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

Answer»

<P>`P=(E_0I_0)/SQRT2`
P=zero
`P=(E_0I_0)/2`
`P=sqrt2E_0I_0`

Solution :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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