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In any AC circuit the emf (e) and the cu...

In any `AC` circuit the emf `(e)` and the current `(i)` at any instant are given respectively by `e= E_(0)sin omega t`
`i=I_(0) sin (omegat-phi)`
The average power in the circuit over one cycle of `AC` is

A

`E _(0) I _(0)`

B

`(E _(0) I _(0))/(2)`

C

`(E _(0) I _(0))/(2) sin phi`

D

`(E _(0) I _(0))/(2) cos phi`

Text Solution

Verified by Experts

`e = E _(0) sin omega t `
`I = I _(0) sin (omega t -phi)`
Avg. power in circuit over one cycle of AC is
`{P} = {ei} = { (E _(0) sin omega t ) [I _(0) sin (omega t - phi)}`
`=E _(0) I _(0) { (sin ^(2) omega t cos phi - sin omega t cos omega t sin phi)}`
` = (1)/(2) E _(0) I _(0) cos phi`
Avg. power can also be determined by
`= (E _(0) I _(0))/( 2) cos phi = (E_(0))/( sqrt2) xx (I_(0))/( sqrt 2 ) xx cos phi`
where `phi` is phase difference between current
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