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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

Respuesta :

skyluke89
The root mean square of the current is given by
[tex]I_{rms} = \frac{I_0}{ \sqrt{2} } [/tex]
where [tex]I_0[/tex] is the maximum value of the current.

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

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