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In A.C. circuit, when E = E0 sin omegat ...

In A.C. circuit, when `E = E_0 sin omegat` supply applied, current is obtained in circuit `I=I_0 sin (omegat-pi/2)` , so power consumed is circuit will be ……

A

`P=(E_0I_0)/sqrt2`

B

P=zero

C

`P=(E_0I_0)/2`

D

`P=sqrt2E_0I_0`

Text Solution

Verified by Experts

The correct Answer is:
B
From equation of E and I, phase difference between their oscillations is `pi/2`
`therefore` Real power `P = E_"rms" I_"rms" cosdelta`
`=E_(rms)I_(rms)xx "cos"pi/2`
`=E I xx 0 [ because E_(rms) = E, I_(rms)=I]`
`therefore` P=0
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