An alternating current is given by the equation

Question:

An alternating current is given by the equation $\mathrm{i}=\mathrm{i}_{1} \sin \omega \mathrm{t}+\mathrm{i}_{2} \cos \omega \mathrm{t}$. The rms current will be

  1. $\frac{1}{\sqrt{2}}\left(i_{1}^{2}+i_{2}^{2}\right)^{\frac{1}{2}}$

  2. $\frac{1}{\sqrt{2}}\left(i_{1}+i_{2}\right)^{2}$

  3. $\frac{1}{2}\left(i_{1}^{2}+i_{2}^{2}\right)^{\frac{1}{2}}$

  4. $\frac{1}{\sqrt{2}}\left(i_{1}+i_{2}\right)$


Correct Option: 1

Solution:

$\mathrm{i}=\mathrm{i}_{1} \sin \omega \mathrm{t}+\mathrm{i}_{2} \sin (\omega \mathrm{t}+90)$

$\mathrm{i}=\sqrt{\mathrm{i}_{1}^{2}+\mathrm{i}_{2}^{2}} \sin (\omega \mathrm{t}+\phi)$

$\mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{0}}{\sqrt{2}}=\frac{\sqrt{\mathrm{i}_{1}^{2}+\mathrm{i}_{2}^{2}}}{\sqrt{2}}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now