Home
Class 12
PHYSICS
An AC source is rated 220 V, 50 Hz. The ...

An AC source is rated `220 V, 50 Hz`. The average voltage is calculated in a time interval of `0.01 s`. It

A

must be zero

B

may be zero

C

is never zero

D

is `(220//sqrt2V`

Text Solution

Verified by Experts

The correct Answer is:
B

may be zero
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    RESONANCE ENGLISH|Exercise EXERCISE SECTION A|4 Videos
  • ALTERNATING CURRENT

    RESONANCE ENGLISH|Exercise SECTION B|15 Videos
  • ALTERNATING CURRENT

    RESONANCE ENGLISH|Exercise EXAMPLE|40 Videos
  • ATOMIC PHYSICS

    RESONANCE ENGLISH|Exercise Advanved level problems|17 Videos

Similar Questions

Explore conceptually related problems

An Ac source is rated 222 V, 60 Hz. The average voltage is calculated in a time interval of 16.67 ms.IL

An AC source is rated at 220V, 50 Hz. The time taken for voltage to change from its peak value to zero is

An AC source is 120 V-60 Hz. The value of voltage after 1/720 s from start will be

The average value of voltage (V) in one time period will be

An ac source is of (200)/(sqrt(2)) V, 50 Hz. The value of voltage after (1)/(600)s from the start is

An ac source is of (200)/(sqrt(2)) V, 50 Hz. The value of voltage after (1)/(600)s from the start is

An AC voltage is given by E=E_(0) sin(2pit)/(T) . Then the mean value of voltage calculated over time interval of T/2 seconds

The peak voltage in a 220 V AC source is

The peak voltage in a 220 V AC source is

The frequency of the wave propagating on a string is 20 Hz . What will be the difference in phase of a particle of the string in a time interval of 0.01 s ?