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# An electric current has both DC and AC components. DC component of 8 A and AC component is given as $I = 6\sin \omega t$. So, the ${I_{rms}}$ value of the resultant current is(A) 9.05 A(B) 8.05 A(C) 11.58 A(D) 13.58 A

Last updated date: 13th Jun 2024
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Hint:Here,we are going to apply the concept of alternating current as well as direct current and determine the resultant current by adding the AC and DC component of the current. Use the formula for the rms value of the current to determine the rms current.

Formula used:
$\Rightarrow I_{rms}^2 = \dfrac{{\int\limits_0^T {{I^2}dt} }}{T}$
Here, ${I_{rms}}$ is the rms value of the resultant current and T is the period.

The resultant value of the current is the sum of AC and DC currents.
$\Rightarrow I = {I_{DC}} + {I_{AC}}$
$\Rightarrow I = 8 + 6\sin \omega t$
The rms value of the resultant current is given as,
$\Rightarrow I_{rms}^2 = \dfrac{{\int\limits_0^T {{I^2}dt} }}{T}$
Here, T is the period.
Substitute the resultant value of the current in the above equation.
$\Rightarrow I_{rms}^2 = \dfrac{{\int\limits_0^T {{{\left( {8 + 6\sin \omega t} \right)}^2}dt} }}{T}$
Solve the above equation further as follows.
$\Rightarrow I_{rms}^2 = \dfrac{{\int\limits_0^T {\left( {64 + 36{{\sin }^2}\omega t + 96\sin \omega t} \right)dt} }}{T}$
Use, ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$.
$\Rightarrow I_{rms}^2 = \dfrac{{\int\limits_0^T {\left( {64 + 36\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right) + 96\sin \omega t} \right)dt} }}{T}$
$\Rightarrow I_{rms}^2 = \dfrac{{\int\limits_0^T {\left( {64 + 18 - 18\cos 2\omega t + 96\sin \omega t} \right)dt} }}{T}$
$\Rightarrow I_{rms}^2 = \dfrac{{\left( {82t + 18\dfrac{{\sin \omega t}}{{2\omega }} + 96\dfrac{{\cos \omega t}}{\omega }} \right)_0^T}}{T}$
$\Rightarrow I_{rms}^2 = \dfrac{{\left( {82T + 18\dfrac{{\sin \omega T}}{{2\omega }} + 96\dfrac{{\cos \omega T}}{\omega }} \right) - \left( {82\left( 0 \right) + 18\dfrac{{\sin \omega \left( 0 \right)}}{{2\omega }} + 96\dfrac{{\cos \omega \left( 0 \right)}}{\omega }} \right)}}{T}$
Use the relation, $\omega = \dfrac{{2\pi }}{T}$ in the above equation. We get,
$\Rightarrow I_{rms}^2 = \dfrac{{\left( {82T + 18\dfrac{{\sin 2\pi }}{{2\omega }} + 96\dfrac{{\cos 2\pi }}{\omega }} \right) - \left( {82\left( 0 \right) + 18\dfrac{{\sin \omega \left( 0 \right)}}{{2\omega }} + 96\dfrac{{\cos \omega \left( 0 \right)}}{\omega }} \right)}}{T}$
$\Rightarrow I_{rms}^2 = \dfrac{{\left( {82T + \dfrac{{96}}{\omega }} \right) - \left( {\dfrac{{96}}{\omega }} \right)}}{T}$
$\Rightarrow I_{rms}^2 = 82$
Therefore,
$\Rightarrow{I_{rms}} = 9.05\,A$
So, the correct answer is option (A).

Note: The integration of $\sin \theta$ is $- \cos \theta$ and integration of $\cos \theta$ is $\sin \theta$. Also, remember $\cos n\pi = {\left( { - 1} \right)^n}$.Also remember that alternating current can be defined as a current that changes its magnitude and polarity at regular intervals of time. It can also be defined as an electrical current which repeatedly changes or reverses its direction opposite to that of Direct Current or DC which always flows in a single direction.