Question 28

$$ \text{An alternating current is given by } I=I_A\sin\omega t + I_B\cos\omega t. \text{ The r.m.s. current will be:} $$

Name: \\text{An alternating current is given by } I=I_A\\sin\\omega t + I_B\\cos\\omega t. \\text{ The r.m.s. current will be:}
Uploaded: 2026-05-16T16:28:23.771251+05:30
Duration: 3 min 24 s
Description: \\text{An alternating current is given by } I=I_A\\sin\\omega t + I_B\\cos\\omega t. \\text{ The r.m.s. current will be:}

Solution

$$I = \sqrt{I_A^2 + I_B^2} \sin(\omega t + \phi) \implies I_0 = \sqrt{I_A^2 + I_B^2}$$

$$I_{\text{r.m.s.}} = \frac{I_0}{\sqrt{2}} \implies I_{\text{r.m.s.}} = \frac{\sqrt{I_A^2 + I_B^2}}{\sqrt{2}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$$

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$$ \text{An alternating current is given by } I=I_A\sin\omega t + I_B\cos\omega t. \text{ The r.m.s. current will be:} $$

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