Alternating Current Test

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Alternating Current Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

At inductance $$1\ H$$ is connected in series with an $$AC$$ source of $$220\ V$$ and $$50\ Hz$$. The inductive resistance (in ohm) is :

A
$$2\pi$$

B
$$50\pi$$

C
$$100\pi$$

D
$$1000\pi$$

Solution

Inductive reactants $$X_{L} = wL$$
$$= 2\pi vL$$
$$= 2\pi \times 50\times 1$$
$$= 100\pi$$.
Question 2
1 / -0

An AC source is connected in parallel with an L-C-R circuit as shown. Let $$\displaystyle { I }_{ S },{ I }_{ L },{ I }_{ C }$$ and $$\displaystyle { I }_{ R }$$ denote the currents through and $$\displaystyle { V }_{ S },{ V }_{ L },{ V }_{ C }$$ and $$\displaystyle { V }_{ R }$$ voltages across the corresponding components. Then :

A
$$\displaystyle { I }_{ S }={ I }_{ L }+{ I }_{ C }+{ I }_{ R }$$

B
$$\displaystyle { V }_{ S }={ V }_{ L }+{ V }_{ C }+{ V }_{ R }$$

C
$$\displaystyle \left( { I }_{ L },{ I }_{ C },{ I }_{ R } \right) <{ I }_{ S }$$

D
$$\displaystyle { I }_{ L },{ I }_{ C }$$ may be greater than $$\displaystyle { I }_{ S }$$

Solution

For parallel RLC circuit,
$$I_S = \sqrt{I_R^2 + (I_L-I_C)^2}$$

In case of resonance, $$I_L = I_C$$
So, $$I_S = I_R$$

Here $$I_L$$ and $$I_C$$ can be much greater than $$I_R$$

Option D is correct.
Question 3
1 / -0

In an LCR circuit having L$$=8$$H, C$$=0.5\mu F$$ and $$R=100\Omega$$ in series, the resonance frequency rad/s is?

A
$$600$$

B
$$200$$

C
$$250/\pi$$

D
$$500$$

Solution

From the formula, $$\omega =\displaystyle\frac{1}{\sqrt{LC}}$$
Given, $$L=8$$H,
$$C=0.5\mu F=0.5\times 10^{-6}F$$
$$\therefore \omega =\displaystyle\frac{1}{\sqrt{8\times 0.5\times 10^{-6}}}$$
$$\Rightarrow \omega =\displaystyle\frac{1}{2\times 10^{-3}}=500$$ rad/s
Question 4
1 / -0

In an $$L-C-R$$ series circuit, the values of $$R, X_{L}$$ and $$X_{C}$$ are $$120\Omega, 180\Omega$$ and $$130\Omega$$, what is the impedance of the circuit?

A
$$120\Omega$$

B
$$130\Omega$$

C
$$180\Omega$$

D
$$330\Omega$$

Solution

Given, $$R = 120\Omega$$
$$X_{L} = 180\Omega$$ and $$ X_{C} = 130\Omega$$
The impedance of $$L-C-R$$ circuit
$$Z = \sqrt {R^{2} + (X_{L} - X_{C})^{2}}$$
$$Z = \sqrt {(120)^{2} + (180 - 130)^{2}}$$
$$Z= 130\Omega$$.
Question 5
1 / -0

A series resonant circuit contains $$L = \dfrac{5}{\pi} mH, C = \dfrac{200}{\pi} \mu F$$ and $$R = 100 \mu$$. If a source of emf $$e = 200 sin 1000 \pi t$$ is applied, then the rms current is:

A
$$2A$$

B
$$200 \sqrt 2 A$$

C
$$100 \sqrt 2 A$$

D
$$1.41 A$$

Solution

Source emf is given as $$e = 200 sin 1000 \pi t$$
Comparing it with $$e = e_o \sin wt$$
We get $$ \omega = 1000 \pi$$
$$\Rightarrow f = \dfrac{\omega}{2 \pi} = 500 Hz$$
Also, $$e_0 = 200 V$$
At resonance $$Z = R$$
So, current flowing $$i_0 = \dfrac{e_o}{R} = \dfrac{200}{100} = 2A$$
Rms value of current $$i_{rms} = \dfrac{i_0}{\sqrt 2} = \dfrac{2}{\sqrt 2} = \sqrt 2 = 1.41 A$$
Question 6
1 / -0

The alternating voltage and current in an electric circuit are respectively given by $$E=100\sin { 100\pi t } ,I=5\sin { 100\pi t } $$. The reactance of the circuit will be :

A
$$1\Omega $$

B
$$0.05\Omega $$

C
$$20\Omega $$

D
Zero

Solution

The general equations of alternating voltage and current in an AC circuit are as follows
$$E={ E }_{ 0 }\sin { \omega t } $$ ....(i)
$$I={ I }_{ 0 }\sin { \omega t } $$ ....(ii)
Given that,
$$E=100\sin { 100\pi t } $$ ....(iii)
$$I=5\sin { 100\pi t } $$ ....(iv)
Comparing equations (ii) and (iv), we get,
Peak value of current, $${ I }_{ 0 }=5A$$
Comparing equations (i) and (iii), we get
Peak value of voltage, $${ E }_{ 0 }=100 V$$
Now, r.m.s. value of voltage is given by
$${ E }_{ rms }=\dfrac { { E }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 100 }{ \sqrt { 2 } } V$$
Similarly, r.m.s. value of current is given by
$${ I }_{ rms }=\dfrac { { I }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 5 }{ \sqrt { 2 } } A$$
Hence, reactance of the circuit is given by
$$Z=\dfrac { { E }_{ rms } }{ { I }_{ rms } } =\dfrac { \dfrac { 100 }{ \sqrt { 2 } } }{ \dfrac { 5 }{ \sqrt { 2 } } } =20\Omega $$
Question 7
1 / -0

In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?

