}} $$

$$ {{Z}_{L}}=50\sqrt{2} $$

Now, the current is

$${{I}_{2}}=\dfrac{20}{50\sqrt{2}}\,A$$

Now, the total current is

$$ I=\sqrt{I_{1}^{2}+I_{2}^{2}} $$

$$ I=\sqrt{{{\left( \dfrac{20}{100\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{20}{50\sqrt{2}} \right)}^{2}}} $$

$$ I=\dfrac{1}{\sqrt{10}}\,A $$

Now, voltage across 100 Ω resistor is

$$ {{V}_{100}}=\dfrac{20}{100\sqrt{2}}\times 100 $$

$$ {{V}_{100}}=10\sqrt{2}\,V $$

Now, voltage across 50 Ω resistor is

$$ {{V}_{50}}=\dfrac{20}{50\sqrt{2}}\times 50 $$

$$ {{V}_{50}}=10\sqrt{2}\,V $$

Hence, the voltage across 50Ω resistor is $$10\sqrt{2}\,V$$
Question 7
1 / -0

The maximum value of a.c. voltage in a circuit is 707V. Its r.m.s. value is

A
70.7 V

B
100 V

C
500 V

D
707 V

Solution

As we know,
$$V_{rms}=\dfrac{V_0}{\sqrt 2}$$

$$V_{rms}=\dfrac{707}{\sqrt 2}=500\ V$$
Question 8
1 / -0

Impedance of the following circuit will be:

A
$$150\ \Omega$$

B
$$200\ \Omega$$

C
$$250\ \Omega$$

D
$$340\ \Omega$$

Solution
Question 9
1 / -0

In the given circuit the potential difference across resistance is $$54$$V and power consumed by it is $$16$$W. If AC frequency is $$60$$ Hz find the value of L.

A
$$1$$H

B
$$2$$H

C
$$5$$H

D
$$4$$H

Solution
Question 10
1 / -0

If a cell of $$200V$$ is connected across an inductor, the current is found to be $$5A$$. When cell is replaced by $$200 V - 100 rad/s$$ supply, r.m.s. value of current becomes $$4A$$. The inductance of inductor is-

A
$$0.2$$ henry

B
$$0.3$$ henry

C
$$0.4$$ henry

D
$$0.5$$ henry