Alternating Current Test

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Alternating Current Test - 35

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Question 1
1 / -0

A series $$AC$$ circuit has a resistance of $$4 \Omega$$ and a reactance of $$3 \Omega$$. The impedance of the circuit is

A
$$5 \Omega$$

B
$$7 \Omega$$

C
$$12/7 \Omega$$

D
$$7/12 \Omega$$

Solution

$$Impedance(Z) = \sqrt{{R}^{2} + {X}^{2}}$$

where, $$R = resistance$$

$$X = reactance$$

Given $$R = 4\Omega$$ and $$X = 3\Omega$$

Substitute values back in equation

$$Z = \sqrt{{4}^{2}+{3}^{2}}$$

$$Z = 5\Omega$$
Question 2
1 / -0

A leaky capacitor $$10\ \Omega/ 60^{\circ}$$ is connected in series with a 10 $$\Omega$$ resistance. Find the overall impedance.

A
$$\displaystyle 10\ \Omega$$

B
$$\displaystyle 10 \sqrt 2\ \Omega$$

C
$$\displaystyle 15\ \Omega$$

D
$$\displaystyle 10\sqrt 3\ \Omega$$

Solution

$$\displaystyle \sqrt{r^2+X^2_c} = 10\ \text{and}\ \dfrac{X_c}{r} = \tan 60$$

$$\displaystyle \therefore \sqrt{r^2+(\sqrt 3r)^2} = 10$$ or $$\displaystyle r = 5 \Omega, X_c = 5\sqrt 3 \Omega$$

$$\displaystyle |Z| = \sqrt{(10+5)^2+5\sqrt 3} = 10\sqrt 3 \Omega$$;

$$\displaystyle \Phi = \tan^{-1} \frac{X_c}{R} = tan^{-1} \left(\dfrac{5\sqrt 3}{15}\right)$$

$$\displaystyle \tan^{-1} \left(\dfrac{1}{\sqrt 3}\right)$$ or $$\displaystyle \Phi = 30^{\circ}$$
Question 3
1 / -0

An $$AC$$ source producing emf $$\displaystyle E = E_0 [cos (100 \pi s^{-1}) t] + E_0 cos [(500 \pi s^{-1})t ]$$ is connected in series with a capacitor and a resistance, the steady state current in the circuit is found to be $$i = i_1 [cos (100 \pi s^{-1}) t + \varphi_1] + i_2 cos [(500 \pi s^{-1})t + \varphi_2]$$

A
$$\displaystyle i_1>i_2$$

B
$$\displaystyle i_1 = i_2$$

C
$$\displaystyle i_1 < i_2$$

D
Insufficient information

Solution
Question 4
1 / -0

If the output is taken across a capacitor in a series RLC circuit then it acts as

A
band-pass filter

B
high-pass filter

C
low-pass filter

D
band reject filter

Solution

Given circuit is one simple low-pass filter circuit consists of a resistor in series with a load, and a capacitor in parallel with the load. The capacitor exhibits reactance, and blocks low-frequency signals, forcing them through the load instead. At higher frequencies the reactance drops, and the capacitor effectively functions as a short circuit.
Question 5
1 / -0

An inductor, a resistor and a capacitor are joined in series with an $$AC$$ source. As the frequency of the source is slightly increased from a very low value, the reactance

A
of the inductor increases

B
of the resistor increases

C
of the capacitor increases

D
of the circuit increases

Solution

Reactance in electriacal and electronic systems is the opposition of a circuit element to a change of electric current or voltage.
Ideally a resistor has zero reactance.

Reactance of inductor of inductance $$L,\quad X_L= L \omega $$. It increases with frequency.

Reactance of capacitor of capacitance $$C,\quad X_C =\dfrac{ 1}{C \omega }$$. It decreases with frequency.

Reactance of circuit $$= L \omega - \dfrac{1}{C \omega} $$. It follows a curve both increasing and decreasing when the frequency increases.
Question 6
1 / -0

$$\displaystyle I = 6\cos wt + 8\sin wt$$ is applied across a resistor of 40 $$\Omega$$. Find the potential difference across the resistor.

A
660 V

B
80 V

C
330 V

D
400 V

Solution

Here $$I=10(\dfrac{6}{10}\cos\omega t+\dfrac{8}{10}\sin \omega t)$$

$$=10cos(\delta- \omega t)$$

$$|I|=10A$$

$$V=IR=400V$$
Question 7
1 / -0

Which of the following plots may represent the reactance of a series $$LC$$ combination?

A
(a)

B
(b)

C
(c)

D
(d)

Solution

Reactance of capacitor, $$X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{\omega C}$$

Reactance of inductor, $$X_L=\omega L$$

Since in series combination, net reactance$$=X_C+X_L=j(\omega L-\dfrac{1}{\omega C})$$

So reactance magnitude tends to infinity as frequency tends to 0 as well as infinity
Question 8
1 / -0

An inductor $$10 \Omega/60^{\circ}$$ is connected to a $$5 \Omega$$ resistance in series. Find net impedance.

A
$$15 \Omega$$

B
$$12 \Omega$$

C
$$13.2 \Omega$$

D
$$18 \Omega$$

Solution

$$\displaystyle 10 = \sqrt{r^2+X^2_L} \ and\ \frac{X_L}{r} = tan 60$$

$$\displaystyle 10 = \sqrt{r^2+(r\sqrt 3)^2} \ or\ r = 5 \Omega, X_L = 5\sqrt 3 \Omega.$$

$$\displaystyle Z = \sqrt{(5+5)^2 + (5\sqrt3)^2} = \sqrt{175} = 13.2 \Omega$$

$$\displaystyle tan\Phi = \frac{X_L}{R+r} = \frac{5\sqrt3}{10}$$

or$$\displaystyle \Phi = tan^{-1} \left(\frac{\sqrt3}{2}\right)$$
Question 9
1 / -0

A capacitor acts as an infinite resistance for

A
$$DC$$

B
$$AC$$

C
$$DC$$ as well as $$AC$$

D
neither $$AC$$ nor $$DC$$

Solution

Capacitors contain at least two electrical conductors separated by a dielectric/insulator and is used to store energy electrostatically between the conductors. It acts like an open circuit and hence acts like an infinite resistance for DC currents.
Question 10
1 / -0

The correct curve representing the variation of capacity reactance $$X_c$$ with frequency $$f$$ is

A

B

C

D

Solution

$$ { X }_{ C }=\dfrac { 1 }{ C2\pi f } $$

$$ {X}_{C} $$ and $$ f $$ are inversely proportional.