Question

You have a special lightbulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.30 A , even for an instant.
What is the largest root-mean-square current you can run through this bulb?

Answer 1

Answer:

about 0.919 amp

Explanation:

$I_r_m_s=\frac{I_m_a_x} {\sqrt{2}}\\ I_r_m_s=\frac{1.3}{\sqrt{2}}\\I_r_m_s\approx 0.919 amp$