Question

You have a special light bulb with a very delicate wire filament. the wire will break if the current in it ever exceeds 1.50 a, even for an instant. what is the largest root-mean-square current you can run through this bulb? 31.2 . a sinusoidal current

Answer 1

The root mean square of the current is given by
$I_{rms} = \frac{I_0}{ \sqrt{2} }$
where $I_0$ is the maximum value of the current.

In our problem, the maximum current allowed without breaking the filament is equal to $I_0=1.50 A$. Therefore, the largest root-mean-square current allowed without breaking the wire is
$I_{rms}= \frac{1.50 A}{ \sqrt{2} }=1.06 A$