Question

An electric current has both A.C and D.C. components. The value of the D.C. component is equal to $12$A. While the A.C component is given as $I=9\sin$wt A. Determine the formula for the resultant current and also calculate the value of $I_{rms}$.

Answer 1

$I=12+9\sin \omega t$

$I_{rms}^2=\frac{{\int }_0^T{I}^{2}dt}{T}$

$=\frac{{\int }_0^T(12+9\sin \omega t{)}^{2}dt}{T}$

$=\frac{{\int }_0^T(144+216\sin \omega t+81{\sin }^2\omega t)dt}{T}$

$=\frac{1}{T}{\int }_0^T(144+216\sin \omega t+\frac{81}{2}(1-\cos 2\omega t\left)\right)dt$

$=\frac{1}{T}{[144t+\frac{216}{\omega }\cos \omega t+\frac{81}{2}t+\frac{81}{4\omega }\sin \omega t]}_0^T$

$I_{rms}^2=\frac{1}{T}.\frac{369T}{2}$

$⟹I_{rms}=\sqrt{\frac{369}{2}}A$

Simply we can see that $I_{rms}=\sqrt{{12}^2+\frac{9^2}{2}}=\sqrt{\frac{369}{2}}A$