Alternating Current: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:

For an alternating current $I = I_0 \sin \omega t$, the rms current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$. Evaluating this integral over one complete cycle gives $I_{rms} = \frac{I_0}{\sqrt{2}}$. Therefore, the relation between rms and peak current is $I_{rms} = I_0/\sqrt{2}$.