Alternating Current — Page 2 | NEET Physics

Q11

In a series LCR circuit, the inductance

L

is

10 ~\mathrm{mH}

, capacitance

C

is

1 ~\mu \mathrm{F}

and resistance

R

is

100 ~\Omega

. The frequency at which resonance occurs is :-

A 15.9 kHz

B 1.59 rad/s

C 1.59 kHz

D 15.9 rad/s

Correct Answer

Option C

Solution

The goal is to find the frequency at which resonance occurs in a series LCR circuit.

To achieve resonance, the inductive reactance (

X_L

) should equal the capacitive reactance (

X_C

). Given are: Inductance,

L = 10 ~\mathrm{mH} = 10 \times 10^{-3} ~\mathrm{H}

Capacitance,

C = 1 ~\mu\mathrm{F} = 1 \times 10^{-6} ~\mathrm{F}

Resistance,

R = 100 ~\Omega

(which does not directly impact the resonance frequency) Resonance occurs when the angular frequency ( $\omega$ ) meets the condition

\omega L = \frac{1}{\omega C}

, leading to the formula for calculating the resonance frequency (

f

) as

f = \frac{\omega}{2\pi} = \frac{1}{2\pi\sqrt{LC}}

. Substituting the given values:

f = \frac{1}{2\pi\sqrt{10 \times 10^{-3} \cdot 1 \times 10^{-6}}} = 1.59 \mathrm{kHz}

Thus, the frequency at which resonance occurs in this LCR circuit is 1.59 kHz.

Q12

A

12 \mathrm{~V}, 60 \mathrm{~W}

lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of

220 \mathrm{~V}

. Assuming the transformer to be ideal, what is the current in the primary winding ?

A 2.7 A

B 3.7 A

C 0.37 A

D 0.27 A

Correct Answer

Option D

Solution

To find the current in the primary winding of an ideal step-down transformer, we first understand that an ideal transformer's power input (primary side) equals its power output (secondary side).

This is because an ideal transformer is assumed to have 100% efficiency, meaning there are no losses in the transformation process.

The power of the lamp (which is connected to the secondary side of the transformer) is given as

60\,\mathrm{W}

, and it operates at

12\,\mathrm{V}

. The voltage on the primary side of the transformer is

220\,\mathrm{V}

. Using the power formula

P = IV

, where

P

is the power in watts,

I

is the current in amperes, and

V

is the voltage in volts, we equate the power on both sides of the transformer because of its ideal nature, i.e.,

P_{\text{primary}} = P_{\text{secondary}}

. Hence,

60\,\mathrm{W} = 220\,\mathrm{V} \times I_{\text{primary}}

. Solving for

I_{\text{primary}}

gives:

I_{\text{primary}} = \frac{60\,\mathrm{W}}{220\,\mathrm{V}} = 0.2727... \approx 0.27\,\mathrm{A}

. Therefore, the current in the primary winding of the transformer is approximately

0.27\,\mathrm{A}

.

Q13

An ac source is connected to a capacitor C. Due to decrease in its operating frequency

A displacement current increases.

B displacement current decreases.

C capacitive reactance remains constant.

D capacitive reactance decreases.

Correct Answer

Option B

Solution

When an AC source is connected to a capacitor, the current (I) depends on the capacitive reactance (X C ), where

X_{C} = \frac{1}{\omega C}

, with $\omega$ representing the angular frequency $\omega = 2\pi f$ , and C denoting the capacitance.

The formula for capacitive reactance shows that it is inversely proportional to the frequency (f) of the operation.

Thus, if the frequency decreases, the capacitive reactance increases because $X_{C} = \dfrac{1}{\omega C}$ and $\omega$ is proportional to frequency.

As capacitive reactance increases, the displacement current (I), which flows through the capacitor, decreases according to the relationship

I = \frac{V}{X_{C}}

, with V being the voltage across the capacitor.

So, among the given choices, the correct interpretation is that the displacement current decreases with a decrease in operating frequency.

Q14

The magnetic energy stored in an inductor of inductance

4 ~\mu \mathrm{H}

carrying a current of

2 \mathrm{~A}

is :

A

4 \mathrm{~mJ}

B

8 \mathrm{~mJ}

C

8 ~\mu \mathrm{J}

D

4 ~\mu \mathrm{J}

Correct Answer

Option C

Solution

The formula for calculating the magnetic energy stored in an inductor is given by:

E = \frac{1}{2} L I^2

where:

E

is the energy stored (in joules, J),

L

is the inductance of the inductor (in henrys, H),

I

is the current flowing through the inductor (in amperes, A). Plugging the given values into the formula:

E = \frac{1}{2} \times 4 \mu H \times (2 A)^2

E =\frac{1}{2} \times 4 \times 10^{-6} H \times 4 A^2

E = 2 \times 10^{-6} H \cdot 4

E = 8 \times 10^{-6} J

Hence, the energy stored in the inductor is

8 \mu J

.

Q15

The net impedance of circuit (as shown in figure) will be :

A

15 \Omega

B

5 \sqrt{5} \Omega

C

25 \Omega

D

10 \sqrt{2} \Omega

Correct Answer

Option B

Solution

To find the net impedance, Z, of the circuit, we need to calculate the inductive reactance (XL), the capacitive reactance (XC), and use the resistance (R) in the circuit.

