Alternating Current Test

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Alternating Current Test - 60

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Question 1
1 / -0

An alternative current, L.R circuit comprises of an inductor, whose reactance $$X_L = 3R$$, where $$R$$ is the resistance of the circuit. If a capacitor, whose reactance $$X_C = R$$ is connected in series then what will be the ratio of the new and the old power factor?

A
$$\sqrt{2}$$

B
$$\dfrac{1}{\sqrt{2}}$$

C
$$2$$

D
$$1$$

Solution
Question 2
1 / -0

At $$t=0$$, an inductor of zero
resistance is joined to a cell of emf `E' through a resistance. The current
increases with a time constant $$\tau$$. The emf across the coil after time
$$t$$ is

A
$$\dfrac{{Et}}{\tau }$$

B
$$E(1 - {e^{ - t/}}\tau )$$

C
$$E{e^{ - \frac{t}{\tau }}}$$

D
$$E{e^{ - \frac{{2t}}{\tau }}}$$

Solution
Question 3
1 / -0

In the adjoining figure the impedance of the circuit will be_

A
$$120\Omega$$

B
$$50\Omega$$

C
$$60\Omega$$

D
$$90\Omega$$

Solution

According to the question,

$$ i_L = \dfrac{90}{30} = 3 \,A $$ , $$ i_C = \dfrac{90}{20} = 4.5 \,A $$

Net current through circuit $$ i = i_C - i_L = 1.5 \,A $$

$$ \therefore \, Z = \dfrac{V}{i} = \dfrac{90}{1.5} = 60\Omega$$
Question 4
1 / -0

An inductance of $$1mH$$, a condenser of $$10 \mu F$$ and resistance of $$50\Omega$$ are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be:-

A
$$100 \Omega$$

B
$$30 \Omega$$

C
$$3.2 \Omega$$

D
$$10 \Omega$$

Solution

Given
$$ \omega L = \dfrac{1}{\omega C} $$ $$ \Rightarrow \, \omega^2 = \dfrac{1}{LC} $$
or $$ \omega = \dfrac{1}{\sqrt{10^{-3} \times 10 \times 10^{-6}}} = \dfrac{1}{\sqrt{10^{-8}}} = 10^4 $$

$$ X_L = \omega L = 10^4 \times 10^{-3} = 10 \, \Omega $$
Question 5
1 / -0

The graph given below depict the dependence of two reactive impedances $$X_1$$ and $$X_2$$ on the frequency of the alternating e.m.f. applied individually to them. We can say that:

A
$$X_1$$ is an inductor and $$X_2$$ is a capacitor

B
$$X_1$$ is a resistor and $$X_2$$ is a capacitor

C
$$X_1$$ is a capacitor and $$X_2$$ is an inductor

D
$$X_1$$ is an inductor and $$X_2$$ is a resistor

Solution

As we know,

$$ X_C = \dfrac{1}{C \times 2 \pi f} $$
hence from the above figure $$X_!$$is a capacitor and

$$ X_L = L \times 2 \pi f $$
hence $$X_2$$ is an inductor.
Question 6
1 / -0

An AC circuit draws $$5 A$$ at $$160 V$$ and the power consumption is $$600 W$$. Then the power factor us :-

A
1

B
0.75

C
0.50

D
Zero

Solution

$$\begin{array}{l} i=5A \\ v=160v \\ p=600w \\ p=V.I\cos \phi \\ \Rightarrow \cos \phi =\frac { 6 }{ 8 } =\frac { 3 }{ 4 } =0.75 \end{array}$$
$$\therefore$$ Option $$B$$ is correct.
Question 7
1 / -0

An inductor of inductance 2.0 mH is connect across a charge capacitor of capacittance 5.0 uF , and the resulting LC circuit is set oscillating at natural frequency. Let Q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found the maximum value of Q is 200 uC .When Q=100uC the value of $$\dfrac{dI}{dt} $$ is found to be $$10^{x}$$ A/s. Find the value of x is then

A
x=-4

B
x=-5

C
x=4

D
x=5

Solution

In an $$Le$$ system $$he$$ charge oscillation between $$I$$ and $$Q$$
Applying $$KVL$$ gives,
$$-L\dfrac {dI}{dt}+\dfrac {9}{c}=0$$ where $$L=2\ mH$$
$$c=5\mu F$$
also, maximum charge $$Q_0=200\ \mu c$$
Thus, total energy of system, $$\mu =\dfrac {Q_0}{2c}=\dfrac {200Z^2}{2\times 5}=4000\ J$$
when $$9=100\ u c$$, we get from $$(1)$$,
$$\dfrac {dI}{dt}dfrac {+9}{Lc}=\dfrac {+100}{2\times 10^{-3}\times 5}=+10^{-4}\ A/s$$
as $$\dfrac {dI}{dt}=10^x$$ in gives $$x=-4$$
Question 8
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A coil has a resistance $$ 10 \Omega $$ and an inductance of 0.4 henry. It is connected to an AC source of $$ 6.5 V , \frac {30} { \pi } Hz. $$ The average power consumed in the circuit, is :

A
$$ \cfrac {5} { 8} W $$

B
$$ \cfrac {4} {3} W $$

C
$$ \cfrac {3} {8} W $$

D
$$ \cfrac {6} {7} W $$

Solution
Question 9
1 / -0

In an electric circuit applied ac emf is e = 20 sin 300t volt and the current is i = 4 sin 300 t ampere. The reactance of the circuit is -

A
5 ohm

B
80 ohm

C
$$1.8 k\Omega $$

D
zero

Solution

Given that,

Emf, $$e=20\sin 300t$$

Current, $$i=4\sin 300t$$

Reactance,

$$ R=\dfrac{emf}{I} $$

$$ R = \dfrac{20}{4}=5\,ohm $$
Question 10
1 / -0

The inductive reactance of an inductive coil with $$\dfrac{1}{\pi}$$ henry and $$50$$Hz:-

A
$$\dfrac{50}{\pi}ohm$$

B
$$\dfrac{\pi}{50}ohm$$

C
100 ohm

D
50 ohm

Solution

As we know,
$$ X_L = 2 \pi \nu L = 2 \times \pi \times 50 \times \dfrac{1}{\pi} = 100 \, \Omega $$