In an ac circuit, the instantaneous values of current and applied voltage are respectively [ i = 2 ( Amp ) sin ( 250 πs - 1 ) t and ε = ( 10 V ) sin [ ( 250 π s – 1 ) t + π 3 ] . Find the instantaneous power drawn from the source at t = 2 3 ms and its average value.

Given that Potential difference of AC source ε = ( 10 V ) sin ( 250 π s – 1 ) t + π 3 Instantaneous values of current i = 2 ( Amp ) sin ( 250 πs - 1 ) t Instantaneous power P ,consumed by the resistance is P = ε i = ( 10 V ) sin ( 250 π s – 1 ) t + π 3 2 ( Amp ) sin ( 250 πs - 1 ) t At t = 2 3 ms Instantaneous power is P = ( 10 V ) sin ( 250 π ) 2 3 × 10 - 3 + π 3 2 ( Amp ) sin 250 π × 2 3 × 10 - 3 ⇒ P = 20 sin π 6 + π 3 sin π 6 ⇒ P = 20 sin π 2 sin π 6 = 10 W Average power consumed in a cycle P avg = ∫ 0 2 π ω P d t 2 π ω ⇒ P avg = ∫ 0 2 π ω ( 10 V ) sin ( 250 π s – 1 ) t + π 3 2 ( Amp ) sin ( 250 πs - 1 ) t d t 2 π ω ∵ P avg = ∫ 0 2 π ω V m sin ω t i m sin ω t + ϕ d t 2 π ω = V m i m 2 cos ϕ ⇒ P avg = 10 cos π 3 = 5 W

A resistor, a capacitor and an inductor ( R = 300 Ω , C = 20 μF , L = 1 . 0 Henry ) are connected in series with an AC source of, E rms = 50 V and v = 50 π Hz . Find (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor.

Given that Resistor R = 300 Ω Capacitor C = 20 μF Inductor L = 1 . 0 henry AC source voltage and frequency E rms = 50 V and ν = f = 50 π Hz . The reactance of the circuit X L = 2 π f L = 2 π 50 π 1 = 100 Ω and X C = 1 2 π f C = 1 2 π 50 π 20 × 10 - 6 = 500 Ω The total reactance of the circuit The reactance of the circuit X = X C – X L = 500 - 100 = 400 Ω Since X L < X C , the circuit is called capacitive Impedance of the circuit Z = R 2 + x 2 = 300 2 + 400 2 = 500 Ω The current i r m s = V r m s Z = 50 500 = 0 . 1 A b The rms potential differences across the capacitor V C = i r m s x C = 0 . 1 × 500 = 50 V The rms potential differences across the resistor V L = i r m s x L = 0 . 1 × 100 = 10 V The rms potential differences across the inductor. V R = i r m s R = 0 . 1 × 300 = 30 V (Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.)

A coil draws a current of 1 . 0 ampere and a power of 100 watt from an A.C. source of 110 volt and 5 22 π Hz . Find the inductance and resistance of the coil.

Given that, Rating of coil, i rms = 1 . 0 A and P = 100 watt . Voltage and frequency of AC mains V rms = 110 volts and f = 5 π 22 Hz Total power is consumed by resistance, so we assume lamp as a resistance for dc circuit. The resistance of the coil P = i 2 R ⇒ 100 = 1 R ⇒ R = 100 Ω For AC circuit Impedance of the circuit i rms = V rms Z ; here i rms = 1 . 0 A and V rms = 110 V So, Z = V rms i rms = 110 1 = 110 Ω (i) For inductance of the inductor, From Z 2 = R 2 + x 2 ⇒ x 2 = Z 2 - R 2 ⇒ x 2 = 110 2 - 100 2 ⇒ x = 10 21 Ω For inductance of the inductor x = 2 π f L ⇒ 10 21 = 2 π 5 22 π L ⇒ L = 21 22 H

HARD

JEE Main/Advance

IMPORTANT

Earn 100

A resistor of resistance $100 Ω$ is connected to an AC source $ε = (12 V) \sin (250 π s^{- 1}) t .$ Find the power consumed by the bulb.

Important Questions on Alternating Current

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 7

In an ac circuit, the instantaneous values of current and applied voltage are respectively

[i = 2 (Amp) \sin (250 {πs}^{- 1}) t

and

ε = (10 V) \sin [(250 π s^{- 1}) t + \frac{π}{3}]

. Find the instantaneous power drawn from the source at

t = \frac{2}{3} ms

and its average value.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 8

The dielectric strength of air is

3.0 \times 10^{6} V m^{- 1}

. A parallel-plate air-capacitor has the area

20 {cm}^{2}

and plate separation

0.10 mm

. Find the maximum RMS voltage of an AC source which can be safely connected to this capacitor.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 9

An electric bulb is designed to consume

55 W

when operated at

110 volts

. It is connected to a

220 V

50 Hz

line through a choke coil in series. What should be the inductance of the coil for which the bulb gets the correct voltage?

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 10

A resistor, a capacitor and an inductor

(R = 300 Ω, C = 20 μF, L = 1.0 Henry)

are connected in series with an AC source of,

E_{rms} = 50 V

and

v = \frac{50}{π} Hz .

Find (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 11

In a series LCR circuit with a source of $50 V, R = 300 Ω$ , frequency $v = \frac{50}{π} Hz$ . The average electric field energy stored in the capacitor and average magnetic field energy stored in the coil are $25 mJ$ and $5 mJ$ respectively. The RMS current in the circuit is $0.10 A$ . Then find:

(a) Capacitance $(C)$ of capacitor

(b) Inductance $(L)$ of inductor.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 12

A $20 volts, 5 watt$ lamp (lamp to be treated as a resistor) is used on AC mains of $200 volts$ and $\frac{50}{π} \sqrt{11}$ c.p.s. Calculate the

(i) capacitance of the capacitor, or inductance of the inductor, to be put in series to run the lamp.

(ii) How much pure resistance should be included in place of the above device so that the lamp can run on its rated voltage.

(iii) which is more economical (the capacitor, the inductor or the resistor).

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 13

A circuit has a resistance of $50 ohms$ and an inductance of $\frac{3}{π}$ henry. It is connected in series with a condenser of $\frac{40}{π} μF$ and AC supply voltage of $200 V$ and $50 cycles per \sec .$ Calculate

(i) the impedance of the circuit,

(ii) the p.d. across inductor coil and condenser.

(iii) Power factor

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 1>Q 14

A coil draws a current of

1.0 ampere

and a power of

100 watt

from an A.C. source of

110 volt

and

\frac{5 \sqrt{22}}{π} Hz

. Find the inductance and resistance of the coil.

A resistor of resistance 100 Ω is connected to an AC source ε = (12V) sin (250π s–1)t. Find the power consumed by the bulb.

Important Questions on Alternating Current

Learn and Prepare for any exam you want !

A resistor of resistance $100 Ω$ is connected to an AC source $ε = (12 V) \sin (250 π s^{- 1}) t .$ Find the power consumed by the bulb.