Question

An alternating current is given by i = i₁ cos ωt + i₂ sin ωt. The rms current is given by

विकल्प

• `(l_1 +l_2)/sqrt2`

• `|i_1 + i_2|/sqrt2` 

• `sqrt(i_1^2 + i_2^2)/2`

• `sqrt(i_1^2+i_2^2)/sqrt2`

Answer 1

`sqrt(i_1^2 + i_2^2)/2`
Given:
i = i₁ cos ωt + i₂ sin ωt
The rms value of current is given by,
`i_{rms} = sqrt(\[\int_0^T  i^2 dt]/\[\int_0^T  dt`
`i = i_1 cos ωt + i_2 sin omegat`

`i_{rms} = sqrt(\[\int_0^T(i_1 cos omegat + i_2 sin omegat)^2 dt)/(\int_0^T dt)`  

`i_rms = sqrt(\[\int_0^T (i_1^2 cos^2 omegat + i_2^2 sin^2 omegat + 2i_1 i_2 sin omegat cos omegat) dt)/\int_0^T)`

`i_{rms} = sqrt(\[\int_0^T (i_1^2 ((cos 2omegat + 1))/2+ i_2^2((1-cos 2omegat))/2 + i_1i_2 sin 2omegat) dt )/[\int_0^T dt`

`[therefore cos^2 omegat = ((cos 2omegat + 1))/2 , sin^2 omegat = ((1 - cos 2omegat))/2 ]`

We know that, T = 2π

Integrating the above expression
\[i_{rms} = \sqrt\frac{{\frac{1}{2} i_1^2\left(\int_0^{2\pi} 1dt + \int_0^{2\pi} cos 2\omega t\ dt\right) + i_2^2\left( \int_0^{2\pi} 1 dt - \int_0^{2\pi} cos 2 \omega t\ dt\right) + i_1 i_2 \int_0^{2\pi} sin 2 \omega t\ dt }}{\int_0^{2\pi} dt} \]

The following integrals become zero
\[\int_0^{2\pi} cos 2 \omega t \ dt  = 0 = \int_0^{2\pi} sin2 \omega  t\]

Therefore, it becomes
\[i_{rms} = \sqrt\frac{{\frac{i_1^2}{2} \left(\int_0^{2\pi} 1dt\right) + \frac{i_2^2}{2}\left(\int_0^{2\pi}1dt\right)}}{\int_0^{2\pi} dt}\]

`i_rms = sqrt((i_1^2/2 xx 2pi + (i_2^2)/2 xx 2pi)/(2pi)`

`⇒i_rms = sqrt((i_1^2) +(i_2^2))/2`