A
$$Resistive$$

B
$$Capacitive$$

C
$$Inductive$$

D
$$None\ of\ these$$

Solution

At resonant frequency
$$X_L=X_C$$ $$\left(\omega L=\displaystyle\frac{1}{\omega C}\right)$$
At frequencies higher than resonance frequencies
$$X_L > X_C$$
i.e., behaviour is inductive.
Question 8
1 / -0

A resistor R, inductor L and a capacitor C are connected in series to an oscillator of frequency v. If the resonant frequency is $$v_r$$, then the current lags behind the voltage, when:

A
$$v = 0$$

B
$$v < v_r$$

C
$$v>v_r$$

D
$$v=v_r$$

Solution

The current will lag behind the voltage when reactance of inductor is more than the reactance of capacitor i.e.,
$$\omega L>\dfrac{1}{\omega c}$$
or $$\omega > \dfrac{1}{\sqrt{LC}}\ or \ 2\pi v>\dfrac{1}{\sqrt{LC}}$$
$$v>\dfrac{1}{2\pi \sqrt{LC}}\ or\ v>v_r$$
where $$v_r$$ is the resonant frequency.
Question 9
1 / -0

In a circuit, $$L,\ C$$ and $$R$$ are connected in series with an alternating voltage source of frequency $$f$$. The current leads the voltage by $$45^o$$. The value of $$C$$ is :

A
$$ \dfrac {1}{\pi f ( 2 \pi f L + R)} $$

B
$$ \dfrac {1}{\pi f ( 2 \pi f L - R)} $$

C
$$ \dfrac {1}{2 \pi f ( 2 \pi f L + R)} $$

D
$$ \dfrac {1}{2 \pi f ( 2 \pi f L - R)} $$

Solution

The phase differenec $$ \phi $$ between current and voltage is given by
$$ \tan \phi = \dfrac {X_L - X_C}{R} $$
$$ \tan 45^o = 1 $$
$$ X_C = X_L - R $$

$$ \dfrac {1}{2 \pi f C } = 2 \pi fL - R $$

$$ C = \dfrac {1}{2 \pi f (2 \pi f L - R)} $$
Question 10
1 / -0

For the given circuit, the natural frequency is given by :

A
$$\frac {1}{2\pi}\sqrt{LC}$$

B
$$\frac {1}{2\pi\sqrt{LC}}$$

C
$$\sqrt{\frac {L}{C}}$$

D
$$\sqrt{\frac {C}{L}}$$

Solution

Natural frequency $$(\omega)$$ is given by
$$f=\dfrac {1}{2\pi} \cdot \dfrac{1}{\sqrt{LC}}$$
For the given circuit $$C_{net}=\dfrac{C}{2}$$ (in series)
$$L_{net}=2L$$ (in series)
$$f=\dfrac {1}{2\pi} \cdot \dfrac{1}{\sqrt{LC}}$$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 50

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Alternating Current Test - 50

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SHARING IS CARING

Question 1

$$2\pi$$

$$50\pi$$

$$100\pi$$

$$1000\pi$$

Solution

Question 2

$$\displaystyle { I }_{ S }={ I }_{ L }+{ I }_{ C }+{ I }_{ R }$$

$$\displaystyle { V }_{ S }={ V }_{ L }+{ V }_{ C }+{ V }_{ R }$$

$$\displaystyle \left( { I }_{ L },{ I }_{ C },{ I }_{ R } \right) <{ I }_{ S }$$

$$\displaystyle { I }_{ L },{ I }_{ C }$$ may be greater than $$\displaystyle { I }_{ S }$$

Solution

Question 3

$$600$$

$$200$$

$$250/\pi$$

$$500$$

Solution

Question 4

$$120\Omega$$

$$130\Omega$$

$$180\Omega$$

$$330\Omega$$

Solution

Question 5

$$2A$$

$$200 \sqrt 2 A$$

$$100 \sqrt 2 A$$

$$1.41 A$$

Solution

Question 6

$$1\Omega $$

$$0.05\Omega $$

$$20\Omega $$

Zero

Solution

Question 7

$$Resistive$$

$$Capacitive$$

$$Inductive$$

$$None\ of\ these$$

Solution

Question 8

$$v = 0$$

$$v < v_r$$

$$v>v_r$$

$$v=v_r$$

Solution

Question 9

$$ \dfrac {1}{\pi f ( 2 \pi f L + R)} $$

$$ \dfrac {1}{\pi f ( 2 \pi f L - R)} $$

$$ \dfrac {1}{2 \pi f ( 2 \pi f L + R)} $$

$$ \dfrac {1}{2 \pi f ( 2 \pi f L - R)} $$

Solution

Question 10

$$\frac {1}{2\pi}\sqrt{LC}$$

$$\frac {1}{2\pi\sqrt{LC}}$$

$$\sqrt{\frac {L}{C}}$$

$$\sqrt{\frac {C}{L}}$$

Solution

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