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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

User Thaerith
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The root mean square of the current is given by

I_(rms) = (I_0)/( √(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)= (1.50 A)/( √(2) )=1.06 A
User EzPizza
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