Since R is given as 10 $\Omega$ , we calculate the reactances as follows: Inductive reactance, $X_L = 2\pi fL$ , where $f$ is the frequency and $L$ is the inductance.

With $f = 50$ Hz and $L$ given as $\dfrac{50}{\pi}$ mH, or $\dfrac{50 \times 10^{-3}}{\pi}$ H, we find $X_L = 2 \pi \times 50 \times \dfrac{50 \times 10^{-3}}{\pi} = 5$ $\Omega$ .

Capacitive reactance, $X_C = \dfrac{1}{2\pi fC}$ , where $C$ is the capacitance.

Given $C = \dfrac{10^3}{\pi}$ μF, or $\dfrac{10^{-3}}{\pi}$ F, results in $X_C = \dfrac{1}{2 \pi \times 50 \times \dfrac{10^{-3}}{\pi}} = 10 \Omega$ .

With these values, the net impedance $Z$ of the circuit can be calculated using the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ Substituting $R = 10 \Omega$ , $X_L = 5 \Omega$ , and $X_C = 10 \Omega$ , we find: $Z = \sqrt{10^2 + (5 - 10)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \Omega$ Therefore, the correct option is $5\sqrt{5} \Omega$ .

Q16

An inductor of inductance 2 mH is connected to a 220 V, 50 Hz ac source. Let the inductive reactance in the circuit is X 1 . If a 220 V dc source replace the ac source in the circuit, then the inductive reactance in the circuit is X 2 . X 1 and X 2 respectively are :

A 0.628

\Omega

, infinity

B 6.28

\Omega

, zero

C 6.28

\Omega

, infinity

D 0.628

\Omega

, zero

Correct Answer

Option D

Solution

We know, for A.C. source

{X_L} = \omega L

= 2\pi f(L)

= 100\pi (2 \times {10^{ - 3}})

= 0.2\pi \,\Omega = 0.628\,\Omega

For D.C. source The inductor behaves as a closed circuit offering no resistance at all (at steady state) as

\omega = 0

(For D.C.) $\therefore$

{X_L} = 0\,\Omega

Q17

A standard filament lamp consumes 100 W when connected to 200 V ac mains supply. The peak current through the bulb will be :

A 2 A

B 0.707 A

C 1 A

D 1.414 A

Correct Answer

Option B

Solution

{P_{avg}} = {V_{rms}} \times {I_{rms}} \times \cos \phi

Here,

\phi = 0^\circ

as current and voltage are in same phase in a resistor.

\Rightarrow 100 = 200 \times {I_{rms}} \Rightarrow {I_{rms}} = {1 \over 2} \Rightarrow {{{I_0}} \over {\sqrt 2 }} = {1 \over 2} \Rightarrow {I_0} = {1 \over {\sqrt 2 }} = 0.707\,A

Q18

The peak voltage of the ac source is equal to

A The value of voltage supplied to the circuit

B The rms value of the ac source

C

\sqrt2

times the rms value of the ac source

D 1/

\sqrt2

times the rms value of the ac source

Correct Answer

Option C

Solution

We know, RMS value of A.C.

{E_{rms}} = {{{E_0}} \over {\sqrt 2 }}

{E_0} = \sqrt 2 {E_{rms}}

Q19

A series LCR circuit with inductance 10 H, capacitance 10

\mu

F, resistance 50

\Omega

is connected to an ac source of voltage, V = 200sin(100t) volt. If the resonant frequency of the LCR circuit is v 0 and the frequency of the ac source is v, then

A

{v_0} = v = 50

Hz

B

{v_0} = v = {{50} \over \pi }

Hz

C

{v_0} = {{50} \over \pi }

Hz,

v = 50

Hz

D

v = 100

Hz,

{v_0} = {{100} \over \pi }

Hz

Correct Answer

Option B

Solution

\omega L = {1 \over {\omega C}}

\omega = {1 \over {\sqrt {LC} }} = {1 \over {\sqrt {10 \times 10 \times {{10}^{ - 6}}} }}

\omega = 100

\omega = 2\pi f \Rightarrow f = {\omega \over {2\pi }}

{v_0} = {f_0} = {{100} \over {2\pi }} = {{50} \over \pi }

Hz,

\omega = 100

v = f = {{100} \over {2\pi }} = {{50} \over \pi }

Q20

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40V, 10V and 40V, respectively. The amplitude of current flowing through LCR series circuit is 10

\sqrt 2

A. The impedance of the circuit is :

A 5

\Omega

B 4

\sqrt 2

\Omega

C 5/

\sqrt 2

\Omega

D 4

\Omega

Correct Answer

Option A

Solution

I 0 = 10

\sqrt 2

A I RMS =

{I \over {\sqrt 2 }}

= 10A

{V_{RMS}} = \sqrt {V_R^2 + {{({V_L} - {V_C})}^2}}

= \sqrt {{{(40)}^2} + {{(40 - 10)}^2}}

= 50 V Impedance,

Z = {{{V_{RMS}}} \over {{I_{RM}}}} = {{50V} \over {10V}} = 5\Omega