The equation of AC voltage is E220sint6voltand the AC current is I10sint6amp The average power dissipated is

EASY

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 18

A

10 Ω

resistance,

5 mH

coil and

10 μ F

capacitor are joined in series. When a variable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency:

EASY

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 19

The

p . d .

across an instrument in an

AC

circuit of frequency

f

is

V

and the current flowing through it is

I

such that

V = 5 \cos (2 π f t) volt

and

I = 2 \sin (2 π f t) amp

. The power dissipated in the instrument is:

EASY

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 20

In an

A . C .

circuit, maximum value of voltage is

423 volt

. Its effective voltage is :

EASY

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 21

Resonance frequency of a circuit is

f

. If the capacitance is made

4

times the initial value, then the resonance frequency will become:

MEDIUM

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 22

The current flowing through a coil is

3 A

and the power consumed is

108 W

. If the AC source is of

120 V, 50 Hz

, the resistance of the circuit is,

EASY

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 23

Calculate the power factor of

L - C - R

circuit at resonance.

MEDIUM

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 24

The turn ratio of transformers is given as

2 : 3

. If the current through the primary coil is

3 A

, calculate the current through load resistance.

EASY

BITSAT

IMPORTANT

Comprehensive Guide to BITSAT Physics. Other applicable Exams - JEE Main, VITEEE, SRM JEE, MHT-CET, K-CET, EAMCET, AMU & Other State Engg. Entrance Exams>Chapter 9 - Alternating Current>Exercise 2>Q 25

A sinusoidal voltage

V_{0} \sin ω t

is applied across a series combination of resistance

R

and capacitance

C

. The amplitude of the current in this circuit is,

The equation of $A C$ voltage is $E = 220 \sin (ω t + \frac{π}{6}) volt$ and the $A . C .$ current is $I = 10 \sin (ω t - \frac{π}{6}) amp$ . The average power dissipated is

Important Questions on Alternating Current

Learn and Prepare for any exam you want !

The equation of AC voltage is E=220sinωt+π6 volt and the A.C. current is I=10sinωt-π6 amp. The average power dissipated is

Important Questions on Alternating Current

Learn and Prepare for any exam you want !

The equation of $A C$ voltage is $E = 220 \sin (ω t + \frac{π}{6}) volt$ and the $A . C .$ current is $I = 10 \sin (ω t - \frac{π}{6}) amp$ . The average power dissipated is